5.13: Comments and solutions

137. Note: The spirit of constructions restricts us to:

• drawing the line joining two known points
• drawing the circle with centre a known point and passing through a known point.

The whole thrust of this first problem is to find some way to “jump” from A (or B) to C. So the problem leaves us with very little choice; AB is given, and A and B are more-or-less indistinguishable, so there are only two possible 'first moves’ - both of which work with the line segment AC (or BC).

Join AC.

Then construct the point X such that ΔACX is equilateral (i.e. use Euclid's Elements, Book I, Proposition 1). Construct the circle with centre A which passes through B; let this circle meet the line AX at the point Y, where either

(i) Y lies on the segment AX (if $\underset{\_}{AB}⩽\underset{\_}{AC}$ ), or

(ii) Y lies on AX produced (i.e. beyond X, if AB > AC).

In each case, AY = AB. Finally construct the circle with centre X which passes through Y. In case (i), let the circle meet the line segment XC at D; in case (ii), let the circle meet CX produced (beyond X) at D.

In case (i), CX = CD + DX; therefore

$$\begin{array}{l}\underset{\_}{CD}=\underset{\_}{CX}-\underset{\_}{DX}\\ =\underset{\_}{AX}-\underset{\_}{YX}\\ =\underset{\_}{AY}=\underset{\_}{AB}.\end{array}$$

In case (ii),

$$\begin{array}{l}\underset{\_}{CD}=\underset{\_}{CX}+\underset{\_}{XD}\\ =\underset{\_}{AX}+\underset{\_}{XY}\\ =\underset{\_}{AY}=\underset{\_}{AB}.\end{array}$$ QED

138.

∠ A X C = ∠ A X B − ∠ C X B                               = ∠ C X D − ∠ C X B ( since the two straight angles ∠ A X B   a n d   ∠ C X D   a r e   e q u a l )                               = ∠ B X D .

QED

139.

AM = AM

MB = MC (by construction of the midpoint M)

BA = CA (given).

$\therefore \Delta AMB\equiv \mathrm{\Delta }AMC$ (by SSS-congruence)

$\therefore \angle AMB=\angle AMC$, so each angle is exactly half the straight angle =BMC. Hence AM is perpendicular to BC. QED

140. Let ABCDEF be a regular hexagon with sides of length 1. Then ΔABC (formed by the first three vertices) satisfies the given constraints, with ∠ABC = 120°.

Let ∠B'C'D be an equilateral triangle with sides of length 2, and with A' the midpoint of B'D. Then ∠A'B'C' satisfies the given constraints with angle ∠A'B'C' = 60°.

141.

(i) Join CO. Then ΔACO is isosceles (since OA = OC) and ΔBCO is isosceles (since OB = OC).

Hence ∠OAC = ∠OCA = x (say), and ∠OBC = ∠OCB = y (say).

So ∠C = x + y (and ∠A +∠B+∠C= x +(x+y)+y = 2(x + y)). QED

(ii) If ∠A+∠B+∠C =2(x+y) is equal to a straight angle, then ∠C= x + y is half a straight angle, and hence a right angle. QED

142.

(i) Draw the circle with centre A and passing through B, and the circle with centre B passing through A. Let these two circles meet at C and D.

Wherever the midpoint M of AB may be, we know from Euclid Book I, Proposition 1 and Problem 139:

that ΔABC is equilateral, and that CM is perpendicular to AB, and that ΔABD is equilateral, and that DM is perpendicular to AB.

Hence CMD is a straight line.

So if we join CD, then this line cuts AB at its midpoint M. QED

(ii) We may suppose that BA ≤ BC.

Then the circle with centre B, passing through A meets BC internally at Aʹ (say).

Let the circle with centre A and passing through B meet the circle with centre Aʹ and passing through B at the point D.

Claim BD bisects ∠ABC.

Proof

BA = BAʹ(radii of the same circle with centre B)

AD = AB (radii of the same circle with centre A)

= AʹB= AʹD (radii of same circle with centre Aʹ)

BD = BD

Hence Δ BAD ≡ Δ BAʹD (by SSS-congruence).

∴ ∠ABD =∠AʹBD. QED

(iii) Suppose first that PA = PB. Then the circle with centre P and passing through A meets the line AB again at B. If we construct the midpoint M of AB as in part (i), then PM will be perpendicular to AB.

Now suppose that one of PA and PB is longer than the other. We may suppose that PA>PB, so B lies inside the circle with centre P and passing through A. Hence this circle meets the line AB again at Aʹ where B lies between A and Aʹ. If we now construct the midpoint M of AAʹ as in part (i), then PM will be perpendicular to AAʹ, and hence to AB. QED

143. Let M be the midpoint of AB.

(a) Let X lie on the perpendicular bisector of AB.

∴ Δ XMA ≡ Δ XMB (by SAS congruence, since XM = XM, ∠XMA =∠XMB, MA = MB)

∴ XA = XB.

(b) If X is equidistant from A and from B, then Δ XMA ≡ Δ XMB (by SSS-congruence, since XM = XM, MA = MB, AX = BX).

∴ ∠XMA =∠XMB, so each must be exactly half a straight angle.

∴ X lies on the perpendicular bisector of AB. QED

144. Let X lie on the plane perpendicular to NS, through the midpoint M.

∴ Δ XMN ≡ Δ XMS (by SAS congruence, since XM = XM, ∠XMN =∠XMS, MN = MS)

∴ XN = XS.

Let X be equidistant from N and from S, then Δ XMN ≡ Δ XMS (by SSS-congruence, since XM = XM, MN = MS, NX = SX).

∴ ∠XMN =∠XMS, so each must be exactly half a straight angle.

∴ X lies on the perpendicular bisector of NS. QED

145. Let M be the midpoint of AC. Join BM and extend the line beyond M to the point D such that MB = MD. Join CD. Then Δ AMB ≡ Δ CMD (by SAS-congruence, since

AM = CM (by construction of the midpoint M)

∠AMB =∠CMD (vertically opposite angles)

MB = MD (by construction)).

∴ ∠DCM =∠BAM.

Now

$$\begin{array}{l}\angle ACX=\angle DCM+\angle DCX\\ >\end{array}$$DCM                 =∠BAM                 =∠A.

Hence∠ACX >∠A.

Similarly, we can extend AC beyond C to a point Y. Let N be the midpoint of BC. Join AN and extend the line beyond N to the point E such that NA = NE.

Join CE.

Then Δ BNA ≡ Δ CNE (again by SAS-congruence).

∴ ∠BCY >∠BCE =∠CBA =∠B. QED

146.

(a) Suppose that AB > AC.

Let the circle with centre A, passing through C, meet AB (internally) at X.

Then Δ ACX is isosceles, so∠ACX =∠AXC.

By Problem 145, ∠AXC >∠ABC.

∴ ∠ACB; (>∠ACX =∠AXC) >∠ABC. QED

(b) Suppose the conclusion does not hold. Then either

(i) AB = AC, or

(ii) AC > AB.

(i) If AB = AC, then Δ ABC is isosceles, so∠ACB =∠ABC – contrary to assumption.

(ii) If AC > AB, then∠ABC >∠ACB (by part (a)) – again contrary to assumption.

Hence, if∠ACB >∠ABC, it follows that AB > AC. QED

(c) Extend AB beyond B to the point D, such that BD = BC.

Then Δ BDC is isosceles with apex B, so∠BDC =∠BCD.

Now

∠ACD =∠ACB +∠BCD >∠BCD =∠BDC.

Hence, by part (b), AD > AC.

By construction,

AD = AB + BD = AB +BC,

so AB +BC > AC. QED

147. Suppose∠C =∠A +∠B, but that C does not lie on the circle with diameter AB. Then C lies either inside, or outside the circle. Let O be the midpoint of AB.

(i) If C lies outside the circle, then OC > OA = OB.

∴ ∠OAC >∠OCA and∠OBC >∠OCB (by Problem 146(a)).

∴ ∠C =∠OCA +∠OCB <∠A +∠B – contrary to assumption.

Hence C does not lie outside the circle.

(ii) If C lies inside the circle, then OC < OA = OB.

∴ ∠OAC < ∠OCA and∠OBC < ∠OCB (by Problem 146(a)).

∴ ∠C =∠OCA +∠OCB >∠A +∠B – contrary to assumption.

Hence C does not lie inside the circle.

Hence C lies on the circle with diameter AB. QED

148. Suppose, to the contrary, that OP is not perpendicular to the tangent at P.

Drop a perpendicular from O to the tangent at P to meet the tangent at Q. Extend PQ beyond Q to some point X. Then ∠OQP and ∠OQX are both right angles. Since Q (≠ P) lies on the tangent, Q lies outside the circle, so OQ > OP. Hence (by Problem 146(a))∠OPQ >∠OQP = ∠OQX – contrary to the fact that∠OQX >∠OPQ (by Problem 145). QED

149. Let Q lie on the line m such that PQ is perpendicular to m.

Let X be any other point on the line m, and let Y be a point on m such that Q lies between X and Y.

Then ∠PQX and ∠PQY are both right angles.

Suppose that PX < PQ.

Then∠PQY =∠PQX < ∠PXQ (by Problem 146(a)), which contradicts Problem 145 (since ∠PQY is an exterior angle of Δ PQX).

Hence PX ≥ PQ as required. QED

150.∠A +∠B +∠C, and∠XCA +∠C are both equal to a straight angle. So∠A +∠B =∠XCA.

151. Join OA, OB, OC. Since these radii are all equal, this produces three isosceles triangles. There are five cases to consider.

(i) Suppose first that O lies on AB. Then AB is a diameter, so ∠ACB is a right angle (by Problem 141), ∠AOB is a straight angle, and the result holds.

(ii) Suppose O lies on AC, or on BC. These are similar, so we may assume that O lies on AC.

Then Δ OBC is isosceles, so∠OBC =∠OCB.

∠AOB is the exterior angle of Δ OBC, so

∠AOB =∠OBC +∠OCB = 2· ∠ACB

(by Problem 150).

(iii) Suppose O lies inside Δ ABC.

Δ OAB, Δ OBC, Δ OBC are isosceles, so let ∠OAB =∠OBA = x, ∠OBC =∠OCB = y, ∠OBC =∠OAC = z.

Then∠ACB = y + z, ∠ABC = x + y, ∠BAC = x + z.

The three angles of Δ ABC add to a straight angle, so 2(x + y + z) equals a straight angle.

Hence, in Δ OBA,

∠AOB = 2(x + y + z) - (∠OAB +∠OBA) = 2(y + z)= 2· ∠ACB.

(iv) Suppose O lies outside Δ ABC with O and B on opposite sides of AC.

Δ OAB, Δ OBC, Δ OBC are isosceles, so let ∠OAB =∠OBA = x, ∠OBC =∠OCB = y, ∠OBC =∠OAC = z.

Then∠ACB = y - z, ∠ABC = x + y, ∠BAC = x - z.

The three angles of Δ ABC add to a straight angle, so 2x + 2y - 2z equals a straight angle.

Hence

2x + 2y - 2z =∠AOB +∠OAB +∠OBA =∠AOB + 2x,

so∠AOB = 2y - 2z = 2· ∠ACB.

(v) Suppose O lies outside Δ ABC with O and B on the same side of AC.

Δ OAB, Δ OBC, Δ OBC are isosceles, so let ∠OAB =∠OBA = x, ∠OBC =∠OCB = y, ∠OBC =∠OAC = z.

Then∠ACB = y + z, ∠ABC = y - x, ∠BAC = z - x.

The three angles of ABC add to a straight angle, so 2y + 2z - 2x equals a straight angle.

Since C lies on the minor arc relative to the chord AB, we need to interpret “the angle subtended by the chord AB at the centre O” as the reflex angle outside the triangle Δ AOB – which is equal to “2x more than a straight angle” , so ∠AOB = 2y + 2z = 2· ∠ACB. QED

152. The chord AB subtends angles at C and at D on the same arc. Similarly BC subtends angles at A and at D on the same arc. Hence (by Problem 151)

∠ACB =∠ADB, and∠BAC =∠BDC.

Hence

∠ADC =∠ADB +∠BDC =∠ACB +∠BAC, so ∠ADC +∠ABC equals the sum of the three angles in Δ ABC. QED

153. Let O be the circumcentre of Δ ABC, and let∠XAB = x.

Then ∠XAO is a right angle, and Δ OAB is isosceles. There are two cases.

(i) If X and O lie on opposite sides of AB, then∠OBA =∠OAB = 90° - x.

∴ ∠AOB = 2x.

∴ ∠ACB = X =∠XAB.

(ii) If X and O lie on the same side of AB, then Y and O lie on opposite sides of AB and of AC. Hence we can apply case (i) (with Y in place of X, AC in place of AB, ∠YAC in place of ∠XAB, and ∠ABC in place of ∠ACB) to conclude that∠YAC =∠ABC. Hence

∠XAB +∠BAC +∠YAC =∠XAB + ∠BAC +∠ABC

are both straight angles, so ∠XAB =∠ACB (since the three angles of Δ ABC also add to a straight angle). QED

154.

(a) (i) Extend AD to meet the circumcircle at X. Then (applying Problem 145 to Δ DXB), the exterior angle∠ADB > ∠DXB = ∠AXB = ∠ACB.

(ii) We are told that the points C, D lie “on the same side of the line AB”. This “side” of the line AB (or “half-plane” ) is split into three parts by the half-lines “AC produced beyond C” and “BC produced beyond C” . There are two very different possibilities.

Suppose first that D lies in one of the two overlapping wedge-shaped regions “between AB produced and AC produced” or “between BA produced and BC produced”. Then either DA or DB cuts the arc AB at a point X (say), and ∠ACB = ∠AXB > ∠ADB (by Problem 145 applied to Δ AXD or to Δ BXD).

The only alternative is that D lies in the wedge shaped region outside the circle at the point C, lying “between AC produced and BC produced”. Then C lies inside Δ ADB, so C lies inside the circumcircle of Δ ADB. Hence part (i) implies that ∠ACB < ∠ADB as required. QED

(b) If D does not lie on the circumcircle of Δ ABC, then it lies either inside, or outside the circle – in which case ∠ADB ≠ ∠ACB (by part (a)).

(c) ∠D is less than a straight angle, so D must lie outside Δ ABC. Moreover, B, D lie on opposite sides of the line AC (since the edges BC, DA do not cross internally, and neither do the edges CD, AB). Consider the circumcircle of Δ ABC, and let X be any point on the circle such that X and D lie on the same side of the line AC. Then ABCX is a cyclic quadrilateral, so ∠ABC and ∠CXA are supplementary. Hence∠CXA =∠CDA =∠D. But then D lies on the circle (by part (b)).

155. Δ OPQ ≡ Δ OPʹQ (by RHS-congruence: ∠OPQ =∠OPʹQ are both right angles, OP = OPʹ, OQ = OQ)

∴ QP = QPʹ, and∠QOP =∠QOPʹ. QED

156.

(a) Let X be any point on the bisector of ∠ABC. Drop the perpendiculars XY from X to AB and XZ from X to CB. Then:∠XYB =∠XZB are both right angles by construction; and∠XBY =∠XBZ since X lies on the bisector of ∠YBZ; hence∠BXY =∠BXZ (since the three angles in each triangle have the same sum; so as soon as two of the angles are equal in pairs, the third pair must also be equal). Hence Δ BXY ≡ Δ BXZ (by ASA-congruence:∠YBX =∠ZBX, BX = BX, ∠BXY =∠BXZ.)

∴ XY= XZ. QED

(b) Suppose X is equidistant from m and from n. Drop the perpendiculars XY from X to m, and XZ from X to n.

Then Δ BXY ≡ Δ BXZ (by RHS-congruence:

∠BYX =∠BZX are both right angles XY = XZ, since we are assuming X is equidistant from m and from n BX = BX).

Hence∠XBY =∠XBZ, so X lies on the bisector of ∠YBZ. QED

157.

(i) Join AC. Then Δ ABC ≡ Δ CDA by ASA-congruence:

∠BAC =∠DCA (alternate angles, since AB || DC)

AC = CA

∠ACB =∠CAD (alternate angles, since CB || DA).

In particular, Δ ABC and Δ CDA must have equal area, and so each is exactly half of ABCD. QED

Note: Once we prove (Problem 161 below) that a parallelogram has the same area as the rectangle on the same base and lying between the same pair of parallels (whose area is equal to “base × height” ), the result in part (i) will immediately translate into the familiar formula for the area of the triangle

$$\begin{array}{cc}& \genfrac{}{}{0.1ex}{}{1}{2}(\text{base}\times \text{height}).\hfill \end{array}$$

(ii) Δ ABC ≡ Δ CDA by part (i). Hence AB = CD, BC = DA; and∠B =∠ABC =∠CDA =∠D.

To show that∠A =∠C, we can either copy part (i) after joining BD to prove that Δ BCD ≡ Δ DAB, or we can use part (i) directly to note that∠A =∠BAC +∠DAC =∠DCA +∠BCA =∠C. QED

(iii) Δ AXB ≡ Δ CXD by ASA-congruence:

∠XAB =∠XCD (alternate angles, since AB || DC)

AB = CD (by part (ii))

∠XBA =∠XDC (alternate angles, since AB || DC).

Hence XA (in Δ AXB) = XC (in Δ CXD), and XB (in Δ AXB) = XD (in Δ CXD). QED

158. We may assume that m cuts the opposite sides AB at Y and DC at Z.

Δ XYB ≡ Δ XZD by ASA-congruence:

∠YXB =∠ZXD (vertically opposite angles)

XB = XD (by Problem 157(iii))

∠XBY =∠XDZ (alternate angles).

Therefore

area(YZCB) = area(Δ BCD) - area(Δ XZD) + area(Δ XYB)

= area(Δ BCD)

$=$12area(ABCD).

QED

159.

(a) Join AC. Then Δ ABC ≡ Δ CDA by SAS-congruence:

BA = DC (given)

∠BAC =∠DCA (alternate angles, since AB || DC)

AC = CA.

Hence∠BCA =∠DAC, so AD || BC as required. QED

(b) Join AC. Then Δ ABC≡ Δ CDA by SSS-congruence:

AB = CD (given)

BC = DA (given)

CA = AC.

Hence∠BAC =∠DCA, so AB || DC; and∠BCA =∠DAC, so BC || AD.

QED

(c)∠A +∠B +∠C +∠D = 2 ∠A + 2 ∠B = 2 ∠A + 2 ∠D are each equal to two straight angles.

∴ ∠A +∠B is equal to a straight angle, so AD || BC; and∠A +∠D is equal to a straight angle, so AB || DC. QED

Note: The fact that the angles in a quadrilateral add to two straight angles is proved in the next chapter. However, if preferred, it can be proved here directly. If we imagine pins lOBCted at A, B, C, D, then a string tied around the four points defines their “convex hull” – which is either a 4-gon (if the string touches all four pins), or a 3-gon (if one vertex is inside the triangle formed by the other three). In the first case, either diagonal (AC or BD) will split the quadrilateral internally into two triangles; in the second case, one of the three ‘edges’ joining vertices of the convex hull to the internal vertex cannot be an edge of the quadrilateral, and so must be a diagonal, which splits the quadrilateral internally into two triangles.

160. AM = MD (by construction of M as the midpoint), and BN = NC.

∴AM = BN (since AD = BC by Problem 157(ii)).

∴ ABNM is a parallelogram (by Problem 159(a)), so MN || AB.

Let AC cross MN at Y.

Then Δ AYM ≡ Δ CYN (by ASA-congruence, since

∠YAM =∠YCN (alternate angles, since AD || BC)

AM = CN

∠AMY =∠CNY (alternate angles, since AD || BC).

Hence AY = CY, so Y is the midpoint of AC –the centre of the parallelogram (where the two diagonals meet (by Problem 157(iii))). QED

161.

Note: In the easy case, where the perpendicular from A to the line DC meets the side DC internally at X, it is natural to see the parallelogram ABCD as the “sum” of a trapezium ABCX and a right angled triangle Δ AXD. If the perpendicular from B to DC meets DC at Y, then Δ AXD ≡ Δ BYC. Hence we can rearrange the two parts of the parallelogram DCBA to form a rectangle XYBA.

However, a general proof cannot assume that the perpendicular from A (or from B) to DC meets DC internally. Hence we are obliged to think of the parallelogram in terms of differences. This is a strategy that is often useful, but which can be surprisingly elusive.

Draw the perpendiculars from A and B to the line CD, and from C and D to the line AB. Choose the two perpendiculars which, together with the lines AB and CD define a rectangle that completely contains the parallelogram ABCD (that is, if AB runs from left to right, take the left-most, and the right-most perpendiculars). These will be either the perpendiculars from B and from D, or the perpendiculars from A and from C (depending on which way the sides AC and BD slope).

Suppose the chosen perpendiculars are the one from B – meeting the line DC at P, and the one from D – meeting the line AB at Q.

Then BP || QD (by Problem 159(c)), so BQDP is a parallelogram with a right angle – and hence a rectangle. Hence BQ = PD, and BP = QD (by Problem 157(ii)).

Δ QAD ≡ Δ PCB (by RHS-congruence), so each is equal to half the rectangle on base PC and height PB. Hence

area(ABCD) = area(rectangle; BQDP)

- area(rectangle on base PC with height PB)

= area(rectangle on base CD with height DQ).

QED

162.

(a) ABCBʹ is a parallelogram, so ABʹ|| CB and ABʹ = BC.

Similarly, BCACʹ is a parallelogram, so ACʹ|| CB and ACʹ= CB.

Hence BʹA = ACʹ, so A is the midpoint of BʹCʹ.

Similarly B is the midpoint of CʹAʹ, and C is the midpoint of AʹBʹ.

(b) Let H be the circumcentre of Δ AʹBʹCʹ – that is, the common point of the perpendicular bisectors of AʹBʹ, BʹCʹ, and CʹAʹ. Then H is a common point of the three perpendiculars from A to BC, from B to CA, and from C to AB.

163. Consider any path from H to V. Suppose this reaches the river at X. The shortest route from H to X is a straight line segment; and the shortest route from X to V is a straight line segment.

If we reflect the point H in the line of the river, we get a point Hʹ, where HHʹ is perpendicular to the river and meets the river at Y (say).

Then Δ HXY ≡ Δ HʹXY (by SAS-congruence, since HY = HʹY, ∠HYX =∠HʹYX, and YX = YX).

Hence HX = HʹX, so the distance from H to V via X is equal to HX + XV = HʹX + XV, and this is shortest when Hʹ, X, and V are collinear.(So to find the shortest route, reflect H in the line of the river to Hʹ, then draw HʹV to cross the line of the river at X, and travel from H to V via X.)

164.

(a) Let Δ PQR be any triangle inscribed in Δ ABC, with P on BC, Q on CA, R on AB (not necessarily the orthic triangle). Let Pʹ be the reflection of P in the side AC, and let Pʹʹ be the reflection of P in the side AB. Then PQ = PʹQ, and PR = PʹʹR (as in Problem 163).

∴ PQ + QR + RP = PʹQ + QR + RPʹʹ.

Each choice of the point P on AB determines the positions of Pʹ and Pʹʹ. Hence the shortest possible perimeter of Δ PQR arises when PʹQRPʹʹ is a straight line. That is, given a choice of the point P, choose Q and R by:

– constructing the reflections Pʹ of P in AC, and Pʹʹ of P in AB;

– join PʹPʹʹand let Q, R be the points where this line segment crosses AC, AB respectively.

It remains to decide how to choose P on BC so that PʹPʹʹ is as short as possible.

The key here is to notice that A lies on both AC and on AB.

∴ AP = APʹ, and AP = APʹʹ, so Δ APʹPʹʹis isosceles.

Also∠PAC =∠PʹAC, and∠PAB =∠PʹʹAB.

∴ ∠PʹAPʹʹ= 2· ∠A.

Hence, for each position of the point P, Δ APʹPʹʹ is isosceles with apex angle equal to 2· ∠A. Any two such triangles are similar (by SAS-similarity).

Hence the triangle Δ APʹPʹʹ with the shortest “base” PʹPʹʹoccurs when the legs APʹ and APʹʹare as short as possible. But AP = APʹ = APʹʹ, so this occurs whenAP is as short as possible – namely when AP is perpendicular to BC.

Since the same reasoning applies to Q and to R, it follows that the required triangle Δ PQR must be the orthic triangle of Δ ABC. QED

(b) Let Δ PQR be the orthic triangle of Δ ABC, with P on BC, Q on CA, R on AB. Let H be the orthocenter of Δ ABC.

∠BPH and ∠BRH are both right angles, so add to a straight angle. Hence (by Problem 154(c)), BPHR is a cyclic quadrilateral. Similarly CPHQ and AQHR are cyclic quadrilaterals.

In the circumcircle of CPHQ, we see that the initial “angle of incidence” ∠CQP =∠CHP. Also∠CHP =∠AHR (vertically opposite angles); and in the circumcircle of AQHR, ∠AHR =∠AQR.

Hence∠CQP =∠AQR, so a ray of light which traverses PQ will reflect at Q along the line QR. Similarly one can show that∠ARQ =∠BRP, so that the ray will then reflect at R along RP; and∠BPR =∠CPQ, so the ray will then reflect at P along PQ. QED

165.

(a) (i) Triangles Δ ABL and Δ ACL have equal bases BL = CL, and the same apex A – so lie between the same parallels. Hence they have equal area (by Problems 157 and 161).

Similarly, Δ GBL and Δ GCL have equal bases BL = CL, and the same apex G – so have equal area.

Hence the differences Δ ABG = Δ ABL - Δ GBL and Δ ACG = Δ ACL - Δ GCL have equal area.

(ii) [Repeat the solution for part (i) replacing A, B, C, L, G by B, C, A, M, G.]

(b) Δ ABL and Δ ACL have equal area (as in (a)(i)). Similarly Δ GBL and Δ GCL have the same area – say x (since BL = CL). Hence Δ ABG and Δ ACG have equal area.

In the same way Δ BCM and Δ BAM have equal area; and Δ GCM and Δ GAM have the same area – say y (since CM = AM). Hence Δ BCG and

Δ BAG have equal area.

But then Δ ABG = Δ ACG = Δ BCG and Δ ACG = Δ AMG + Δ CMG = 2y, Δ BCG = Δ BLG + Δ CLG = 2x. Hence X = y, so Δ AMG, Δ CMG, Δ CLG, Δ BLG all have the same area x, and Δ ABG has area 2x.

The segment GN divides Δ ABG into two equal parts (Δ ANG andΔ BNG), so each part has area x.

Hence Δ CAG + Δ ANG has the same area (3x) as Δ CAN. Hence ∠CGN is a straight angle, and the three medians AL, BM, CN all pass through the point G.

Note: At first sight, the ‘proof’ of the result in (b) using vectors seems considerably easier.(If A, B, C have position vectors a, b, c respectively, then L has the position vector $\frac{1}{2}$(b + c), and M has position vector $\frac{1}{2}$(c + a), and it is easy to see that AL and BM meet at G with position vector $\frac{1}{3}$(a + b + c). One can then check directly that G lies on CN, or notice that the symmetry of the expression $\frac{1}{3}$(a + b + c) guarantees that G is also the point where BM and CN meet.

The inscrutable aspect of this ‘proof’ lies in the fact that all the geometry has been silently hidden in the algebraic assumptions which underpin the unstated axioms of the 2-dimensional vector space, and the underlying field of real numbers. Hence, although the vector ‘proof’ may seem simpler, the two different apprOAChes cannot really be compared.

166.

(a) AN = BN (by construction of N as the midpoint of AB)

∠ANM =∠BNMʹ (vertically opposite angles)

NM = NMʹ (by construction).

∴ Δ ANM ≡ Δ BNMʹ (by SAS-congruence). QED

(b) BMʹ = AM = CM.

∠NAM =∠NBMʹ (since Δ ANM ≡ Δ BNMʹ)

∴ BMʹ || MA (i.e. BMʹ || CM). QED

(c) BMʹMC is a quadrilateral with opposite sides CM, BMʹ equal and parallel.

∴ BMʹMC is a (by Problem 158(a)). QED

167. Since A and B are interchangeable in the result to be proved, we may assume that A is the point on the secant that lies between P and B.

In order to make deductions, we have to create triangles – so join AT and BT. This creates two triangles: Δ PAT and Δ PTB, in which we see that:

∠TPA =∠BPT,

∠PTA =∠PBT (by Problem 153),

∴ ∠PAT = ∠PTB (since the three angles in each triangle add to a straight angle).

Hence Δ PAT ~ Δ PTB (by AAA-similarity).

∴ PT: PB= PA: PT, orPA × PB= PT². QED

168. Since A, B are interchangeable in the result to be proved, and C, D are interchangeable, we may assume that A lies between P and B, and that C lies between P and D.

(i) Let the tangent from P to the circle touch the circle at T.

Then Problem 167 guarantees that PA × PB= PT².

Replacing the secant PAB by PCD shows similarly that PC × PD= PT².

Hence PA ×PB = PC × PD. QED

(ii) Note: The first proof is so easy, one may wonder why anyone would ask for a second proof. The reason lies in Problem 169 – which looks very much like Problem 168, but with P inside the circle. Hence, when we come to the next problem, the easy apprOACh in (i) will not be available, so it is worth looking for another proof of 168 which has a chance of generalizing.

Notice that AC is a chord which links the two secants PAB and PCD. So join AD and CB.

Then Δ PAD ~ Δ PCB (by AAA-similarity: since∠APD =∠CPB, and∠PDA =∠PBC).

∴ PA: PC = PD: PB, orPA × PB= PC × PD. QED

169. Join AD and CB.

Then Δ PAD ~ Δ PCB (by AAA-similarity: since

∠APD =∠CPB (vertically opposite angles)

∠PDA =∠PBC (angles subtended by chord AC on the same arc)

∠PAD =∠PCB (angles subtended by a chord BD on the same arc)).

∴ PA: PC= PD: PB, orPA × PB= PC × PD. QED

170.

(a) If a = b, then ABCD is a parallelogram (by Problem 159(a)).

∴ AD = BC (by Problem 157(ii)).

∴ AM = BN, so ABNM is a parallelogram (by Problem 159(a)).

∴ MN = AB has length a, and MN || AB (by Problem 160).

If a ≠ B, then a < B, or B < a. We may assume that a < b.

Construct the line through b parallel to AD, and let this line meet DC at Q.

Then ABQD is a parallelogram, so DQ = AB, and AD = BQ. Hence QC has length B - a.

Construct the line through M parallel to QC (and hence parallel to BA), and let this meet BQ at P, and BC at Nʹ.

Then ABPM and MPQD are both parallelograms.

∴ MP = AB has length a, and

BP = AM = MD = PQ.

Now Δ BPNʹ ~ Δ BQC (by AAA-similarity, since PNʹ || QC); so

BP: BQ = BNʹ: BC = 1: 2.

Hence Nʹ = n is the midpoint of BC, MN || BC, and MN has length

$\begin{array}{cc}& a+\genfrac{}{}{0.1ex}{}{b-a}{2}=\genfrac{}{}{0.1ex}{}{a+b}{2}.\hfill \end{array}$

(b) Suppose first that a = b. Then ABCD is a parallelogram (by Problem 159(a)), so the area of ABCD is given by a × d ( “(length of base) × height” ). (by Problem 161). Hence we may suppose that a < b.

Solution 1: Extend AB beyond B to a point X such that BX = DC (so AX has length a + b).

Extend DC beyond C to a point Y such that CY = AB (so DY has length a + b).

Clearly ABCD and YCBX are congruent, so each has area one half of area(AXYD).

Now AX || DY, and AX = DY, so AXYD is a parallelogram (by Problem 159(a)).

Hence AXYD has area “(length of base)× height” (by Problem 161), so ABCD has area $\frac{a+b}{2}$ × d.

Solution 2: [We give a second solution as preparation for Problem 171.]

Now a < B implies that∠BAD +∠ABC is greater than a straight angle.

[Proof. The line through B parallel to AD meets DC at Q, and ABQD is a parallelogram.

Hence DQ = AB, so Q lies between D and C, and

∠ABC =∠ABQ +∠QBC >∠ABQ.

∴ ∠BAD +∠ABC >∠BAD +∠ABQ, which is equal to a straight angle.]

So if we extend DA beyond A, and CB beyond B, the lines meet at X, where X, D are on opposite sides of AB. Then Δ XAB ~ Δ XDC (by AAA-similarity), whence corresponding lengths in the two triangles are in the ratio AB: DC = a: b. In particular, if the perpendicular from X to AB has length h, then $\frac{h}{h+d}=\frac{a}{b}$, so h(b - a) = ad.

Now

area(ABCD) = area(Δ XDC) - area(Δ XAB)

= $$12b(h+d)−12ah

= $\frac{1}{2}bd+\frac{1}{2}h(b-a)$

= $\frac{1}{2}bd+$12ad

= $\left(\frac{a+b}{2}\right)d$.

171. Note: The volume of a pyramid or cone is equal to

$\frac{1}{3}$ × (area of base) ×height.

There is no elementary general proof of this fact. The initial coefficient of $\frac{1}{3}$ arises because we are “adding up” , or integrating, cross-sections parallel to the base, whose area involves a square x², where x is the distance from the apex (just as the coefficient $\frac{1}{2}$ in the formula for the area of a triangle arises because we are integrating linear cross-sections whose size is a multiple of x – the distance from the apex). Special cases of this formula can be checked – for example, by noticing that a cube ABCDEFGH of side s, with base ABCD, and upper surface EFGH, with E above D, F above A, and so on, can be dissected into three identical pyramids – all with apex E: one with base ABCD, one with base BCHG, and one with base ABGF. Hence each pyramid has volume $\frac{1}{3}$s³, which may be interpreted as

$\frac{1}{3}\times (areaofbase=s^2)\times (height=s)$.

To obtain the frustum of height d, a pyramid with height h (say) is cut off a pyramid with height h + d.

∴ volume(frustum) = $\left[\frac{1}{3}\times b^2\times (h+d)\right]-\left[\frac{1}{3}\times a^2\times h\right]$

$=\left[\frac{1}{3}\times b^2\times d\right]+\left[\frac{1}{3}\times (b^2-a^2)\times h\right]$.

Let N be the midpoint of BC, and let the line AN meet the upper square face of the frustum at M.

Let the perpendicular from the apex A to the base BCDE meet the upper face of the frustum at Y and the base BCDE at Z.

ThenΔ AYM ~ Δ AZN (by AAA-similarity), so AY: AZ = YM: ZN.

∴$\frac{h}{h+d}=\frac{a}{b},$soh(b−a)=ad.

∴ $volume(frustum)=$13b2d+13(b2−a2)h

$=$13b2d+13(b+a)ad

$=\frac{1}{3}\left(b^2+ab+a^2\right)d$.

172. Construct the line through A which is parallel to AʹBʹ, and let it meet the line BBʹ at P. Similarly, construct the line through B which is parallel to BʹCʹ and let it meet the line CCʹ at Q.

Then Δ ABP ~ Δ BCQ (by AAA-similarity), so AB: BC = AP: BQ.

Now AAʹBʹP is a parallelogram, so AP = AʹBʹ, and BBʹCʹQ is a parallelogram, so BQ = BʹCʹ.

∴ AB: BC = AʹBʹ: BʹCʹ. QED

173.

(a) Suppose AB has length a and CD has length b. Problem 137 allows us to construct a point X such that AX = CD, so AX has length b. Now construct the point Y where the circle with centre A and radius AX meets BA produced (beyond A). Then YB has length a + b.

If a > B, let Z be the point where the circle with centre A and passing through X meets the segment AB internallyThen ZB has length a - b.

(b) Use Problem 137 to construct a point U (not on the line CD) such that DU = XY, and a point V on DU (possibly extended beyond U) such that DV = AB. Hence DU has length 1 and DV has length a.

Construct the line through V parallel to UC and let it meet the line DC at W. Then Δ DVW ~ Δ DUC, with scale factor DW: DC = DV: DU = a: 1. Hence DW has length ab.

To construct a line segment of length $\frac{a}{b}$ construct U, V as above. Then let the circle with centre D and radius DU meet CD at Uʹ. Let the line through Uʹ parallel to CV meet DV at X. The Δ DUʹX ~ Δ DCV, with scale factor DX: DV = DUʹ: DC = 1: B, so DX has length $\frac{a}{b}$.

(c) Use Problem 137 to construct a point G so that AG = XY = 1. Then draw the circle with centre A passing through G to meet BA extended at H. Hence HA = 1, AB = a.

Construct the midpoint M of HB; draw the circle with centre M and passing through H and B.

Construct the perpendicular to HB at the point A to meet the circle at K.

Then Δ HAK ~ Δ KAB, so AB: AK = AK: AH. Hence AK = $\sqrt{a}$.

174. The key to this problem is to use Problem 158: a parallelogram(and hence any rectangle) is divided into two congruent pieces by any straight cut through the centre. If A is the centre of the rectangular piece of fruitcake, and B is the centre of the combined rectangle consisting of “fruitcake plus icing” , then the line AB gives a straight cut that divides both the fruitcake and the icing exactly in two.

175.

(a) The minute hand is pointing exactly at “7” , but the hour hand has moved $\frac{7}{12}$ of the way from “1” to “2” : that is, $\frac{7}{12}$ of 30°, or 17$\frac{1}{2}$°. Hence the angle between the hands is 162$\frac{1}{2}$°.

The same angle arises whenever the hands are trying to point in opposite directions, but are off-set by $\frac{7}{12}$ of 30°, or 17$\frac{1}{2}$°. This suggests that instead of “35 minutes after 1” we should consider “35 minutes before 11” , or 10:25.

(b) The two hands coincide at midnight. The minute hand then races ahead, and the hands do not coincide again until shortly after 1:00 – and indeed after 1:05.

More precisely, in 60 minutes, the minute hand turns through 360°, so in x minutes, it turns through 6x°. In 60 minutes the hour hand turns through 30°, so in x minutes, the hour hand turns through $\frac{1}{2}$x°.

The hands overlap when

$$$$12x ≡6x (mod360),

or when $\frac{11}{2}$x is a multiple of 360.

This occurs when X = 0 (i.e. at midnight); then not until

X = $\frac{720}{11}=65\frac{5}{11}$,

and the time is 5$\frac{5}{11}$ minutes past 1: that is, after 1$\frac{1}{11}$ hours.

It occurs again 1$\frac{1}{11}$ hours, or 65$\frac{5}{11}$ minutes, later – namely at 10$\frac{10}{11}$ minutes past 2, and so on.

Hence the phenomenon occurs at midnight, and then 11 more times until noon (with noon as the 12th time; and then 11 more times until midnight – and hence 23 times in all (including both midnight occurrences).

(c) If we add a third hand (the ‘second hand'), all three hands coincide at midnight. In x minutes, the second hand turns through 360x°.

We now know exactly when the hour hand and minute hand coincide, so we can check where the second hand is at these times. For example, at 5$\frac{5}{11}$ minutes past 1, the second hand has turned through (360 × 65$\frac{5}{11}$)°, and

360 × 65$\frac{5}{11}$ ≡ 360 × $\frac{5}{11}($mod360),

so the second hand is nowhere near the other two hands.

The k^th occasion when the hour hand and minute hands coincide occurs at k × 1$\frac{1}{11}$ hours after midnight, when the two hands point in a direction $(\frac{}{}$360k11)° clockwise beyond “12” . At the same time, the second hand has turned through (360 × 65$\frac{5}{11}$ × k)°, and

$360\times 65\frac{5}{11}\times k\equiv 360\times \frac{5}{11}\times k($mod360),

or five times as far round, and these two are never equal $($mod360).

176.

Note: One of the things that makes it possible to calculate distances exactly here is that the angles are all known exactly, and give rise to lots ofright angled triangles.

The rotational symmetry of the clockface means we only have to consider segments with one endpoint at 12 o'clock (say A). The reflectional symmetry in the line AG means that we only have to find AB, AC, AD, AE, AF, and AG.

Clearly AG = 2. If O denotes the centre of the clockface, then AD is the hypotenuse of an isosceles triangle Δ OAD with legs of length 1, so AD = $\sqrt{2}$.

Δ ACO is isosceles (OA = OC = 1) with apex angle∠AOC = 60°, so the triangle is equilateral. Hence AC = OA = 1.

It follows that ACEGIK is a regular hexagon, so AE = $\sqrt{3}$ (if OC meets AE at X, then Δ ACX is a 30-60-90 triangle, and so is half of an equilateral triangle, whence AX = $\frac{\sqrt{3}}{2})$.

It remains to find AB and AF.

Let OB meet AC at Y. Then OY = $\frac{\sqrt{3}}{2}$ (since Δ OYA is a 30-60-90 triangle).

∴ BY= 1 - $\frac{\sqrt{3}}{2}$, AY = $\frac{1}{2}$, so AB = $\sqrt{2-\sqrt{3}}$.

Finally, AB subtends∠AOB = 30° at the centre, whence∠OAB =∠OBA = 75° and∠AGB = 15°.

∴ ∠ABG = 90° so we may apply Pythagoras’ Theorem to Δ ABG to find BG =AF = $\sqrt{2+\sqrt{3}}$.

177.

$\sqrt{1}$ is the distance from (0, 0, 0) to (1, 0, 0).

$\sqrt{2}$ is the distance from (0, 0, 0) to (1, 1, 0).

$\sqrt{3}$ is the distance from (0, 0, 0) to (1, 1, 1).

$\sqrt{4}$ is the distance from (0, 0, 0) to (2, 0, 0).

$\sqrt{5}$ is the distance from (0, 0, 0) to (2, 1, 0).

$\sqrt{6}$ is the distance from (0, 0, 0) to (2, 1, 1).

$\sqrt{7}$ cannot be realized as a distance between integer lattice points in 3D.

$\sqrt{8}$ is the distance from (0, 0, 0) to (2, 2, 0).

$\sqrt{9}$ is the distance from (0, 0, 0) to (3, 0, 0).

$\sqrt{10}$ is the distance from (0, 0, 0) to (3, 1, 0).

$\sqrt{11}$ is the distance from (0, 0, 0) to (3, 1, 1).

$\sqrt{12}$ is the distance from (0, 0, 0) to (2, 2, 2).

$\sqrt{13}$ is the distance from (0, 0, 0) to (3, 2, 0).

$\sqrt{14}$ is the distance from (0, 0, 0) to (3, 2, 1).

$\sqrt{15}$ cannot be realized as a distance between integer lattice points in 3D.

$\sqrt{16}$ is the distance from (0, 0, 0) to (4, 0, 0).

$\sqrt{17}$ is the distance from (0, 0, 0) to (4, 1, 0).

Note: The underlying question extends Problem 32:

Which integers can be represented as a sum of three squares?

This question was answered by Legendre (1752–1833):

Theorem. A positive integer can be represented as a sum of three squares if and only if it is not of the form 4^a(8b + 7).

178.

(a) Let AD and CE cross at X. Join DE. Then∠AXC =∠EXD (vertically opposite angles). Hence∠CAD +∠ACE =∠ADE +∠CED, so the five angles of the pentagonal star have the same sum as the angles of Δ BED. Hence the five angles have sum $\pi$ radians.

(b) Start with seven vertices A, B, C, D, E, F, G arranged in cyclic order.

(i) Consider first the 7-gonal star ADGCFBE. Let GD and BE cross at X and let GC and BF cross at Y. Join DE, BG. As in part (a),

∠BGC +∠GBF =∠BFC +∠GCF,

and

∠BGD +∠GBE =∠BED +∠GDE,

so the angles in the 7-gonal star have the same sum as the angles in Δ ADE. Hence the seven angles have sum $\pi$ radians.

(ii) Similar considerations with the 7-gonal star ACEGBDF show that its seven angles have sum 3$\pi$ radians.

Note: Notice that the three possible “stars” (including the polygon ABCDEFG) have angle sums $\pi$ radians, 3$\pi$ radians, and 5$\pi$ radians.

(c) if n = 2k + 1 is prime, we may join A to its immediate neighbour B (1-step), or to its second neighbor C (2-step), … , or to its k^th neighbour (k-step), so there are k different stars, with angle sums

$$\begin{array}{cc}& (n-2)\pi ,(n-4)\pi ,(n-6)\pi ,\dots \mathrm{,3}\pi ,\pi \hfill \end{array}$$

respectively.

If n is not prime, the situation is slightly more complicated, since, for each divisor m of n, the km-step stars break up into separate components.

179.

(a) (i) Let the other three pentagons be CDRST, DEUVW, EAXYZ.

At A we have three angles of 108°, so∠NAX = 36°.

Δ ANX is isosceles (AN = AB = AE = AX), so ∠AXN = 72°.

Hence∠AXY +∠AXN = 180°, so Y, X, N lie in a straight line.

Similarly M, N, X lie in a straight line.

Hence M, N, X, Y lie on a straight line segment MY of length 1 + YN = 2 + NX.

(ii) In the same way it follows that MP passes through L and Q, that PS passes through O and T, etc. so that the figure fits snugly inside the pentagon MPSVY, whose angles are all equal to 108°. Moreover, Δ ANX ≡ Δ BQL (by SAS-congruence), so XN = LQ, whence MP = YM. Similarly MP = PS = SV = VY, so MPSVY is a regular pentagon.

(iii) In the regular pentagon EAXYZ the diagonal EY || AX.

Moreover XAC is a straight line, and ∠ACB +∠NBC = 180°, so AC || NB.

Hence YE || NB, and YE = NB = $(n-2)\pi ,(n-4)\pi ,(n-6)\pi ,\dots ,3\pi ,\pi$ (the Golden Ratio $\tau$), so YEBN is a parallelogram.

Hence YN = EB = $\frac{1+\sqrt{5}}{2}$, so YM = 1 + ${\tau }^2$ .

(b)(i)

Δ MPY ≡ Δ PSM ≡ Δ SVP ≡ Δ VYS ≡ Δ YMV

(by SAS-congruence), so

YP = MS = PV = SY = VM

Also∠PMS =∠MPY = 36°;

∴ Δ BMP has equal base angles, and so is isosceles.

Hence BM =BP, and∠MBP = 108° =∠ABC.

Similarly∠AMY =∠AYM = 36°.

∴ Δ AMY ≡ Δ BPM (by ASA-congruence), so

AY = AM = BM = BP.

Δ MAB ≡ Δ PBC – by ASA-congruence:

∠BPC = 108° -∠BPM -∠CPS = 36° =∠AMB,

MA = PB, and ∠PBC =∠MBA =∠MAB.

Hence AB = BC.

Continuing round the figure we see that

AB = BC = CD = DE = EA

and that

∠A =∠B =∠C =∠D =∠E.

(ii) Extend DB to meet MP at L, and extend DA to meet MY at N. Then 36° =∠DBC =∠LBM (vertically opposite angles). Hence Δ LBM has equal base angles and so is isosceles: LM = LB. Similarly NM = NA. Now Δ LBM ≡Δ NAM (by ASA-congruence, since MA = MB), so LM = LB = NA = NM.

In the regular pentagon ABCDE we know that∠DBC = 36°; and in Δ LBM, ∠LBM =∠DBC (vertically opposite angles), so ∠MLB = 108°. Hence∠ BLP = 72° =∠LBP, so Δ PLB is isosceles: PL = PB. In the regular pentagon MPSVY, Δ PMA is isosceles, so PM = PA. ∴ LM = PM -PL = PA - PB = BA.

Hence ABLMN is a pentagon with five equal sides. It is easy to check that the five angles are all equal.

(iii) We saw in (i) that the five diagonals of MPSVY are equal. We showed in Problem 3 that each has length $\tau ={\tau }^2$, and that MPDY is a rhombus, so DY = PM = 1. Similarly SE = 1.

Hence SD = $\tau$- 1, and DE = SE - SD = 2 - $\tau$ = ${\left({\tau }^2\right)}^{-1}$ .

180.

(a) Since the tiles fit together “edge-to-edge” , all tiles have the same edge length. The number k of tiles meeting at each vertex must be at least 3 (since the angle at each vertex of a regular n-gon =$\left(1-\frac{2}{n}\right)\pi \text{<}\pi )$, and can be at most 6 (since the smallest possible angle in a regular n-gon occurs when n = 3, and is then $\frac{\pi }{3}$ ).

We consider each possible value of k in turn.

– If k = 6, then n = 3 and we have six equilateral triangles at each vertex.

– If k = 5, then we would have $\left(1-\frac{2}{n}\right)\pi =\frac{}{}$2π5,son=103 is not an integer.

– If k = 4, then n = 4 and we have four squares at each vertex.

– If k = 3, then n = 6 and we have three regular hexagons at each vertex.

Hence n = 3, or 4, or 6.

(b)(i) If k = n = 4, it is easy to form the vertex figure. It may seem ‘obvious’ that it continues “to infinity” ; but if we think it is obvious, then we should explain why: (choose the scale so that the common edge length is “1” ; then let the integer lattice points be the vertices of the tiling, with the tiles as translations of the unit square formed by (0, 0), (1, 0), (1, 1), (0, 1)).

(ii) If k = 6, n = 3, let the vertices correspond to the complex numbers p + q$\omega$ , where p, q are integers, and where $\omega$ is a complex cube root of 1

(that is, a solution of the equation $\omega$³ = 1 ≠ $\omega$, or $\omega$² + $\omega$ + 1 = 0), with the edges being the line segments of length 1 joining nearest neighbours (at distance 1).

(iii) If k = 3, n = 6, take the same vertices as in (ii), but eliminate all those for which p + q ≡ 0 (mod 3), then let the edges be the line segments of length 1 joining nearest neighbours.

181.

(a)(i) As before, the number k of tiles at each vertex lies between 3 and 6. However, this time k does not determine the shape of the tiles. Hence we introduce a new parameter: the number of t of triangles at each vertex, which can range from 0 up to 6. The derivations are based on elementary arithmetic, for which it is easier to work with angles in degrees.

* If t = 6, then the vertex figure must be 3⁶.

* If t = 5, the remaining gap of 60° could only take a sixth triangle, so this case cannot occur.

* If t = 4, we are left with angle of 120°, so the only possible vertex figure is 3⁴.6.

* If t = 3, then we are left with an angle of 180°, so the only possible configurations are 3³.4² (with the two squares together), or 3².4.3.4 (with the two squares separated by a triangle).

* If t = 2, we are left with an angle of 240°, which cannot be filled with 3 or more tiles (since the average angle size would then be at most 80°, and no more triangles are allowed), so the only possible vertex figures are 3².4.12, 3.4.3.12, 3².6², 3.6.3.6 (since 3².5.n or 3.5.3.n would require a regular n-gon with an angle of 132°, which is impossible).

Note: The compactness of the argument based on the parameter t is about to end. We continue the same apprOACh, with the focus shifting from the parameter tto a new parameter s– namely the number of squares in each vertex figure.

* Suppose t= 1. We are left with an angle of 300°.

If s = 0, the 300° cannot be filled with 3 or more tiles (since then the average angle size would be at most 100°, and no squares can be used), so there are exactly two additional tiles. Since each tile has angle $\le$ 180°, we cannot use a hexagon, so the smallest tile has at least 7 sides; and since the average of the two remaining angle sizes is 150°, the smallest tile has at most 12 sides. It is now easy to check that the only possible vertex figures are 3.7.42, 3.8.24, 3.9.20, 3.10.15, 3.12².

If s = 1, we would be left with an angle of 210°, which would require two larger tiles with average angle size 105°, which is impossible.

If s = 2, the only possible vertex figures are 3.4².6, or3.4.6.4.

Clearly we cannot have s = 3 (or we would be left with a gap of 30°); and t = 1, s > 3 is also impossible.

* Hence we may assume that t = 0, in which case s is at most 4.

If s = 4, then the vertex figure is 4⁴.

If s = 3, then the remaining gap could only take a fourth square, so this case does not occur.

If s = 2, we are left with an angle of 180°, which cannot be filled.

If s = 1, we are left with an angle of 270°, so there must be exactly two additional tiles and the only possible vertex figures are 4.5.20, 4.6.12, or 4.8² (since a regular 7-gon would leave an angle of 141${\frac{3}{7}}^{\circ }$ ).

* Hence we may assume that t = s = 0, and proceed using the parameter f– namely the number of regular pentagons. Clearly f is at most 3, and cannot equal 3 (or we would leave an angle of 36°).

If F = 2, we are left with an angle of 144°, so the only vertex figure is 5².10.

If F = 1, we are left with an angle of 252°, which requires exactly two further tiles, whose average angle is 126°; but this forces us to use a hexagon – leaving an angle of 132°, which is impossible.

* Hence we may assume that t = s = F = 0. So the smallest possible tile is a hexagon, and since we need at least 3 tiles at each vertex, the only possible vertex figure is 6³.

Hence, the simple minded necessary condition (namely that the vertex figure should have no gaps) gives rise to a list of twenty-one possible vertex figures:

3⁶, 3⁴.6, 3³.4², 3².4.3.4, 3².4.12, 3.4.3.12, 3².6², 3.6.3.6, 3.7.42, 3.8.24, 3.9.20, 3.10.15, 3.12², 3.4².6, 3.4.6.4, 4⁴, 4.5.20, 4.6.12, 4.8², 5².10, 6³.

(ii) Lemma. The vertex figures 3².4.12, 3.4.3.12, 3².6², 3.7.42, 3.8.24, 3.9.20, 3.10.15, 3.4².6, 4.5.20, 5².10 do not extend to semi-regular tilings of the plane.

Proof. Suppose to the contrary that any of these vertex figures could be realized by a semi-regular tiling of the plane. Choose a vertex B and consider the tiles around vertex B.

In the first eight of the listed vertex figures we may choose a triangle T = ABC, which is adjacent to polygons of different sizes on the edges BA (say) and BC.

In the two remaining vertex figures, there is a face T = ABC·swith an odd number of edges, which has the same property – namely that of being adjacent to an a-gon on the edge BA (say) and a b-gon on the edge BC with a ≠ b.(For example, in the vertex figure 3².4.12, T = ABC is a triangle, and the faces on BA and on BC are – in some order – a 3-gon and a 4-gon, or a 3-gon and a 12-gon.)

In each case let the face T be a p-gon.

If the face adjacent to T on the edge BA is an a-gon, and that on edge BC is a b-gon, then the vertex figure symbol must include the sequence “$\dots$ a.p.b$\dots$” .

If we now switch attention from vertex B to the vertex A, then we know that A has the same vertex figure, so must include the sequence “$\dots$ a.p.b$\dots$” , so the face adjacent to the other edge of T at A must be a b-gon. As one traces round the edges of the face T, the faces adjacent to

T are alternately a-gons and b-gons – contradicting the fact that T has an odd number of edges.

Hence none of the listed vertex figures extends to a semi-regular tiling of the plane. QED

(b) It transpires that the remaining eleven vertex figures

3⁶, 3⁴.6, 3³.4², 3².4.3.4, 3.6.3.6, 3.12², 3.4.6.4, 4⁴, 4.6.12, 4.8², 6³

can all be realized as semi-regular tilings(and one– namely 3⁴.6 – can be realized in two different ways, one being a reflection of the other).

In the spirit of Problem 180(b), one should want to do better than to produce plausible pictures of such tilings, by specifying each one in some canonical way. We leave this challenge to the reader.

182. [We construct a regular hexagon, and take alternate vertices.]

Draw the circle with centre O passing through A. The circle with centre A passing through O meets the circle again at X and Y.

The circle with centre X and passing through A and O meets the circle again at B; and the circle with centre Y and passing through A and O, meets the circle again at C.

Then Δ AOX, Δ AOY, Δ XOB, Δ YOC are equilateral, so∠AXB = 120° =∠AYC, and ∠XAB = 30° =∠YAC.

Hence Δ AXB ≡ Δ AYC, so AB = AC.Δ ABC is isosceles so∠B =∠C, with apex angle∠BAC =∠XAY -∠XAB -∠YAC = 60°; hence Δ ABC is equiangular and so equilateral.

183.

(a) Draw the circle with centre O and passing through A.

Extend AO beyond O to meet the circle again at C.

Construct the perpendicular bisector of AC, and let this meet the circle at B and at D.

Then BA = BC (since the perpendicular bisector of AC is the locus of points equidistant from A and from C); similarly DA = DC.

Δ OAB and Δ OCB are both isosceles right angled triangles, so ∠ABC is a right angle (or appeal to “the anglesubtended on the circle by the diameter AC” ). Similarly ∠A, ∠C, ∠D are right angles, so ABCD is a rectangle with BA = BC, and hence a square.

Note: This construction starts with the regular 2-gon AC inscribed in its circumcircle, and doubles it to get a regular 4-gon, by constructing the perpendicular bisectors of the “two sides” to meet the circumcircle at B and at D.

(b) Erect the perpendiculars to AB at A and at B.

Then draw the circles with centre A and passing through B, and with centre B and passing through A.

Let these circles meet the perpendiculars to AB (on the same side of AB) at D and at C.

Then AD = BC, and AD || BC, so ABCD is a parallelogram (by Problem 159(a)), and hence a (being a parallelogram with a right angle), and so a square (since AB = AD).

184.

(a)(i) First construct the regular 3-gon ACE with circumcentre O. Then construct the perpendicular bisectors of the three sides AC, CE, EA, and let these meet the circumcircle at B, D, F.

Note: Here we emphasise the general step from inscribed regular n-gon to inscribed regular 2n-gon – even though this may seem perverse in the case of a regular 3-gon (since we constructed the inscribed regular 3-gon in Problem 182 by first constructing the regular 6-gon and then taking alternate vertices).

(ii) First construct the regular 4-gon ACEG with circumcentre O. Then construct the perpendicular bisectors of the four sides AC, CE, EG, GA, and let these meet the circumcircle at B, D, F, H.

(b)(i) Construct an equilateral triangle ABO.

Then draw the circle with centre O and passing through A and B. Then proceed as in 182.

(ii) Extend AB beyond B, and let this line meet the circle with centre B and passing through A at X.

Now construct a square BXYZ on the side BX as in 183(a), and let the diagonal BY meet the circle at C.

Construct the circumcentre O of Δ ABC (the point where the perpendicular bisectors of AB, BC meet).

Construct the next vertex D as the point where the circle with centre O and passing through A meets the circle with centre C and passing through B. The remaining points E, F, G, H can be found in a similar way.

185.

(a)(i) There are various ways of doing this – none of a kind that most of us might stumble upon. Draw the circumcircle with centre O and passing through A. Extend the line AO beyond O to meet thecircle again at X.

Construct the perpendicular bisector of AX, and let this meet the circle at Y and at Z. Construct the midpoint M of OZ, and join MA.

Let the circle with centre M and passing through A meet the line segment OY at the point F.

Finally let the circle with centre A and passing through F meet the circumcircle at B. Then AB is a side of the required regular 5-gon.(The vertex C on the circumcircle is then obtained as the second meeting point of the circumcircle with the circle having centre B and passing through A. The points D, E can be found in a similar way.)

The proof that this construction does what is claimed is most easily accomplished by calculating lengths.

Let OA = 2. Then OF = $\sqrt{5}$ - 1, so AF = $\sqrt{10-2\sqrt{5}}$ = AB. It remains to prove that this is the correct length for the side of a regular pentagon inscribed in a circle of radius 2. Fortunately the work has already been done, since Δ OAB is isosceles with apex angle equal to 72°. If we drop a perpendicular from O to AB, then we need to check whether it is true that

$$\begin{array}{cc}& \sin 36°=\genfrac{}{}{0.1ex}{}{\sqrt{10-2\sqrt{5}}}{4}.\hfill \end{array}$$

But this was already shown in Problem 3(c).

(ii) To construct a regular 10-gon ABCDEFGHIJ, first construct a regular 5-gon ACEGI with circumcentre O; then construct the perpendicular bisectors of the five sides, and so find B, D, F, H, J as the points where these bisectors meet the circumcircle.

(b)(i) The first move is to construct a line BX through B such that∠ABX = 108°. Fortunately this can be done using part (a), by temporarily treating B as the circumcentre, drawing the circle with centre B and passing through A, and beginning the construction of a regular 5-gon AP·s inscribed in this circle.

Then∠ABP = 72°; so if we extend the line PB beyond B to X, then ∠ABX = 108°. Let BX meet the circle with centre B through A at the point C. Then BA = BC and ∠ABC = 108°, so we are up and running.

If we let the perpendicular bisectorsof AB and BC meet at O, then the circle with centre O and passing through A also passes through B and C (and the yet to be lOBCted points D and E). The circle with centre C and passing through B meets this circle again at D; and the circle with centre A and passing through B meets the circle again at E.

(ii) To construct the regular 10-gon ABCDEFGHIJ, treat B as the point O in (a)(i) and construct a regular 5-gon AXCYZ inscribed in the circle with centre B and passing through A.

Then∠ABC = 144°, and BA = BC, so C is the next vertex of the required regular 10-gon. We may now proceed as in (a)(ii) to first construct the circumcentre O of the required regular 10-gon as the point where the perpendicular bisectors of AB and BC meet, then draw the circumcircle, and finally step off successive vertices D, E, $$\text{.}\text{.}\text{.}$$ of the 10-gon around the circumcircle.

186. The number k of faces meeting at each vertex can be at most five (since more would produce an angle sum that is too large). And k ≥ 3 (in order to create a genuine corner.

• If k = 5, then the vertex figure must be 3⁵ (or the angle sum would be>360°).
• If k = 4, then the vertex figure must be 3⁴ (or the angle sum would be too large).
• If k = 3, the angle in each of the regular polygons must be<120°, so the only possible vertex figures are 5³, 4³, and 3³.

187. The respective midpoints have coordinates:

of AB: $\left(\frac{1}{2},\frac{1}{2},0\right)$; of AC: $\left(\frac{1}{2},0,\frac{1}{2}\right)$; of AD: $\left(1,\frac{1}{2},\frac{1}{2}\right)$; of BC: $\left(0,\frac{1}{2},\frac{1}{2}\right)$; of BD: $\left(\frac{1}{2},1,\frac{1}{2}\right)$ ; of CD: $\left(\frac{1}{2},\frac{1}{2},1\right)$ .

∴ PQ = $\frac{1}{\sqrt{2}}$ = PR = PS = PT = QR = RS = ST = TQ.

188.

(a) There are infinitely many planes through the apex A and the base vertex B. Among these planes, the one perpendicular to the base BCD is the one that passes through the midpoint M of CD. Let the perpendicular from A to the base, meet the base BCD at the point X, which must lie on BM. Let AX have length h. To find h we calculate the area of Δ ABM in two ways.

First, BM has length $\sqrt{3}$, so area(Δ ABM) = $\frac{1}{2}\left(\sqrt{3}\times h\right)$

Second, Δ ABM is isosceles with base AB and apex M, so has height $\sqrt{2}$.

∴ area(Δ ABM)= $\frac{1}{2}(2\times \sqrt{2})$ = $\sqrt{2}$.

If we now equate the two expressions for area(Δ ABM), we see that h = $\sqrt{\frac{8}{3}}$.

(b)(i) When constructing a regular octahedron (whether using card, or tiles such as Polydron®) one begins by arranging four equilateral triangles around a vertex such as A. This ‘vertex figure’ is not rigid: though we know that BC = CD = DE = EB, there is no a priori reason why the four neighbours B, C, D, E of A should form a square, or a rhombus, or a planar quadrilateral. We show that these four neighbours lie in a single plane:B, C, D, E are all distance 2 from A and distance 2 from F, so (by Problem 144) they must all lie in the plane perpendicular to the line AF and passing through the midpoint X of AF. Moreover, Δ ABX ≡ Δ ACX (by RHS-congruence, since AB = AC = 2, AX = AX, and∠AXB =∠AXC are both right angles). Hence XB = XC, so B, C lie on a circle in this plane with centre X. Similarly Δ ABX ≡ Δ ADX ≡ Δ AEX, so BCDE is a cyclic quadrilateral (and a rhombus), and hence a square – with X as the midpoint of both BD and CE.

(ii) Let M be the midpoint of BC and N the midpoint of DE.

Then NM and AF cross at X and so define a single plane. In this plane, Δ ANM ≡ Δ FMN (by SSS-congruence, since AN= FM = $\sqrt{3}$, NM = MN, MA = NF = $\sqrt{3}$); hence ∠ANM =∠FMN, so AN || MF.

Similarly, if P is the midpoint of AE and Q is the midpoint of FC, then Δ DPQ ≡ Δ BQP, so DP || QB. Hence the top face DEA is parallel to the bottom face BCF, so the height of the octahedron sitting on the table is equal to the height of Δ FMN. But this triangle has sides of lengths 2, $\sqrt{3}$, $\sqrt{3}$, so this height is exactly the same as the height h in part (a).

189.

(i) Let L = $\left(\frac{1}{2},0,0\right)$andM= $\left(\frac{1}{2},\frac{1}{2},0\right)$. Then L lies on the line ST and M is the midpoint of NP.

Δ LSM is a right angled trianglewith legs of length LS = A, LM = $\frac{1}{2}$, so MS = $\sqrt{a^2+\frac{1}{4}}$.

Δ MNS is a right angled trianglewith legs of length MN = $\frac{1}{2}$ - A, MS = $\sqrt{a^2+\frac{1}{4}}$.

HenceNS = $\sqrt{2a^2-a+\frac{1}{2}}$.

Similarly NU = $\sqrt{2a2-a+\frac{1}{2}}$.

Let Lʹ = $0,\frac{1}{2}$, 0\right) and Mʹ = $\left(0,\frac{1}{2},\frac{1}{2}\right)$. Then Mʹ is the midpoint of WX and Lʹ lies immediately below Mʹ on the line joining (0, 0, 0) to (0, 1, 0). In the right angled triangleΔ LʹNMʹ we find NMʹ = $\sqrt{a^2+\frac{1}{4}}$, and in the right angled triangle Δ MʹWN we then find NW = $\sqrt{2a^2-a+\frac{1}{2}}$. Similarly NX= $\sqrt{2a^2-a+\frac{1}{2}}$.

(ii) NP = NS precisely when 1 - 2a = $\sqrt{2a^2-a+\frac{1}{2}}>0$; that is, whena = $\frac{3-\sqrt{5}}{4}$, so all edges of the polyhedron have length$\frac{\sqrt{5}-1}{2}=\tau -1$.

Note: The rectangle NPRQ is a “1 by $\tau -1$” rectangle, and 1: $\tau$ - 1 = $\tau$: 1. Hence the regular icosahedron can be constructed from three congruent, and pairwise perpendicular, copies of a “Golden rectangle” .

190. We mimic the classification of possible vertex figures for semi-regular tilings.

We are assuming that the angles meeting at each vertex add to<360°, so the number k of faces at each vertex lies between 3 and 5. Because faces are regular, but not necessarily congruent, k does not determine the shape of the faces. Hence we let t denote the number of triangles at each vertex, which can range from 0 up to 5.

• If t = 5, then the vertex figure must be 3⁵.
• If t = 4, the remaining polygons have angle sum<120°, so the possible vertex figures are 3⁴, 3⁴.4, and 3⁴.5.
• If t = 3, then the remaining polygons have angle sum<180°, so the only possible vertex figures are 3³, and 3³.n (for any n >3).
• If t = 2, then the remaining polygons have angle sum< 240°, so there are at most 2 additional faces in the vertex figure (since if there were 3 or more extra faces, the average angle size would then be at most 80°, with no more triangles allowed). If there is just 1 additional face, we get the vertex figure 3².n for any n>3. So we may assume that there are 2 additional faces – the smallest of which must then be a 4-gon or a 5-gon.

If the next smallest face is a 4-gon, then we get the possible vertex figures

3².4² and 3.4.3.4, 3².4.5 and 3.4.3.5, 3².4.6 and 3.4.3.6, 3².4.7 and 3.4.3.7, 3².4.8 and 3.4.3.8, 3².4.9 and 3.4.3.9, 3².4.10 and 3.4.3.10, 3².4.11 and 3.4.3.11.

If the next smallest face is a 5-gon, then we get the possible vertex figures

3².5² and 3.5.3.5, 3².5.6 and 3.5.3.6, 3².5.7 and 3.5.3.7.

Note: Before proceeding further it is worth deciding which among the putative vertex figures identified so far seem to correspond to semi-regular polyhedra – and then to prove that these observations are correct.

• The vertex figure 3⁵ corresponds to the regular icosahedron.
• The vertex figure 3⁴ corresponds to the regular octahedron; 3⁴.4 corresponds to the snubcube; 3⁴.5 corresponds to the snub dodecahedron – which comes in left-handed and right-handed forms.
• The vertex figure 3³ corresponds to the regular tetrahedron, and 3³.n (for any n>3) corresponds to the n-gonal antiprism.
• The vertex figure 3².n for any n > 3 does not seem to arise.
• The vertex figure 3².4² does not seem to arise; 3.4.3.4 corresponds to the cuboctahedron; 3².4.5 and 3.4.3.5, 3².4.6 and 3.4.3.6, 3².4.7 and 3.4.3.7, 3².4.8 and 3.4.3.8, 3².4.9 and 3.4.3.9, 3².4.10 and 3.4.3.10, 3².4.11 and 3.4.3.11 do not seem to arise.
• The vertex figure 3².5² does not seem to arise, whereas 3.5.3.5 corresponds to the icosidodecahedron; 3².5.6 and 3.5.3.6, 3².5.7 and 3.5.3.7 do not seem to arise.

To avoid further proliferation of spurious ‘putative vertex figures’ we inject a version of the Lemma used for tilings somewhat earlier than we did fortilings, and then apply the underlying idea to eliminate other spurious possibilities as they arise.

Lemma. The vertex figures

3².4², 3².n (n>3), 3².4.5, 3.4.3.5, 3².4.6, 3.4.3.6, 3².4.7, 3.4.3.7, 3².4.8, 3.4.3.8, 3².4.9, 3.4.3.9, 3².4.10, 3.4.3.10, 3².4.11, 3.4.3.11, 3².5², 3².5.6, 3.5.3.6, 3².5.7, 3.5.3.7

do not arise as vertex figures of any semi-regular polyhedron.

Proof outline. Each of these requires that the vertex figure of any vertex B includes a tile T = ABC·swith an odd number of edges, for which the edge BA is adjacent to an a-gon, the edge BC is adjacent to a b-gon (where a ≠ b), and where the subsequent faces adjacent to T are forced to alternate –a-gon, b-gon, a-gon, $\dots$ – which is impossible. QED

For the rest we introduce the additional parameter s to denote the number of 4-gons in the vertex figure.

Suppose t= 1. Then the remaining polygons have angle sum< 300°, so there are at most 3 additional faces in the vertex figure (since if there were 4 or more extra faces, the average angle size would be<75°, with no more triangles allowed).

If there are 3 additional faces, the average angle size is<100°, so s > 0.

• If s > 1, then the possible vertex figures are 3.4³ (which corresponds to the rhombicuboctahedron), 3.4² (which corresponds to the 3-gonal prism), 3.4².5 (which is impossible as in the Lemma) and 3.4.5.4 (which corresponds to the small rhombicosidodecahedron).
• If s = 1, then the remaining faces have angle sum<210°, so there can only be one additional face, and every 3.4.n (n > 4) is impossible as in the Lemma.
• If s = 0, then the remaining faces have angle sum < 300°, so there are exactly two other faces and the smallest face has < 12 edges, so the only possible vertex figures are

  – 3.5.n (4 < n), which is impossible as in the Lemma;

  – 3.6², which corresponds to the truncated tetrahedron;

  – 3.6.n (6 < n), which is impossible as in the Lemma;

  – 3.7.n (6 < n), which is impossible as in the Lemma;

  – 3.8², which corresponds to the truncated cube;

  – 3.8.n (8 < n), which is impossible as in the Lemma;

  – 3.9.n (8 < n), which is impossible as in the Lemma;

  – 3.10², which corresponds to the truncated dodecahedron;

  – 3.10.n (10 < n), which is impossible as in the Lemma;

  – 3.11.n (10 < n), which is impossible as in the Lemma.

Thus we may assume that t = 0 – in which case, s<4.

• If s = 3, then the only possible vertex figure is 4³, which corresponds to the cube.
• If s = 2, then the remaining faces have angle sum<180°, so there is exactly one additional face, and every 4².n (n>4) corresponds to the n-gonal prism.
• If s = 1, then the remaining faces have angle sum<270°, so there are at most 2 other faces with the smallest face having<8 edges, so the only possible vertex figures are 4.5.n (for 4<n<20), 4.6², 4.6.n (for 6<n<12), 4.7², 4.7.n (for 7<n<10). Among these 4.5.n is impossible as in the Lemma; 4.6² corresponds to the truncated octahedron; 4.6.n with n odd is impossible as in the Lemma; 4.6.8 corresponds to the great rhombicuboctahedron; 4.6.10 corresponds to the great rhombicosidodecahedron; 4.7.n is impossible as in the Lemma.
• If s = 0, there must be exactly three faces at each vertex, and the smallest must be a 5-gon, so the only possible vertex figures are 5³, which corresponds to the regular dodecahedron; 5².n (for n>5), which is impossible as in the Lemma; 5.6², which corresponds to the truncated icosahedron; or 5.6.7, which is impossible as in the Lemma.

191.

(a)(i)∠AXD = 100°, so∠ADB = 40°. In Δ ABD we then see that∠ABD = 40°, so Δ ABD is isosceles with AB = AD. Δ ABC is isosceles with a base angle∠BAC = 60°, so Δ ABC is equilateral. Hence∠CBD = 20°.

(ii) Δ ABC is equilateral, so AC = AB = AD. Hence Δ ADC is isosceles, so∠ACD = 70° =∠ADC, whence∠BDC = 70° -∠ADB = 30°.

(b)(i) As before ∠AXD = 100°, so∠ADB = 40°.

In Δ ABD we then see that∠ABD = 30°, so Δ ABD is not isosceles (as it was in (a)).

Δ ABC is isosceles, so∠BCA = 70°, whence∠CBD = 10°.

In Δ XCD, ∠CXD = 80°, so∠BDC +∠ACD = 100°, but there is no obvious way of determining the individual summands: ∠BDC and ∠ACD.

(ii) No lengths are specified, so we may choose the length of AC. The point B then lies on the perpendicular bisector of AC, and∠CAB = 70° determines the lOBCtion of B exactly. The line AD makes an angle of 40° with AC, and BD makes an angle of 80° with AC, so the lOBCtion of D is determined. Hence, despite our failure in part (i), the angles are determined.

192.

(i) If P lies on CB, then PC = b cos C, AP = B sin C, and in the right angled triangle Δ APB we have:

c² = (b sin C)² + (a - B cos C)²

= a² + b²(sin² C + cos² C) - 2ab cos C

= a² + b² - 2ab cos C.

If P lies on CB extended beyond B, then PC = B cos C, AP = B sin C as before, and in the right angled triangle Δ APB we have:

c² = (b sin C)² + (b cos C -a)²

= (bsin C)² + (a - bcos C)²

= a² + b²(sin² C + cos² C) - 2abcos C

= a² + b² - 2abcos C.

(ii) If P lies on BC extended beyond C, then PC = bcos ∠ACP= -bcos C, AP = bsin ∠ACP = bsin C, and in the right angled triangle Δ APB we have:

c² = (bsin C)² + (a + PC)²

= (bsin C)² + (a - bcos C)²

= a² + b²(sin² C + cos² C) - 2abcos C

= a² + b² - 2abcos C.

193. There are many ways of doing this – once one knows the Sine Rule and Cosine Rule. If we let ∠BDC = y and ∠ACD = z, then one route leads to the identity cos(z - 10°) = 2 sin 10° · sin z, from which it follows that z = 80°, y = 20°.

194.

(a) The angle between two faces, or two planes, is the angle one sees “end-on” – as one looks along the line of intersection of the two planes. This is equal to the angle between two perpendiculars to the line of intersection – one in each plane. If M is the midpoint of BC, then Δ ABC is isosceles with apex A, so the median AM is perpendicular to BC; similarly Δ DBC is isosceles with apex D, so the median DM is perpendicular to BC.

Δ MAD is isosceles with apex M, MA = MD = $\sqrt{3}$, AD= 2, so we can use the Cosine Rule to conclude that 2² = 3 + 3 - 2· 3· cos(∠AMD), whence cos(∠AMD) = $\frac{1}{3}$.

(b) cos(∠AMD) = $\frac{1}{3}$, and $\frac{1}{3}$ < $\frac{1}{2}$, so∠AMD > 60°; hence we cannot fit six regular tetrahedra together so as to share an edge.

Since ∠AMD is acute, ∠AMD < 90°, so we can certainly fit four regular tetrahedra together with lots of room to spare.

We can now appeal to “trigonometric tables” , or a calculator, to see that

$\arccos \left(\frac{1}{3}\right)\text{<}1.24$,

that is1.24 radians, so five tetrahedra use up less than 6.2 radians – which is less than 2$\pi$. Hence we can fit five regular tetrahedra together around a common edge with room to spare (but not enough to fit a sixth).

195.

(a) The angle between the faces ABC and FBC is equal to the angle between two perpendiculars to the common edge BC. Since the two triangles are isosceles with the common base BC, it suffices to find the angle between the two mediansAM and FM.

In Problem 188 we saw that BCDE is a square, with sides of length 2. If we switch attention from the opposite pair of vertices A, F to the pair C, E, then the same proof shows that ABFD is a square with sides of length 2. Hence the diagonal AF = 2$\sqrt{2}$.

Now apply the Cosine Rule to Δ AMF to conclude that:

$\left(2\sqrt{2}\right)$² = 3 + 3 - 2· 3 · cos(∠AMF),

so it follows that cos(∠AMF) = -$\frac{1}{3}$.

(b) cos(∠AMF) = -$\frac{1}{3}$ < 0, so∠AMF > 90°; hence we cannot fit four regular octahedra together so as to share a common edge. Moreover, -$\frac{1}{2}$ < -$\frac{1}{3}$, so∠ AMD < 120°; hence we can fit three octahedra together to share an edge with room to spare (but not enough room to fit a fourth).

196. The angle ∠AMD = $\arccos \left(\frac{1}{3}\right)$ in Problem 194 is acute, and the angle∠AMF = $\arccos \left(-\frac{1}{3}\right)$ in Problem 195 is obtuse. Hence these angles are supplementary; so the regular tetrahedron fits exactly into the wedge-shaped hole between the face ABC of the regular octahedron and the table.

197.

(i) AB = $\sqrt{2}$ = BC = CA = AW = BW = CW. Hence the four faces ABC, BCW, CWA, WAB are all equilateral triangles, so the solid is a regular tetrahedron (or, more correctly, the surface of the solid is a regular tetrahedron).

(ii) AC = $\sqrt{2}$ = CD = DA = AX = CX = DX. Hence the four faces ACD, CDX, DXA, XAC are equilateral triangles, so the solid is a regular tetrahedron.

(iii) AD = $\sqrt{2}$ = DE = EA = AY = DY = EY. Hence the four faces ADE, DEY, EYA, YAD are equilateral triangles, so the solid is a regular tetrahedron.

(iv) AE = $\sqrt{2}$ = EB = BA = AZ = EZ = BZ. Hence the four faces AEB, EBZ, BZA, ZAE are equilateral triangles, so the solid is a regular tetrahedron.

(v) We get another four regular tetrahedra– such as FBCP, where P = (1, 1, -1).

(vi) ABCDEF is a regular octahedron (or, more correctly, the surface of the solid is a regular octahedron).

Note: The six vertices (± 1, 0, 0), (0, ± 1, 0), (0, 0, ± 1) span a regular octahedron, with a regular tetrahedron fitting exactly on each face. The resulting compound star-shaped figure is called the stellated octahedron. Johannes Kepler (1571–1630) made an extensive study of polyhedra and this figure is sometimes referred to as Kepler's ‘stella octangula'. It is worth making in order to appreciate the way it appears to consist of two interlocking tetrahedra.

198. Let the unlabeled vertex of the pentagonalface ABW-V be P. In the pentagon ABWPV the edge AB is parallel to the diagonal VW. Hence ABWV is an isosceles trapezium. The sides VA and WB (produced)meet at S in the plane of the pentagon ABWPV.

Δ SAB has equal base angles, so SA = SB.

Hence SV = SW.

Similarly BC || WX, and WB and XC meet at some point Sʹ on the line WB.

Now Δ SʹBC ≡ Δ SAB, so SʹB = SA= SB. Hence Sʹ = S, and the lines VA, WB, XC, YD, ZE all meet at S.

Since VW || AB, we know that Δ SAB ~ Δ SVW, with scale factor

$\tau :1$ = VW: AB = SV: SA.

If SA = x, then X + 1: X = $\tau :1$, so X = $\tau$.

Let M be the midpoint of AB and let O denote the circumcentre of the regular pentagon ABCDE.

Δ OAB is isosceles, so OM is perpendicular to AB, and the required dihedral angle between the two pentagonal faces is ∠OMP.

It turns out to be better to find not the dihedral angle ∠OMP, but its supplement: namely the angle ∠SMO.

From Δ OAM we see that

OA= $\frac{1}{2\sin {36}^{°}}$,

and we know that SA = $\tau =$2cos 36°. One can then check that OS = $\cot$ 36°.

Similarly, in Δ OAM we have

OM = $\frac{\cot {36}^{°}}{2}$.

Hence in Δ OMS we have

tan(∠SMO) = 2.

Hence the required dihedral angle is equal to $\pi -\arctan 2\approx$ 116.56°.

199.

(a)(i) Let M be the midpoint of AC. The angle between the two faces is equal to ∠BMD.

In Δ BMD, we have BM = DM = $\sqrt{3}$, BD = 2$\tau$ = 1 + $\sqrt{5}$. So the Cosine Rule in Δ BMD gives:

$6+2\sqrt{5}=3+3$ - 2· 3· cos(∠BMD),

so cos(∠BMD) = -$\frac{\sqrt{5}}{3}$,

∠$BMD=\arccos \left(-\frac{\sqrt{5}}{3}\right)=\pi -\arccos \left(\frac{\sqrt{5}}{3}\right)\approx {138.19}^{°}$.

(ii) $\frac{\sqrt{5}}{3}>\frac{1}{2}$; hence∠BMD>$\frac{}{}$2π3, so we can fit two copies along a common edge, but not three.

(b)(i) Now let M denote the midpoint of BC, and suppose that OM extended beyond M meets the circumcircle at V. Then BV is an edge of the regular 10-goninscribed in the circumcircle. The circumradius OB of BCDEF (which is equal to the edge length of the inscribed regular hexagon) is

OB= $\frac{1}{\sin {36}^{°}}$;

and ∠MBV = 18°,

so BV= $\frac{1}{\cos {18}^{°}}$.

It is easiest to use the converse of Pythagoras’ Theorem, and to write everything first in terms of cos 36°, then (since $\tau$ $=2\cos 36$°, so $\cos {36}^{°}=\frac{1+\sqrt{5}}{4})$ write everything in terms of $\sqrt{5}$.

If we use sin² 36° = 1 - cos² 36°, and 2cos² 18° - 1 = cos 36° = $\frac{1+\sqrt{5}}{4}$, then

BV² + OB² = ${\left(\frac{1}{\cos {18}^{°}}\right)}^2+\left(\frac{1}{\sin {36}^{°}}\right)$²

= $\frac{8}{5+\sqrt{5}}+\frac{8}{5-\sqrt{5}}$

= $\frac{80}{20}$ = 2²

= AB².

Hence, in the right angled triangle Δ AOB, we must have OA = BV as claimed.

(ii) Miraculously no more work is needed. Let the vertex at the ‘south pole’ be L, and let the pentagon formed by its five neighbours be GHIJK. It helps if we can refer to the circumcircle of BCDEF as the ‘tropic of Cancer', and to the circumcircle of GHIJK as the ‘tropic of Capricorn'.

The pentagon GHIJK is parallel to BCDEF, but the vertices of the southern pentagon have been rotated through $\frac{\pi }{5}$ relative to BCDEF, so that G (say) lies on the circumcircle of the pentagon GHIJK, but sits directly below the midpoint of the minor arc BC. Let X denote the point on the ‘tropic of Capricorn’ which is directly beneath B. Then Δ BXG is a right angled triangle with BG = AB = 2, and XG = BV is equal to the edge length of a regular 10-gon inscribed in the circumcircle of GHIJK. Hence, by the calculation in (i), BX is equal to the edge length of the regular hexagon inscribed in the same circle – which is also equal to the circumradius.

200. A necessary condition for copies of a regular polyhedron to “tile 3D (without gaps or overlaps)” is that an integral number of copies should fit together around an edge. That is, the dihedral angle of the polyhedron should be an exact submultiple of 2$\pi$. Only the cube satisfies this necessary condition.

Moreover, if we take as vertices the points (p, q, r) with integer coordinates p, q, r, and as our regular polyhedra all possible translations of the standard unit cube having opposite corners at (0, 0, 0) and (1, 1, 1), then we see that it is possible to tile 3D using just cubes.

201. The four diameters form the four edges of a square ABCD of edge length 2.

The protruding semicircular segments on the left and right can be cut off and inserted to exactly fill the semicircular indentations above and below.

Hence the composite shape has area exactly equal to 2² = 4 square units.

202.

(a) If the regular n-gon is ACEG·sand the regular 2n-gon is ABCDEFG·s, then AC = s_n = s and AB = s_2n = t. If M is the midpoint of AC, AM= $\frac{s}{2}$ and

MB = $1-\sqrt{(1-\left(\frac{s}{2}\right)=\left(\frac{s}{2}\right)}$².

$$\begin{array}{ccc}\therefore t^2\hfill & ={AB}^2\hfill & \hfill \text{(1)}\\ \hfill & ={\left(\genfrac{}{}{0.1ex}{}{s}{2}\right)}^2+{[1-\sqrt{1-{\left(\genfrac{}{}{0.1ex}{}{s}{2}\right)}^2}]}^2\hfill \\ \hfill & =1+1-2\sqrt{1-{\left(\genfrac{}{}{0.1ex}{}{s}{2}\right)}^2}\hfill \\ \hfill & =2-\sqrt{4-s^2}.\hfill \end{array}$$

(b) Put s = s₂= 2 in (1) to get

t = s₄= $\sqrt{2}$. Then put s = s₄= $\sqrt{2}$ in (1) to get t = s₈= $\sqrt{2-\sqrt{2}}$.

(c) Put t = s₆= 1 in (1) to get s = s₃= $\sqrt{3}$. Then put s = s₆= 1 to get

t = s₁₂= $\sqrt{2-\sqrt{3}}$.

(d) Put s = s₅= $\frac{\sqrt{10-2\sqrt{5}}}{2}$ to get

$t={}_{}$s10=3−52.

203.

(a) (i)

n = 3: p₃ = 3$\sqrt{3}\times r$

n = 4: p₄ = $4\sqrt{2}\times r$

n = 5: p₅ = $\frac{5\sqrt{10-2\sqrt{5}}}{2}\times r$

n = 6: p₆ = 6 $\times r$

n = 8: p₈ = 8$\sqrt{2-\sqrt{2}}\times r$

n = 10: p₁₀ = 5$\sqrt{6-2\sqrt{5}}\times r$

n = 12: p₁₂ = 12$\sqrt{2-\sqrt{3}}\times r$.

(ii)

$$\begin{array}{ccc}{c}_3=5.19\cdots \hfill & <\hfill & c_4=5.65\cdots \hfill \\ \hfill & <\hfill & c_5=5.87\cdots \hfill \\ \hfill & <\hfill & c_6=6\hfill \\ \hfill & <\hfill & c_8=6.12\cdots \hfill \\ \hfill & <\hfill & c_{10}=6.18\cdots \hfill \\ \hfill & <\hfill & c_{12}=6.21\cdots .\hfill \end{array}$$

(b)(i)

n = 3: P₃ = 6$\sqrt{3}\times r$

n = 4: P₄ = 8 $\times r$

n = 5: P₅ = 10$\sqrt{5-2\sqrt{5}}\times r$

n = 6: P₆ = 4$\sqrt{3}\times r$

n = 8: P₈ = 8$\left(2\sqrt{2}-2\right)\times r$

n = 10: P₁₀ = $4\sqrt{25-10\sqrt{5}}\times r$

n = 12: P₁₂ = $12\left(4-2\sqrt{3}\right)\times r$.

(ii)

$$\begin{array}{ccc}{C}_3=10.39\cdots \hfill & >\hfill & C_4=8\hfill \\ \hfill & >\hfill & C_5=7.26\cdots \hfill \\ \hfill & >\hfill & C_6=6.92\cdots \hfill \\ \hfill & >\hfill & C_8=6.62\cdots \hfill \\ \hfill & >\hfill & C_{10}=6.49\cdots \hfill \\ \hfill & >\hfill & C_{12}=6.43\cdots .\hfill \end{array}$$

(c) Let O be the centre of the circle of radius r. Let A, B lie on the circle with∠AOB = 30°.

Let M be the midpoint of AB – so that Δ OAB is isosceles, with apex O and height h = OM.

Let Aʹ lie on OA produced, and let Bʹ lie on OB produced, such that OAʹ = OBʹ and AʹBʹ is tangent to the circle.

Then Δ OAB ~ Δ OAʹBʹ with Δ OAʹBʹ larger than Δ OAB, so the scale factor $\frac{1}{h}=2\sqrt{2-\sqrt{3}}>1$.

Hence ${}_{}$P12=22−3×p12>p12,soC12>c12.

204. (i) $\pi r$ (ii) $$π2r (iii) $\theta r$

205.

(a) (i) Note: This could be a long slog. However we have done much of the work before: when the radius is 1, the most of the required areas were calculated exactly back in Problem 3 and Problem 19.

Alternatively, the area of each of the n sectors is equal to $$12sinθ, where $\theta$ is the angle subtended at the centre of the circle, and the exact values of the required trig functions were also calculated back in Chapter 1.

$a_3=\frac{3\sqrt{3}}{4}\times$ r²

$a_4=2\times$ r²

$a_5=\frac{5}{8}\sqrt{10+2\sqrt{5}}\times$ r²

$a_6=\frac{3\sqrt{3}}{2}\times$ r²

$a_8=2\sqrt{2}\times$ r²

${}_{}$a10=5410−25× r²

${}_{}$a12=3× r²

(ii)

$$\begin{array}{ccc}{d}_3=1.29\cdots \hfill & <\hfill & d_4=2\hfill \\ \hfill & <\hfill & d_5=2.37\cdots \hfill \\ \hfill & <\hfill & d_6=2.59\cdots \hfill \\ \hfill & <\hfill & d_8=2.82\cdots \hfill \\ \hfill & <\hfill & d_{10}=2.93\cdots \hfill \\ \hfill & <\hfill & d_{12}=3.\hfill \end{array}$$

(b)(i) Note: This could also be a long slog. However we have done much of the work before. But notice that, when the radius is 1, the area of each of the n sectors is equal to half the edge length times the height (= 1); so if r=1, then the total area is numerically equal to “half the perimeter P_n” .

$A_3=3\sqrt{3}\times$ r²

$A_4=4\times$ r²

$A_5=5\sqrt{5-2\sqrt{5}}\times$ r²

$A_6=2\sqrt{3}\times$ r²

$A_8=8\left(\sqrt{2}-1\right)\times$ r²

${}_{}$A10=225−105× r²

${}_{}$A12=12(2−3)× r²

(ii)

$$\begin{array}{ccc}{D}_3=5.19\cdots \hfill & >\hfill & D_4=4\hfill \\ \hfill & >\hfill & D_5=3.63\cdots \hfill \\ \hfill & >\hfill & D_6=3.46\cdots \hfill \\ \hfill & >\hfill & D_8=3.31\cdots \hfill \\ \hfill & >\hfill & D_{10}=3.24\cdots \hfill \\ \hfill & >\hfill & D_{12}=3.21\cdots .\hfill \end{array}$$

(c) Let O be the centre of the circle of radius r. Let A, B lie on the circle with∠AOB = 30°.

Let M be the midpoint of AB – so that Δ OAB is isosceles, with apex O and height h = OM.

Let Aʹ lie on OA produced, and let Bʹ lie on OB produced, such that OAʹ = OBʹ and AʹBʹ is tangent to the circle.

Then Δ OAB ~ Δ OAʹBʹ with Δ OAB contained in Δ OAʹBʹ, so the scale factor $\frac{1}{h}=2\sqrt{2-\sqrt{3}}>1$.

Hence 4(2 - $\sqrt{3}$)· $a_{12}=A_{12}>a_{12}$,

so $D_{12}>d_{12}$.

206. The rearranged shape (shown in Figure 9 is an “almost rectangle” , where OAforms an “almost height” and OA = r. Half of the 2n circular arcs such as AB form the upper “width” , and the other half form the “lower width” , so each of these “almost widths” is equal to half the perimeter of the circle –namely $\pi$ r.

Figure 9: Rectification of a circle.

Hence the area of the rearranged shape is very close to

$$r×πr=π r².

207.

(a) Cut along a generator, open up and lay the surface flat to obtain: a 2$\pi$ r by h rectangle and two circular discs of radius r.

Hence the total surface area is

2$\pi$ r²+ 2$\pi$ rh = $$2πr(r+h).

(b) The lateral surface consists of n rectangles, each with dimensions s_n by h (where s_n is the edge length of the regular n-gon), and hence has area P_nh = 2${\Pi }_{n}rh$.

Adding in the twoend discs (each with area $$12Pnr) then gives total surface area $$2Πnr(r+h).

208.

(a) (i) $$12π r²; (ii) $$14π r²; (iii) $\frac{\theta }{2}$r².

(b) The sector has two radii of total length 2r. Hence the circular must have length $(\pi -2)r$, and so subtends an angle $\pi$ - 2 at the centre, so has area $\frac{\pi -2}{2}$r².

209.

(a) Focus first on the sloping surface. If we cut along a “generator” (a straight line segment joining the apex to a point on the circumference of the base), the surface opens up and lays flat to form a sector of a circle of radius l. The outside arc of this sector has length 2$\pi$ r, so the sector angle at the centre is equal to $\frac{r}{l}$· 2$\pi$, and hence its area is $\frac{r}{l}$· $\pi$ l² = $\pi$ rl.

Adding the area of the base gives the total surface area of the cone as

$\pi r(r+l)=\frac{1}{2}$· $$2πr(r+l).

(b) Let M be the midpoint of the edge BC and let l denote the ‘slant height’ AM.

Then the area of the n sloping faces is equal to $\frac{1}{2}{P}_n$· l, while the area of the base is equal to $\frac{1}{2}{P}_n$· r.

Hence the surface area is precisely$\frac{1}{2}{P}_n$$(r+l)$.

210.

(a) If AB is an edge of the inscribed regular n-gon ABCD$\mathrm{...}$, and O is the circumcentre, then∠AOB = $\frac{}{}$2πn, so AB = 2r$\sin \frac{\pi }{n}$. Hence the required ratio is equal to $\frac{}{}$sinπnπn, which tends to 1 as n tends to $\infty$.

(b) If AB is an edge of the circumscribed regular n-gon ABCD$\mathrm{...}$, and O is the circumcentre, then∠AOB = $\frac{}{}$2πn, so AB = 2r$\tan \frac{\pi }{n}$. Hence the required ratio is equal to $\frac{}{}$tanπnπn, which tends to 1 as n tends to $\infty$.

211.

(a) If AB is an edge of the inscribed regular n-gon ABCD$\mathrm{...}$, and O is the circumcentre, then∠AOB = $\frac{}{}$2πn, so area(Δ OAB) = $$12r2sin2πn. Hence the required ratio is equal to $\frac{}{}$sin2πn2πn, which tends to 1 as n tends to $\infty$.

(b) If AB is an edge of the circumscribed regular n-gon ABCD$\mathrm{...}$, and O is the circumcentre, then∠AOB = $\frac{}{}$2πn, so area(Δ OAB) = r²$$tanπn. Hence the required ratio is equal to $\frac{}{}$tanπnπn, which tends to 1 as n tends to $\infty$.

212.

(a) area(P₂) = area(P₁) + area(P₂ - P₁) > area(P₁).

(b) This general result is clearly related to the considerations of the previous section. But it is not clear whether we can really expect to prove it with the tools available. So it has been included partly in the hope that readers might come to appreciate the difficulties inherent in proving such an “obvious” result.

In the end, any attempt to prove it seems to underline the need to use “proof by induction” – for example, on the number of edges of the inner polygon. This method is not formally treated until Chapter 6, but is needed here.

• Suppose the inner polygon P₁ = ABC has just n= 3 edges, and has perimeter p₁.

Draw the line through A parallel to BC, and let it meet the (boundary of the) polygon P₂ at the points U and V. The triangle inequality (Problem 146(c)) guarantees that the length UV is less than or equal to the length of the compound path from U to V along the perimeter of the polygon P₂ (keeping on the opposite side of the line UV from B and C). So, if we cut off the part of P₂ on the side of the line UV opposite to B and C, we obtain a new outer convex polygon P₃, which contains P₁, and whose perimeter is no larger than that of P₂.

Now draw the line through B parallel to AC, and let it meet the boundary of P₃ at points W, X. If we cut off the part of P₃ on the side of the line WX opposite to A and C, we obtain a new outer convex polygon P₄, which contains P₁, and whose perimeter is no larger than that of P₃.

If we now do the same by drawing a line through C parallel to AB, and cut off the appropriate part of P₄, we obtain a final outer convex polygon P₅, which contains the polygon P₁, and whose perimeter is no larger than that of P₄– and hence no larger than that of the original outer polygon P₂.

All three vertices A, B, C of P₁ now lie on the boundary of the outer polygon P₅, so the triangle inequality guarantees that AB is no larger than the length of the compound path along the boundary of P₅ from A to B (staying on the opposite side of the line AB from C). Similarly BC is no larger than the length of the compound path along the boundary of P₅ from B to C; and CA is no larger than the length of the compound path along the boundary of P₅ from C to A.

Hence the perimeter p₁ of the triangle P₁ is no larger than the perimeter of the outer polygon P₅, whose perimeter was no larger than the perimeter of the original outer polygon P₂. Hence the result holds when the inner polygon is a triangle.

• Now suppose that the result has been proved when the inner polygon is a k-gon, for some k ≥ 3, and suppose we are presented with a pair of polygons P₁, P₂ where the inner polygon P₁ = ABCD·sis a convex (k+1)-gon.

Draw the line m through C parallel to BD. Let this line meet the outer polygon P₂ at U and V. Cut off the part of P₂ on the opposite side of the the line UCV to B and D, leaving a new outer convex polygon P with perimeter no greater than that of P₂. We prove that the perimeter of polygon P₁ is less than that of polygon P – and hence less than that of P₂. Equivalently, we may assume that UCV is an edge of P₂.

Translate the line m parallel to itself, from m to BD and beyond, until it reaches a position of final contact with the polygon P₁, passing through the vertex X (and possibly a whole edge XY)of the inner polygon P₁. Let this final contact line parallel to m be mʹ.

Since P₁ is convex and k ≥qslant 3, we know that X is different from B and from D. As before, we may assume that mʹ is an edge of the outer polygon P₂. Cut both P₂ and P₁ along the line CX to obtain two smaller configurations, each of which consists of an inner convex polygon inside an outer convex polygon, but in which

• each of the inner polygons has at most k edges, and

• in each of the smaller configurations, the inner and outer polygons both share the edge CX.

Then (by induction on the number of edges of the inner polygon) the perimeter of each inner polygon is no larger than the perimeter of the corresponding outer polygon; so for each inner polygon, the partial perimeter running from C to X (omitting the edge CX) is no larger than the partial perimeter of the corresponding outer polygon running from C to X (omitting the edge CX). So when we put the two parts back together again, we see that the perimeter of P₁is no larger than the perimeter of P₂.

Hence the result holds when P₁is a triangle; and if the result holds whenever the inner polygon has k ≥qslant 3 edges, it also holds whenever the inner polygon has (k+ 1) edges.

It follows that the result holds whatever the number of edges of the inner polygon may be. QED

213.

(a) Join PQ. Then the lines y = B and X = d meet at R to form the right angled triangle PQR.

Pythagoras’ Theorem then implies that (d - a)² + (e - b)² = PQ².

(b) Join PQ. The points P = (a, B, c) and R = (d, e, c) lie in the plane z = c. If we work exclusively in this plane, then part (a) shows that

PR² = (d - a)² + (e - b)².

QR = | F - c|, and Δ PRQ has a right angle at R.

Hence

PQ² = PR² + RQ² = (d - a)² + (e - b)² + (f - c)².

214.

(a) area(Δ ABC) = $\frac{}{}$bc2, area(Δ ACD) = $\frac{}{}$cd2, area(Δ ABD) = $\frac{}{}$bd2.

(b) [This can be a long algebraic slog. And the answer can take very different looking forms depending on how one proceeds. Moreover, most of the resulting expressions are not very pretty, and are likely to incorporate errors.

One way to avoid this slog is to appeal to the fact that the modulus of the vector product DB × DC is equal to the area of the parallelogram spanned by DB

and DC – and so is twice the area of Δ BCD.]

(c) However part (b) is apprOAChed, it is in fact true that area(Δ $BCD{)}^2={(\frac{}{}}^{}$bc2)2+ ${(\frac{}{}}^{}$cd2)2+(bd2)2,

so that

area(Δ ABC)² + area(Δ ACD)² + area(Δ ABD)² = area(Δ BCD)².

215.

(a) b and c are indeed lengths of arcs of great circles on the unit sphere: that is, arcs of circles of radius 1 (centred at the centre of the sphere). However, back in Chapter 1 we used the ‘length’ of such circular arcs to define the angle (in radians) subtended by the arc at the centre. So b and c are also angles (in radians).

(b) The only standard functions of angles are the familiar trig functions (sin, cos).

(c) If c = 0, then the output should specify that a = B, so c should have no effect on the output. This suggests thatwe might expect a formula that involves “adding sin c” or “multiplying by cos c” .

Similarly when B = 0, the output should give a= c, so we might expect a formula that involved “adding sin b” or “multiplying by cos b” .

In general, we should expect a formula in which b and c appear interchangeably (since the input pair(b, c) could equally well be replaced by the input pair (c, b) and should give the same output value of a).

(d)(i) If BC runs along the equator and∠B =∠C = $\frac{\pi }{2}$, then BA and CA run along circles of longitude, so A must be at the North pole. Since A is a right angle, it follows that a = B = c = $\frac{\pi }{2}$.(This tends to rule out the idea that the formula might involve “adding sin B and adding sin c” .)

(ii) Suppose that∠B = $\frac{\pi }{2}$. Since we can imagine AB along the equator, and since there is a right angle at A, it follows that AC and BC both lie along circles of longitude, and so meet at the North pole. Hence C will be at the North pole, so a = B = $\frac{\pi }{2}$.

The inputs to any spherical version of Pythagoras’ Theorem are then B = $\frac{\pi }{2}$, and c. And c is not constrained, so every possible input value of c mustlead to the same output a = $\frac{\pi }{2}$. This tends to suggest that the formula involves some multiple of a product combining “cos b” with some function ofc. And since the inputs “b” and “c” must appear symmetrically, we might reasonably expect some multiple of “cos B · cos c” .

216.

(a) Δ OABʹ has a right angle at A with∠AOBʹ =∠AOB = c. Hence ABʹ = tan c. Similarly ACʹ= tan b.

Hence BʹCʹ²= tan² B + tan² c.

(b) In Δ OABʹ we see that OBʹ = sec c. Similarly OCʹ = sec b. We can now apply the Cosine Rule to Δ OBʹCʹ to obtain:

$$\begin{array}{l}{\tan }^{2}b+{\tan }^{2}c={\sec }^{2}b+{\sec }^{2}c-2\sec b·\sec c·\cos \left(\angle B^{\prime }O{C}^{\prime }\right)\\ ={\sec }^{2}b+{\sec }^{2}c-2\sec b·\sec c·\cos a.\end{array}$$

Hence cos a = cos B · cos c. QED

217. Construct the plane tangent to the sphere at A. Extend OB to meet this plane at Bʹ, and extend OC to meet the plane at Cʹ.

Δ OABʹ has a right angle at A with∠AOBʹ =∠AOB = c. Hence ABʹ = tan c. Similarly ACʹ = tan b.

Hence BʹCʹ²= tan² B + tan² c - 2· tan B · tan c· cos A.

In Δ OABʹ we see that OBʹ = sec c. Similarly OCʹ = sec b.

We can now apply the Cosine Rule to Δ OBʹCʹ to obtain:

$$\begin{array}{l}{\tan }^{2}b+{\tan }^{2}c-2·\tan b·\tan c·\cos A\\ ={\sec }^{2}b+{\sec }^{2}c-2\sec b·\sec \left(\angle B^{\prime }O{C}^{\prime }\right)\\ ={\sec }^{2}b+{\sec }^{2}c-2\sec b·\sec c·\cos a.\end{array}$$

Hence cos a = cos b· cos c + sin b· sin c · cos A. QED

218. We show that $\frac{}{}$sinbsina=sinB (where B denotes the angle ∠ABC at the vertex B).

Construct the plane T which is tangent to the sphere at B. Let O be the centre of the sphere; let OA

produced meet the plane T at Aʹʹ, and let OC produced meet the plane T at Cʹʹ.

Imagine BA positioned along the equator; then BAʹʹ is horizontal; AC lies on a circle of longitude, so AʹʹCʹʹis vertical. Hence∠Cʹʹ BAʹʹ=∠B, and∠BAʹʹ Cʹʹis a right angle; so sin $B=\frac{A\prime \prime C\prime \prime }{BC\prime \prime }$.

Δ OAʹʹCʹʹ is right angled at Aʹʹ; and∠AʹʹOCʹʹ = b; so sin B = $\frac{A\prime \prime C\prime \prime }{OC\prime \prime }$.

Δ OBCʹʹ is right angled at B; and∠BOCʹʹ = a; so sin a = $\frac{BC\prime \prime }{OC\prime \prime }$.

Hence$\frac{\sin b}{\sin a}$ = sin B depends only on the angle at B, so $\frac{}{}$sinbsina=sinb′sina′=sinB.

219. We show that

$\frac{}{}$sinasinA=sinbsinB.

Position the triangle (or rather “rotate the sphere” ) so that AB runs along the equator, with AC leading into the northern hemisphere.

(i) If ∠A is a right angle, then

$\frac{}{}$sinbsina=sinB=sinBsinA

by Problem 218. Hence

$\frac{}{}$sinasinA=sinbsinB.

The same is true if ∠B is a right angle.

(ii) If ∠A and ∠B are both less than a right angle, one can draw the circle of longitude from C to some point X on AB. One can then apply Problem 218 to the two triangles Δ CXA and Δ CXB (each with a right angle at X). Let x denote the length of the CX. Then $\frac{}{}$sinxsinb=sinA,andsinxsina=sinB.

Hence sin B · sin A = sin X = sin a · sin B, so

$\frac{}{}$sinasinA=sinbsinB.

(iii) If ∠A (say) is greater than a right angle, the circle of longitude from C meets the line BA extended beyond A at a point X (say). If we let CX= x, then one can argue similarly using the triangles Δ CXA and Δ CXB to get $\frac{}{}$sinxsina=sinB,andsinxsinb=sinA, whence

$\frac{}{}$sinasinA=sinbsinB.

220. Let X = (x, y) be an arbitrary point of the locus.

(i) Then the distance from X to m is equal to y, so setting this equal to XF gives the equation:

y² = x² + (y - 2a)²,

or

$x^2=4a(y-a)$.

(ii) If we change coordinates and choose the line y = a as a new x-axis, then the equation becomes x² = 4aY. The curve is then tangent to the new x-axis at the (new) origin, and is symmetrical about the y-axis.

221.

(a) Choose the line AB as x-axis, and the perpendicular bisector of AB as the y-axis. Then A = (-3, 0) and B = (3, 0). The point X = (x, y) is a point of the unknown locus precisely when

(x + 3)² + y² = XA² = (2· XB)² = 2²((x - 3)² + y²)

that is, when

(x - 5)² + y² = 4².

This is the equation of a circle with centre (5, 0) and radius r = 4.

(b) Choose the line AB as x-axis, and the perpendicular bisector of AB as the y-axis.

If F = 1, the locus is the perpendicular bisector of AB.

We may assume that F > 1 (since if F < 1, then BX: AX = f^-1: 1 and f^-1 > 1, so we may simplyswap the labelling of A and B).

Now A = (-b, 0) and B = (b, 0), and the point X = (x, y) is a point of the unknown locus precisely when

(x + b)² + y² = XA² = (f· XB)² = f²[(x - b)² + y²]

that is, when

x²(f² - 1) - 2bx(f² + 1) + y²(f² - 1) + b²(f² - 1) = 0

${\left(x-\frac{b(f^2+1)}{f^2-1}\right)}^2+y^2={\left(\frac{2fb}{f^2-1}\right)}^2.$

This is the equation of a circle with centre $\left(\frac{b(f^2+1)}{f^2-1},0\right)$ and radius r = $\frac{2fb}{f^2-1}$.

222.

(a) Choose the line AB as x-axis, and the perpendicular bisector of AB as the y-axis.

Then A = (-c, 0) and B = (c, 0). The point X = (x, y) is a point of the unknown locus precisely when

2a - $2a=AX+BX=\sqrt{{(x+c)}^2+y^2}+\sqrt{{(x-c)}^2+y^2}$,

that is, when

$$2a-\sqrt{{(x+c)}^2+y^2}=\sqrt{{(x-c)}^2+y^2}$$

∴ 4a² - 4a$\sqrt{{(x+c)}^2+y^2}=\sqrt{{(x-c)}^2+y^2}$ $\sqrt{{(x+c)}^2+y^2}+[{(x+c)}^2$

∴ a² + cx = a (x + c)² + y²

∴ (a² - c²)x²+ a²y² = a²(a² - c²)

Setting $\genfrac{}{}{0.1ex}{}{c}{a}$ = e then yields the equation for the locus in the form:

$\frac{}{}$x2a2+y2a2(1−e2)=1.

Note: In the derivation of the equation we squared both sides (twice). This may introduce spurious solutions. So we should check that every solution (x, y) of the final equation satisfies the original condition.

(b) The real number e< 1 is given, so we may set the distance from F to m be $\frac{a}{e}\left(1-e^2\right)$. Choose the line through F and perpendicular to m as x-axis. To start with, we choose the line m as y-axis and adjust later if necessary.

Hence F has coordinates $\left(\frac{a}{e}\left(1-e^2\right),0\right)$, and the point X = (x, y) is a point of the unknown locus precisely when

${\left(x-\frac{a}{e}\left(1-e^2\right)\right)}^2+y^2={(ex)}^2$,

which can be rearranged as

$\left(1-e^2\right)x^2-2\frac{a}{e}(1-e^2)x+{\left(\frac{a}{e}\right)}^2{(1-e^2)}^2+y^2=0$,

and further as

(1 - e²)$\left[x^2-2\frac{a}{e}x+{\left(\frac{a}{e}\right)}^2\right]+y^2={\left(\frac{a}{e}\right)}^2(1-e^2)-{\left(\frac{a}{e}\right)}^2{(1-e^2)}^2$

= $\left(\frac{a}{e}\right)$²(e² - e⁴)

= a²(1 - e²)

∴${\left(x-\frac{a}{e}\right)}^2+\frac{y^2}{1-e^2}=a^2$.

If we now move the y-axis to the line $x=\frac{a}{e}$ the equation takes the simpler form:

$\frac{}{}$x2a2+y2a2(1−e2)=1.

(c) This was done in the derivations in the solutions to parts (a) and (b).

223.

(a) The triangle inequality shows that, if AX > BX, then AB + BX ≥qslant AX; hence the locus is non-empty only when a $⩽$ c. If a = c, then X must lie on the line AB, but not between A and B, so the locus consists of the two half-lines on AB outside AB. Hence we may assume that a < c.

Choose the line AB as x-axis, and the perpendicular bisector of AB as y-axis.

Then A = (-c, 0) and B = (c, 0). The point X = (x, y) is a point of the unknown locus precisely when

$2a=|<i><u>AX</u></i>-<i><u>BX</u></i>|=\left|\sqrt{{(x+c)}^2+y^2}-\sqrt{{(x-c)}^2+y^2}\right|$.

If AX > BX, we can drop the modulus signs and calculate as in Problem 222.

$2a+\sqrt{{(x-c)}^2+y^2}=\sqrt{{(x+c)}^2+y^2}$.

∴ $$4a2+4a(x−c)2+y2+(x−c)2+y2=(x+c)2+y2

∴ $a^2-cx=-a\sqrt{{(x-c)}^2+y^2}$

∴ $(c^2-a^2)x^2-a^2{y}^2=a^2(c^2-a^2)$

Setting $\genfrac{}{}{0.1ex}{}{c}{a}$ = e (> 1), then yields the equation for the locus in the form:

$\frac{x^2}{a^2}-\frac{y^2}{a^2(e^2-1)}=1$.

Note: In the derivation of the equation we squared both sides (twice). This may introduce spurious solutions. So we should check that every solution (x, y) of the final equation satisfies the original condition. In fact, the squaring process introduces additional solutions in the form of a second branch of the locus, corresponding precisely to points X where AX < BX.

(b) The real number e> 1 is given, so we may set the distance from F to m be $\frac{a}{e}$(e² - 1). Choose the line through F and perpendicular to m as x-axis. To start with, we choose the line m as y-axis and adjust later if necessary.

Hence F has coordinates $\left(\frac{a}{e}(e^2-1),0\right)$$\left(\frac{a}{e}(e^2-1),0\right)$, and the point X = (x, y) is a point of the unknown locus precisely when

${\left(x-\frac{a}{e}(e^2-1)\right)}^2+y^2={(ex)}^2$

∴ (e^{2 - 1})x² + $\frac{}{}$2ae(e2−1)x−y2=(ae)2(e2−1)2

∴ $(e^2-1)[x^2+$2aex+(ae)2 ]−y2=(ae)2(e2−1)2+(ae)2(e2−1)

= ${\left(\frac{a}{e}\right)}^2(e^4-e^2)$

∴ (e^{2 - 1})$[x^2+$2aex+(ae)2 ]−y2=a2(e2−1)

∴ ${\left(x+\frac{a}{e}\right)}^2-\frac{y^2}{e^2-1}=a^2$

If we now move the y-axis to the line X = -$\frac{a}{e}$ the equation takes the simpler form:

$\frac{x^2}{a^2}-\frac{y^2}{a^2(e^2-1)}=1$.

(c) This was done in derivations in the solutions to parts (a) and (b).

224.

(a) When z = k is a constant, the equation reduces to that of a circle

x² + y² = (rk)² in the plane z = k. When the cutting plane is the

xy-plane “z = 0” , the circle has radius 0, so is a single point.

(b) (i) A vertical plane through the apex cuts the cone in a pair of generators crossing at the apex.

(ii) If the cutting plane through the apex is less steep than a generator, then it cuts the cone only at the apex.

If the cutting plane through the apex is parallel to a generator, then it cuts the cone in a generator – a single line (the next paragraph indicates that this line may be better thought of as a pair of “coincident” lines).

What happens when the cuttingplane through the apex is steeper than a generator may not be intuitively clear. One way to make sense of this is to treat the cross-section as the set of solutions of two simultaneous equations – one for the cone, and the other for the plane (say y = nz, with n < r). This leads to the equation

$x^2=(r^2-n^2)z^2,y=nz$,

with solution set

$x=\pm z\sqrt{r^2-n^2},xy=nz$

which specifies a pair of lines crossing at the apex.

A slightly easier way to visualize the cross-section in this case is to let the apex of the double cone be A, and to let X be any other point of the cross-section. Then the line AX is a generator of the cone, so lies on the cone's surface. But A and X also lie in the cutting plane – so the whole line AX must lie in the cutting plane. Hence the cross-section contains the whole line AX.

(c)(i) If the cutting plane passes through the apex and is parallel to a generator of the cone, then we saw in (b) that the cross-section is simply the generator itself.

(ii) Thus we assume that the cutting plane does not pass through the apex, and may assume that it cuts the bottom half of the cone. If V is the point nearest the apex where the cutting plane meets the cone, then the cross-section curve starts at V and becomes wider as we go down the cone. Because the plane is parallel to a generator, the plane never cuts the “other side” of the bottom half of the cone, so the cross section never “closes up” – but continues to open up wider and wider as we go further and further down the bottom half of the cone.

Let S be the sphere, which is inscribed in the cone above the cutting plane, and which is tangent to the cutting plane at F. Let C be the circle of contact between S and the cone. Let A be the apex of the cone, and let the apex angle of the cone be equal to 2$\theta$. Let X be an arbitrary point of the cross-section. To illustrate the general method, consider first the special case where X = V is the “highest” point of the cross-section. The line segment VA is tangent to the sphere S, so crosses the circle C at some point M. Now VF lies in the cutting plane, so is also tangent to the sphere S at the point F. Any two tangents to a sphere from the same exterior point are equal, so it follows that VM = VF. Moreover, VM is exactly equal to the distance from V to the line m (since

• firstly the line m lies in the horizontal plane through C and so is on the same horizontal level as M, and

• secondly the shortest line VM* from V to m, runs straight up the cutting plane, which is parallel to a generator – so the angle∠MVM*= 2 $\theta$, whence VM* = VM = VF).

Now let X be an arbitrary point of the cross-sectional curve, and use a similar argument. First the line XA is always a generator of the cone, so is tangent to the sphere S, and crosses the circle C at some point Y. Moreover, XF is also tangent to the sphere. Hence XY = XF. It remains to see that XY is equal to the distance XY* from X to the closest point Y* on the line m – since

• firstly the two points Y and Y* both lie on the same horizontal level (namely the horizontal plane through the circle C), and

• secondly both make the same angle $\theta$ with the vertical.

Hence the cross-sectional curve is a parabola with focus F and directrix m.

(d) (i) If the cutting plane is less steep than a generator, the cross-section is a closed curve. If V and W are the highest and lowest points of intersection of the plane with the cone, then the cross-section is clearly symmetrical under reflection in the line VW. Intuitively it is tempting to think that the lower end near W should be ‘fatter’ than the upper part of the curve (giving an egg-shaped cross-section). This turns out to be false, and the correct version was known to the ancient Greeks, though the error was repeated in many careful drawings from the 14th and 15th centuries (e.g. Albrecht Dürer (1471–1528)).

(ii) The derivation is very similar to that in part (c), and we leave the reader to reconstruct it.

An alternative apprOACh is to insert a second sphere S’ below the cutting plane, and inflate it until it makes contact with the cone along a circle C’ while at the same time touching the cutting plane at a point Fʹ. If X is an arbitrary point of the cross-sectional curve, then XA is tangent to both spheres, so meets the circle C at some point Y and meets the circle C’ at some point Yʹ. Then Y, X, Yʹ are collinear. Moreover, XY = XF (since both are tangents to the sphere S from the point X), and XYʹ = XFʹ (since both are tangents to the sphere S’ from the point X), so

XF + XFʹ = XY + XYʹ = YYʹ

But YY’ is equal to the slant height of the cone between the two fixed circles C and C', and so is equal to a constant k. Hence, the focus-focus specification in Problem 222 shows that the cross-section is an ellipse.

(e)(i) If the cutting plane is steeper than a generator, the cross-section cuts both the bottom half and the top half of the cone to give two separate parts of the cross-section. Neither part “closes up” , so each part opens up more and more widely.

If V is the highest point of the cross-section on the lower half of the cone, and W is the lowest point of the cross-section on the upper half of the cone, then it seems clear that the cross-section is symmetrical under reflection in the line VW. But it is quite unclear that the two halves of the cross-section are exactly congruent (though again it was known to the ancient Greeks).

(ii) The formal derivation is very similar to that in part (c), and we leave the reader to reconstruct it.

An alternative apprOACh is to copy the alternative in (d), and to insert a second sphere S'in the upper part of the cone, on the same side of the cutting plane as the apex, inflate it until it makes contact with the cone along a circle C’ while at the same time touching the cutting plane at a point Fʹ. If

X is an arbitrary point of the cross-sectional curve, then XA is tangent to both spheres, so meets the circle C at some point Y and meets the circle C’ at some point Yʹ. Then Y, X, Yʹ are collinear. Moreover, X, F, and Fʹ all lie on the cutting plane. Now XY = XF (since both are tangents to the sphere S from the point X), and XY’ = XFʹ (since both are tangents to the sphere S’ from the point X). If X is on the upper half of the cone, then

XF -XFʹ = XY- XYʹ = YYʹ.

But YYʹ is equal to the slant height of the cone between the two fixed circles C and C', and so is equal to a constant k. Hence, the focus-focus specification in Problem 223 shows that the cross-section is a hyperbola.

225.

(a)(i) 2¹; (ii) 1 = 2⁰

(b) (i) 2²; (ii) 1 = 2⁰; (iii) 4 = 2 \times 2⁰ + 2¹

(c) (i) 2³; (ii) 1 = 2⁰; (iii) 12 = 2 \times 4 + 2²; (iv) 6 = 2 \times 2⁰ + 4

(d) (i) 2⁴; (ii) 1 = 2⁰; (iii) 32 = 2 \times 12 + 2³; (iv) 24 = 2 \times 6 + 12; (v) 8 = 2 \times 2⁰ + 6

226.

(c) (i)If you look carefully at the diagram shown here you should be able to see not only the upper and lower 3D-cubes, but also the four other 3D-cubes formed by joining each 2D-cube in the upper 3D-cube to the corresponding 2D-cube in the lower3D-cube.

Note: Once we have the 3D-cube expressed in coordinates, we can specify precisely which planes produce which cross-sections in Problem 38. The plane X + y + z = 1 passes through the three neighbours of (0, 0, 0) and creates an equilateral triangular cross-section. Any plane of the form z = c (where c is a constant between 0 and 1) produces a square cross-section. And the plane X + y + z = $\frac{3}{2}$ is the perpendicular bisector of the line joining (0, 0, 0) to (1, 1, 1), and creates a regular hexagon as cross-section.

227.

(a) View the coordinates as (x, y, z). Start at the origin (0, 0, 0) and travel round 3 edges of the lower 2D-cube “z = 0” to the point (0, 1, 0). Copy this path of length 3 on the upper 2D-cube “z = 1” (from (0, 0, 1) to (0, 1, 1). Then join (0, 0, 0) to (0, 0, 1) and join (0, 1, 0) to (0, 1, 1). The result

(0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 1, 1), (1, 1, 1), (1, 0, 1), (0, 0, 1) (and back to (0, 0, 0))

has the property that exactly one coordinate changes when we move from each vertex to the next. This is an example of a Gray code of length 3.

Note: How many such paths/circuits are there in the 3D-cube?We can certainly count those of the kind described here. Each such circuit has a “direction” : the 12 edges of the 3D-cube lie in one of 3 “directions” , and each such circuit contains all four edges in one of these 3 directions. Moreover this set of four edges can be completed to a circuit in exactly 2 ways. So there are 3 × 2 such circuits. In 3D this accounts for all such circuits. But in higher dimensions the numbers begin to explode (in the 4D-cube there are 1344 such circuits).

(b) View the coordinates as (w, x, y, z). Start at the origin (0, 0, 0, 0) and travel round the 8 vertices of the lower 3D-cube “z = 0” to the point (0, 0, 1, 0). Then copy this path on the upper 3D-cube “z = 1” from (0, 0, 0, 1) to (0, 0, 1, 1). Finally join (0, 0, 0, 0) to (0, 0, 0, 1) and join (0, 0, 1, 0) to (0, 0, 1, 1). The result

(0, 0, 0.0), (1, 0, 0, 0), (1, 1, 0, 0), (0, 1, 0, 0), (0, 1, 1, 0), (1, 1, 1, 0),

(1, 0, 1, 0), (0, 0, 1, 0) (0, 0, 1, 1), (1, 0, 1, 1), (1, 1, 1, 1), (0, 1, 1, 1),

(0, 1, 0, 1), (1, 1, 0, 1), (1, 0, 0, 1), (0, 0, 0, 1) (and back to (0, 0, 0, 0))

has the property that exactly one coordinate changes when we move from each vertex to the next. This is an example of a Gray code of length 4.

Note: The general construction in dimension n + 1 depends on the previous construction in dimension n, so makes use of mathematical induction (see Problem 262 in Chapter 6).