18.6: Method of complex coordinates
The following problem illustrates the method of complex coordinates.
Let \(\triangle OPV\) and \(\triangle OQW\) be isosceles right triangles such that
\(\measuredangle VPO = \measuredangle OQW=\dfrac{\pi}{2}\)
and \(M\) be the midpoint of \([VW]\) . Assume \(P\) , \(Q\) , and \(M\) are distinct points. Show that \(\triangle PMQ\) is an isosceles right triangle.
- Solution
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Choose the complex coordinates so that \(O\) is the origin; denote by \(v, w, p, q, m\) the complex coordinates of the remaining points respectively.
Since \(\triangle OPV\) and \(\triangle OQW\) are isosceles and \(\measuredangle VPO=\measuredangle OQW=\dfrac{\pi}{2}\) , 18.3.2 and 18.5.1 imply that
\(\begin{aligned} v-p&=i\cdot p, & q-w&=i\cdot q.\end{aligned}\)
Therefore
\(\begin{aligned} m&=\tfrac12\cdot(v+w)= \\ &=\tfrac{1+i}2\cdot p+\tfrac{1-i}2\cdot q.\end{aligned}\)
By a straightforward computations, we get that
\(p-m=i\cdot (q-m).\)
In particular, \(|p-m|=|q-m|\) and \(\arg\dfrac{p-m}{q-m}=\dfrac{\pi}{2}\) ; that is, \(PM=QM\) and \(\measuredangle QMP =\dfrac{\pi}{2}\) .
Consider three squares with common sides as on the diagram. Use the method of complex coordinates to show that
\(\measuredangle EOA+\measuredangle EOB+\measuredangle EOC=\pm\dfrac{\pi}{2}.\)
- Hint
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We can choose the complex coordinates so that the points \(O, E, A, B\), and \(C\) have coordinates \(0, 1, 1 + i, 2 + i\), and \(3 + i\) respectively.
Set \(\measuredangle EOA = \alpha\), \(\measuredangle EOB = \beta\), and \(\measuredangle EOC = \gamma\). Note that
\(\begin{array} {rcl} {\alpha + \beta + \gamma} & \equiv & {\arg (1 + i) + \arg (2 + i) + \arg (3 + i) \equiv} \\ {} & \equiv & {\arg [(1 + i) \cdot (2 + i) \cdot (3 + i)] \equiv \arg [10 \cdot i] = \dfrac{\pi}{2}} \end{array}\)
Note that these three angles are acute and conclude that \(\alpha + \beta + \gamma = \dfrac{\pi}{2}\).
Check the following identity with six complex cross ratios:
\((u,w;v,z)\cdot(u',w';v',z')=\frac{(v,w';v',w)\cdot(z,u';z',u)}{(u,v';u',v)\cdot(w,z';w',z)}.\)
Use it together with Theorem 18.5.1 to prove that if \(\square UVWZ\) , \(\square UVV'U'\) , \(\square VWW'V'\) , \(\square WZZ'W'\) , and \(\square ZUU'Z'\) are inscribed, then \(\square U'V'W'Z'\) is inscribed as well.
- Hint
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The identity can be checked by straightforward computations.
By Theorem 18.5.1 , five from six cross ratios in the this identity are real. Therefore so is the sixth cross ratio; it remains to apply the theorem again.
Suppose that points \(U\) , \(V\) and \(W\) lie on one side of line \((AB)\) and \(\triangle UAB\sim \triangle BVA \sim \triangle ABW\) . Denote by \(a\) , \(b\) , \(u\) , \(v\) , and \(w\) the complex coordinates of \(A\) , \(B\) , \(U\) , \(V\) , and \(W\) respectively.
- Show that \(\dfrac{u-a}{b-a}=\dfrac{b-v}{a-v}=\dfrac{a-b}{w-b}=\dfrac{u-v}{w-v}\) .
- Conclude that \(\triangle UAB\sim \triangle BVA \sim \triangle ABW\sim \triangle UVW\) .
- Hint
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Use 3.7 and 3.10 to show that \(\angle UAB, \angle BVA\), and \(\angle ABW\) have the same sign. Note that by SAS we have that
\(\dfrac{AU}{AB} = \dfrac{VB}{VA} = \dfrac{BA}{BW}\) and \(\measuredangle UAB = \measuredangle BVA =\measuredangle ABW.\)
The later means that \(|\dfrac{u - a}{b - a}| = |\dfrac{b - v}{a - v}| = |\dfrac{a - b}{w - b}|\), and \(\arg \dfrac{b - a}{u - a} = \arg \dfrac{a - b}{w - b}\). It implies the first two equalities in
\[\dfrac{b - a}{u - a} = \dfrac{a - v}{b - v} = \dfrac{w - b}{a -b} = \dfrac{w - v}{u - v}.\]
the last equality holds since
\(\dfrac{(b - a) + (a -v) + (w - b)}{(u - a) + (b - v) + (a -b)} = \dfrac{w - v}{u - v}.\)
To prove (b), rewrite 18.6.1 using angles and distances between the points and apply SAS.