19.3: Constructions with a set-square

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A set-square is a construction tool shown on the picture — it can produce a line thru a given point that makes the angles $\tfrac\pi2$ or $\pm\tfrac\pi4$ to a given line and it can be also used as a ruler; that is, it can produce a line thru a given pair of points.

Exercise $\PageIndex{1}$

Trisect a given segment with a set-square.

Hint: Note that with a set-square we can construct a line parallel to given line thru the given point. It remains to modify the construction in Exercise 14.2.1.

Let us consider set-square constructions. Following the same lines as in the previous section, we can define set-square constructible numbers and prove the following analog of Theorem 19.2.1:

Theorem $\PageIndex{1}$

Theorem Assume that the initial configuration of a geometric construction is given by the points $A_1=(0,0)$, $A_2=(1,0)$, $A_3=(x_3,y_3),\dots,A_n=(x_n,y_n)$. Then a point $X=(x,y)$ can be constructed using a set-square construction if and only if both coordinates $x$ and $y$ can be expressed from the integer numbers and $x_3$, $y_3$, $x_4$, $y_4,\dots,x_n$, $y_n$ using the arithmetic operations "$+$", "$-$", "$\cdot$", and "$/$" only.

Let us apply this theorem to show the impossibility of some constructions with a set-square.

Note that if all the coordinates $x_3,y_3,\dots,x_n,y_n$ are rational numbers, then the theorem above implies that with a set-square, one can only construct the points with rational coordinates. A point with both rational coordinates is called rational, and if at least one of the coordinates is irrational, then the point is called irrational.

Exercise $\PageIndex{2}$

Show that an equilateral triangle in the Euclidean plane has at least one irrational point.

Conclude that with a set-square, one cannot construct an equilateral triangle with a given base.

Hint: Assume that two vertices have rational coordinates, say $(a_1, b_1)$ and $(a_2, b_2)$. Find the coordinates of the third vertex. Use that the number $\sqrt{3}$ is irrational to show that the third vertex is an irrational point.

Exercise \(\PageIndex{1}\)

Theorem \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)