
# 19.3: Constructions with a set-square


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.