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Mathematics LibreTexts

19.3: Constructions with a set-square

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    截屏2021-03-02 上午10.14.12.png

    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.


    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.


    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.