23. Construct a Triangle Given Two Sides and the Inradius

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This Demonstration draws a triangle given two side lengths and and the inradius (the radius of the inscribed circle). This construction involves solving a cubic and is not possible with a ruler and compass.


Let the third side length be and let the area of be . Define the semiperimeter . Since equals the sum of the areas of the three triangles with apex at the incenter, .

Using that together with Heron's formula (Wolfram MathWorld), gives , which is a cubic equation for . The equations for and are shown at the bottom of the graphic.


Contributed by: Izidor Hafner (November 2017)
Open content licensed under CC BY-NC-SA



The case , , gives , where . Substitute to get .

Since the leading coefficient of the equation is , a rational root would have to be an integer that divides 32. But no integer , , , , , , , , , is a root of the equation. Therefore the last equation in (and so also the first in ) has no rational solutions.

According to the theorem on p. 42 of [1], none of the roots can be constructed by ruler and compass, but the roots can be constructed using a marked ruler [1, p. 134].


[1] G. E. Martin, Geometric Constructions, New York: Springer, 1998.

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