Maths Olympiad Prep

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, 2015

Geometry Difficulty 3.5 AMC 10/12 Prove it Slovenia

We inscribe a regular octagon in a square with side of length aa, so that 4 sides of the octagon lie on the sides of the square. Express the side length of the inscribed octagon in terms of aa.

Solution

Denote by xx the side length of the inscribed octagon. The four triangles that are formed at the vertices of the square are isosceles right-angled triangles. Since their hypotenuse is of length xx, their legs are of length x2\frac{x}{\sqrt{2}}. Thus
a=x+2x2=x+x2=x(2+1). a = x + 2\frac{x}{\sqrt{2}} = x + x\sqrt{2} = x(\sqrt{2} + 1).
From this we deduce

Figure 1
x=a2+1=a(2−1)2−1=2a−a. x = \frac{a}{\sqrt{2} + 1} = \frac{a(\sqrt{2} - 1)}{2 - 1} = \sqrt{2}a - a.

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