Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Slovenia

Let $ABCDEF$ be a hexagon with $\angle BAF = 150^\circ$ , $\angle ACB = \angle ADC = 90^\circ$ and $|AC| = |BC|$ . Assume also that the triangle $ABC$ is similar to the triangle $ADE$ and the triangle $BCD$ is similar to the triangle $DEF$ . Find the ratio of the lengths of the segments $AB$ and $AF$ .

Solution

Let $|AB| = a$ . Since $ABC$ is an isosceles right triangle with the apex at $C$ , we have $|AC| = \frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}$ and $\angle ABC = \angle BAC = \frac{\pi}{4}$ . The triangles $ADE$ and $ABC$ are similar, so $\angle AED = \angle ACB = \frac{\pi}{2}$ and $|DE| = |AE|$ .

$\angle ACD = \angle BCD - \frac{\pi}{2} = \angle DEF - \frac{\pi}{2} = \angle AEF$

$\frac{|AC|}{|CD|} = \frac{|BC|}{|CD|} = \frac{|DE|}{|EF|} = \frac{|AE|}{|EF|}$

$150^\circ = \angle BAF = \angle BAC + \angle CAD + \angle CAE + \angle EAF = \frac{\pi}{2} + 2\angle CAD$

we deduce that $\angle CAD = 30^\circ$ . The triangle $CAD$ is one half of an equilateral triangle and $|AD| = \frac{|AC|\sqrt{3}}{2} = \frac{a\sqrt{6}}{4}$ . Since $ADE$ is an isosceles right triangle, we have $|AE| = \frac{|AD|}{\sqrt{2}} = \frac{a\sqrt{3}}{4}$ . Finally, since $AEF$ is one half of an equilateral triangle, we get $|AF| = \frac{|AE|\sqrt{3}}{2} = \frac{3a}{8}$ . Hence, $\frac{|AB|}{|AF|} = \frac{8}{3}$ .

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