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Philippine Mathematical Olympiad Qualifying Stage

Geometry Difficulty 4.5 AIME Prove it Philippines

Problem:

Let $\triangle ABC$ be an isosceles triangle with a right angle at $A$ , and suppose that the diameter of its circumcircle $\Omega$ is $40$ . Let $D$ and $E$ be points on the arc $BC$ not containing $A$ such that $D$ lies between $B$ and $E$ , and $AD$ and $AE$ trisect $\angle BAC$ . Let $I_{1}$ and $I_{2}$ be the incenters of $\triangle ABE$ and $\triangle ACD$ respectively. The length of $I_{1}I_{2}$ can be expressed in the form $a+b \sqrt{2}+c \sqrt{3}+d \sqrt{6}$ , where $a, b, c$ , and $d$ are integers. Find $a+b+c+d$ .

Solution

Solution:

Let $O$ be the center of $\Omega$ . Note that $\angle OBD = \angle CBD = \frac{\pi}{3}$ , so $\triangle OBD$ is an equilateral triangle. Thus, $BD = BO = \frac{40}{2} = 20$ . This implies that $EC = DE = BD = 20$ .

Clearly, $I_{1}$ and $I_{2}$ lie on the segments $AD$ and $AE$ respectively. It is well-known that $DI_{1} = DB$ , so $DI_{1} = 20$ . Similarly, $EI_{2} = 20$ .

Now, note that $\angle EDI_{1} = \angle EDC + \angle CDA = 30^{\circ} + 45^{\circ} = 75^{\circ}$ . Similarly, $\angle DEI_{2} = 75^{\circ}$ .

Looking at the isosceles trapezoid $I_{1}I_{2}ED$ , the length of $I_{1}I_{2}$ must then be $DE - DI_{1} \cos 75^{\circ} - EI_{2} \cos 75^{\circ} = 20 - 40 \cos 75^{\circ} = 20 - 40\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) = 20 - \sqrt{6} + \sqrt{2}$ , and so $a + b + c + d = 20 + 1 + 0 + (-1) = 20$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.