Library / /1160 of 1394

, 2023

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $\angle BAC > 90^{\circ}$ . Let $D$ be the foot of the perpendicular from $A$ to side $BC$ . Let $M$ and $N$ be the midpoints of segments $BC$ and $BD$ , respectively. Suppose that $AC = 2$ , $\angle BAN = \angle MAC$ , and $AB \cdot BC = AM$ . Compute the distance from $B$ to line $AM$ .

Solution

Solution:

Extend $AM$ to meet the circumcircle of $\triangle ABC$ at $X$ . Then, we have $\triangle ABM \sim \triangle CXM$ , which implies that $\frac{CX}{CM} = \frac{AB}{AM}$ . Using the condition $AB \cdot BC = AM$ , we get that $CX = \frac{1}{2}$ .

Now, the key observation is that $\triangle ANB \sim \triangle ACX$ . Thus, if we let $Y$ be the reflection of $X$ across point $C$ , we get that $\angle AYC = 90^{\circ}$ . Thus, Pythagorean's theorem gives $AY = \sqrt{AC^2 - CY^2} = \sqrt{15/4}$ and $AX^2 = \sqrt{AY^2 + XY^2} = \sqrt{19/4}$ .

Finally, note that the distance from $B$ and $C$ to line $AM$ are equal. Thus, let $H$ be the foot from $C$ to $AM$ . Then, from $\triangle XCH \sim \triangle XAY$ , we get that
$CH = AY \cdot \frac{CX}{AX} = \frac{\sqrt{15}}{2} \cdot \frac{1/2}{\sqrt{19}/2} = \frac{\sqrt{285}}{38}$

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