HMMT February

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Triangle $ABC$ with $\angle BAC > 90^{\circ}$ has $AB = 5$ and $AC = 7$ . Points $D$ and $E$ lie on segment $BC$ such that $BD = DE = EC$ . If $\angle BAC + \angle DAE = 180^{\circ}$ , compute $BC$ .

Solution

Solution:

Let $M$ be the midpoint of $BC$ , and consider dilating about $M$ with ratio $-\frac{1}{3}$ . This takes $B$ to $E$ , $C$ to $D$ , and $A$ to some point $A'$ on $AM$ with $AM = 3A'M$ . Then the angle condition implies $\angle DAE + \angle EA'D = 180^{\circ}$ , so $AD A' E$ is cyclic. Then by power of a point, we get
$\frac{AM^2}{3} = AM \cdot A'M = DM \cdot EM = \frac{BC^2}{36}$
But we also know $AM^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4}$ , so we have $\frac{2AB^2 + 2AC^2 - BC^2}{12} = \frac{BC^2}{36}$ , which rearranges to $BC^2 = \frac{3}{2}(AB^2 + AC^2)$ . Plugging in the values for $AB$ and $AC$ gives $BC = \sqrt{111}$ .

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