HMMT November, 2022

Geometry Difficulty 4.3 AIME Find the answer United States

Problem:
Let $\triangle ABC$ be an isosceles right triangle with $AB = AC = 10$ . Let $M$ be the midpoint of $BC$ and $N$ the midpoint of $BM$ . Let $AN$ hit the circumcircle of $\triangle ABC$ again at $T$ . Compute the area of $\triangle TBC$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Note that since quadrilateral $BAC T$ is cyclic, we have
$\angle BTA = \angle BCA = 45^\circ = \angle CBA = \angle CTA$
Hence, $TA$ bisects $\angle BTC$ , and $\angle BTC = 90^\circ$ . By the angle bisector theorem, we then have
$\frac{BT}{TC} = \frac{BN}{NC} = \frac{1}{3}.$
By the Pythagorean theorem on right triangles $\triangle TBC$ and $\triangle ABC$ , we have
$10\, BT^2 = BT^2 + TC^2 = AB^2 + AC^2 = 200$
so $BT^2 = 20$ . Note that the area of $\triangle TBC$ is
$\frac{BT \cdot TC}{2} = \frac{3 \cdot BT^2}{2}$
so our answer is then
$\frac{3}{2} \cdot BT^2 = \frac{3}{2} \cdot 20 = 30$

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