HMMT November

Geometry Difficulty 4.9 AIME Find the answer

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

Note that since quadrilateral $BACT$ 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 $10BT^{2}=BT^{2}+TC^{2}=AB^{2}+AC^{2}=200$ 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 $\frac{3}{2} \cdot BT^{2}=\frac{3}{2} \cdot 20=30$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.