HMMT November

Geometry Difficulty 4.8 AIME Find the answer

Consider triangle $A B C$ with side lengths $A B=4, B C=7$ , and $A C=8$ . Let $M$ be the midpoint of segment $A B$ , and let $N$ be the point on the interior of segment $A C$ that also lies on the circumcircle of triangle $M B C$ . Compute $B N$ .

Your answer

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

Solution

Let $\angle B A C=\theta$ . Then, $\cos \theta=\frac{4^{2}+8^{2}-7^{2}}{2 \cdot 4 \cdot 8}$ . Since $A M=\frac{4}{2}=2$ , and power of a point gives $A M \cdot A B=A N \cdot A C$ , we have $A N=\frac{2 \cdot 4}{8}=1$ , so $N C=8-1=7$ . Law of cosines on triangle $B A N$ gives $B N^{2}=4^{2}+1^{2}-2 \cdot 4 \cdot 1 \cdot \frac{4^{2}+8^{2}-7^{2}}{2 \cdot 4 \cdot 8}=17-\frac{16+15}{8}=15-\frac{15}{8}=\frac{105}{8}$ so $B N=\frac{\sqrt{210}}{4}$ .

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.