HMMT November

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Consider triangle $ABC$ with side lengths $AB = 4$ , $BC = 7$ , and $AC = 8$ . Let $M$ be the midpoint of segment $AB$ , and let $N$ be the point on the interior of segment $AC$ that also lies on the circumcircle of triangle $MBC$ . Compute $BN$ .

Solution

Solution:

Answer: $\frac{\sqrt{210}}{4}$ OR $\frac{\sqrt{105}}{2 \sqrt{2}}$

Let $\angle BAC = \theta$ . Then,
$\cos \theta = \frac{4^2 + 8^2 - 7^2}{2 \cdot 4 \cdot 8}.$
Since $AM = \frac{4}{2} = 2$ , and power of a point gives $AM \cdot AB = AN \cdot AC$ , we have
$AN = \frac{2 \cdot 4}{8} = 1,$
so $NC = 8 - 1 = 7$ .

Law of cosines on triangle $BAN$ gives
$BN^2 = 4^2 + 1^2 - 2 \cdot 4 \cdot 1 \cdot \frac{4^2 + 8^2 - 7^2}{2 \cdot 4 \cdot 8} = 17 - \frac{16 + 15}{8} = 17 - \frac{31}{8} = \frac{105}{8}$
so
$BN = \frac{\sqrt{210}}{4}.$

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