Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Consider triangle ABCABC with side lengths AB=4AB = 4, BC=7BC = 7, and AC=8AC = 8. Let MM be the midpoint of segment ABAB, and let NN be the point on the interior of segment ACAC that also lies on the circumcircle of triangle MBCMBC. Compute BNBN.

Solution

Solution:

Answer: 2104\frac{\sqrt{210}}{4} OR 10522\frac{\sqrt{105}}{2 \sqrt{2}}

Let ∠BAC=θ\angle BAC = \theta. Then,
cos⁡θ=42+82−722⋅4⋅8. \cos \theta = \frac{4^2 + 8^2 - 7^2}{2 \cdot 4 \cdot 8}.
Since AM=42=2AM = \frac{4}{2} = 2, and power of a point gives AM⋅AB=AN⋅ACAM \cdot AB = AN \cdot AC, we have
AN=2⋅48=1, AN = \frac{2 \cdot 4}{8} = 1,
so NC=8−1=7NC = 8 - 1 = 7.

Law of cosines on triangle BANBAN gives
BN2=42+12−2⋅4⋅1⋅42+82−722⋅4⋅8=17−16+158=17−318=1058 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=2104. 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.