Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

$ABC$ is a triangle with $AB = 15$ , $BC = 14$ , and $CA = 13$ . The altitude from $A$ to $BC$ is extended to meet the circumcircle of $ABC$ at $D$ . Find $AD$ .

Solution

Solution:

Answer: $\boxed{\dfrac{63}{4}}$

Let the altitude from $A$ to $BC$ meet $BC$ at $E$ . The altitude $AE$ has length $12$ ; one way to see this is that it splits the triangle $ABC$ into a $9$ - $12$ - $15$ right triangle and a $5$ - $12$ - $13$ right triangle; from this, we also know that $BE = 9$ and $CE = 5$ .

Now, by Power of a Point, $AE \cdot DE = BE \cdot CE$ , so $DE = (BE \cdot CE) / AE = (9 \cdot 5) / (12) = 15 / 4$ . It then follows that $AD = AE + DE = 12 + 15/4 = 63/4$ .

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