HMMT November

Geometry Difficulty 4.7 AIME Find the answer

$A B C$ is a triangle with $A B=15, B C=14$ , and $C A=13$ . The altitude from $A$ to $B C$ is extended to meet the circumcircle of $A B C$ at $D$ . Find $A D$ .

Your answer

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

Solution

Let the altitude from $A$ to $B C$ meet $B C$ at $E$ . The altitude $A E$ has length 12 ; one way to see this is that it splits the triangle $A B C$ into a $9-12-15$ right triangle and a $5-12-13$ right triangle; from this, we also know that $B E=9$ and $C E=5$ . Now, by Power of a Point, $A E \cdot D E=B E \cdot C E$ , so $D E=(B E \cdot C E) / A E=(9 \cdot 5) /(12)=15 / 4$ . It then follows that $A D=A E+D E=63 / 4$ .

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