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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Triangle $ABC$ obeys $AB = 2AC$ and $\angle BAC = 120^\circ$ . Points $P$ and $Q$ lie on segment $BC$ such that
$\begin{aligned} AB^2 + BC \cdot CP &= BC^2 \\ 3AC^2 + 2BC \cdot CQ &= BC^2 \end{aligned}$

Find $\angle PAQ$ in degrees.

Solution

Solution:

Find $\angle PAQ$ in degrees.

Answer: $40^\circ$

We have $AB^2 = BC(BC - CP) = BC \cdot BP$ , so triangle $ABC$ is similar to triangle $PBA$ .

Also, $AB^2 = BC(BC - 2CQ) + AC^2 = (BC - CQ)^2 - CQ^2 + AC^2$ , which rewrites as $AB^2 + CQ^2 = BQ^2 + AC^2$ .

We deduce that $Q$ is the foot of the altitude from $A$ .

Thus, $\angle PAQ = 90^\circ - \angle QPA = 90^\circ - \angle ABP - \angle BAP$ .

Using the similar triangles, $\angle PAQ = 90^\circ - \angle ABC - \angle BCA = \angle BAC - 90^\circ = 40^\circ$ .

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