HMMT February, 2024

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle. Let $X$ be the point on side $\overline{AB}$ such that $\angle BXC = 60^{\circ}$ . Let $P$ be the point on segment $\overline{CX}$ such that $BP \perp AC$ . Given that $AB = 6$ , $AC = 7$ , and $BP = 4$ , compute $CP$ .

Solution

Solution:

Construct parallelogram $BPCQ$ . We have $CQ = 4$ , $\angle ACQ = 90^{\circ}$ , and $\angle ABQ = 120^{\circ}$ . Thus, $AQ = \sqrt{AC^{2} + CQ^{2}} = \sqrt{65}$ , so if $x = CP = BQ$ , then by Law of Cosine, $x^{2} + 6x + 6^{2} = 65$ . Solving this gives the answer $x = \sqrt{38} - 3$ .

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