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2025/26 AHSMC Part 1 Problem 16

Problem 16 of 16

PP is a point inside the triangle △ABC\triangle ABC. We know that ∠BAP=30∘\angle BAP = 30^\circ, ∠CAP=20∘\angle CAP = 20^\circ, and ∠ACP=50∘\angle ACP = 50^\circ and that the length of the segment APAP is equal to the length of the segment BCBC. Find the angle ∠BCP\angle BCP.

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Solution

Let x=∠BCPx = \angle BCP. Then ∠APC=110∘\angle APC = 110^\circ and ∠ABC=80∘−x\angle ABC = 80^\circ - x. The law of sines in △APC\triangle APC and △ABC\triangle ABC gives AP=AC⋅sin⁡50∘sin⁡110∘,BC=AC⋅sin⁡50∘sin⁡(80∘−x). AP = AC \cdot \frac{\sin 50^\circ}{\sin 110^\circ}, \qquad BC = AC \cdot \frac{\sin 50^\circ}{\sin(80^\circ-x)}. Since AP=BCAP = BC we get sin⁡110∘=sin⁡(80∘−x)\sin 110^\circ = \sin(80^\circ-x). The angles cannot be equal (as 80∘−x<80∘<110∘80^\circ-x < 80^\circ < 110^\circ), so they are supplementary: 110∘+(80∘−x)=180∘110^\circ + (80^\circ-x) = 180^\circ, hence x=10∘x = 10^\circ.

Alternatively, reflecting PP over ACAC to QQ makes ABCQABCQ an isosceles trapezoid, whence ∠ABC=70∘\angle ABC = 70^\circ and ∠BCP=10∘\angle BCP = 10^\circ. The answer is (B).

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