Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

In triangle ABCABC, ∠ABC=50∘\angle ABC = 50^{\circ} and ∠ACB=70∘\angle ACB = 70^{\circ}. Let DD be the midpoint of side BCBC. A circle is tangent to BCBC at BB and is also tangent to segment ADAD; this circle intersects ABAB again at PP. Another circle is tangent to BCBC at CC and is also tangent to segment ADAD; this circle intersects ACAC again at QQ. Find ∠APQ\angle APQ (in degrees).

Solution

Solution:

Suppose the circles are tangent to ADAD at E,FE, F, respectively; then, by equal tangents, DE=DB=DC=DF⇒E=FDE = DB = DC = DF \Rightarrow E = F (as shown). So, by the Power of a Point Theorem, AP⋅AB=AE2=AF2=AQ⋅AC⇒AP/AQ=AC/AB⇒△APQ∼△ACBAP \cdot AB = AE^{2} = AF^{2} = AQ \cdot AC \Rightarrow AP / AQ = AC / AB \Rightarrow \triangle APQ \sim \triangle ACB, giving ∠APQ=∠ACB=70∘\angle APQ = \angle ACB = 70^{\circ}.

Figure 1

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