Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Right triangle $X Y Z$ has right angle at $Y$ and $X Y = 228$ , $Y Z = 2004$ . Angle $Y$ is trisected, and the angle trisectors intersect $X Z$ at $P$ and $Q$ so that $X, P, Q, Z$ lie on $X Z$ in that order. Find the value of $(P Y + Y Z)(Q Y + X Y)$ .

Solution

Solution:

The triangle's area is $(228 \cdot 2004) / 2 = 228456$ . All the angles at $Y$ are $30$ degrees, so by the sine area formula, the areas of the three small triangles in the diagram are $Q Y \cdot Y Z / 4$ , $P Y \cdot Q Y / 4$ , and $X Y \cdot P Y / 4$ , which sum to the area of the triangle. So expanding $(P Y + Y Z)(Q Y + X Y)$ , we see that it equals
$4 \cdot 228456 + X Y \cdot Y Z = 6 \cdot 228456 = 1370736$

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