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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Soviet Union

Problem:

The quadrilateral $ABCD$ is inscribed in a fixed circle. It has $AB$ parallel to $CD$ and the length $AC$ is fixed, but it is otherwise allowed to vary. If $h$ is the distance between the midpoints of $AC$ and $BD$ and $k$ is the distance between the midpoints of $AB$ and $CD$ , show that the ratio $h/k$ remains constant.

Solution

Solution:

Let the center of the circle be $O$ and its radius be $R$ . Let $\angle AOB = 2x$ (variable) and let $\angle AOC = 2y$ (fixed). Then $AC = 2R \sin y$ . We find $\angle COD = 180^\circ - x - 2y$ , so $2h = 2R \sin x + 2R \sin(x + 2y)$ . Angle $ACD = x + y$ , so $k = AB \sin(x + y) = 2R \sin(x + y) \sin y$ . Hence the ratio $h / k = (\sin x + \sin(x + 2y)) / (2 \sin(x + y) \sin y)$ . We have $\sin x = \sin(x + y - y) = \sin(x + y) \cos y - \cos(x + y) \sin y$ , and $\sin(x + 2y) = \sin((x + y) + y) = \sin(x + y) \cos y + \cos(x + y) \sin y$ , so $(\sin x + \sin(x + 2y)) = 2 \sin(x + y) \cos y$ . Hence $h / k = \cot y$ , which is constant.

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