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#1 h Yesterday at 8:21 AM

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Let $\omega$ be a circle inscribed inside a rhombus $ABCD$ . Let $P$ and $Q$ be variable points on $AB$ and $AD$ respectively, such as $PQ$ is always the tangent line to $\omega$ .
Prove that for any position of $P$ and $Q$ the area of triangle $\triangle CPQ$ is the same.

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#2 h Yesterday at 9:44 AM

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An interesting concept.

First, gleefully ignore configuration issues by assuming $\omega$ lies outside $\triangle APQ$ . The other case is analogous to this.

Now note that the distance from $C$ to $PQ$ is the distance from $A$ to $PQ$ plus two times the radius of $\omega$ . Thus, the area of $CPQ$ is simply that amount multiplied by $\frac{PQ}{2}$ ; this results in $[CPQ] = [APQ] + 2[OPQ]$ where $O$ is the center of $\omega$ . But then this is precisely the area of $\triangle ADE$ where $D, E$ are the foot of the altitudes from $O$ to $AB$ , $AD$ . This area is obviously constant, so we are done.

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