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Points $P$, $Q$, $R$ and $S$ are chosen on the sides of parallelogram $ABCD$, so that $P$ is on line $AB$, $Q$ is on line $BC$, $R$ is on line $CD$, $S$ is on line $DA$, and $AP=BQ=CR=DS=\frac{1}{3}(AB)$. Compute the ratio of the area of $PQRS$ to the area of $ABCD$.

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    I meant to comment, but forgot for a while. Sorry! There is something wrong with the problem. I can change the parallelogram and get different ratios. I am sure that with a little work you can also. My guess is that there was a transcription error, and we should have $BQ=(1/3)BC$, and similar things for the others. Then it is straightforward to give a general solution.2011-10-28

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A nice trick for solving problems like this is that the desired ratio must be the same for any parallelogram with that construction, so we can choose a convenient one and just solve the special case.

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Let ABCD be a square of side length 3; then PQRS is a square inside of ABCD excluding four right triangles in the corners, each of which has legs length 1 and 2. The area of ABCD is 9, and the area of the excluded triangles is $4(\frac{1}{2}*1*2) = 4$, so the area of PQRS is 5. Therefore the ratio of the area of PQRS to the area of ABCD is $\frac{5}{9}$.

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    PQRS is wholly contained in ABCD, so "the intersection of PQRS and ABCD" is the same thing as "PQRS".2011-10-27