# Interscholastic Mathematics League Senior B Division #2

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 $ABCD$.

• Um, is this an ongoing competition? It would be nice to be assured that answers here will be to late to be of any help with cheating. – hmakholm left over Monica Oct 27 '11 at 0:54
• Yes this competition has already been completed. No one in my class knew a fast shortcut to answer this problem, so that's why I came here. ;) – Tayyeb Din Oct 27 '11 at 0:56
• 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. – André Nicolas Oct 28 '11 at 19:45

## 1 Answer

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.

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 $\frac{5}{9}$.

• Thank you! But the question asked for the ratio of PQRS to the area of ABCD, not the interestion of PQRS and ABCD to ABCD. The answer was 5:9. – Tayyeb Din Oct 27 '11 at 0:43
• PQRS is wholly contained in ABCD, so "the intersection of PQRS and ABCD" is the same thing as "PQRS". – hmakholm left over Monica Oct 27 '11 at 0:59