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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=1/3 AB. Compute the ratio of the area of PQRS to the area of ABCD.

The special case where ABCD is a square is solved here: Interscholastic Mathematics League Senior B Division #2

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marked as duplicate by Jonas Meyer, mixedmath Jan 1 '15 at 5:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

if you really want someone to answer your question, you should take the time to write it out carefully. Posting a link to another question and vaguely asking someone to work out a special case is not helpful. Please specify exactly what you're looking for. – Bill Cook Oct 28 '11 at 2:10
So this is an exact duplicate? – The Chaz 2.0 Oct 28 '11 at 2:16
I'm not sure -- might be. It's hard to tell what Victor is looking for. – Bill Cook Oct 28 '11 at 2:18
@BillCook - i am trying to solve that problem without using the rectangle – Victor Oct 28 '11 at 2:47
up vote 1 down vote accepted

Let $\alpha = AB$ and $\beta = AD$ be the side lengths of the original parallelogram. If you combine the corner triangles obtained from the construction of $PQRS$ then $PBQ + SDR = PBRD$, a parallelogram with side lengths $\frac{\alpha}{3}$ and $\frac{2\alpha}{3}$. Analogously, the other two triangles gives $APS + RCQ = AQCR$ with side lengths $\frac{\alpha}{3}$ and $\beta - \frac{\alpha}{3}$. The area of a parallelogram is given by

$$A = \alpha\beta\sin\theta$$

where $\theta$ is an angle of the parallelogram. A key observation here is that the angles of the two smaller parallelograms formed are equal to the original. In particular, $PBRD$ contains $\angle ABC$ and $AQCR$ contains $\angle BCD$. Note that this would be enough to reduce the construction to an arbitrary rectangle but not to a square. It turns out that this ratio will depend on the ratio of the side lengths.

Taking the ratio, we have

$$\frac{A_{PQRS}}{A_{ABCD}} = \frac{\alpha\beta\sin\theta - \frac{2\alpha^2}{9}\sin\theta - \frac{\alpha}{3}\left(\beta - \frac{\alpha}{3}\right)\sin\theta}{\alpha\beta\sin\theta}$$

canceling $\alpha$ and $\sin\theta$ and simplifying gives us

$$\frac{A_{PQRS}}{A_{ABCD}} = \frac{\beta - \frac{2\alpha}{9} - \frac{\beta}{3} + \frac{\alpha}{9}}{\beta} = \frac{2}{3} - \frac{1}{9}\cdot\frac{\alpha}{\beta}$$

In the case $\alpha = \beta$ this gives us

$$\frac{A_{PQRS}}{A_{ABCD}} = \frac{5}{9}$$

as required.

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