Points $P$, $Q$, $R$, and $S$ are taken on sides $AB$, $BC$, $CD$, and $DA$ of square $ABCD$ respectively, so that $\frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = \frac{1}{n}$. Then the ratio of the area of $PQRS$ to the area of $ABCD$ is

Since $\frac{AP}{PB} = \frac{1}{n}$, the parts are in ratio $1 : n$.

Let $AP = 1$ and $PB = n$, so the side of the square is:

$AB = 1 + n$

The same split applies to every side of the square.

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Look at $\triangle BPQ$:

$PB = n$, $BQ = 1$, and $\angle B = 90°$ (corner of the square)

By the Pythagorean theorem:

$PQ = \sqrt{n^2 + 1}$

By symmetry, every corner triangle has legs $n$ and $1$ with a $90°$ angle between them, so:

$PQ = QR = RS = SP = \sqrt{n^2 + 1}$

All four sides of $PQRS$ are equal, and since each corner triangle is identical, all four angles of $PQRS$ are equal. Four equal angles summing to $360°$ means each angle $= 90°$.

So $PQRS$ is a square.

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Since both $PQRS$ and $ABCD$ are squares:

$\dfrac{\text{Area of } PQRS}{\text{Area of } ABCD} = \dfrac{(\text{side of } PQRS)^2}{(\text{side of } ABCD)^2}$

$= \dfrac{(\sqrt{n^2 + 1})^2}{(n + 1)^2}$

$= \dfrac{n^2 + 1}{(n + 1)^2}$

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Sanity check: let $n = 1$ (midpoints of each side)

$\dfrac{1 + 1}{(1 + 1)^2} = \dfrac{2}{4} = \dfrac{1}{2}$

This matches the well-known result that connecting midpoints of a square gives a square with half the area. ✅