Problem 13 · Perimeter of a shaded region
Three rectangles tilted 45° along the diagonal of a square.
Integer answer, at most 4 digitsIn the figure, $ABCD$ is a square while $EFGH$, $HIKL$ and $LMNP$ are just rectangles. Side $AB$ of the square measures $16\sqrt{2}$ cm and segment $ID$ measures $6\sqrt{2}$ cm. What is the perimeter of the shaded region, in centimetres?
Reasoned solution
Key idea: all sides of the three rectangles make $45°$ with the sides of the square, and points $H$ and $L$ lie on the diagonal $AC$. The perimeter of the two small rectangles does not depend on their exact position!
The diagonal of the square measures $AC = 16\sqrt{2}\cdot\sqrt{2} = 32$ cm.
Central rectangle $HIKL$ (side $HL$ on the diagonal). We have $AI = AD - ID = 16\sqrt{2} - 6\sqrt{2} = 10\sqrt{2}$. Side $AD$ makes $45°$ with the diagonal and $IH \perp AC$, so triangle $AIH$ is an isosceles right triangle:
The same argument at vertex $C$ (with $K$ on side $DC$ and $KL \perp AC$, and $KL = IH = 10$ as opposite sides) gives $CL = KL = 10$. Therefore
Rectangle $EFGH$ ($F$ on side $AB$, $H$ on the diagonal). Sides $FE$ and $FG$ make $45°$ with $AB$, so the vertical “depth” from $F$ down to $H$ equals $(FE+FG)\cdot\frac{\sqrt{2}}{2}$. But that depth is the distance from $H$ to side $AB$, which equals $AH\cdot\frac{\sqrt{2}}{2} = 5\sqrt{2}$. Hence
wherever point $F$ happens to be.
Rectangle $LMNP$ ($N$ on side $BC$, $L$ on the diagonal): the same argument with the distance from $L$ to side $BC$, which equals $CL\cdot\frac{\sqrt{2}}{2} = 5\sqrt{2}$, gives perimeter $20$.
The three rectangles touch only at the points $H$ and $L$, so the perimeter of the shaded region is the sum: