Problem 487

The rectangle in Figure A has dimensions $7\times8$, and thus has perimeter $2\times7+2\times8$ or 30.
Solution 1

We begin by labelling Figure B as shown.

The width of Figure B is $PU=7$.

However, the width is also equal to $QR+ST$.

Since $QR=3$, then $ST=7-3=4$.

Similarly, $PQ+RS=UT=8$ and since $RS=1$, then $PQ=7$.

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The perimeter of Figure B is $PQ+QR+RS+ST+TU+UP=7+3+1+4+8+7=30$.

Solution 2

The answer in Solution 1, 30, is equal to the answer inpart (a). Why?

Consider sliding $RS$ horizontally left to $QV$, and sliding $QR$ vertically down to $VS$, as shown.

Since $QRSV$ is a rectangle, then $PVTU$ is a rectangle.

[[IMAGE1]]

Further, since $RS=QV$ and $QR=VS$, then the perimeter of Figure B is equal to the perimeter of rectangle $PVTU$.

Since $PV=UT=8$ and $VT=PU=7$, the perimeter of rectangle $PVTU$ is 30 (this is the rectangle from part (a)).

Therefore, the perimeter of Figure B is 30.
Following the approach in Solution 2 above, the perimeter of Figure C is equal to the perimeter of a rectangle with side lengths $k+4$ and $k+2$.

Since the perimeter of Figure C is 56, then $2(k+4)+2(k+2)=56$ or $2k+8+2k+4=56$ and so $4k=44$ or $k=11$.

(Alternatively, we could have determined that the two missing lengths in Figure C are each equal to $k$, and then found the perimeter of Figure C, $4k+12$, by adding the lengths of the six segments.)
Each of the segments having lengths 4 or 7 may be “pushed out” to show that the perimeter of Figure D is equal to the perimeter of a square with side length $8n+1$.

That is, the perimeter of Figure D is $4(8n+1)=32n+4$.

We require the largest integer $n$ for which $32n+4<1000$.

Solving this inequality, we get $32n<996$ or $n<\frac{996}{32}$ and so $n<31.125$.

Thus, the largest integer $n$ for which the perimeter of Figure D is less than 1000 is 31.