# Problem of the Week Problem D and Solution From Square to Hexagon

## Problem

A square piece of paper, $$PQRS$$, has side length $$40$$ cm. The page is grey on one side and white on the other side. Point $$M$$ is the midpoint of side $$PQ$$ and point $$N$$ is the midpoint of side $$PS$$.

The paper is folded along $$MN$$ so that $$P$$ touches the paper at the point $$P'$$.

Point $$T$$ lies on $$QR$$ and point $$U$$ lies on $$SR$$ such that $$TU$$ is parallel to $$MN$$, and when the paper is folded along $$TU$$, the point $$R$$ touches the paper at the point $$R'$$ on $$MN$$.

What is the area of hexagon $$NMQTUS$$?

Here are some known properties of the diagonals of a square that may be useful:

• the diagonals are equal in length; and

• the diagonals right bisect each other; and

• the diagonals bisect the corner angles.

## Solution

To determine the area of hexagon $$NMQTUS$$, we will subtract the area of $$\triangle PMN$$ and the area of $$\triangle TRU$$ from the area of square $$PQRS$$.

Since $$M$$ and $$N$$ are the midpoints of $$PQ$$ and $$PS$$, respectively, we know $$PM=\frac{1}{2}(PQ)=20$$ cm and $$PN=\frac{1}{2}(PS)=20$$ cm. Therefore, $$PM=PN=20$$ and $$\triangle PMN$$ is an isosceles right-angled triangle. It follows that $$\angle PNM=\angle PMN=45\degree$$.

After the first fold, $$P$$ touches the paper at $$P'$$. $$\triangle P'MN$$ is a reflection of $$\triangle PMN$$ in the line segment $$MN$$. It follows that $$\angle P'MN=\angle PMN=45\degree$$ and $$\angle P'NM=\angle PNM=45\degree$$. Therefore, $$\angle PMP'=\angle PNP'=90\degree$$. Since all four sides of $$PMP'N$$ are equal in length and all four corners are $$90\degree$$, $$PMP'N$$ is a square.

Since $$\angle MPP'=\angle MPR=45\degree$$, the diagonal $$PP'$$ of square $$PMP'N$$ lies along the diagonal $$PR$$ of square $$PQRS$$. Let $$O$$ be the intersection of the two diagonals of square $$PMP'N$$. It is also the intersection of $$MN$$ and $$PR$$. (We will show later that this is in fact $$R'$$, the point of contact of $$R$$ with the paper after the second fold.)

The length of the diagonal of square $$PMP'N$$ can be found using the Pythagorean Theorem. $PP'=\sqrt{(PM)^2+(MP')^2}=\sqrt{20^2+20^2}=\sqrt{800}=\sqrt{400}\sqrt{2}=20\sqrt{2}$

Thus, $$PO=\frac{1}{2}(PP')=\frac{1}{2}(20\sqrt{2})=10\sqrt{2}$$ cm.

In the last two steps of calculating $$PP'$$, we simplified the radical. We will do this quite often in the solution. Here is the process to simplify radicals, for students who may not be familiar with this:

• Find the largest perfect square that divides into the radicand (the number under the root symbol). In this case, 400 is the largest perfect square that divides 800.

• Rewrite the radicand as the product of the perfect square and the remaining factor. In this case, we get $$\sqrt{400\times 2}$$.

• Take the square root of the perfect square. In this case, we get $$20\sqrt{2}$$.

Since $$TU$$ is parallel to $$MN$$, it follows that $$\angle RTU=\angle RUT=45\degree$$ and $$\triangle TRU$$ is an isosceles right-angled triangle with $$TR=RU$$.

When $$\triangle TRU$$ is reflected in the line segment $$TU$$ with $$R'$$ being the image of $$R$$, a square, $$TRUR'$$, is created. We will not present the argument here because it is very similar to the argument presented for $$PMP'N$$. Since $$\angle TRR'=\angle TRP=45\degree$$, $$RR'$$ lies along the diagonal $$PR$$. Also, $$R'$$ lies on $$MN$$. This means that $$R'$$ and $$O$$ are the same point and so $$PR' = PO = 10\sqrt{2}$$ cm.

The length of the diagonal of square $$PQRS$$ can be calculated using the Pythagorean Theorem. $PR=\sqrt{(PQ)^2+(QR)^2}=\sqrt{40^2+40^2}=\sqrt{3200}=\sqrt{1600}\sqrt{2}=40\sqrt{2}$

The length of $$RR'$$ equals the length of $$PR$$ minus the length of $$PR'$$. $RR'= PR - PR' = 40\sqrt{2}-10\sqrt{2}=30\sqrt{2}$

But $$RR'=TU$$, so $$TU=30\sqrt{2}$$ cm. Let $$TR=RU=x$$. Then, using the Pythagorean Theorem in $$\triangle TRU$$, \begin{aligned} (TR)^2+(RU)^2&=(TU)^2\\ x^2+x^2&=(30\sqrt{2})^2\\ x^2+x^2&=900\times 2\\ 2x^2&=1800\\ x^2&=900\end{aligned} And since $$x>0$$, this gives $$x=30$$ cm. We now have enough information to calculate the area of hexagon $$NMQTUS$$. \begin{aligned} \text{Area }NMQTUS&=\text{Area }PQRS-\text{Area }\triangle PMN-\text{Area }\triangle TRU\\ &=PQ\times QR-\frac{PM\times PN}{2}-\frac{TR\times RU}{2}\\ &=40\times 40-\frac{20\times 20}{2}-\frac{30\times 30 }{2}\\ &=1600-\frac{400}{2}-\frac{900}{2}\\ &=1600-200 - 450\\ &=950\end{aligned} Therefore, the area of hexagon $$NMBPQD$$ is $$950\text{ cm}^2$$.