# Problem of the Week Problem E and Solution Fold Once

## Problem

A rectangular piece of paper, $$PQRS$$, has $$PQ = 30$$ cm and $$PS = 40$$ cm. The paper has grey lines on one side and is plain white on the other.

The paper is folded so that the two diagonally opposite corners $$P$$ and $$R$$ coincide. This creates a crease along line segment $$AC$$, with $$A$$ on $$PS$$ and $$C$$ on $$QR$$.

Determine the length of $$AC$$.

## Solution

Since $$PQRS$$ is a rectangle, all angles inside $$PQRS$$ are $$90^\circ$$. After the fold, $$R$$ coincides with $$P$$. Label the point that $$S$$ folds to as $$D$$. The angle at $$D$$ is the same as the angle at $$S$$. Since $$PQRS$$ is a rectangle, $$\angle PSR=90^\circ$$ and $$SR = 30$$, and it follows that $$\angle PDA=90^\circ$$ and $$PD = 30$$.

Let $$a$$ represent the length of $$QC$$ and $$b$$ represent the length of $$AS$$.
Then $$CR=QR - QC=40-a$$ and $$PA=PS-AS=40-b$$.

Since $$S$$ folds to $$D$$, it follows that $$AD=AS=b$$.
Since $$R$$ folds to $$P$$, it follows that $$PC=CR=40-a$$.

All of the information is recorded on the following diagram.

Since $$\triangle PQC$$ is a right-angled triangle, we can use the Pythagorean Theorem to find $$a$$. \begin{aligned} QC^2+PQ^2&=PC^2\\ a^2+30^2&=(40-a)^2\\ a^2+900&=1600-80a+a^2\\ 80a&=700\\ a&=\dfrac{35}{4}\end{aligned}

Since $$\triangle PDA$$ is a right-angled triangle, we can use the Pythagorean Theorem to find $$b$$. \begin{aligned} AD^2+PD^2&=PA^2\\ b^2+30^2&=(40-b)^2\\ b^2+900&=1600-80b+b^2\\ 80b&=700\\ b&=\dfrac{35}{4}\end{aligned}

Therefore, $$a=b=\dfrac{35}{4}$$ cm.

We still need to find the length of the crease. From $$A$$ drop a perpendicular to $$QR$$ intersecting $$QR$$ at $$B$$. Then $$ABRS$$ is a rectangle. It follows that $$AB=SR=30$$ and $$BR=AS=b$$.
Also, $$CB=QR-QC-BR=40-a-b=40-\dfrac{35}{4}-\dfrac{35}{4}=\dfrac{90}{4}=\dfrac{45}{2}$$ cm.

Using the Pythagorean Theorem in $$\triangle CAB$$, \begin{aligned} AC^2&=AB^2+CB^2\\ &=30^2+\left(\dfrac{45}{2}\right)^2\\ &=900+\dfrac{2025}{4}\\ &=\dfrac{5625}{4}\end{aligned} Since $$AC >0$$, it follows that $$AC=\dfrac{75}{2} = 37.5$$ cm. Therefore, the length of the crease is $$\dfrac{75}{2}$$ cm.