# Problem of the Week Problem E and Solution Shape Building

## Problem

Sina drew square $$ABCD$$ with side length $$6$$ cm on a piece of paper and passed the paper to Theo. Theo drew a circle on top of the square so that the circle passes through $$A$$ and $$D$$, and the circle is tangent to side $$BC$$ at point $$P$$.

Determine the radius of the circle.

Note: You may find the following known result about circles useful:

If a line is tangent to a circle, then the perpendicular to that line at the point of tangency passes through the centre of the circle.

## Solution

Let $$O$$ be the centre of the circle and $$r$$ be the radius. Construct line segment $$PQ$$ perpendicular to $$CB$$ with $$Q$$ on side $$AD$$ of the square. Since $$CB$$ is tangent to the circle with point of tangency $$P$$, $$PQ$$ must pass through the centre of the circle, $$O$$. Therefore, $$PO=r$$.

Since $$PQ\perp BC$$, $$PQ\parallel AB,$$ and $$PQ=AB=6$$, then $$QO=PQ-PO=6-r$$. Since $$A$$ and $$D$$ are on the circle, $$AO=DO=r$$.

Using the Pythagorean Theorem, $$AQ^2=AO^2-QO^2=r^2-(6-r)^2$$ and $$DQ^2=DO^2-QO^2=r^2-(6-r)^2$$. Therefore, $$AQ^2=DQ^2$$ and $$AQ=DQ$$ follows. Since $$AQ=DQ$$ and $$AQ+QD=AD=6$$, we can substitute to obtain $$AQ+AQ=2AQ=6$$ or $$AQ=3$$.

Using the Pythagorean Theorem in $$\triangle AQO$$, \begin{aligned} AO^2&=AQ^2+QO^2\\ r^2&=3^2+(6-r)^2\\ r^2&=9+36-12r+r^2\\ 12r&=45\\ r&=\frac{45}{12}\\ r&=3.75 \end{aligned} Therefore, the radius of the circle is $$3.75$$ cm.