#
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.