UKMT Maclaurin 2023 - Q3

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Olympiad
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UKMT Maclaurin

Published on Mon Mar 20 2023. Views

My solution to the third question of the UKMT Maclaurin 2023 paper.

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UKMT Maclaurin 2023 Question 1 Solution

Problem Statement

ABCD is a square and X is a point on the side DA such that the semicircle with diameter CX touches the side AB. Find the ratio AX:XD.

My Solution

A diagram like the following can be drawn:

Let

$r = XO = OC$ be the radius of the semicircle
the length of the square be $2$

Now, if we were to construct the line $OQ$ perpendicular to $DC$ through $O$ and $PO$ perpendicular to $AD$ :

We obtain two congruent triangles, namely $\Delta XOP$ and $\Delta OCQ$ , since they have equal angles and a hypotenuse of the same length (the radius).

Hence, $PO = DQ = QC$ . So $Q$ is the midpoint of $DC$ .

As the side length of the square is $2$ , $DQ = QC = 1$ .

By Pythagoras' theorem:

\begin{aligned} OQ = \sqrt{r^2 - 1^2} \end{aligned}

As $TQ$ is equal to the side length of the square: $TO + OQ = 2$ , hence:

\begin{aligned} r + \sqrt{r^2 - 1} &= 2 \\ \sqrt{r^2 - 1} &= 2 - r \\ r^2 - 1 &= 4 - 4r + r^2 \\ 4r &= 5 \\ r &= \frac{5}{4} \end{aligned}

Thus, the length of the diameter $XC = 2r = \frac{5}{2}$ .

By Pythagoras' theorem:

\begin{aligned} XD &= \sqrt{XC^2 - DC^2} \\ &= \sqrt{\frac{25}{4} - \frac{16}{4}} = \frac{3}{2} \\ \\ \therefore AX &= 2 - XD = \frac{1}{2} \end{aligned}

The final answer is:

\begin{aligned} AX:XD = 1 : 3 \end{aligned}