You may also like

Five Circuits, Seven Spins

A circular plate rolls inside a rectangular tray making five circuits and rotating about its centre seven times. Find the dimensions of the tray.

Mach Attack

Have you got the Mach knack? Discover the mathematics behind exceeding the sound barrier.

Construct the Solar System

Make an accurate diagram of the solar system and explore the concept of a grand conjunction.

Set Square

Age 16 to 18 Challenge Level:

The locus of the point $P$ is a straight line as the vertices $Q$ and $R$ slide along the walls.

Sue Liu of Madras College, St Andrew's sent this solution to the problem. Without loss of generality we can let the length of $QR$ be 1 unit, and take a coordinate system with the origin at $O$ and axes along $OR$ and $OQ$.

If $\angle PQR = \alpha$, where $0 < \alpha < 90^\circ$, then $PQ = \cos \alpha$ and $PR = \sin \alpha.$ Let $\angle QRO = \theta $ where $0 \leq \theta \leq 90^\circ$. Then, from the right angled triangles $PSQ$ and $PTR$, we have $\angle PRT = \angle QPS = \alpha - \theta$, and hence we can write down the coordinates of the point $P$.

\begin{eqnarray} x &=& \cos \alpha \cos (\alpha - \theta) \\ y &=& \sin \alpha \cos (\alpha - \theta). \end{eqnarray}
We see that $${y\over x}= {\sin \alpha \cos (\alpha - \theta)\over \cos \alpha \cos (\alpha - \theta)}= \tan \alpha.$$ and so $P$ lies on the straight line $y = x\tan \alpha$.

The position $(x,y)$ depends only on $cos(\alpha - \theta)$, $\alpha$ being a constant, and $\theta$ a variable. The distance of the point $P$ from $O$ is given by $$OP^2 = x^2 + y^2 = \cos^2(\alpha - \theta)(\cos^2\alpha + \sin^2\alpha) = \cos^2(\alpha - \theta).$$ Hence $OP = \cos(\alpha - \theta)$ which is a maximum when $\cos(\alpha - \theta) = 1$, that is when $\alpha = \theta$. This occurs when $OQPR$ is a rectangle as shown in the diagram.

We get an even simpler method of solution by using the fact that the angles $QOR$ and $QPR$ are both 90 degrees so that $OQPR$ is a cyclic quadrilateral with $PR$ as a chord. We have $\angle POR = \angle PQR = \alpha$ because these two angles are subtended by the same chord of the circle. This shows that $\angle POR$ is constant and hence that the locus of $P$ is the straight line $y = x \tan \alpha.$

What can you say about the locus of $P$ if the triangle $PQR$ is not a right angled triangle?