Set Square
A triangle PQR, right angled at P, slides on a horizontal floor with Q and R in contact with perpendicular walls. What is the locus of P?
Problem
A triangle PQR, right angled at P, slides on a horizontal floor
with Q and R in contact with perpendicular walls. What is the locus
of P?
Getting Started
You might look for a cyclic quadrilateral.
We have constructed this in geogebra to help
Created with GeoGebra
NOTES AND BACKGROUND
You might like to download your own free copy of GeoGebra from the link above and draw this dynamic diagram for yourself. You will find it easy to get started on GeoGebra with the Quickstart guide for beginners.
Student Solutions
The locus of the point $P$ is a straight line as the vertices $Q$ and $R$ slide along the walls.
Image
| 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 $P$. |
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?