Recently, I've asked for help from my teacher, as well as gathering help from other people. At the moment, what troubles me is how to even start to find the maximum and minimum points without a calculator, other than that there is some symmetry in the question, so if $PC/PD$ is a maximum where point $P$ is one side of the line, then if point $P$ was on the opposite side of the line $AB$, then it would be the minimum instead. As well as this, given that this was once an Olympiad question, I haven't found any simple solutions to the problem without the use of a computer or calculator to maximise and minimise it. From some research, the only other source of help I had found were hints from The Mathematical Olympiad Handbook- an Introduction to Problem Solving based on the First 32 British Mathematical Olympiads 1965-1996
If anyone has solutions to the question without using computation, it would be much appreciated. Preferably, simplify it and show each step, as this is meant to be a question aimed at students, and I just want to see a solution in full and see the process.
Here is the question written here in full:
(1990 British Mathematical Olympiad Round 1 – Question 2)
$ABCD$ is a square and $P$ is a point on the line $AB.$ Find the maximum and minimum values of the ratio $PC/PD$, showing that these occur for the points $P$ given by $$AP \times BP=AB^2$$
Best Answer
This can be answered in a geometrical way, if you remember that the locus of points $P$ in a given ratio $r$ to two points $C$ and $D$ is a circle. If $a=CD$, then this circle has a radius given by $$ R={ar\over1-r^2} $$ (I'm supposing $r<1$, in the other case just reflect everything about the perpendicular bisector of $CD$) and its center $O$ lies on line $CD$, at a distance $$ x={ar^2\over1-r^2} $$ from $C$. When $r$ is minimum then $R$ is minimum too. Hence point $P$ on line $AB$ corresponding to a minimum of $r$ is the tangency point when that circle is tangent to $AB$, that is when $R=a$. This is equivalent to $$ {r\over1-r^2}=1 $$ which also entails $$ x(a+x)=a^2, $$ as it was to be proved.