Ptolemy's theorem for olympiads

This is part of a series of articles based on my high school research paper from 2021 covering nonstandard approaches to solving olympiad problems. The problems from the Latvian Open Mathematics Olympiad mentioned in the article can be found in the LU NMS archive.

Ptolemy’s theorem is a somewhat rare, but generally very simple statement in geometry. Let’s quickly introduce ourselves to this formula and check out one problem where it can be used.

Ptolemy’s theorem

Ptolemy’s theorem is typically presented as follows:

$ABCD$ is a cyclic quadrilateral if and only if $AD\cdot BC+AB\cdot CD=AC\cdot BD.$

So, if $ABCD$ is cyclic, then the sum of the products of the opposite side lengths equals the product of the diagonal lengths. And conversely: if the sum of the products of the opposite side lengths equals the product of the diagonal lengths, then the rectangle must be cyclic.

Olympiad example

Latvian Open Mathematics Olympiad 2016/2017, Form 10, Problem 3.
Given rectangle $ABCD,$ a circle is drawn through $A$ which intersects $AB$, $AC$ and $AD$ at points $P$, $Q$ and $R$ respectively. Prove that \[AB\cdot AP+AD\cdot AR=AC\cdot AQ.\]

In the problem, we’re pretty much given the cyclic quadrilateral $APQR$ right away, and the equation we wish to prove looks a lot like Ptolemy’s theorem. Thus, it feels logical that we might make use of \[QR\cdot AP+QP\cdot AR=PR\cdot AQ.\] On top of that, the second factor in each product in this equation matches with the equation we want to prove, so we could first investigate relationships between $QR,QP,PR$ and $AB,AD,AC.$

Since $APQR$ is cyclic, \[\angle RQP=180^\circ-\angle RAP=180^\circ-90^\circ=90^\circ.\] Then, $\triangle ABC\sim\triangle RQP$ (AA) because $\angle ABC=\angle RQP=90^\circ$ and $\angle BAC=\angle QRP$, since both angles span the chord $PQ.$ The similarity implies $\frac{AB}{QR}=\frac{AC}{PR}.$

And $\triangle CDA\sim\triangle RQP$ (AA) because $\angle CDA=\angle RQP=90^\circ$ and $\angle CAD=\angle RPQ$, since both angles span the chord $QR.$ The similarity implies $\frac{AD}{PQ}=\frac{AC}{PR}.$

Call $\frac{AC}{PR}$ the similarity coefficient $k.$ Combining the equations, we get \[\frac{AB}{QR}=\frac{AC}{PR}=\frac{AD}{PQ}=k,\] hence \[QR=\frac{AB}k,~PR=\frac{AC}k,~PQ=\frac{AD}k.\]

Lastly, since $APQR$ is cyclic, Ptolemy’s theorem gives \[QR\cdot AP+QP\cdot AR=PR\cdot AQ.\] Plugging in the expressions found earlier gives \[\frac{AB}k\cdot AP+\frac{AD}k\cdot PQ=\frac{AC}k\cdot AQ,\] and multiply both sides by $k$ gives the equation we sought to prove, \[{AB}\cdot AP+{AD}\cdot PQ={AC}\cdot AQ.\]

Final thoughts

I’ve always been under the impression that the newest olympiad participants tend to find geometry in particular to be the most challenging. Ptolemy’s theorem is a relatively simple idea, so this statement is worth memorising, just in case it turns out to be useful in a problem. However, the most important thing, as usual, is to practice geometry problems, for example, from archives - that helps the most.