Intersections of the Nine-Point Circle and (BHC)

In this article, we shall explore some properties of the intersections two highly celebrated circles in the world of Olympiad geometry. Although the individual circles are thoroughly explored and considered mostly ‘well-known’, the configuration involving their intersections is still relatively unexplored, with only two notable examples appearing on an Olympiad to date, as far as I am aware.

We shall follow the notation given in the above diagram, with most points carrying their traditional labels.

A summary of the key results pertaining to this configuration,

First, we can consider the inversion centered at with radius followed by a reflection across point . This swaps with and with . Hence, and as well as and are inverses of each other.
- This immediately implies that points $P,H$ and $S$ and points $Q,H$ and $R$ are collinear.
- Further, this yields $HS \cdot HP = HF \cdot HC$ which indicates that $PCSF$ , and similarly $QBRE$ is cyclic.
- Since it is well known that $PQDM$ is cyclic, this also implies that $(ARS)$ passes through $G$ .
Next, look at the inversion centered at with radius . This swaps with and with . Hence, points and all invert to themselves.
- Hence, $AP=AQ=AR=AS$ so quadrilateral $PQRS$ is cyclic with center $A$ .
- Moreover, since $AP=AQ$ , the quadrilateral $APA'Q$ is a kite and is thus harmonic. Projecting through $G$ onto line $\overline{EF}$ immediately implies that $(X, HM \cap EF; PQ)=-1$ which is sufficient to conclude that the polar of $X$ with respect to $(ABC)$ passes through the intersection of lines $EF$ and $HM$ which solves this problem by Kudo3105.
Finally, look at the inversion centered at with radius . This swaps with and with (remember ). Hence, this swaps points and and similarly swaps points and .
- We now have that points $M,R$ and $P$ and points $M,S$ and $Q$ lie on the same lines. This combined with our first observations finishes 2016 Canada MO Problem 5.
- From our first observation that points $P,H$ and $S$ are collinear, it follows that quadrilateral $GQMR$ and similarly quadrilateral $GPMS$ are cyclic.
- Further, we have $MP \cdot MR = MH_a \cdot MA = MQ \cdot MS$ so quadrilaterals $PRH_aA$ and $QSH_aA$ are both cyclic.
- Similarly, $MP \cdot MR = MH \cdot MG = MQ \cdot MS$ which implies that quadrilaterals $PRHG$ and $QSHG$ are both cyclic.
- Finally, $MP \cdot MR = MD \cdot MX = MQ \cdot MS$ which implies that quadrilaterals $PRDX$ and $QSDX$ are also both cyclic.
- Also worth noting is that by Brokard’s Theorem on cyclic quadrilateral $PRSQ$ we have that $\triangle XHM$ is self polar with respect to $(PQRS)$ .
Using both the above main results we can then show with a little angle chasing that quadrilaterals and are cyclic.
- Considering the aforementioned inversion at $M$ , this implies that quadrilaterals $XSH_aP$ and $XRH_aQ$ are cyclic.
By Radical Center Theorem on circles $(ABC)$ , $(BHC)$ and $(PQRS)$ it follows that lines $PQ , RS$ and $BC$ concur at $X$ .
A straightforward angle chase now allows us to show that lies on and similarly lies on , which solves 2024 German MO Problem 3.
- From our previous observation that $CP = CQ$ it follows that $\triangle UEC \sim \triangle VFB$ and hence $UV \parallel EF$ . An alternate proof is to apply Pascal’s Theorem on concyclic hexagon $PQBH_bH_cC$ .
- Furthermore, by Reim’s Theorem it hence follows that quadrilateral $BCUV$ is cyclic.
Using harmonics it is not hard to see that lies on the line through and .
- This then allows us to show that $\overline{DT}$ is the internal $\angle LDN-$ bisector proving 2025 Brazil MO Problem 5.