14th European Girls' Mathematical Olympiad (EGMO 2025)

Problem 3. Let ABC be an acute triangle. Points B, D, E, and C lie on a line in this order and satisfy BD = DE = EC. Let M and N be the midpoints of AD and AE, respectively. Suppose triangle ADE is acute, and let H be its orthocentre. Points P and Q lie on lines BM and CN , respectively, such that D, H, M , and P are concyclic and pairwise different, and E, H, N , and Q are concyclic and pairwise different. Prove that P , Q, N , and M are concyclic.
The orthocentre of a triangle is the point of intersection of its altitudes.

SOLUTION 2

Let $B'$ be the reflection of $B$ over $M$ , and let $C'$ be the reflection of $C$ over $N$ .
It is clear that $A$ is the midpoint of $B'C'$ , and $B'C' \parallel MN \parallel BC$ .

Claim: $\angle HB'C' = \angle HC'B' = 90^\circ - \angle DAE$

Proof:
Note that $M, A, B', E$ are concyclic and $M, A, C', D$ are also concyclic.
The rest follows from simple angle chasing. $\blacksquare$

Since $P, M, H, D$ are concyclic, we have:
$\angle HPB' = \angle HPM = \angle MDH = 90^\circ - \angle DAE$ Similarly,
$\angle HQC' = 90^\circ - \angle DAE$
By the claim, we know:
$\angle HPB' = \angle HC'B' \quad \text{and} \quad \angle HQC' = \angle HB'C'$ So, the points $B', C', P, Q, H$ lie on a circle.

Finally, notice that $MN \parallel B'C'$ .
Applying Reim's Theorem yields that $M, N, P, Q$ are concyclic.

SOLUTION 1

Let $\{S\}=(DMH)\cap(ENH)$ . We prove that $SH, BM,CN$ concur, and the conclusion follows.
We prove that the intersection point is in fact the point that divides the $A$ -median in a $\frac{2}{3}$ ratio.
Let $\{P'\}=BM\cap AC$ , $\{Q'\}=CN\cap AB$ , $\{T\}=BP'\cap CQ'$ and $\{R\}=AT\cap BC$ .
$\boldsymbol{Claim:}$ $\frac{AT}{TR}=\frac{2}{3}$ .
$\boldsymbol{Proof:}$ Applying Menelaus' Theorem in $\triangle ADC$ we get that $\frac{AP'}{P'C}=\frac{1}{3}$ , and similarly $\frac{AQ'}{Q'B}=\frac{1}{3}$ as well. Then, by Ceva, $BR=CR$ , and applying Menelaus' Theorem again, in $\triangle ARC$ , the claim follows.

Now, let $\{T'\}=SH\cap AR$ , $O$ be the center of $(ADE)$ , $O'$ be the center of $(AMN)$ and $\{L\}=OR\cap SH$ .
$\boldsymbol{Claim:}S,H,O'$ are collinear.
$\boldsymbol{Proof:}$ Since $D,S,H,M$ and $E,S,H,N$ are concyclic, we have that $\angle MSH=\angle NSH=90\textdegree - \angle DAE$ , which quickly yields that $M,S,N,O'$ are concyclic. Since $O'M=O'N$ , $SO'$ is the bisector of $\angle SMN$ -- but so is $SH$ . Therefore, $S,H,O'$ are collinear.
All that's left for the problem's conclusion to be obtained is to find that $T=T'$ , i.e. $\frac{AT'}{T'R}=\frac{2}{3}$ .
By homothety, $O'$ is the midpoint of $AO$ , and since $AH\parallel OL$ , then $AHOL$ is a parallelogram and $OL=AH$ . By Menelaus' Theorem in $\triangle ARO$ , $\frac{AT'}{T'R}=\frac{OL}{LR}=\frac{AH}{AH+\frac{AH}{2}}=\frac{2}{3}$ .
So $T=T'$ , meaning that $SH, BM$ and $CN$ concur in $T$ , and by power of point we have that $P,M,N,Q$ are concyclic.

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