Problem A. 805. (September 2021)
A. 805. In acute triangle \(\displaystyle ABC\) the feet of the altitudes are \(\displaystyle A_1\), \(\displaystyle B_1\) and \(\displaystyle C_1\) (with the usual notations on sides \(\displaystyle BC\), \(\displaystyle CA\) and \(\displaystyle AB\), respectively). The circumcircles of triangles \(\displaystyle AB_1C_1\) and \(\displaystyle BC_1A_1\) intersect the circumcircle of triangle \(\displaystyle ABC\) at points \(\displaystyle P\ne A\) and \(\displaystyle Q\ne B\), respectively. Prove that lines \(\displaystyle AQ\), \(\displaystyle BP\) and the Euler line of triangle \(\displaystyle ABC\) are either concurrent or parallel to each other.
Submitted by Géza Kós, Budapest
(7 pont)
Deadline expired on October 11, 2021.
Solution of Tashi Diaconescu
For a triangle \(\displaystyle XYZ\) let \(\displaystyle \mathcal{C}_{XYZ}\) denote its circumcircle. For a circle \(\displaystyle \Gamma\) and a point \(\displaystyle X\) let \(\displaystyle \rho(X,\Gamma)\) be the power of point \(\displaystyle X\) with respect to circle \(\displaystyle \Gamma\).
Let \(\displaystyle H\) and \(\displaystyle O\) be the orthocenter and the circumcenter of triangle \(\displaystyle ABC\), respectively.
Claim:
\(\displaystyle H\) lies on the radical axis of the circles \(\displaystyle \mathcal{C}_{AOQ}\) and \(\displaystyle \mathcal{C}_{BOP}\).
Proof: Let \(\displaystyle HP \cap \mathcal{C}_{BOP}\setminus \{P\}=\{M\}\), \(\displaystyle HQ \cap \mathcal{C}_{AOQ} \setminus\{Q\}=\{N\}\). Let \(\displaystyle A'\) and \(\displaystyle B'\) be the antipodes of the points \(\displaystyle A\) and \(\displaystyle B\), respectively, in \(\displaystyle \mathcal{C}_{ABC}\). Since \(\displaystyle \angle HPA=\angle HC_1A=90^\circ=\angle A'PA\), it follows that \(\displaystyle A'\in PH\). Similarly \(\displaystyle B'\in QH\).
Since
\(\displaystyle \angle HMO=\angle PMO=\angle PBO=\angle PBB'=\angle PA'B'=\angle HA'B', \)
it follows that \(\displaystyle MO\parallel A'B'\). Similarly \(\displaystyle NO\parallel A'B'\). Hence \(\displaystyle MN\parallel A'B'\).
By Thales's theorem in \(\displaystyle \triangle HA'B'\) it follows that \(\displaystyle \dfrac{HM}{HA'}=\dfrac{HN}{HB'}\), which implies
$$\begin{align*} \rho (H,\mathcal{C}_{AOQ})&= HQ\cdot HN=\dfrac{HN}{HB'}\cdot (HQ\cdot HB')= \dfrac{HN}{HB'}\cdot \rho(H,\mathcal{C}_{ABC})\\ &=\dfrac{HM}{HA'}\cdot (HP\cdot HA')=HP\cdot HM=\rho(H,O_{BOP}), \end{align*}$$which concludes the proof of the claim.
Now, we have two cases.
Case 1: \(\displaystyle AQ\nparallel BP\). Let \(\displaystyle \{K\}=AQ\cap BP\). Let \(\displaystyle \Psi\) be the inversion respect to \(\displaystyle \mathcal{C}_{ABC}\).
Note that for each \(\displaystyle X\in \mathcal{C}_{ABC}\): \(\displaystyle \Psi(X)=X\). Hence \(\displaystyle \Psi(AQ)=\mathcal{C}_{AOQ}\) and \(\displaystyle \Psi(BP)={BOP}\), which implies \(\displaystyle K^*=\Psi(K)\in \mathcal{C}_{AOQ}\, \cap
\,\mathcal{C}_{BOP}\setminus\{O\}\).
From this and the claim it follows that \(\displaystyle H\in OK^*\). Because \(\displaystyle K\in OK^*\) it follows that \(\displaystyle O\), \(\displaystyle H\), \(\displaystyle K^*\) and \(\displaystyle K\) are collinear, which implies that lines \(\displaystyle AQ\), \(\displaystyle BP\) and the Euler line \(\displaystyle HO\) of triangle \(\displaystyle ABC\) are concurrent.
Case 2: \(\displaystyle AQ\parallel BP\) (see the second figure). In this case \(\displaystyle AQBP\) is an isosceles trapezium and the circumcenters of \(\displaystyle \triangle AOQ\) and \(\displaystyle \triangle BOP\) lies on the (common) perpendicular bisector of the segments \(\displaystyle AQ\) and \(\displaystyle BP\), which implies that \(\displaystyle \mathcal{C}_{AOQ}\) and \(\displaystyle \mathcal{C}_{BOP}\) are tangent at \(\displaystyle O\). From the claim it follows that \(\displaystyle HO\) is tangent to \(\displaystyle \mathcal{C}_{BOP}\) at \(\displaystyle O\), which implies
\(\displaystyle \angle HOP=\angle OBP=\angle OPB. \)
Hence \(\displaystyle HO\parallel BP\parallel AQ\), and we are done.
Remark: As in the official solution, Pascal's theorem in cyclic hexagon \(\displaystyle AQB'BPA'\) quickly solves the problem.
Statistics:
12 students sent a solution. 7 points: Bán-Szabó Áron, Bognár 171 András Károly, Diaconescu Tashi, Lovas Márton, Molnár-Szabó Vilmos, Móra Márton Barnabás, Nádor Benedek, Seres-Szabó Márton, Sztranyák Gabriella, Varga Boldizsár. 6 points: Török Ágoston. 1 point: 1 student.
Problems in Mathematics of KöMaL, September 2021