Let $ABC$ be an acute scalene triangle such that $AB <AC$. The midpoints of sides $AB$ and $AC$ are $M$ and $N$, respectively. Let $P$ and $Q$ be points on the line $MN$ such that $\angle CBP = \angle ACB$ and $\angle QCB = \angle CBA$. The circumscribed circle of triangle $ABP$ intersects line $AC$ at $D$ ($D\ne A$) and the circumscribed circle of triangle $AQC$ intersects line $AB$ at $E$ ($E \ne A$). Show that lines $BC, DP,$ and $EQ$ are concurrent. Nicolás De la Hoz, Colombia
Problem
Source: 2020 IberoAmerican p1
Tags: geometry, concurrency, concurrent
jbaca
18.11.2020 03:41
Solution. Angle equalities imply that $BMQC$ and $BPNC$ are isosceles trapezoids, thus symmetric with respect to the perpendicular bisector of $\overline{BC}$. To finish, define $L$ as the midpoint of the latter. Therefore $$\measuredangle LQC=\measuredangle BML=\measuredangle BAC=\measuredangle EAC=\measuredangle EQC;\quad \measuredangle LPB=\measuredangle CNL=\measuredangle CAB=\measuredangle DAB=\measuredangle DPB$$leading to the required result. $\blacksquare$
Packito
18.11.2020 04:15
This problem was proposed by Colombia by me (Nicolás De la Hoz).
L567
28.05.2021 09:33
Let the perpendicular bisector of $BC$ meet $AB, AC$ at $X, Y$. Then, obviously $B,P,Y$ and $C,Q,X$ are collinear. Let $XY$ meet $BC$ at $Z$, the midpoint of $BC$. I claim $Z$ is the concurrency point. This is because $\angle ZQC = \angle ZMB = \angle BAC = \angle EAC = \angle EQC$ and so $E,Z,Q$ are collinear. Similarly, $D,P,Z$ are collinear and so $DP, EQ, BC$ are concurrent. $\blacksquare$
The second intersections of $(AQC), (APB)$ lies on the line $MN$. To see this, let $MN$ meet $(ACQ)$ at $R$. Then, $\angle ARP = \angle ARQ = \angle ACQ = \angle ABC - \angle ACB = \angle ABP$ and so $R$ lies on $(ABP)$ too
Math_Is_Fun_101
01.10.2021 18:10
Let $A'$ be the point on $(ABC)$ such that $A'ACB$ is an isosceles trapezoid. Note that $\angle CBA' = \angle ACB = \angle CBP$, so $P$ lies on $\overline{BA'}$. Then, since $\overline{MP}\parallel\overline{AA'}$ and $M$ is the midpoint of $\overline{AB}$, we have $P$ is the midpoint of $\overline{BA'}$. Now, from the cyclic quads we have \[ \angle PDC = \angle PDA =\angle PBA = \angle A'BA = \angle A'CA, \]so $\overline{DP}\parallel\overline{CA'}$. Since $P$ is the midpoint of $\overline{BA'}$, this implies the intersection of lines $\overline{BC}$ and $\overline{DP}$ is the midpoint of $\overline{BC}$. Analogously, the intersection of lines $\overline{EQ}$ and $\overline{BC}$ is the midpoint of $\overline{BC}$, so the three lines concur. $\blacksquare$
Mahdi_Mashayekhi
15.12.2021 10:58
Let K be midpoint of BC. We will prove this is where our lines meet. QCBM is isosceles trapezoid ---> ∠KQC = ∠KMB = ∠CAB = ∠CQE ---> Q,K,E are collinear PBCN is isosceles trapezoid ---> ∠BPK = ∠KNC = ∠BAC = 180-∠DAB = 180-∠DPB ---> K,P,D are collinear we're Done.
rstenetbg
29.12.2023 18:18
Let $K$ be the midpoint of $BC$. We will show that $K\in PD$. In a similar way we could show that $K\in QE$. Indeed, let $KP\cap AC = D'$. Notice that $$\angle NMK = \angle ACB=\angle CBP=\angle KBP.$$Hence, $PMBK$ is cyclic. Therefore, $\angle CAB = \angle BMK =\angle BPK$ and thus $PBAD'$ is cyclic. However, we know that $ADPB$ is cyclic. This means that $D\equiv D'$, so we are done.