Let $ABC$ ($BC > AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD, OCD$ and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q, P$ and $K$, respectively, such that $P \in (AK)$ and $K \in (PQ)$. The line $PD$ intersects the circumcircle of triangle $BKQ$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $BQ$. Prove that $TB = TQ$.
Problem
Source: 2019 Balkan MO Shortlist G5 - BMO
Tags: geometry, circumcircle, equal segments
08.01.2021 10:32
Bump! Here's my progress. It's sufficient to prove that -$OPKT$ is cyclic, or -$BC$ and $KT$ are parralel Also I found that -$KBC$ is isosceles -$TQ$ and $TB$ should be tangents to $(BCD)$ (not proven) -triangles $DQB$ and $ABC$ are similar -$DQ=DC$ Remark: In my diagram (the above statements are based on it) the point order on $AC$ turns out to be $C, A, P, Q, K$. Hopefully this is not a problem since the statement still seems true.
04.02.2021 21:42
Here is an outline. Let $X$ be the intersection of $C$ symmedian of $\triangle ABC$ and $(ABC)$ and let $Y$ be the intersection of $D$ symmedian of $DQB$ and $(DQB)$. I want to prove $T$ is the intersection of tangents to $(DQB)$ at $Q$ and $B$ and what will imply the problem. By chasing $K$ lies on the circumcircle of $Q$, $B$ and the circumcenter of $(DQB)$. Now we have to prove $D-P-Y$ collinear. Let $P'$ be the intersection of $DY$ and $CA$, we have to prove $CXP'D$ cyclic. By chasing $DC=DQ$ and by tangents $DC=DX$ so $DX=DQ$. Also $\triangle BCA$ and $\triangle BQD$ are similar by and $B$ is the center of spiral similarity. This spiral similarity also carries $X$ to $Y$, so it carries $CX$ to $DY$ from where $C-X-Y$ are collinear. By chasing you can prove $DQ$ is tangent to $(YP'Q)$. From there $P'YBA$ is cyclic. By some more chasing we conclude $CXP'D$ is cyclic as desired.
26.03.2021 15:07
Can anybody post full solution pls?
29.08.2021 09:02
$\textbf{Claim:}$ $P$ is on the $D$-symmedian of $QDB$. Proof: $S=DP \cap (DQCB), M$ is the midpoint of $AB$. We have $\angle DCA= \angle DQA=\angle B \implies BD$ is the angle bisector of $CBQ$. $DP \cdot DS=DC^2 \implies \angle PCS=\angle DCS-\angle DCP=\angle DPC-\angle B=\angle CMD-\angle B=\angle MCB \implies CS$ is the $C$ symmedian of $ACB \implies C(D,S;A,B)=-1 \implies (D,S;Q,B)=-1 \implies P$ is on the $D$-symmedian of $QDB$.$\square$ If $T'$ is the intersection of perpendicular bisector of $QB$ and the line $DP$ ($T'Q, T'B$ are tangents) $180-2\angle C=\angle QT'B$, $\angle CKB=180-\angle AOB=180-2\angle C \implies KQBT'$ is a cyclic quadrilateral $\implies T=T'$ as desired.
07.01.2022 17:36
This problem was proposed by me, here is the solution I submitted (at a TST use later on the successful people approached similarly to @SerdarBozdag). Denote by $X$ the midpoint of $AB$. Then the pentagon $PXOCD$ is inscribed in the circle with diameter $OD$, hence $\angle PXA = \angle PXD = \angle PCD = \angle QCD = \angle QBA$ (the latter is due to $QBCD$ being cyclic). We deduce that $PX \parallel QB$ and that $P$ is the midpoint of $AQ$. Now let $T_1$ be the midpoint of the arc $BQ$, not containing $K$, from the circumcircle of $\triangle BKQ$. Due to $\angle DPO = 90^{\circ}$, it suffices to show that $\angle OPT_1 = 90^{\circ}$ -- indeed, $T\equiv T_1$ and $TB = TQ$ would follow. Denote by $Y$ the midpoint of $BQ$. Then $\angle OXB = \angle T_1YB = 90^{\circ}$ и $\angle XBO = \frac{\angle AKB}{2} = \frac{\angle BT_1Q}{2}=\angle BT_1Y$ and thus $\triangle OXB \sim \triangle BYT_1$. Consequently, $$\frac{OX}{XP}=\frac{OX}{BY}=\frac{XB}{T_1Y}=\frac{PY}{T_1Y}$$which along with $\angle PXB = \angle PYB$ and $\angle OXB = \angle T_1YB$ gives $\triangle OXP \sim \triangle PYT_1$. In conclusion, $$\angle OPT_1 = \angle XPY + \angle XPO + \angle YPT_1 = \angle PXA + \angle XPO + \angle XOP = 90^{\circ}. $$
26.02.2022 16:23
Let $DP$ meet circle $DBC$ at $X$. we will prove $DX$ is symmedian of $QDB$. Note that $QDB$ and $ACB$ are similar and $\angle QDX = \angle ACX$ so we can instead prove $CX$ is symmedian of $ACB$. Claim1 : $CX$ is symmedian of $ACB$. Proof : Let $M$ be midpoint of $AB$. Note that $\angle OCD = \angle 90 = \angle OMD$ so $OMDC$ is cyclic so $M$ is intersection of $OCD$ with $AB$. $\angle ACX = \angle QCX = \angle QDP = \angle DPC - \angle DQC = \angle DMC - \angle DBC = \angle MCB$. Now we have $T$ lies on $D$ - symmedian of $DQB$ so for proving $TB = TQ$ we need to prove $\angle QTB = \angle 180 - 2\angle QDB$. we have $ \angle QTB = \angle BKA = \angle 180 - \angle AOB = \angle 180 - 2\angle ACB = \angle 180 - 2\angle QDB$. we're Done.
10.12.2022 00:25
26.10.2024 06:26
How am I so washed at geometry So we claim that $T$ is the intersection of the tangents from $B$ and $Q$ to $(BCDQ)$. Reverse reconstruct $T$ to be as such. Notice that $\measuredangle QTB=\measuredangle TQB+\measuredangle QBT=2\measuredangle QDB=2\measuredangle QCB=2\measuredangle ACB=\measuredangle AOB=\measuredangle AKB=\measuredangle QKB$ as desired. So it suffices to show now that $D-P-T$. We now let $H=(BCDQ)\cap\overline{DP}\neq D$ and $H’=(BCDQ)\cap\overline{DT}\neq D$. It suffices to show that $H=H’$. However notice that $(DH’;QB)=-1$ so it suffices to show that $(DH;QB)=-1$ as well. First let $M$ be the midpoint of $AB$. Clearly $M\in(OCDP)$. By Reim’s theorem then $\overline{MP}\parallel\overline{QB}$. Furthermore $\measuredangle DCQ=\measuredangle CBA=\measuredangle CBD=\measuredangle CQD$, so $CD=DQ$, and thus the tangent at $D$ to $(BCDQ)$ is parallel to $CQ$. Now $-1=(M\infty_{AB};AB)\stackrel{\infty_{BQ}}{=}(P\infty_{AC};AQ)\stackrel{D}{=}(HD;BQ)=(DH;QB)$, thus done. (I missed the clean symmedian solution by similar triangles, oh well )