Let $\omega$ be the circumcircle of the triangle $ABC$ with $\angle B = 3\angle C$. The internal angle bisector of $\angle A$, intersects $\omega$ and $BC$ at $M$ and $D$, respectively. Point $E$ lies on the extension of the line $MC$ from $M$ such that $ME$ is equal to the radius of $\omega$. Prove that circumcircles of triangles $ACE$ and $BDM$ are tangent. Proposed by Mehran Talaei - Iran
Problem
Source: 2023 IGO Intermediate P3
Tags: geometry, tangent circles
VicKmath7
18.01.2024 18:47
Let $O$ be the circumcenter of triangle $ABC$, let $T=AB \cap (BDM)$ and let $\angle ABC=3\varphi$, $\angle ACB=\varphi$. Hence, $\angle ADB=\angle TBM=\angle TDM=90^{\circ}-\varphi$, so $\angle BDT=2\varphi$ and $\angle ATD=\varphi$. Thus, the triangles $\triangle ATD$ and $\triangle ACD$ are congruent, so $\angle DTC=\angle DCT=\varphi$. Since $AO=OM=ME$ and $\angle AOM=\angle OME=180^{\circ}-2\varphi$, we obtain that $AOME$ is an isosceles trapezoid, which implies that $\angle AEM=2\varphi=\angle ATC$, i.e. that $ACET$ is cyclic. Take a point $R$ on the tangent of $(BDM)$ at $T$; we have that $\angle RTB=\angle TDB=\angle ACT$, so $RT$ touches $(ACE)$ as well and we are done.
NO_SQUARES
18.01.2024 20:04
An important step towards solving the problem will be the realization that the desired point of contact actually lies on the straight line $AB$.
X.Allaberdiyev
19.01.2024 20:13
Let $C'$ be the reflection of $C$ through $AM$. Then these two circles touch each other at point $C'$, which can be shown just using angle chasing(after finding that $AE$ is perpendicular to $BC$).
EquationTracker
30.01.2024 18:43
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Let $\angle C=\alpha$, $\angle B=3\alpha$, $\frac{1}{2}\angle A=\theta$. Then $\theta = 90^{\circ}-2\theta$. Let $R$ be the circumradius of $\Delta ABC$. Let $O,O_1,O_2$ be the circumcenters of triangles $\triangle ABC,\triangle BDM,\triangle ACE$. Let $H$ be a point on $AB$ such that $MB=MH$. \[\angle MHB = \angle MBH = \angle ACM = \alpha + \theta = \angle ADB\implies BDMH\text{ cyclic}\]\[\angle MHA = \angle MCA,\ \angle MAH=\angle MAC\implies \triangle AMH\cong\triangle AMC\implies AH=AC\implies \angle AHC = 90^{\circ}-\theta\]$\angle OME = 180^{\circ}-(90^{\circ}-\theta)=90^{\circ}+90^{\circ}-2\alpha=180^{\circ}-2\alpha$, $\angle AOM = 2\angle ACM = 2\alpha+2\theta = 180^{\circ}-2\alpha$ and $AO=OM=ME=R$, $\triangle AOM\cong\triangle OME\implies AM=OE$. Combining all these things gives us, $AOME$ is an isosceles trapezoid. Thus, \[\angle AEC = \angle OMC = 90^{\circ}-\theta = \angle AHC\implies ACEH\text{ cyclic}\] Finally, \[\angle O_1HB = 90^{\circ}-\angle BMH = 90^{\circ}-2\alpha=\theta = 90^{\circ}-(90^{\circ}-\theta) = 90^{\circ}-\angle ACH = \angle O_2HA\implies O_1,O_2,H\text{ are colinnear}\]
sami1618
16.02.2024 19:14
Let $F\neq M$ be the point along line $AB$ such that $MB=MF$. We claim that $F$ is the desired tangency point. For the solution let $\angle ACB=x$ Claim 1: The points $B$, $D$, $M$, and $F$ are concyclic This is simple angle chasing, $180-\angle BFM=180-\angle FBM=\angle ABM=\angle ADC=\angle BDM$. Claim 2: $DC=DF$ Let $A'$ be the reflection of $A$ about the perpendicular bisector of $BC$. It is not hard to show that cyclic quadrilaterals $DBFM$ and $BAA'M$ are similar. Now: $$DF=A'B\cdot \frac{DM}{MB}=AC\cdot \frac{DM}{MB}=DC$$Claim 3: $CE=EA$ Consider isosceles triangle $OME$. We know that $\angle OME=180-2x$ so $\angle OEM=x$. In addition, $\angle ACE=90-x$ so we have $OE$ is perpendicular to $AC$. But $OA=OC$, so $CE=EA$. Claim 4: The points $A$, $C$, $E$, and $F$ are concyclic We already know that $\angle AEC=2x$ so we need to show that $AEC=2x$. But using the fact that $\angle AFD=x$ and $DC=DF$ this follows quickly. Claim 5: $\widehat{AF}=\widehat{BF}=4x$ This is trivial by angle chasing and this implies the result.
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