Three equal circles have a common point $M$ and intersect in pairs at points $A, B, C$. Prove that that $M$ is the orthocenter of triangle $ABC$.
Problem
Source: JBMO Shortlist 2003
Tags: JBMO, geometry
RagvaloD
13.10.2017 13:11
Let centers of circles are $X,Y,Z$ and $AM$ intersect $BC$ at $D$. Easy to prove, that $XM \parallel CZ \parallel AY;YM \parallel AX \parallel BZ;ZM \parallel XC \parallel YB$ $\angle CAM=\frac{\angle CXM}{2},\angle ACM= \frac{\angle AXM}{2}, \angle DCM=\frac{\angle MZB}{2}=\frac{180-\angle AXC}{2}$ and so $180-\angle ADC = \angle CAM+\angle ACM+\angle DCM = \frac{\angle CXM+\angle AXM+180-\angle AXC}{2}=90 \to angle ADC=90$ so $AD$ is altitude of $ABC$. Same way we can prove, that $BM,CM$ are altitudes of $ABC$ so $M$ is the orthocenter of triangle $ABC$
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sunken rock
14.10.2017 17:19
In our mathematical literature this problem is called 'the 5 lei coin problem'. Best regards, sunken rock
Gomes17
14.10.2017 18:09
Let $J_A = AM \cap BC$, etc. Perform an inversion around $M$ with radius $r$. Since the three circles (call the circle opposite to $A$ of $\omega_A$, etc) have equal radius, we have that the distance $M-\omega'$ is equal for $A,B,C$, hence $M$ is the incenter of triangle $A'B'C'$.
Now notice that $J'_A=(AM \cap BC)'=A'M \cap (MB'C')$ and, since line $A'M-A'J'_A$ is the $\angle B'A'C'$ angle bissector and $(B'MC'J'_A)$ is cyclic, we find that $J'_A$ is the $A'- \text{excenter}$, hence
$$\angle J'_AB'M=90^{\circ} \Rightarrow \angle AJ_AB=90^{\circ}$$hence proved.
Math_legendno12
24.06.2023 13:48
We show <MCB + <ABC =90 by converting to length of arcs. This is done as the circles have equal radius, Thus we get MC perpendicular to AB and by similar conditions we obtain that M is orthocenter of ABC.