Let $C_{1}$ and $C_{2}$ be concentric circles, with $C_{2}$ in the interior of $C_{1}$. From a point $A$ on $C_{1}$, draw the tangent $AB$ to $C_{2}$ $(B \in C_{2})$. Let $C$ be the second point of intersection of $AB$ and $C_{1}$,and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects $C_{2}$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$. This question is taken from Mathematical Olympiad Challenges , the 9-th exercise in 1.3 Power of a Point.
Problem
Source:
Tags: ratio, circumcircle, geometry unsolved, geometry
frenchy
07.06.2010 13:43
A very nice problem. This is an approach worked out by my friend. From the power of the point theorem and since $D$ is the middle of $AB$ with $AB=AC$(because $OM\perp AC$ and $OA=OC$) we get that $AB^2=AE\cdot AF=AD\cdot AC$. From the above we obtain that the quadrilateral $EDCF$ is cyclic. So $M$ is the center of circle passing through points $E,D,F,C$. So $M$ is the middle of $DC$, from which $\frac{AM}{MC}=\frac{5}{3}$. That's it.
littletush
05.11.2011 07:49
frenchy wrote: A very nice problem. This is an approach worked out by my friend. From the power of the point theorem and since $D$ is the middle of $AB$ with $AB=AC$(because $OM\perp AC$ and $OA=OC$) we get that $AB^2=AE\cdot AF=AD\cdot AC$. From the above we obtain that the quadrilateral $EDCF$ is cyclic. So $M$ is the center of circle passing through points $E,D,F,C$. So $M$ is the middle of $DC$, from which $\frac{AM}{MC}=\frac{5}{3}$. That's it. very nice solution the key is to find that $D,E,F,C$ are concyclic. I have second-killed it in the exactly same way~
Cassius
15.05.2012 20:51
We have that $AD \cdot AC=AB^2=AE \cdot AF$ since $AD$ is one quarter of $AC$, so $DFEC$ is cyclic; by definition the center of its circumscribed circle is $M$, which is also therefore the midpoint of $DC$. So $\frac {AM} {MC}=\frac 5 3$.
aksshatkhanna
02.02.2018 20:52
same in usamo 1998