Let us denote the midpoint of $AB$ with $O$. The point $C$, different from $A$ and $B$ is on the circle $\Omega$ with center $O$ and radius $OA$ and the point $D$ is the foot of the perpendicular from $C$ to $AB$. The circle with center $C$ and radius $CD$ and $\omega$ intersect at $M$, $N$. Prove that $MN$ cuts $CD$ in two equal segments.
Problem
Source: MDA TST for egmo 2017, problem 2
Tags: geometry
Medjl
26.04.2017 14:52
Snakes wrote: If $M$ and $N$ are the intersection points of the circle with center $C$ and radius $CD$ . can you explain??
Snakes
26.04.2017 14:56
@above, edited. Thnx for spotting the mistake.
Medjl
26.04.2017 15:25
can someone post figure ?
Snakes
26.04.2017 22:16
Anyone? This problem isn't hard.
aryanna30
26.04.2017 23:16
Very easy by power of the point$G$(the intersection of $CD,EF$) in circles $(EDF),(ABE)$
SBLNuclear17
01.02.2023 17:54
You can do this without much problem by some analytic geometry. $O(0,0)$ $A(-1,0)$ $B(1,0)$ $C(cos(a),sin(a))$ $D(cos(a),0)$ $D: (x-cos(a))^{2}+(y-sin(a))^{2}=sin^{2}(a)$ We can get that MN is: $2xcos(a)+2ysin(a)=1+cos^{2}(a)$ Which clearly passes through ($cos(a),\frac{sin(a)}{2}$) which is the midpoint of CD.
eauna002
01.02.2023 20:08
Let $X$ be the intersection of $CD$ and $MN$. Invert over the circle $(MND)$, and observe $CX^{*} = 2CD = \frac{CD^2}{CX}$, and the result follows immediately.