Copy paste the proof of EGMO problem 4.31

A homothety takes $\Gamma_1$ to $\Omega$ and $C$ to $X$, which is the midpoint of the arc $\overarc{AB}$ not containing $D$. Similarly, a homothety takes $\Gamma_2$ to $\Omega$ and $E$ to $Y$, which is the midpoint of the arc $\overarc{AB}$ not containing $F$ (and containing $D$). Hence $XY$ is a diameter of $\Omega$.

I think there is an easy proof with euclidian Geometry.

mricaomi12 wrote: I think there is an easy proof with euclidian Geometry. is that pythagorean geometry?

See my solution here

Ghoshadi wrote: See my solution here : https://mathsupportweb.wordpress.com/2017/10/08/geometry-problems-of-rmo17-east-region/ Be it east or west, RMO paper is the same.

Oh I didn't know that before. In past years they use to set different papers for different regions.

This is pure angle chasing proof and nothing else

Kayak wrote: Copy paste the proof of EGMO problem 4.31 EGMO 4.31 MEANS

Just apply Archimedes Lemma 2 times and that's it.

Kayak wrote: Ghoshadi wrote: See my solution here : https://mathsupportweb.wordpress.com/2017/10/08/geometry-problems-of-rmo17-east-region/ Be it east or west, RMO paper is the same. The Maharashtra and Goa region paper was different. I do not know the Mumbai region paper though.

Mumbai region paper was same as the rest of the papers

arulxz wrote: Mumbai region paper was same as the rest of the papers No.

Ghoshadi wrote: See my solution here Thanks

Sorry for reviving but can someone post a solution involving homothey of circle on one side to the other.

Ghoshadi wrote: arulxz wrote: Mumbai region paper was same as the rest of the papers No. Except maharashtra

thewitness wrote: Let \(\Omega\) be a circle with a chord \(AB\) which is not a diameter. \(\Gamma_{1}\) be a circle on one side of \(AB\) such that it is tangent to \(AB\) at \(C\) and internally tangent to \(\Omega\) at \(D\). Likewise, let \(\Gamma_{2}\) be a circle on the other side of \(AB\) such that it is tangent to \(AB\) at \(E\) and internally tangent to \(\Omega\) at \(F\). Suppose the line \(DC\) intersects \(\Omega\) at \(X \neq D\) and the line \(FE\) intersects \(\Omega\) at \(Y \neq F\). Prove that \(XY\) is a diameter of \(\Omega\) .

We can also prove it by using mixtilinear circles. We can show that X and Y are the midpoints of respective arcs and the solution follows.

Mate007 wrote: We can also prove it by using mixtilinear circles. We can show that X and Y are the midpoints of respective arcs and the solution follows. I agree

@above Unnecessary Bump

Suppose $O$ be the center of $\Omega$ and $O_1,O_2$ be the centers of $\Gamma_1 ,\Gamma_3$ respectively .At first suppose $X,YO$ are not coliner. Let, $OY$ met $AB$ at $N$ $OX$ met $ AB $at $M$ We want to show $M=N$ to show contradiction . Note that $\angle AEF =\angle YEB$.And $\angle FEO_2 =\angle ETO_2 =90-\angle AEF$. Again we have $\angle YEN =\angle AEF$ which means $\triangle YEN$ is right angled so we get $OY\perp AB$ . Now come back to smaller circle .$\angle CDB =\angle XDM$ and $\angle XMD=\angle O_1DC =\angle O_1CD =\angle OXD = 90-\angle CBD$. So $OY\perp AB$ . So,M,N are concurrent.

thewitness wrote: Let \(\Omega\) be a circle with a chord \(AB\) which is not a diameter. \(\Gamma_{1}\) be a circle on one side of \(AB\) such that it is tangent to \(AB\) at \(C\) and internally tangent to \(\Omega\) at \(D\). Likewise, let \(\Gamma_{2}\) be a circle on the other side of \(AB\) such that it is tangent to \(AB\) at \(E\) and internally tangent to \(\Omega\) at \(F\). Suppose the line \(DC\) intersects \(\Omega\) at \(X \neq D\) and the line \(FE\) intersects \(\Omega\) at \(Y \neq F\). Prove that \(XY\) is a diameter of \(\Omega\) . We prove Archimedes lemma : If in a circle $\omega$, a circle $\Omega$ is tangent to a chord $AB$ and $\omega$ at $C, D$ respectively, then $CD$ passes through midpoint of arc $AB$ which doesn't contain $C$. Proof : Take homothety from $\Omega \mapsto \omega$. We see that $X, Y$ are midpoints of a minor arc and major arc which share a chord. This means $XY$ is diameter. Hence proved

If i somehow prove that AB<XY Then we can XY is diameter Cause AB was choosen arbitrary And XY is longest chord