The original version of the problem, was not correct according to it's figure, the equality of the ratios was not correct even if AB<AC, as I used geogebra to check it that's why it was changed above, here is the original translation: Quote: Let $ABC$ be a triangle with $AB>AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IC }{IB}$

Using Geogebra I checked that there are two right statements of the problem, as the the equality is true when one fraction is reversed: 1. Let $ABC$ be a triangle with $AB<AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ 2. Let $ABC$ be a triangle with $AB>AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ and only the first one mathes the figure given. It sounds unbelievable, to have proposed a wrong problem in an Math Olympiad in China !!!

parmenides51 wrote: Using Geogebra I checked that there are two right statements of the problem, as the the equality is true when one fraction is reversed: 1. Let $ABC$ be a triangle with $AB<AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ 2. Let $ABC$ be a triangle with $AB>AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ and only the first one mathes the figure given. It sounds unbelievable, to have proposed a wrong problem in an Math Olympiad in China !!! The problem is valid for any triangle to have the point $D$ on the segment $BA$ we need to take $BA<AC$ otherwise $D$ will be on the half-line $[BA)$ but beyond $A$ but that doesn't change the result

here is my solution

Dear Mathlinlers, 1. DE goes through the midpoint of the arc BAC 2. DE // BI 3. BED and CIB are similar and we are done... Sincerely Jean-Louis

Dear Mathlinkers, I forgot to give this link http://jl.ayme.pagesperso-orange.fr/Docs/La%20ponctuelle%20(MI).pdf p 23-25. Sincerely Jean-Louis

parmenides51 wrote: Using Geogebra I checked that there are two right statements of the problem, as the the equality is true when one fraction is reversed: 1. Let $ABC$ be a triangle with $AB<AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ 2. Let $ABC$ be a triangle with $AB>AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ and only the first one mathes the figure given. It sounds unbelievable, to have proposed a wrong problem in an Math Olympiad in China !!! ？？？

sqing wrote: parmenides51 wrote: Using Geogebra I checked that there are two right statements of the problem, as the the equality is true when one fraction is reversed: 1. Let $ABC$ be a triangle with $AB<AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ 2. Let $ABC$ be a triangle with $AB>AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ and only the first one mathes the figure given. It sounds unbelievable, to have proposed a wrong problem in an Math Olympiad in China !!! ？？？ https://artofproblemsolving.com/community/c6h602657p10667541 https://artofproblemsolving.com/community/c1h1674953p10667539

jayme wrote: Dear Mathlinlers, 1. DE goes through the midpoint of the arc BAC 2. DE // BI 3. BED and CIB are similar and we are done... Sincerely Jean-Louis Where does $1$ come from?

the main part of the proof is to prove that to prove that $DE \parallel BI$: let $K=AB\cap CI$ we have $ \frac{EK}{EI}= \frac{EI}{EC}= \frac{KI}{IC}$ but from Menelaus we get $\frac{DK}{DB} = \frac{IK}{IC}$ so $\frac{DK}{DB} = \frac{EK}{EI}$ hence $\frac{ED}{EI}=\frac{ED}{EK}.\frac{EK}{EI}=\frac{ED}{EK}.\frac{IK}{IC}$but $\frac{IK}{EK}=\frac{IB}{ED}$ so $\frac{ED}{EI}=\frac{IB}{IC}$ RH HAS

Yes,proving $DE||BI$ is the main struggle, though it's easy to prove that fact using barycentric coordinates: Let $ABC$ be the reference triangle. We obtain $D=(a:b-c:0)$, $E=\bigg(a:b:-\frac{c^2}{a+b}\bigg)$. Suppose $P=ED\cap BI$. We find out $P=(a:-c-a:c)$, so it's the point in infinity, as we wanted. $\square$

Let $I_C$ be the $C$-excenter of $\triangle ABC$, $I'$ be the reflection of $I$ over $M$, and $P$ be the midpoint of $\overline{CF}$. Note that $BICI'$ is a parallelogram, so it suffices to show that $DE/EI=IC/CI'$, or $\overline{DE}\parallel\overline{CI'}$. [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair A=(-43.2,51.3), B=(-60,-30), C=(60,-30), M=(0,-30), D=(-48,27.9), E=(-65.7,13.6), F=(-52,8.8), I=(-24.2,-0.8), O=(0,0), X=(24.2,-59.2), Y=(-107.2,27.9), P=(4,-10.6); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(I); dot(O); dot(P); dot(M); dot(X); dot(Y); draw(circumcircle(A,B,C), linewidth(0.5)); draw(A--B--C--cycle, linewidth(0.5)); draw(M--D, linewidth(0.5)); label("$A$", A, (-0.5,1)); label("$B$", B, SW); label("$C$", C, SE); label("$M$", M, (-0.5,-1)); label("$E$", E, (-1,-0.5)); draw(C--E, linewidth(0.5)); draw(E--Y, linewidth(0.4)+dashed); draw(B--I, linewidth(0.5)); draw(D--E, linewidth(0.5)); draw(I--X--C, linewidth(0.4)+dashed); label("$D$", D, (-0.6,0.6)); label("$I$", I, (0,-1.5)); label("$I'$", X, (-1,-0.5)); draw(circumcircle(A,I,B), linewidth(0.7)+dashed); draw(Y--D, linewidth(0.4)+dashed); label("$I_C$", Y, (-1,0)); label("$F$", F, SW); label("$P$", P, (0.5,-1.5)); draw(M--P, linewidth(0.4)+dashed); [/asy][/asy] To begin, observe that $(C,F;I,I_C)=-1$; it follows by a well-known property of harmonic bundles that $$\frac{I_CI}{IC}=\frac{FI}{IP}=\frac{DI}{IM}.$$Using this in conjunction with the fact that $II_C=2IE$ and $II'=2IM$ yields that $EI/II_C=DI/II'$, which immediately implies the desired conclusion.

Dear Mathlinkers, the one 1 (#8) of my comes from the Pascal's theorem... Sincerely Jean-Louis

That helps a lot.