This is the most difficult problem of IMO1982. It is problem 2 The following table shows the distribution of scores on each individual question. Score 0 1 2 3 4 5 6 7 Question 1 04 18 08 11 20 00 04 54 Question 2 80 09 08 02 01 03 08 08 Question 3 31 01 36 08 09 02 05 27 Question 4 36 04 02 14 08 06 05 44 Question 5 17 03 03 01 03 03 07 82 Question 6 46 12 11 09 08 07 06 20 http://www.srcf.ucam.org/~jsm28/imo-scores/1982/

I remember reading somewhere that this was the most difficult, but I didn't want to scare people off It took me some time to solve it (meaning thet it took me A LOT of time), but I don't think it's THAT nasty... Hint: I think a problem posted on this forum before might help (I'm talking about the proof for Feuerbach's theorem, which Lagrangia posted in the 'Geometry' section; it might be in the 'Solved Problems' section now ) EDIT: Sketch of a solution: An angle chase will show that $S_iS_j\|a_k$ for all permutations $(i,j,k)$ of $(1,2,3)$, so the triangles $S_1S_2S_3,M_1M_2M_3$ are homothetic, meaning that the lines connecting the corresponding points, namely $S_iM_i$, must be concurrent. P.S. The concurrence point is the Feuerbach point of $ABC$, i.e. the point where the nine-point circle touches the incircle.

The Additional Theorem in http://www.mathlinks.ro/Forum/viewtopic.php?t=14927 post #2 is stronger than the problem: it shows that the lines $M_{1}S_{1}$, $M_{2}S_{2}$ and $M_{3}S_{3}$ pass through the Feuerbach point of triangle $A_{1}A_{2}A_{3}$. Darij

Dear Mathlinkers, you can see a synthetic proof on http://perso.orange.fr/jl.ayme vol. 9 Le triangle reflechi p. 10. Sincerely Jean-Louis

PROOF SKETCH: 1.$S_1,S_2,S_3$ all lie on the incircle. 2.$S_1S_2 \parallel M_1M_2,S_2S_3 \parallel M_2M_3,S_3S_1 \parallel M_3M_1$. 3.$\triangle{M_1M_2M_3} \sim \triangle{S_1S_2S_3}$. 4.$M_1S_1,M_2S_2,M_3S_3$ concur at the centre of homothety.

For the proof of Sayantan da in a bit detail. See http://www.math.ust.hk/excalibur/v9_n4.pdf

Rename the vertices so that $A_1=A,A_2=B$ and $A_3=C$. Let $I$ be the incenter of $\triangle ABC$. Claim 1: $S_1, S_2, S_3$ all lie on the incircle. Proof: By construction, $IS_1$ is the reflection of $IT_1$ with respect to the $A$ angle bisector since $I$ lies on the angle bisector of $A$. So, $IS_1=IT_1$ so $S_1$ lies on the incircle. Similarly, $S_2$ and $S_3$ lie on the incircle. $\Box$ Note that $I$ is the circumcenter of $\triangle S_1S_2S_3$ because of equal radii due to Claim 1. Claim 2: $\angle S_1IS_2=2\angle C.$ Proof: In cyclic quadrilateral $T_1IT_2C$, $\angle T_1IT_2=180^{\circ}-\angle C$. Let $D$ be the point where the $A$ angle bisector intersects $BC$. Due to the construction of $S_1$, $\angle S_1IT_1=2\angle DIT_1. $ But, $\angle ADC=180^{\circ}-\angle C-\frac{1}{2}\angle A$ so $\angle DIT_1=\frac{1}{2}\angle A+\angle C-90^{\circ}.$ Thus, $$\angle S_1IT_1=\angle A+2\angle C-180^{\circ}.$$Similarly, $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Therefore, $\angle S_1IS_2=\angle S_1IT_1+\angle T_1IT_2+\angle T_2IS_2=2\angle C.$$\Box$ Hence, we use Claim 2 to get $\angle S_1S_3S_2=\frac{1}{2}\angle S_1IS_2=\angle C$ and we similarly get $\angle S_3S_1S_2=\angle A$ and $\angle S_1S_2S_3=\angle B.$ Claim 3: Let $X_1$ be the midpoint of $S_2S_3.$ Then, $T_1IX_1$ is collinear. Proof: We already derived that $\angle T_1IT_2=180^{\circ}-\angle C$ and $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Recall that a radius is perpendicular to the midpoint of a chord so $S_2S_3\perp IX_1.$ Then, $\angle X_1IS_2=90^{\circ}-\angle X_1S_2I=90^{\circ}-(90^{\circ}-\angle A)=\angle A.$ Thus, $$\angle T_1IX_1=\angle X_1IS_2+\angle S_2IT_2+\angle T_2IT_1=\angle A+\angle B+2\angle C-180^{\circ}+180^{\circ}-\angle C=180^{\circ},$$as desired.$\Box$ Then, $T_1IX_1\perp S_2S_3$ because a diameter is perpendicular to a chord at its midpoint and because $T_1IX_1$ is part of a diameter, by Claim 3. Hence, $S_2S_3\parallel BC.$ But, $M_2M_3\parallel BC$ so $S_2S_3\parallel M_2M_3$. Similarly, we get $S_1S_2\parallel M_1M_2$ and $S_1S_3\parallel M_1M_3$ so $\triangle M_1M_2M_3\sim \triangle S_1S_2S_3$ with corresponding parallel sides. Thus, there exists a homothety that maps $M_1M_2M_3$ to $S_1S_2S_3$ and $M_1S_1, M_2S_2,$ and $M_3S_3$ intersect at the center of the homothety. $\blacksquare$

Here's a proof I found. The distance from $S_2$ to ${A_2}{A_3}$ is equal to the distance from $T_2$ to ${A_1}{A_3}$ which can be written as ${A_1}{T_2}\centerdot\sin\angle{A_2}{A_1}{A_3}$ Similarly, The distance from $S_3$ to ${A_2}{A_3}$ can be written as ${A_1}{T_3}\centerdot\sin\angle{A_2}{A_1}{A_3}$ And we have ${A_1}{T_2}={A_1}{T_3}$ Therefore ${S_2}{S_3}\parallel{A_2}{A_3}\parallel{M_3}{M_2}$ So $\triangle{M_3}{M_2}{M_1}$ is homothetic to $\triangle{S_3}{S_2}{S_1}$, which implies the concurrency of ${M_1}{S_1},{M_2}{S_2},{M_3}{S_3}$ An interesting fact about this proof is that we don't even need to prove that ${S_I}$ lies on the incircle.