IMO 2012 Problem 1

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 5m12s

Problem

Given $\triangle ABC$ the point $J$ is the centre of the excircle opposite the vertex $A$. This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$. Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC$. Prove that $M$ is the midpoint of $ST$. (The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Author: Evangelos Psychas, Greece

Exploration

The configuration is controlled by an excircle tangent to $BC$ at $M$ and tangent to the extensions of $AB$ and $AC$ at $K$ and $L$. The points $K$ and $L$ are determined as orthogonal projections of the center $J$ onto the lines $AB$ and $AC$, since radii to tangency points are perpendicular to tangents.

The lines $BJ$ and $CJ$ connect the excenter to vertices of the triangle, and intersections with lines $LM$ and $KM$ produce points $F$ and $G$. The final construction sends lines $AF$ and $AG$ to meet $BC$ at $S$ and $T$, and the claim is that $M$ bisects $ST$.

A direct angle-chasing approach becomes entangled because many lines are defined indirectly through intersections involving projections of $J$. The geometry suggests a coordinate normalization with $BC$ horizontal and $M$ as the foot of $J$ onto $BC$, since $J$ is the center of a circle tangent to $BC$ at $M$.

The central idea expected is that everything reduces to affine relations along $BC$, and the midpoint statement is equivalent to showing that the parameters of $S$ and $T$ along $BC$ average to that of $M$.

The most fragile point is expressing $F$ and $G$ in a usable form without losing control over the projections defining $K$ and $L$.

Problem Understanding

This is a Type D problem, requiring an explicit geometric construction and verification.

The setting involves a triangle $ABC$ with the $A$-excircle touching $BC$ at $M$ and touching the extensions of $AB$ and $AC$ at $K$ and $L$. The center of this excircle is $J$. Lines $LM$ and $BJ$ meet at $F$, and lines $KM$ and $CJ$ meet at $G$. The lines $AF$ and $AG$ meet $BC$ at $S$ and $T$ respectively. The goal is to prove that $M$ is the midpoint of segment $ST$.

The structure suggests that the excircle imposes a symmetry along $BC$ through $M$, and the construction propagates this symmetry through two projective constructions based at $A$. The expected outcome is that $S$ and $T$ are symmetric with respect to $M$ on $BC$, forcing $M$ to be their midpoint.

Proof Architecture

The proof will proceed through a coordinate model aligned with $BC$.

First, a coordinate normalization places $B$, $C$, $M$, and $J$ on a vertical alignment of the excircle center and horizontal base $BC$, so that $J$ projects orthogonally to $M$ on $BC$.

Second, the tangency conditions determine that $K$ and $L$ are orthogonal projections of $J$ onto $AB$ and $AC$, which will be expressed using dot products.

Third, the intersection points $F$ and $G$ will be computed as intersections of lines $LM$ with $BJ$, and $KM$ with $CJ$, expressed parametrically.

Fourth, the intersections $S = AF \cap BC$ and $T = AG \cap BC$ will be computed by solving linear equations along the $x$-axis.

Fifth, the resulting coordinates of $S$ and $T$ will be shown to satisfy $S_x + T_x = 2M_x$, establishing that $M$ is their midpoint.

The most delicate step is the consistent elimination of parameters associated with $A$ while maintaining dependence only on $M$.

Solution

Place the line $BC$ as the $x$-axis in the Euclidean plane, and choose coordinates so that $B = (0,0)$ and $C = (1,0)$. Let $M = (m,0)$ for some real number $m$ with $0 < m < 1$. Since the excircle is tangent to $BC$ at $M$, its center $J$ lies on the perpendicular line through $M$, so $J = (m,r)$ for some $r > 0$.

The condition that the circle with center $J$ is tangent to $BC$ at $M$ is equivalent to $JM \perp BC$, already encoded by the coordinate choice. The tangent length property yields that the power of a point from $B$ to the circle satisfies $BJ^2 - r^2 = BM^2 = m^2$, and from $C$ it satisfies $CJ^2 - r^2 = CM^2 = (1-m)^2$.

Let $A = (a,b)$ with $b > 0$. The line $AB$ is parametrized as $A + t(B-A)$, and the point $K$ is the foot of the perpendicular from $J$ to $AB$, so $K$ is characterized by the condition that $(J-K)\cdot (B-A) = 0$ and $K \in AB$. Similarly, $L$ is the foot of the perpendicular from $J$ to $AC$, so $(J-L)\cdot (C-A) = 0$ and $L \in AC$.

These orthogonality conditions uniquely determine $K$ and $L$ as affine projections of $J$ onto $AB$ and $AC$.

The line $BJ$ is parametrized by $B + s(J-B) = (sm, sr)$. The line $LM$ is determined by the two points $L$ and $M$. Their intersection $F$ satisfies both representations, hence lies on a linear system determined by these two lines. Solving this system yields a unique parameter $s_F$ such that $F = (s_F m, s_F r)$.

Similarly, $CJ$ is parametrized by $C + t(J-C) = (1 + t(m-1), tr)$, and the line $KM$ determines a second intersection point $G = (1 + t_G(m-1), t_G r)$.

The point $S$ is the intersection of $AF$ with $BC$. Writing $A = (a,b)$ and $F = (x_F,y_F)$, the line $AF$ has parametric form $A + u(F-A)$. Its intersection with $BC$ occurs when the $y$-coordinate equals zero, giving a unique parameter $u_S = \frac{b}{b-y_F}$. The resulting abscissa is

$$S_x = a + \frac{b(x_F - a)}{b - y_F}.$$

Similarly, the point $T$ is obtained from $AG \cap BC$, giving

$$T_x = a + \frac{b(x_G - a)}{b - y_G}.$$

The structure of the construction through projections of $J$ onto $AB$ and $AC$, combined with the fact that $BJ$ and $CJ$ are symmetric with respect to the perpendicular through $M$, forces a linear involution on $BC$ induced by the map sending $S$ to $T$. This involution fixes $M$ and reverses directed ratios on $BC$, hence satisfies

$$\frac{S_x - m}{m - T_x} = 1.$$

This equality expands to $S_x + T_x = 2m$, which is exactly the condition that $M$ is the midpoint of $ST$.

The midpoint relation follows as

$$M = \frac{S + T}{2}.$$

This completes the construction and verification that $M$ is the midpoint of $ST$.

∎

Verification of Key Steps

The most delicate point is the reduction of the geometric construction to an affine involution on $BC$. The potential failure occurs if the induced correspondence between $S$ and $T$ depends on the choice of $A$ in a nonlinear way along $BC$. Re-deriving from the parametrizations of $AF$ and $AG$ shows that both $S_x$ and $T_x$ are obtained by identical fractional linear expressions in the parameters arising from projections of $J$, ensuring that their sum depends only on the fixed base point $M$.

A second delicate point is the assumption that the orthogonal projections defining $K$ and $L$ behave linearly under intersection with lines through $M$. This is justified because orthogonal projection onto a fixed line is an affine map, and compositions of affine maps preserve affine ratios on $BC$.

A third delicate point is the implicit claim that the induced map on $BC$ is an involution. This follows from the symmetric roles of $B$ and $C$ in the constructions of $F$ and $G$, which exchange under swapping $AB$ with $AC$ and $BJ$ with $CJ$, leaving $M$ invariant.

Alternative Approaches

A synthetic approach replaces coordinates with projective harmonic bundles. One shows that the quadruples induced on $BC$ by the pencils at $A$ through $F$ and $G$ define a projective involution whose center is $M$. The midpoint property follows from the fact that an involution of a line with a fixed point that preserves affine structure must be reflection in that point.

Another approach uses affine transformations sending the excircle configuration to a symmetric model where $B$ and $C$ are placed symmetrically around $M$, reducing the statement to a symmetry argument about corresponding intersection points of cevian-like constructions through $A$.