Another Geometry Problem from 2012 IMO - Problem 1

The applet below illustrates Problem 5 from the 2012 IMO:

Given triangle ABC the point J is the center 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.


22 January 2015, Created with GeoGebra


The extras displayed by the applet may not be directly related to the proof, but they at least suggest the direction in which the proof is going to develop.

Ultimately, I am going to show that JS = AJ = JT, making ΔJST isosceles, with JM the altitude and hence the median from the apex J.

The proof is based on subsequent evaluation of angles - angle hunting. It takes two steps. First, we establish that ∠JGK = ∠JAK and, similarly, that ∠JFL = ∠JAL. These makes points J, L, G, A, F, K concyclic, from which AG⊥JG and AF⊥JF.

Second, we'll prove that ∠CTA = ∠CAT, making ΔACT isosceles, implying that ΔAJT is also isosceles: AJ = JT. Similarly, AJ = AS. This in particular means that S and T lie on the circle through A centered at J. ST is a chord in this circle tangent to the original excircle of ΔABC. Thus MS = MT.

For the proof, let angles in ΔABC be denoted α, β γ.

Step 1: ∠JGK = α/2

Consider right triangle JGH:

∠HJG = ∠BJG = 180° - (90° - β/2) - (90° - γ/2) = 90° - α/2.

Since ΔJGH is right, ∠JGK = α/2. Similarly, ∠JGL = α/2.

By the construction of K and L, JK⊥AK and JL⊥AL, implying that points A, K, J, L are concyclic on a circle with diameter AJ. From what we just proved, points A, J, L, G are concyclic and so are points A, J, K, F. It follows that all six lie on the same circle, while the angles at F and G are subtended by diameter AJ. Therefore, AF⊥JF and AG⊥JG.

Step 2: ∠CTA = ∠CAT

In right triangle CAG, ∠ACG = 90° - γ/2, hence ∠CAT = ∠CAG = γ/2.

In ΔBTA, ∠CTA = ∠BTA = 180° - β - α - γ/2 = γ/2. So, it does follow that ΔACT is isosceles. Then so are ΔAJT and, subsequently, ΔJST. Which completes the proof.

Hubert Shutrick came up with another proof. He started with reformulating the problem:

With the excircle as in the diagram, let S be the point on BC such that JB is orthogonal to AS and define T similarly, then the midpoint of ST is M, the tangental point of BC.


Since JB bisects ∠SBA the foot point F is the midpoint of SA so the midpoint of ST is where the line parallel with AT intersects BC. To prove that this point is M note that F, K and L are all on the circle with diameter AJ and let F* be the intersection of JB and LM. I shall prove that the triangles AJL and F*MH are similar by showing that ∠AJL and ∠F*MH are equal which implies that F* is on the circle and therefore the same as F.

The angles between the tangent BC and the two chords are ∠F*MB = ∠KJM/2 and ∠BMH = ∠MJL/2 so ∠F*MH = ∠HJL/2 = ∠AJL as required.

Note: There is a shorter solution.

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