Proving the inradius sum up of two triangle from segment ABC is least from the inradius triangle ABC (Solution Geometry Shortlist IMO 2014)
Let ABC be a triangle. The points K, L, and M lie on the segments BC, CA, and AB, respectively, such that the lines AK, BL, and CM intersect in a common point. Prove that it is possible to choose two of the triangles ALM, BMK, and CKL whose inradii sum up to at least the inradius of the triangle ABC.
, , .
By Ceva’s theorem, abc = 1, so we may, without loss of generality, assume that . Then at least one of the numbers b or c is not greater than 1. Therefore at least one of the pairs (a, b), (b, c) has its ﬁrst component not less than 1 and the second one not greater than 1. Without loss of generality, assume that and .
Therefore, we obtain and , or equivalently
The ﬁrst inequality implies that the line passing through M and parallel to BC intersects the segment AL at a point X (see Figure 1). Therefore the inradius of the triangle ALM is not less than the inradius r1 of triangle AMX.
Similarly, the line passing through M and parallel to AC intersects the segment BK at a point Y , so the inradius of the triangle BMK is not less than the inradius r2 of the triangle BMY . Thus, to complete our solution, it is enough to show that , where r is the inradius of the triangle ABC. We prove that in fact r1 + r2 = r.
Since MX || BC, the dilation with centre A that takes M to B takes the incircle of the triangle AMX to the incircle of the triangle ABC. Therefore
, and similarly .
Adding these equalities gives r1 + r2 = r, as required.
Alternatively, one can use Desargues’ theorem instead of Ceva’s theorem, as follows: The lines AB, BC, CA dissect the plane into seven regions. One of them is bounded, and amongst the other six, three are two-sided and three are three-sided. Now define the points , , and R=AB\cap KL$ (in the projective plane). By Desargues’ theorem, the points P, Q, R lie on a common line l. This line intersects only unbounded regions. If we now assume (without loss of generality) that P, Q and R lie on l in that order, then one of the segments PQ or QR lies inside a two-sided region. If, for example, this segment is PQ, then the triangles ALM and BMK will satisfy the statement of the problem for the same reason.
Also Read : Geometry shortlist problem in IMO 2014