Proving the inradius sum up of two triangle from segment ABC is least from the inradius triangle ABC (Solution Geometry Shortlist IMO 2014)

Problem 2:

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.

Solution.

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Denote

a=\frac{BK}{KC}, b=\frac{CL}{LA}, c=\frac{AM}{MB}.

By Ceva’s theorem, abc = 1, so we may, without loss of generality, assume that a\geqslant 1. 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 1\leqslant a and b\leqslant 1.

Therefore, we obtain bc\leqslant 1 and 1\leqslant ca, or equivalently

\frac{AM}{MB}\leqslant \frac{LA}{CL}\text{ and }\frac{MB}{AM}\leqslant \frac{BK}{KC}

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 r_1+r_2\geqslant r, 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

\frac{r_1}{r_2}=\frac{AM}{AB}, and similarly \frac{r_2}{r}=\frac{MB}{AB}.

Adding these equalities gives r1 + r2 = r, as required.

Comment.

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 P=BC\cap LM, Q=CA\cap MK, 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

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