Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

Circle $k$ of radius $r$ is inscribed in $\triangle ABC$ . Tangent lines of $k$ , that are parallel to sides $AB$ , $BC$ and $CA$ , intersect other sides of $\triangle ABC$ at points $M, N$ ; $P, Q$ and $L, T$ ( $P, T \in AB$ , $L, N \in BC$ and $M, Q \in AC$ ). Denote by $r_1, r_2, r_3$ radii of circles inscribed in triangles $MNC, PQA$ and $LTB$ , respectively. Prove that $r_1 + r_2 + r_3 = r$ .

Fig. 28

Solution

Since all these triangles are similar, we get
$\frac{r_1 + r_2 + r_3}{r} = \frac{p_1 + p_2 + p_3}{p}$
It is not hard to see that (fig. 28)
$\begin{aligned} 2p_1 &= CM + CN + MN = \\ &= CM + CN + MZ + ZN = \\ &= CM + MX + CN + NY = 2p - (AX + AB + BY) = 2p - 2c. \end{aligned}$
Thus,
$\frac{r_1 + r_2 + r_3}{r} = \frac{p_1 + p_2 + p_3}{p} = \frac{p-a+p-b+p-c}{p} = \frac{3p-2p}{p} = 1.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.