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Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

Circle kk of radius rr is inscribed in △ABC\triangle ABC. Tangent lines of kk, that are parallel to sides ABAB, BCBC and CACA, intersect other sides of △ABC\triangle ABC at points M,NM, N; P,QP, Q and L,TL, T (P,T∈ABP, T \in AB, L,N∈BCL, N \in BC and M,Q∈ACM, Q \in AC). Denote by r1,r2,r3r_1, r_2, r_3 radii of circles inscribed in triangles MNC,PQAMNC, PQA and LTBLTB, respectively. Prove that r1+r2+r3=rr_1 + r_2 + r_3 = r.

Figure 1
Fig. 28

Solution

Since all these triangles are similar, we get
r1+r2+r3r=p1+p2+p3p \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)
2p1=CM+CN+MN==CM+CN+MZ+ZN==CM+MX+CN+NY=2p−(AX+AB+BY)=2p−2c. \begin{aligned} 2p_1 &= CM + CN + MN = \\ &= CM + CN + MZ + ZN = \\ &= CM + MX + CN + NY = 2p - (AX + AB + BY) = 2p - 2c. \end{aligned}
Thus,
r1+r2+r3r=p1+p2+p3p=p−a+p−b+p−cp=3p−2pp=1. \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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