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Spring Mathematical Competition

Geometry Difficulty 6.3 National Olympiad Prove it Bulgaria

Problem:

Let $A A'$ , $B B'$ and $C C'$ be the angular bisectors of a triangle $A B C$ with incenter $I$ . The segments $C I$ and $A' B'$ meet at $D$ and the midpoints of the segments $A I$ and $B I$ are denoted by $M$ and $N$ , respectively.

a) If $a = B C$ , $b = A C$ and $c = A B$ , find the ratio $C D : D I$ .

b) If $K = A C \cap \overrightarrow{C' M}$ and $L = B C \cap \overrightarrow{C' N}$ , prove that $D$ is the incenter of $\triangle K L C$ .

Solution

Solution:

a)
We have $C A' = \frac{a b}{b + c}$ and $\frac{A I}{I A'} = \frac{A C}{C A'} = \frac{b + c}{a}$ . By the Menelaus theorem for $\triangle A I C$ and the line $B' A'$ we have
$\frac{C D}{D I} \cdot \frac{I A'}{A' A} \cdot \frac{A B'}{B' C} = 1 \Rightarrow \frac{C D}{D I} = \frac{a + b + c}{a} \cdot \frac{a}{c} = \frac{a + b + c}{c}$

b)
By the Menelaus theorem for $\triangle A I C$ and the line $K C'$ we get
$\frac{A M}{M I} \cdot \frac{I C'}{C' C} \cdot \frac{C K}{K A} = 1$
whence $\frac{C K}{K A} = \frac{a + b + c}{c}$ . Analogously, $\frac{C L}{L B} = \frac{a + b + c}{c}$ .

Let $h$ be the homothety with center $C$ and ratio $\frac{a + b + c}{a + b + 2c}$ . Then $h(A) = K$ , $h(B) = L$ , $h(C) = C$ and it follows from a) that $h(I) = D$ . Therefore $D$ is the incenter of $\triangle K L C$ .

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