Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National Olympiad Prove it Bulgaria

Problem:

In an acute $\triangle ABC$ the altitudes $AA_1$ ( $A_1 \in BC$ ) and $BB_1$ ( $B_1 \in AC$ ) are drawn, $I$ is the incenter and the line $CI$ meets $AB$ at $L$ . It is known that $I$ lies on the circumcircle of $\triangle A_1B_1C$ .

a) Prove that $L$ is the center of excircle of $\triangle A_1B_1C$ tangent to the side $A_1B_1$ .

b) If $CI = 2IL$ , find $\text{ACB}$ .

Solution

Solution:

a) Since $\text{CB 1A 1 = CIA 1}$ and the quadrilateral $ABA_1B_1$ is cyclic, we have $\text{CIA 1 = CB 1A 1 = ABC}$ . This shows that the quadrilateral $LBA_1I$ is cyclic. Then
$\text{LA 1B = LIB = ICB + IBC = 1 2 ( ACB + ABC) = 1 2 ( ACB + A 1B 1C ) = 1 2 BA 1B 1}$

Hence $A_1L$ is the bisector of $\text{BA 1B 1}$ , which completes the proof.

b) If $J$ is the midpoint of $CI$ , then we have $CJ = JI = IL$ . But $IL = IA_1$ from a) and we conclude that $JLA_1$ is a right triangle. Then $A_1J$ is the bisector of $\text{B 1A 1C}$ and therefore $J$ is the incenter of $\triangle A_1B_1C$ . On the other hand, we have $\triangle A_1B_1C \sim \triangle ABC$ , whence $\frac{A_1C}{AC} = \frac{CJ}{CI} = \frac{1}{2}$ and $\text{ACB = 60}$ .

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