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Bulgarian Mathematical Competitions

Geometry Difficulty 6.4 National Olympiad Prove it Bulgaria

Problem:

The bisectors of $\angle A$ , $\angle B$ and $\angle C$ of $\triangle ABC$ meet its circumcircle at points $A_1$ , $B_1$ and $C_1$ , respectively. Set $AA_1 \cap CC_1 = I$ , $AA_1 \cap BC = N$ and $BB_1 \cap A_1C_1 = P$ . Denote by $O$ the circumcenter of $\triangle IPC_1$ and let $OP \cap BC = M$ . If $BM = MN$ and $\angle BAC = 2 \angle ABC$ , find the angles of $\triangle ABC$ .

Solution

Solution:

Set $\angle CAB = \alpha$ , $\angle ABC = \beta$ and $\angle BCA = \gamma$ . Then
$\text{IPC 1 = 1 2 ( BA 1 + C 1B 1 ) = 1 2 ( BA 1 + AC 1 + AB 1 ) = 1 2 ( + + ) = 90 .}$
Hence $O$ is the midpoint of the segment $IC_1$ and it follows that

$\text{IOP = 2 IC 1P = CA 1 = .}$
We also have $\angle CC_1B = \alpha$ and therefore $OP \parallel C_1B$ . Since $C_1O = OI$ and $BM = MN$ , we conclude that $IN \parallel C_1B$ , i.e. $\angle CIA_1 = \alpha$ . On the other hand,
$\text{CIA 1 = 1 2 ( CA 1 + AC 1 ) = 1 2 ( + ) = 90 - 2 .}$
This implies that $\alpha = 90^\circ - \frac{\beta}{2} = \frac{\alpha + \gamma}{2}$ , i.e. $\alpha = \gamma$ . Since $\alpha = 2\beta$ , we obtain that $\alpha = \gamma = 72^\circ$ and $\beta = 36^\circ$ .

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