Difference between revisions of "2023 CMO Problems/Problem 5"

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Revision as of 05:18, 25 May 2024

In an acute triangle $\triangle A B C, K$ is a point on the extension of $B C$. Through $K$, draw lines parallel to $A B$ and $A C$, denoted as $K P$ and $K Q$ respectively, such that $B K=B P$ and $C K=C Q$. Let the circumcircle of $\triangle K P Q$ intersect $A K$ at point $T$. Prove: (1) $\angle B T C+\angle A P B=\angle C Q A$; (2) $A P \cdot B T \cdot C Q=A Q \cdot C T \cdot B P$.

Solution 1

~xiaohuangya|szm

See Also

2023 CMO(CHINA) (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All CMO(CHINA) Problems and Solutions