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, 2007

Geometry Difficulty 4.4 AIME Prove it JBMO

Problem:

Let $M$ be an interior point of the triangle $A B C$ with angles $\text{B A C = 70}$ and $\text{A B C = 80}$ . If $\text{A C M = 10}$ and $\text{C B M = 20}$ , prove that $A B = M C$ .

Solution

Solution:

Let $O$ be the circumcenter of the triangle $A B C$ . Because the triangle $A B C$ is acute, $O$ is in the interior of $\triangle A B C$ . Now we have that $\text{A O C = 2 A B C = 160}$ , so $\text{A C O = 10}$ and $\text{B O C = 2 B A C = 140}$ , so $\text{C B O = 20}$ . Therefore $O \equiv M$ , thus $M A = M B = M C$ . Because $\text{A B O = 80 - 20 = 60}$ , the triangle $A B M$ is equilateral and so $A B = M B = M C$ .

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