Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National Olympiad Prove it United States

Problem:

In triangle $\triangle A B C$ , we have marked points $A_{1}$ on side $B C$ , $B_{1}$ on side $A C$ , and $C_{1}$ on side $A B$ so that $A A_{1}$ is an altitude, $B B_{1}$ is a median, and $C C_{1}$ is an angle bisector. It is known that $\triangle A_{1} B_{1} C_{1}$ is equilateral. Prove that $\triangle A B C$ is equilateral too.

(Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median $B B_{1}$ we know that $B_{1}$ is the midpoint of side $A C$ in $\triangle A B C$ .)

Solution

Solution:

Let equilateral $\triangle A_{1} B_{1} C_{1}$ have sides of length $s$ . Since $\triangle A A_{1} C$ is right with midpoint $B_{1}$ on the hypotenuse $A C$ ,
$B_{1} A = B_{1} C = B_{1} A_{1} = s = B_{1} C_{1},$
so points $A$ , $C_{1}$ , $A_{1}$ , and $C$ lie on a circle centered at $B_{1}$ . Therefore, $\triangle A C C_{1}$ is also right with hypotenuse $A C$ . In other words $C C_{1}$ is an altitude, but since it was an angle bisector, we conclude $A C = B C$ .

In particular, $C_{1}$ is the midpoint of $A B$ . But since $\triangle A A_{1} B$ is also right, the median $A_{1} C_{1}$ is half of the hypotenuse $A B$ . In other words,
$A B = 2 A_{1} C_{1} = 2s = A C$
completing the proof.

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