# 2021 USAMO Problems/Problem 6

## Problem 6

Let be a convex hexagon satisfying , , , andLet , , and be the midpoints of , , and . Prove that the circumcenter of , the circumcenter of , and the orthocenter of are collinear.

## Solution 1

Let , , and be the midpoints of , , and , , and be the midpoints of , , and . Also, let be the orthocenter of . Note that we can use parallel sides to see that , , and are collinear. Thus we have by midlines. Applying this argument cyclically, and noting the condition , , , , , , all lie on a circle concentric with .

Next, realize that basic orthocenter properties imply that the circumcenter of is the orthocenter of , and likewise the circumcenter of is the orthocenter of .

The rest is just complex numbers; toss on the complex plane so that the circumcenter of is the origin. Then we have Note that from the above we have , so is the midpoint of segment . In particular, , , and are collinear, as required.

~ Leo.Euler

## Solution 2

We construct two equal triangles, prove that triangle is the same as medial triangle of both this triangles. We use property of medial triangle and prove that circumcenters of constructed triangles coincide with given circumcenters.

Denote Then

Denote

Similarly we get

The translation vector maps into is

so is midpoint of and Symilarly is the midpoint of and is the midpoint of and is the midpoint of

Similarly is the midpoint of is the midpoint of

Therefore is the medial triangle of

is translated on

It is known (see diagram) that circumcenter of triangle coincide with orthocenter of the medial triangle. Therefore orthocenter of is circumcenter of translated on

It is the midpoint of segment connected circumcenters of and

According to the definition of points quadrangles and are parallelograms. Hence Power of points A,C, and E with respect circumcircle is equal, hence distances between these points and circumcenter of are the same. Therefore circumcenter coincide with circumcenter

Similarly circumcenter of coincide with circumcenter of

**vladimir.shelomovskii@gmail.com, vvsss**

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