In the right triangle ABC, the medians AA1, BB1 and CC1 intersect at point O. Points K, M and N are the midpoints

In the right triangle ABC, the medians AA1, BB1 and CC1 intersect at point O. Points K, M and N are the midpoints of the segments AO, BO and CO, respectively. Find the perimeter of hexagon A1MC1KB1N if AB = a.

Since in a regular triangle the medians are the height, bisector and median and are divided by the intersection point in a ratio of 2 to 1, then AO / A1O = BO / B1O = CO / C1O = 2/1.

Therefore, OC1 = ОА1 = ОВ1 = OK = OM = ON = x, and the resulting hexagon is equilateral.

Therefore, its perimeter is: P = 6 * b, where b is the side of the hexagon.

From the right-angled triangle A1BO, A1B = a / 2, angle A1BO = 30º, since AA1 = BB1 – bisectors, medians and heights.

This means that the angle MOA1 = 60º.

And since OM = OA1, then the MOA1 triangle is equilateral, then b = x.

By the theorem of sines OA1 = x = A1B * tg (OBA1) = a / 2 * tg30º = a / 2 * √3 / 3 = a√3 / 6.

The perimeter of A1MC1KB1N is:

P = 6 * x = 6 * a√3 / 6 = a√3.