Perimeter of inscribed hexagon > In a circle of radius 3 the equilateral triangle ABC is inscribed, and the points X, Y and Z are diametrically opposite to A, B and C (respect) . Find the perimeter of the hexagon AZBXCY. The radius is $d/2 = 3$, the arcs AB, BC, CA are all $120$. I got that the length of the chords, AB, BC, CA is: $3$ each and also that, $P = 6(x)$ where $x$ is $AZ$ for example. But I got: $P = 6\sqrt{3}$ which is incorrect? Help?

**Solution (you have to draw a diagram to follow along):**

Since X is diametrically opposite to A, and ABC is equilateral (which means AB=AC), we can prove that AX bisects ∠BAC. Let's look at triangles AXB and AXC.

Because AX is a diameter, we know that ∠XBA and ∠XCA are both right angles. We also know that ∠BXA and ∠CXA are congruent because they intercept the same chord, and therefore the same arc. If 2 angles of 2 triangles are congruent, the third pair of angles is also congruent. Thus, we get that ∠BAX≅∠CAX. We can see that both angles are equal to 30 degrees.

Now, since ∠BAX and ∠CAX are 30 degrees, the arcs are both 60 degrees. That means that the central angles to those arcs are also 60 degrees. This proves that hexagon AZBXCY is indeed regular.

Since the radius of the circle is 3, each side of the regular hexagon also has a length of 3 (each sector of the hexagon is an equilateral triangle). This means that the perimeter of the hexagon is **18 units.**