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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?

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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.

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  • $\begingroup$ (+1) But I don't understand, just because $r=3$, how can you say the side length $=3$ as well? $\endgroup$ – Amad27 Oct 18 '15 at 16:40
  • $\begingroup$ If the center of the circle is O, then triangle OAZ and all other "sectors" of the hexagon are equilateral. The radius is a side of the equilateral triangle; a side of the hexagon is also a side of the equilateral triangle. Therefore, the 2 segments are congruent. $\endgroup$ – Jed Oct 18 '15 at 17:08
  • $\begingroup$ etc.usf.edu/clipart/43400/43447/6c2_43447_sm.gif This might help. $\endgroup$ – Jed Oct 18 '15 at 17:10
  • $\begingroup$ I thought it only said $1$ equilateral traingle though? (Great answer, accepted) $\endgroup$ – Amad27 Oct 18 '15 at 17:13
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You may try to show that the hexagon is regular. Therefore each of the sides is equal to the radius of the circumcircle.

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  • $\begingroup$ Wait, how do you know each side is equal to the radius? (+1) $\endgroup$ – Amad27 Oct 18 '15 at 16:18
  • $\begingroup$ @Amad27: Uniting the circumcenter with each vertex, we have six isosceles triangles $($since two sides are always the radius$)$, with an angle of $\dfrac{360^\circ}6=60^\circ$. $\endgroup$ – Lucian Oct 19 '15 at 15:20
  • $\begingroup$ @Lucian, Jed's answer has $6$ equilateral triangles though, the problem only says there is one? $\endgroup$ – Amad27 Oct 19 '15 at 16:14
  • $\begingroup$ @Amad27: $(-1)$ for not thinking before speaking. $\endgroup$ – Lucian Oct 19 '15 at 16:26
  • $\begingroup$ @Lucian. What? I am bad at Geometry, which is why I am speaking what I am thinking of the problem. How was he allowed to replace the one equilateral triangle with $6$? $\endgroup$ – Amad27 Oct 19 '15 at 16:37

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