An equilateral triangle with one side 6cm is inscribed in a circle with its three vertices on the circumference. What is the diameter of the circle?

(In reply to Trigonometry? by levik)

I will use levik's notation. In it, the vertices are A, B, and C, the center of the circle is O, and M is a midpoint on segment AB. The legnth of the hypotenuse (OA) is the radius of the circle, since O is the center of the circle and A is on the circle.

Angle CAB (which is angle CAM) is a 60-degree angle, because this is an equilateral triangle. Segment OA bisects this angle, and thus angle OAM is 30 degrees. The cosine of OAM is the length of the adjacent leg (segment OA) divided by the hypotenuse (segment AM).

The cosine of a 30 degree angle is sqrt(3)/2. Using r for the hypotenuse OA (the radius), and 3 cm for the length of AM gives:
sqrt(3)/2 = (3 cm)/r, which solving for r yields:
r = 6/sqrt(3) cm = 2*sqrt(3) cm

The diameter is twice the radius, so d = 2*r = 4*sqrt(3) cm, which is about 6.93 cm (TomM's answer).

I suspect Half-Mad is incribing the circle inside of the triangle, instead of the triangle inside of the circle, since the diameter must be at least 6 cm (the length of a side).

Posted by Ender on 2002-06-04 03:41:39