Properties of Equilateral Triangles in Circles
If there is an equilateral triangle in a circle, would the midpoint of any of the 3 sides be half the radius?
e.g if the radius was 6 and at the midpoint of the triangle (call it B) would center to B be 3 and then B to circle be 3 as well?
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$\begingroup$Yes. If you're familiar with construction using compass and straight edge, one of the easiest ways to construct an equilateral triangle is to draw two circles where each circle's centre lies on the other circle's edge. Drawing a line between the two intersection points and then from each intersection point to the point on one circle farthest from the other creates an equilateral triangle. You can see from this construction that the side of the equilateral triangle between intersection points is equidistant from each centre, proving that the side is halfway between the circle's centre and its edge.
Yes. Draw a picture : From the circle's center draw a radius to a vertex and a line to the midpoint of a side with that vertex at one extreme. Then you get a $\;30-60-90\;$ little triangle and thus the line from center to the midpoint is opposite to the angle of $\;30^\circ\;$ and is thus half the hypotenuse which is the radius.
A drawing could probably help a lot here, but unfortunately I've no idea how to do that here.
Ok, now I've learned! in the drawing below, $\;AM\;$ is a median to $\;BC\;$ in circle $\;O\;$ . Observe that triangle $\;\Delta OMC\;$ is $\;30-60-90\;$ , and $\;OM\;$ is opposite to angle $\;30^\circ\;$ , thus
$$|OM|=\frac{|OC|}2$$
If the inner triangle is equilateral, then the angle $y = 60^o$, and since the OBA is a right triangle, the angle $x = 30^o$
If "B to the circle" in the question is $|BA|$, then it is $3\sqrt{3}$, (not 3).
If "B to the center" in the question is $|BO|$, then it is 3.
$\endgroup$ 13 $\begingroup$No. In that case, the angle of triangle must be $45^{\circ}$ because is isosceles.
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