SOLUTION: Equilateral triangle ABC has a perimeter of 12. the circle with center O and radius 1 is inscribed inside triangle ABC. What is the distance between point G and C? Point G is locat

Question 1119150: Equilateral triangle ABC has a perimeter of 12. the circle with center O and radius 1 is inscribed inside triangle ABC. What is the distance between point G and C? Point G is located at the the edge of the circle and point C is located at the vertice of the triangle. A picture is helpful in this case, but I don't know where I can upload a picture. Let me know how I can do that.
I tried to solve by recognizing all the angles of the triangle are 60 degrees. So if we split one in half that's 30 degrees and makes a 30-60-90 triangle. Knowing the perimeter is 12 that means each side of the triangle is 4. So using the 30-60-90 rule I said x=2 since it's half of one side, so that means the other side is 4 and the other side is 2sqrt3. I divided 2sqrt3 by 3 and got 1.15 but that was not an answer choice listed.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

There is something wrong with the statement of the problem. As you found, the radius of a circle inscribed in an equilateral triangle of side length 4 is 2sqrt3 divided by 3; but the problem says the radius is 1.

Your definition of point G being "on the edge of the circle" is not precise.

If G were a point on the circle where it is tangent to a side of the triangle, then clearly the distance between C and G would be half the side length of the triangle, which is 2. So that is probably not where point G is.

The only other reasonable place for point G is a point on the circle closest to a vertex of the triangle. In that case, since the altitude of the triangle is 2sqrt3 and the radius of the circle is 2sqrt3 divided by 3, the distance from C to G is also 2sqrt3 divided by 3.

So if you ignore the misinformation that the radius of the circle is 1, your answer of 2sqrt3 divided by 3 for the distance from C to G is correct.
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