Circle Inscribed In an Equilateral Triangle

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In yesterday's article about multiple figures in GMAT geometry, I posed a question:

What if a circle is inscribed in an equilateral triangle? ... If I gave you the area of the circle, you have enough information to find, say, the perimeter of the triangle.

To see how the figures are related, click here for a diagram. (Drawing isn't my strong suit, but I think you'll get the idea despite the lopsided circle.)

The Relationship

The key insight here is that the center of the two figures (O) is the same point. Thus, each of the lines from an interior angle of the triangle to the center of the circle bisects that angle. Take, for instance, BO. It bisects angle CBA, so CBO and OBA are each half of the original measure of angle CBA.

Then, if you draw the remaining line (OP) of the right triangle OPB, you have a 30-60-90 triangle. Since we know the ratio of the sides of such a triangle (x : xr3 : 2x), we have the relationship between the radius of the circle (OP) and half of one side of the triangle (PB). (Note: "xr3" means "x times root 3.")

Solving the Problem

I didn't provide an exact value for the area of the circle yesterday, so let's do that now. If the area of the circle is 9pi, what is the perimeter of the triangle?

As we've seen, OP is the radius of circle O. If circle O has an area of 9pi, the radius is 3. Using the 30-60-90 ratio, if x = 3, xr3 = 3r3. BP is the xr3 side of the triangle, so BP = 3r3. Since that's half of the base of the triangle, the length of one side is 2(3r3) = 6r3.

Finally, the perimeter of an equilateral triangle is three times the length of one side, or 3(6r3) = 18r3.



About the author: Jeff Sackmann has written many GMAT preparation books, including the popular Total GMAT Math, Total GMAT Verbal, and GMAT 111. He has also created explanations for problems in The Official Guide, as well as 1,800 practice GMAT math questions.

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