# Manhattan GMAT Challenge Problem of the Week - 11 Mar 10

by , Mar 11, 2010

Welcome back to this week's Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

## Question

A circle is inscribed in an equilateral triangle, such that the two figures touch at exactly 3 points, one on each side of the triangle. Which of the following is closest to the percent of the area of the triangle that lies within the circle?

(A) 50%

(B) 55%

(C) 60%

(D) 65%

(E) 70%

## Solution

First, draw a picture of the circle and triangle and make up good labels for all points. We should add one more point: the center of the circle, which will also be the center of the triangle. Let's call that point O.

In order to deal with the area of the circle, we'll need a radius (with circles, it should always be your first instinct to draw in a radius). We'll put in OZ. Now, because both equilateral triangles and circles are regular, symmetrical figures, and because they're lined up perfectly in this scenario, we can take advantage of "symmetry arguments." For instance, it should be obvious that each of the points X, Y, and Z are the midpoints of their respective sides. You don't need to prove this, nor do you want to under time constraints. Rather, on the exam you should look for regular, symmetrical figures to behave symmetrically, and points that seem like midpoints can actually be proven to be just that. Likewise, we can draw in AO, and we know that this line cuts angle BAC perfectly in half. If you've drawn the figure carefully, this conclusion should be apparent. (We almost have the Deathly Hallows here, for you Harry Potter fans). Now, since any angles in an equilateral triangle is 60, angle AOZ = 60/Z = 30. Moreover, angle AZO is 90 (a radius touches any tangent line, at 90).

Thus, we know AOZ is 60, and we have a 30-60-90 triangle. Let's redraw that triangle by itself: Now, let's make up a convenient radius. Say OZ = 1. We quickly get the dimensions of triangle AOZ: We can now compute the areas of each figure. First, the circle is [pmath]pi r^2 = pi(1)^2 = pi[/pmath]. The triangle is a little harder, but we're almost there. Since AZ = [pmath]sqrt{3}[/pmath], and Z is the midpoint of AC, we know that AC = [pmath]2 sqrt{3}[/pmath]. If you know the formula for the area of an equilateral triangle in terms of its side length, you can plug in right now or you can figure out the height. Let's take the latter approach.

Since we know AO = 2, and OY = 1, we know that AY = 3 (again, we don't need or want a rigorous proof that they form a straight line. Regular figures behave nicely!) And again by symmetry, this tells us that BZ = 3 as well: Now we can calculate the area of the triangle, which equals ()bh = ()( )(3) = . (Incidentally, the area of an equilateral triangle in terms of its side length s is , which in this case gives )

Finally, we compute the ratio of the areas: . We approximate: [pmath]pi[/pmath] 3.1 and [pmath]sqrt{3}[/pmath] 1.7, so we get (Excel gives the ratio as 60.46% to a couple more decimal places, so our approximation didn't create much of an error).

The correct answer is (C) 60%.