Manhattan GMAT Challenge Problem of the Week - 19 May 2010

by Michael, May 19, 2010

Welcome back to this weeks 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

In a rectangular coordinate system, point A has coordinates (d, d), where d > 0. Point A and the origin form the endpoints of a diameter of circle C. What fraction of the area of circle C lies within the first quadrant?

A.

B.

C.

D.

E.

Answer

First, draw the coordinate plane and add in point A, which has coordinates (d,d). Since d is greater than 0, this point lies in the first quadrant, as shown:

Notice that the origin, the point (0, d), the point (d, 0), and point A (d,d) form the corners of a square:

Now, we draw the line segment between (0,0) and (d,d). This is the diameter of a circle, which we also draw:

Notice that the answer does not depend on d. All that matters is that the circle contains a square. The fraction of the circle's area in the first quadrant will be the same, no matter what. Thus, we can drop d as a variable and create a new variable r for the radius of the circle. It will be easier to compute areas in terms of r (which will also cancel out in the end).

The area of the circle is just [pmath]pi r^2[/pmath].

The fraction we are looking for is this:

So we need the shaded area of the circle, which consists of the square ( ) and 2 of the four "leaves of the table" ( ).

The square's area can be found this way:

Now the area of the 4 leaves ( ) is the area of the circle ( ) minus the area of the square ( ). So the area of the 4 leaves is [pmath]pi r^2-2r^2=(pi-2)r^2[/pmath].

This means that the area of 2 leaves is .

The whole shaded area is thus .

Finally, the desired fraction is .

The correct answer is D.

Special Announcement: If you like the explanations we give for our Challenge Problems, try our new Official Guide Companion , available now. It has detailed answer explanations to every quant question in the Official Guide.