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Several ways to solve this problem [#permalink]
18 Dec 2005, 16:26
Thanks for posting the figure. It helps to visulize the problem.
Note that you are looking at an isoceles triangle with one vertex at the center of the circle and two equal sides equal to 1 (the radius of the circle).
The problem boils down to finding the length of the third side of the triangle (say X) to maximize the area.
The area of the triangle is 1/2 ab SinC where a and b two sides of the triangle and C is the included angle between a and b.
In this case, a = b = 1.
Then, the area is 1/2 1.1.SinC = 1/2 SinC.
Note that the value of SinC is between 0 and 1 and the maximum value is 1 happens at C = 90 degrees.
Hence, the Maximum area = 1/2 * 1 = 1/2 = 0.5
Let X be the third side. Then, by dropping a perpendicular from the center of the circle to the third side, you note that it bisects the side (because the triangle is isoceles).
So, the area of the given triangle = 2 (1/2) * X/2 * (sqrt(1-x^2/4))
So, the problem boils down to finding X to maximize X * sqrt(4-X*X).
Note that X is between 0 and 2. The maximum value occuring at X = sqrt(2).
Then, the area of the triangle = 1/2
Let X is the angle made by the radius of the circle to the line segment AB where A and B are the two other sides of the triangle.
Then, AB = 2 CosX and OD = sinX where OD = the length of the perpendicular from Origin O to the side AB.
The triangle area = 2 * (1/2) * CosX * sinX = CosX * SinX = 1/2 (sin2X).
The maximum value of (cosX * sinX) or (sin2X) occurs at X = 45 degrees.
Therefore, the triangle area is 1/2 * 1 = 0.5
Method IV: (the simplest - which was also explained by the previous reply to some extent)
I believe GMAT does not require you to know the above trigonometric formulae or calculating maxima or minima of a triagnometric functions.
So, if you look at finding a solution some other technique..
Let A and B are the points of the triangle. Now, using symmetry, find out points C and D on the other half of the circle (note that A and B stay on the same side of the semi-circle). Now, ABCD form a sqaure (due to symmetry) and AB=BC=CD=DA. The diagonal AC = AO+OC=1+1=2.
If X is the side of the square, X . sqrt(2) = 2, then, X = 2/sqrt(2) = sqrt(2).
The area of the sqaure is sqrt(2) * sqrt(2) = 2 units and the area of the triangle is 1/4*2 = 1/2 (because there are four triangles in the square with equal area - once agian due to symmetry).