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What is the greates possible area of a triangular region

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What is the greates possible area of a triangular region [#permalink] New post 18 Dec 2005, 12:36
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What is the greates possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?
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 [#permalink] New post 18 Dec 2005, 12:55
Nice question, I'll try do solve

I think the biggest area will be spanned by a 1:1:sgrt2 triangle. Since the radius is 1 the area should be 0,5.
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Several ways to solve this problem [#permalink] New post 18 Dec 2005, 16:26
Hi,

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.

Method I:

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

Method II:

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

Method III:

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).

-Srinivas.

Let A and B are the points
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If you would have to guess.. [#permalink] New post 18 Dec 2005, 16:59
Yes, indeed, if the answer choices were given, I would pick a number between 0 and 0.8, and here is the explanation...

Note that the area of circle with radius 1 is equal to (pi * 1 *1) = pi = approximately 3.1417.

Now, if there are 4 such triangles (due to symmetry), the area of each triangle can at most be pi/4 which is approximately 0.8.

-Srinivas (mathguru).
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 [#permalink] New post 18 Dec 2005, 22:37
GMATland's method:
Looks lengthy but in fact very simple.


Hi Mathguru: I'm really much impressed by the way you use that trigonometric formula :wink:
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 [#permalink] New post 19 Dec 2005, 11:50
What is the greates possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

a) sq rt 3/4

b) 1/2

c) TT/4

c) 1

e) sq rt 2
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 [#permalink] New post 19 Dec 2005, 11:51
the oa is 1/2. thanks guys
  [#permalink] 19 Dec 2005, 11:51
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