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by pm » Wed Jul 20, 2011 11:16 am
Points X, Y, and Z lie on circle C, and line segment XY passes through the center of the circle. If the area of circle C is 18pi, what is the greatest possible perimeter of triangle XYZ?



(A)18



(B)3r2 + 18



(C)6r2 + 6



(D)6r2 + 12



(E)9r2
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by MBA.Aspirant » Wed Jul 20, 2011 11:46 am
pm wrote:Points X, Y, and Z lie on circle C, and line segment XY passes through the center of the circle. If the area of circle C is 18pi, what is the greatest possible perimeter of triangle XYZ?



(A)18



(B)3r2 + 18



(C)6r2 + 6



(D)6r2 + 12



(E)9r2

IMO D) 12 + 6√2

XY passes througth the circle's origin, implies that triangle XYZ is a right-angled triangle

radius = √18 = 3√2

diameter = 6 √2

since this is a right triangle and the hypotenuse has a √2 in it, then the sides are 1:1:√2 ratio

If you want to prove that it's 45:45:90, you can watch this video: https://www.youtube.com/watch?v=b0U1NxbRU4w

so sides are 6 : 6 : 6√2

perimeter = 6+6+6√2 = 12 + 6√2
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by knight247 » Wed Jul 20, 2011 11:53 am
Vertices X and Y of the triangle lie on the diameter and z lies elswhere on the circumference. Which clearly indicates that triangle XYZ is inscribed inside a semicircle. Making it a Right Angled Triangle with XY as the hypotenuse.

Given that

PiR^2=18Pi
R=3r2
D=6r2=XY

Perimeter is the sum of all sides of a triangle. By now, most ppl may have narrowed down to option C or D. Now, in a triangle the sum of any two sides has to be greater than the 3rd side. Between C and D D is the only option that satisfies this condition. Hence D
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by pm » Wed Jul 20, 2011 11:57 am
OA D
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