Triangle incribed in Circle

Geometry Level 3

The area of equilateral triangle inscribed in the circle x 2 + y 2 − 4 x + 2 y + 1 = 0 x^2+y^2-4x+2y+1=0 is of the form a √ b c \frac{a√b}{c} Find a + b + c a+b+c


The answer is 7.

Sanjeet Raria by Sanjeet Raria , 27, India

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1 solution

Sanjeet Raria
Sep 23, 2014

We know that radius of the circle x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2+y^2+2gx+2fy+c=0 is R = g 2 + f 2 − c R=\sqrt{g^2+f^2-c} Hence radius of the given circle is R = 4 + 1 − 1 = 2 R=\sqrt{4+1-1}=2 Now area of equilateral triangle with side a A = √ 3 4 a 2 A=\frac{√3}{4}a^2 Also from sine rule a = 2 R sin ⁡ ∠ A = √ 3 R a=2R\sin\angle A=√3R Hence A = 3 √ 3 4 4 = 3 √ 3 A=\frac{3√3}{4}4=3√3 Answer is 3 + 3 + 1 = 7 3+3+1=\boxed7

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You must give some restrictions since $$\frac{3\sqrt{3}}{1}=\frac{6\sqrt{3}{2}$$

Alfa Claresta - 2 years, 5 months ago

Sine rule will be 2Rsin(A/2)

NAYAN CHAUHAN - 6 years, 7 months ago

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