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JEE Advanced 2014 · Paper 2, Mathematics question 47

In a triangle the sum of two sides is xx and the product of the same two sides is yy. If x2−c2=yx^2-c^2=y, where cc is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

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Answer and worked solution

Official answer: B

Find the angle. Let the sides be a,ba,b with a+b=xa+b=x, ab=yab=y. Then
x2−c2=y  ⟹  a2+b2+2ab−c2=ab  ⟹  a2+b2−c2=−ab,x^2-c^2=y\implies a^2+b^2+2ab-c^2=ab\implies a^2+b^2-c^2=-ab,
so cos⁡C=−12\cos C=-\frac12 and C=120°C=120°.

In-radius and circum-radius.

  • Δ=12absin⁡C=34y\Delta=\tfrac12ab\sin C=\dfrac{\sqrt3}4y and s=x+c2s=\dfrac{x+c}2, so r=Δs=3y2(x+c)r=\dfrac\Delta s=\dfrac{\sqrt3y}{2(x+c)}.
  • R=c2sin⁡C=c3R=\dfrac{c}{2\sin C}=\dfrac c{\sqrt3}.

Ratio.
rR=3y2(x+c)⋅3c=3y2c(x+c).\frac rR=\frac{\sqrt3y}{2(x+c)}\cdot\frac{\sqrt3}c=\frac{3y}{2c(x+c)}.

(B)

Topics:Properties of triangles
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