In a triangle, the sum of two sides is and the product of the same two sides is . If , where is the third side of the triangle, then the ratio of the inradius to the circumradius of the triangle is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
Let the two sides be a and b, so a+b = x and ab = y.
Given: x² - c² = y → (a+b)² - c² = ab → a² + 2ab + b² - c² = ab → a² + ab + b² = c².
By the cosine rule: c² = a² + b² - 2ab·cos C, so:
a² + ab + b² = a² + b² - 2ab·cos C → ab = -2ab·cos C → cos C = -1/2 → C = 120°.
Now compute inradius/circumradius = r/R.
Using r = (Area)/s and R = c/(2 sin C):
sin C = sin 120° = √3/2, so Area = (1/2)ab·sin C = (√3/4)y.
Semi-perimeter s = (a+b+c)/2 = (x+c)/2.
r = Area/s = (√3y/4)/((x+c)/2) = √3y/(2(x+c)).
R = c/(2 sin C) = c/(2·√3/2) = c/√3.
r/R = [√3y/(2(x+c))] / [c/√3] = [√3y/(2(x+c))]·(√3/c) = 3y/(2c(x+c)).
This matches option 2.
Given: x² - c² = y → (a+b)² - c² = ab → a² + 2ab + b² - c² = ab → a² + ab + b² = c².
By the cosine rule: c² = a² + b² - 2ab·cos C, so:
a² + ab + b² = a² + b² - 2ab·cos C → ab = -2ab·cos C → cos C = -1/2 → C = 120°.
Now compute inradius/circumradius = r/R.
Using r = (Area)/s and R = c/(2 sin C):
sin C = sin 120° = √3/2, so Area = (1/2)ab·sin C = (√3/4)y.
Semi-perimeter s = (a+b+c)/2 = (x+c)/2.
r = Area/s = (√3y/4)/((x+c)/2) = √3y/(2(x+c)).
R = c/(2 sin C) = c/(2·√3/2) = c/√3.
r/R = [√3y/(2(x+c))] / [c/√3] = [√3y/(2(x+c))]·(√3/c) = 3y/(2c(x+c)).
This matches option 2.
[JEE Advanced 2014]