Single correct option
In triangle ABC, AB = 13, BC = 14 and CA = 15. What is the distance between the incentre and the circumcentre of the triangle?
- A. 1
- B. 65/64
- C. √65/4
- D. √65/8
Show answer and solution
Answer: (D) √65/8
s = (13 + 14 + 15)/2 = 21. Area = √(21 × 8 × 7 × 6) = 84.
Inradius r = Area/s = 84/21 = 4.
Circumradius R = abc/(4 × Area) = (13 × 14 × 15)/336 = 65/8.
Euler's formula: OI² = R(R − 2r) = (65/8)(65/8 − 8) = (65/8)(1/8) = 65/64.
So OI = √65/8 (about 1.008).
65/64 is OI², a trap for those who forget the square root.
Coordinate check: B(0,0), C(14,0), A(5,12) gives I = (6, 4) and O = (7, 33/8); distance² = 1 + 1/64 = 65/64 ✓.
Inradius r = Area/s = 84/21 = 4.
Circumradius R = abc/(4 × Area) = (13 × 14 × 15)/336 = 65/8.
Euler's formula: OI² = R(R − 2r) = (65/8)(65/8 − 8) = (65/8)(1/8) = 65/64.
So OI = √65/8 (about 1.008).
65/64 is OI², a trap for those who forget the square root.
Coordinate check: B(0,0), C(14,0), A(5,12) gives I = (6, 4) and O = (7, 33/8); distance² = 1 + 1/64 = 65/64 ✓.
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