prmo 2012 question 14
14. \( O \) and \( I \) are the circumcentre and incentre of \( \triangle ABC \) respectively. Suppose \( O \) lies in the interior of \( \triangle ABC \) and \( I \) lies on the circle passing through \( B, O, \) and \( C \). What is the magnitude of \( \angle BAC \) in degrees?
We know, because \( I \) is the incenter, \[ \angle IBD = \frac{B}{2}, \quad \angle ICD = \frac{C}{2} \] \[ \Rightarrow \angle BIC = \pi - \left( \frac{B}{2} + \frac{C}{2} \right) = \pi - \left( \frac{\pi - A}{2} \right) = \frac{\pi}{2} + \frac{A}{2} \] But, \( \angle BIC = \angle BOC \), because chord \( BC \) subtends both angles at the center. Also, \( \angle BOC = 2A \), because the angle subtended by a chord at the center \( (O) \) is double the angle subtended by the chord on any part of the circle (passing through \( \triangle ABC \)) in the major sector. \[ \frac{\pi}{2} + \frac{A}{2} = 2A \] Which gives, \[ \angle A = \frac{\pi}{3} = 60^\circ \] \[ \Rightarrow \angle BAC = \frac{\pi}{3} = 60^\circ \]
You can apply a trick here. If you can construct a triangle which satisfies the conditions given in the question, you can answer using that.
Here if you construct an equilateral triangle, O will lie in the interior of the triangle and I will lie on the circle passing through O,B,C since in the equilateral triangle O and I are same. So the answer will simply be 60 degrees.
Here if you construct an equilateral triangle, O will lie in the interior of the triangle and I will lie on the circle passing through O,B,C since in the equilateral triangle O and I are same. So the answer will simply be 60 degrees.
We know, because \( I \) is the incenter, \[ \angle IBD = \frac{B}{2}, \quad \angle ICD = \frac{C}{2} \] \[ \Rightarrow \angle BIC = \pi - \left( \frac{B}{2} + \frac{C}{2} \right) = \pi - \left( \frac{\pi - A}{2} \right) = \frac{\pi}{2} + \frac{A}{2} \] But, \( \angle BIC = \angle BOC \), because chord \( BC \) subtends both angles at the center. Also, \( \angle BOC = 2A \), because the angle subtended by a chord at the center \( (O) \) is double the angle subtended by the chord on any part of the circle (passing through \( \triangle ABC \)) in the major sector. \[ \frac{\pi}{2} + \frac{A}{2} = 2A \] Which gives, \[ \angle A = \frac{\pi}{3} = 60^\circ \] \[ \Rightarrow \angle BAC = \frac{\pi}{3} = 60^\circ \]
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