Math IOQM

 In a right triangle if inradius  = r, hypotenuse  = c and legs are a,b then:  r = (a + b - c)/2 Proof: From the center of the incircle, the perpendiculars are dropped on the 2 legs (a,b). It creates a square and splits the base in 2 parts r, a-r and similarly the other leg is split into r,b-r. Since 2 tangents from one point are equal => c = a - r + b - r => r = (a + b - c)/2 Circumradius of a right triangle = c/2 where c is hypotenuse. Proof: Easy by co-ordinate geometry. A = (0,0) B = (a,0) C = (0,b) Midpoint of BC = a/2,b/2 OA = OB = OC = sqrt(a^2/4 + b^2/4) => O is the circumcenter and OA = Hypotenuse/2  = Circumradius.

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? 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. Detailed solution copied from here : 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 c...