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Prove that the distance between the circumcenter and the orthocenter of triangle ABC is R sqrt(1 -8 cos A cos B cos C) |
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Answer» Solution :Let O and H be the circumcenter and the orthocenter, respectively. If OF is the PERPENDICULAR to AB, we have `angleOAF = 90^(@) - angle AOF = 90^(@) -C` Also, `angleHAL = 90^(@) -C` Hence, `angle OAH = A - angle OAF - angle HAL` `= A -2 (90^(@) -C)` `= A + 2C -180^(@)` `= A + 2C -(A + B +C) = C -B` Also, `OA = R and HA = 2R cos A` Now in `Delta AOH` `OH^(2) = OA^(2) + HA^(2) - 2OA HA cos (angle OAH)` `=R^(2) + 4R^(2) cos^(2) A - 4R^(2) cos A cos (C -B)` `= R^(2) + 4R^(2) cos A [cos A - cos (C -B)]` `=R^(2) -4R^(2) cos A [cos (B + C) + cos (C - B)]` `= R^(2) -8R^(2) cos A cos B cos C` Hence, `OH = R sqrt(1-8 cos A cos B cos C)` |
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