Similar to skinny triangle problem
Geometry
Level
3
Let ABC be a triangle such that AB = 24. A circle having its center on AB is tangent to AC and BC at E and F, respectively. Given that AE = 3 and BF = 1, find the perimeter of the triangle ABC .
The answer is 168.
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1 solution
As the circle is tangent to both AC and BC, equating its radius 3 tan A = tan B 3 Sin A Cos B = Sin B Cos A 4 Sin A Cos B = Sin A Cos B + Sin B Cos A = Sin (A + B) = Sin C Hence 4 a Cos B = c.. From the point C let the two tangents to the circle CE and CF be equal to x each. Then a = 1 + x and b = x + 3. Then the perimeter is 24 + (1 + x) + (x + 3) = 2x +28. Now x is found to be equal to 70 by solving 2 a c Cos B = c^2 + a^2 - b^2 where Cos B = c/(4a) from above. Answer = 2(70) + 28 = 168.