Twin Circles
Triangle ABC is inscribed in a circle so that BC is a diameter. Circle is inscribed in ABC , and circle is tangent to both segment AB and minor arc AB so that the two points of tangency lie on the same diameter of . If and are congruent, then the extended ratio BC : AC : AB can be written a : b : c , where a , b , and c are positive coprime integers. Find .
The answer is 30.
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1 solution
because Γ2 and Γ3 are congruent, we have: R - 2r = AC/2. Let R,r be diameter of circumcircle and inscribed circle of triangle ABC. Alternatively, BC = 2R and r = tan(A/2) . (AB+AC-BC)/2 Therefore, BC - √2( AB + AC - BC) = AC => BC - √2 [√(BC^2 - AC^2) + AC - BC ] = AC BC/AC can be written x, we have: x - √2 [√(x^2 -1) + 1 - x = 1 Solve this equation, we have x = 13/5. So, a : b : c = 13 :5 :12. a + b + c = 13 + 5 + 12 = 30 The answer is 30