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Solution of the Week #418 - Fourth Circle

May 22, 2023 Elliott Line

Because three of the circles are tangent at the same point, their centres lie on a straight line. That being the case, if we draw lines connecting all of the circle centres, we can form two triangles which both use the angle ‘x’: the 90,120,70 triangle and the (R+70),120,R triangle.

We can use the law of cosines which states:

 Cos(a) = (B^2+C^2-A^2)/(2BC), (where side A is opposite angle a)

 If we do so first for the 90,120,70 triangle, we find that cos(x) = 22/27.

 Now if we do the same for the large triangle, we already know cos(x), so:

 22/27 = (140R + 19300)/(240R + 16800)

 22(240R + 16800) = 27(140R + 19300)

 1500R = 151500

 R = 101

 So therefore the radius of the fourth circle is 101.

← Solution of the Week #419 - Minimum TriangleSolution of the Week #417 - Triple Duality →

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