Talk about calculations, here are some thoughts, which maybe helpful. Given a triangle ABC, let O_A, O_B, O_C be the centers of the three circles satisfied the conditions, which are cloest to A, B and C respectively, and let R_A, R_B, R_C be the radius of the three circles. Let O be the center of the circle inscribe ABC, with radius R. On edge AB, let D, E, F be the tangent points of the circles centered at O_A, O and O_B respectively. It is easy to prove that |AE| = 1/2 (b+c-a), |BE| = 1/2(c+a-b) and |AD|/|AE| = R_A/R |BF|/|BE| = R_B/R; Also, notice that |DF|^2+(R_A-R_B)^2 = (R_A+R_B)^2 Hence we got one equation about R_A, R_B: (c-(R_A/R*1/2*(b+c-a)-(R_B/R*1/2*(c+a-b)))^2 = 4*R_A*R_B Similarly, we get two other equations on BC and AC. Notice that, by Helen's formula(?), the area of the triangle is sqrt(s*(s-a)*(s-b)*(s-c)), where s = 1/2*(a+b+c), hence R = sqrt(s*(s-a)*(s-b)*(s-c)) / 2s In short, we have three unkowns R_A, R_B, R_C and three equations..... Some one cares to solve for them? Unfortunately, even we get the solution, it may not be obvious how to make it using ruler/compass. Just think about the messy formula: R = sqrt(s*(s-a)*(s-b)*(s-c)) / 2s and how easy you can make R out of a,b,c using ruler/compass..... |
Establishing the equations is relatively easy. But solving them is more difficult. It envolves a degree 8 polynomial. (see http://mathworld.wolfram.com/MalfattiCircles.html ) And the solutions must be represented in a way that they can be constructed using ruler and compass. |
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