几何作图难题

constant

过节了，大家都吃好的，动点脑筋，免得长胖。

难度：+++++

在已知三角形内作三个两两相切的圆，并且每个圆都与两条边相切。

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constant

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. www.ddhw.com

QL

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.www.ddhw.com 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.www.ddhw.com 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.....   www.ddhw.com

constant

  表达式必须是能用尺规作出来的，例如平方根，已知角的三角函数等。

QFT

  仅给出半径的表达式，还是需要用圆规和直尺做出相应的长度？