M是△ABC边AB上的任意一点,r1,r2,r分别是△AMC,△BMC,△ABC的内切圆半径,q1,q2,q分别是上述三角形在∠ABC内部的旁切圆半径,证明r1除以q1乘以r2除以q2等于r除以q

Question

M是△ABC边AB上的任意一点,r1,r2,r分别是△AMC,△BMC,△ABC的内切圆半径,q1,q2,q分别是上述三角形在∠ABC内部的旁切圆半径,证明r1除以q1乘以r2除以q2等于r除以q

Answer 1

此题是不是抄错了?△AMC不可能在∠ABC内部做旁切圆.原题应该是在∠ACB内.
在∠ACB内时：
△ABC的三边长为BC=a,AC=b,AB=c
sin(A)/a=sin(B)/b=sin(C)/c=1/n
r/q = (a+b-c) / (a+b+c)
=[ n * sin(A) + n * sin(B) - n * sinc(C) ] / [ n * sin(A) + n * sin(B) + n * sinc(C) ]
=[ sin(A) + sin(B) - sinc(C) ] / [ sin(A) + sin(B) + sinc(C) ]
={ sin(B+C) + [ sin(B) - sinc(C) ] } / { sin(B+C) + [ sin(B) + sinc(C) ] }
= [ 2 * sin((B+C)/2) * cos((B+C)/2) + 2 * cos((B+C)/2) * sin((B-C)/2 ] / [ 2 * sin((B+C)/2) * cos((B+C)/2) + 2 * sin((B+C)/2) * cos((B-C)/2 ]
= [ cos((B+C)/2) / sin((B+C)/2) ] * { [ sin((B+C)/2) + sin((B-C)/2) ] / [ cos((B+C)/2) + cos((B-C)/2 ] }
= [ cos((π-A)/2) / sin((π-A)/2)] * { [ 2 * sin(B/2) * cos(C/2) ] / [ 2 * cos(B/2) * cos(C/2) ] }
= tan(A/2) * tan(B/2)
同理
r1/q1 = tan(A/2) * tan(∠AMC/2)
r2/q2 = tan(B/2) * tan(∠BMC/2)
因为∠AMC/2 + ∠BMC/2 = π/2,所以tan(∠AMC/2) * tan(∠BMC/2) = 1
所以
(r1/q1) * (r2/q2) = [ tan(A/2) * tan(∠AMC/2) ] * [ tan(B/2) * tan(∠BMC/2) ]
= tan(A/2) * tan(B/2)
= r/q