✔ 最佳答案
1. 設CE交BE於F,則∠DFE=90度-0.5A ,a=BC=√6, b=AC=√7, c=AB=√5
AE=bsin(C/2), AD=csin(B/2)
由∆ADE知DE^2
=[bsin(C/2)]^2+[csin(B/2)]^2- 2bcsin(B/2)sin(C/2)cos(90度-0.5A)
={b^2(1-cosC)+c^2(1-cosB)-2bc[cos(0.5(B-C))-cos(0.5(B+C)]cos(0.5(B+C))}/2
=[b^2+c^2-b^2cosC-c^2cosB-bc(cosB+cosC-1+cosA)]/2
=0.5[b^2+c^2-b(bcosC+ccosB)-c(ccosb+bcosc)+bc-(b^2+c^2-a^2)/2]
=[2b^2+2c^2-2ba-2ca+2bc-(b^2+c^2-a^2)]/4
=[b^2+c^2+a^2-2ab-2ca+2bc]/4
=[(b+c-a)/2]^2
so, DE=(b+c-a)/2=(√5+√7-√6)/2
2.設p=log(a), q=log(b), r=log(c)
log_a(27)+...+log_c(27)=log_abc(27), 換底得1/p+1/q+1/r=1/(p+q+r)
so, (p+q+r)(pq+qr+pr)=pqr
then (p+q)(q+r)(p+r)=0(乘開即知)
hence p+q=0 or q+r=0 or p+r=0
設p+q=0(其餘同理), then log(a)+log(b)=0, ab=1, b=1/a
then (abc)^2-abc(a+b+c)+(ab+bc+ca)
=c^2-c(a+1/b +c)+(1+c/a+ca) = 1
Ans: 1
2010-05-24 13:05:41 補充：
謝謝樓上的指教!
2010-05-24 17:26:21 補充：
呵呵!大家的眼力都很好,謝謝更正!
