Let ABC be an arbitrary triangle.

Let A_B and A_C be the orthogonal projections of A onto the internal bisectors of angles B and C respectively.

Let B_C and B_A be the orthogonal projections of B onto the internal bisectors of angles C and A respectively.

Let C_A and C_B be the orthogonal projections of C onto the internal bisectors of angles A and B respectively.

Prove that |A_BA_C| + |B_CB_A| + |C_AC_B| = s,

where s is the semiperimeter of triangle ABC.

Hi!

Let be I the intersection ppoint of the interior bisector!

AAc=b*sin(C/2) ; AAb=c*sin(B/2) and AcAAb angle is 90-A/2

Cos theorem in AcAAb triangle give us :

(AcAb)^2=(b*sin(C/2))^2+(c*sin(B/2)^2-2*b*c*sin(A/2)*sin(B/2)*sin(C/2).

I use now the identity cos(A)+cos(B)+cos(C)=1+4*sin(A/2)sin(B/2)sin(C/2) and after replacing sin(C/2)^2 with (1-cos(C))/2  and sin(B/2)^2 with (1-cos(B))/2 i found that

2*(AcAb)^2 = b^2+c^2+b*c-b^2*cos(C)-c^2*cos(B)-b*c*(cos(B)+cos(C)) - b*c*cos(A)

but a=c*cos(B)+b*cos(C) and b*c*cos(A)=(b^2+c^2-a^2)/2 and after replacing and make some calculus i found that

4*(AcAb)^2 = b^2+c^2+a^2+2*b*c-2*a*b-2*a*c = (b+c-a)^2

Therefore

|AcAb|=(-a+b+c) /2

If we make the similar calculus for BcBa and CaCb we found

|BcBa| =(a-b+c) /2

|CaCb| = (a+b-c) /2

So |AcAb| + |BcBa| + |CaCb| = (a+b+c) /2 = s.

Nice puzzle!

Posted by Chesca Ciprian on 2008-01-07 15:49:37