# Homework Help: Incentre of A Triangle (3D)-Quite Urgent

1. Mar 2, 2009

### ritwik06

1. The problem statement, all variables and given/known data

The coordinates of a the vertices of a triangle are given:
A(x1,y1,z1)
B(x2,y2,z2)
C(x3,y3,z3)
and the sides opposite each vertex are a,b,c respectively. Find the coordinates of the incentre.

3. The attempt at a solution
I have been frequently using the formula
X=(ax1+bx2+cx3)/(a+b+c)
Y=(ay1+by2+cy3)/(a+b+c)
Z=(az1+bz2+cz3)/(a+b+c)
for the incenter. But I simply cannot prove the result.

I calculated the in radius= Area/semi-perimeter (a lengthy process) and then tried to prove it by just taking x,y coordinates. But I failed.

http://img502.imageshack.us/img502/9529/triangle.jpg [Broken]

I also know the ratio in which the sides are divided by angle bisectors. For example side a is divied in the ratio s-b:s-c

Somebody please give me the proof for this!! I tried mathworld
http://mathworld.wolfram.com/Incenter.html

Actually I am a bit worried because of my exam which is day after tomorrow. Its not that I have not prepared anything. The only thing is that I get nervous whenever whenever I get stuck on a question b4 exams. Therefore I seeking the proof at present. I promise you that I will work it out myself as soon as my exams are over.

I shall be really glad if someboy finds me a link to this proof(using 3D coordinates). Or just gives me a guidelines on how to prove it. The problem with me is that I cannot sit on the internet for very long so please help me as soon as possible. I hope this is not against the rules of the forum.
regards,
Ritwik

Last edited by a moderator: May 4, 2017
2. Mar 2, 2009

### tiny-tim

Good luck!

Hi Ritwik! Thanks for the PM.

Here's two ways of doing it:

i] if the tangent points are d e and f, then (a - d).(b - c) = (b - e).(c - a), so find the intersection

ii] if the tangent points are at a + p(b - a) etc, and the incentre is at x, then (x - a - p(b - a))2 = (x - b - q(c - b))2 = (x - c - r(c - b))2

(and there's probably some other really neat ways, including one involving areas, as you suggested)

3. Mar 2, 2009

### ritwik06

Re: Good luck!

Hi,
First of all, I wish to say that I have no words to thank you. I am really indebted to you. Thank you very very much for the help.

1)a,b,c,d,e,f are position vectors, right?
D is opposite A, E opp B and F opp C.

But I dont see through why (a - d).(b - c) = (b - e).(c - a)
this holds?
Will line AD pass through incentre??? Similarly, will BE?
Please elaborate! I havent got a clue. I would be highly obliged if you could spare some of ur time to get a rough paint picture if you could manage, please.

2) I assumed l,m,n as the coordinates of X
and continued as u told:

But I have 6 unknowns and 2 equations.
Here is what I got on solving:
q2a2+2q(l(x2-x3)+m(y2-y3)+n(z2-z3))=r2a2+2r(l(x2-x3)+m(y2-y3)+n(z2-z3))=p2c2+2p(l(x1-x2)+m(y1-y2)+n(z1-z2))

l,m,n are unknown. p,q,r are also unknown. And there are only 2 equations. What to do now?

Thanks once again

regards,
Ritwik

Last edited: Mar 2, 2009
4. Mar 2, 2009

### tiny-tim

Hi Ritwik!
ooh, sorry …1) isn't the incentre at all (maybe it's the orthocentre?) …

i was in a hurry to get out this morning, and i didn't draw a diagram

Actually, looking at your formula, I now see there's an easy way to check it.

Your formula has the incentre at X, where (using a b c as the lengths, and A B C as the vectors, so (A - B)2 = c2 etc):

(a+b+c)X = aA + bB + cC,

so (a+b+c)(X - A).(A - B) = b(A - B)2 + c(A - B)(A - C),

so (a+b+c)(X - A).(A - B)/c = bc + (A - B)(A - C),

= (a+b+c)(X - A).(A - C)/b,

so X - A bisects the angle between A - B and A - C