This is kind of stressing. I saw this problem in a a book and i can''t seem to solve it. There is no need for complicated math on this (said the authors). Given a triangle with vertices A, B and C and a point D inside the triangle, prove that dist(A,D) + dist(D,C) < dist(A,B) + dist(B,C) This is very simple. If you draw it in a piece of paper, it makes perfectly sence. The problem i''m having it''s demonstrating it mathematicaly. I always end up with sin or cos in the middle on the inequation. Can you help? NOTE: Don''t flame me. This aint a homework i''m supposed to do. I just got curious with it, and since i''m not getting the solution after several trials, i ask for help. Thanks in advance.

dist(A,B) + dist(B,C)-dist(A,D) - dist(D,C)>0

f(Xd,Yd)=dist(A,B) + dist(B,C)-dist(A,D) - dist(D,C)
is special case of general function:
f(X,Y)=dist(A,B) + dist(B,C)-dist(A,P) - dist(P,C)
where P is any point P=P(X,Y)

Xa,Ya,Xb,Yb,Xc,Yc are constants,
and dist(A,B)=(Xa-Xb)^2+(Ya-Yb)^2
and dist(A,P)=(Xa-X)^2+(Ya-Y)^2 and so on.

You are looking on behaviuor of f(X,Y)and you have to prove that f(X,Y) is greater than zero,when X=Xd and Y=Yd(when P is D).


[edited by - AlexanderCZ on November 11, 2002 10:23:11 AM]

Yes, it works. You''re absolutely right.

My problem was that i convinced myself that there was a solution to this using pure geometry, therefore not even loking at other possible good solutions.

Pretty stupid of me not to consider that method, hem?

Thanks AlexanderCZ

This problem can be made very simple by applying the triange inequality twice.
Tri Ineq says: AB + BC > AC, or, the sum of any two sides must be longer than the third. Apply this to triangle ABC and to triangle ADC.
In ABC:
AB + BC > AC
In ACD:
AD + CD > AC
Subtract the two to get AB + BC - AD - CD > 0, move the two negative terms to the right hand side and you get the desired result of AB + BC > AD + CD QED

Brendan

Brilliant! I knew there was a simple way of doing it by aplying simple geometry.

Just one little touch:
A>Z and B>Z -> A-B>0 : This can be false.

for example: A=5; B=6 or A=B. (This later case of A=B can be eliminated because the point is said to be inside the triangle)

We indeed know that point D is inside the triangle, so in fact
AD + DC < AB + BC. But it''s like if we''re using what we want to demonstrate in the demonstration itself.

But at least i am convinced this is the way. Maybe with one little touch here or there. Time for me to take a piece of paper and a pencil (again).

Thanks to both of you

Wolverine-
You make very valid points indeed. However, I believe (haven''t tried it, plus I have my own work to do tonight ) that if you apply the inequality, you can show that the original sides must be longer than the ones formed with point D in the middle, and therefore, you do not need to worry about the false inequality. Because it is in the boundary, it is clear that this is true, however, I too would like to see a more rigorous proof of it (if one is needed) You''re welcome for the help, its a good question.
Brendan

This is how it''s done (it would help if you make a sketch while following the argument):

Draw a line from A to D. It cuts BC in a point that we will call E.

We will first show that d(A,B)+d(B,C)>d(A,E)+d(E,C) (1)

A triangle inequality applied to ABE gives us
d(B,E)+d(A,B)>d(A,E) => d(A,B)>d(A,E)-d(B,E)

Now we start we the left side of (1) and apply this
d(A,B)+d(B,C)>d(A,E)-d(B,E)+d(B,C)=d(A,E)+d(E,C),
because d(E,C)=d(B,C)-d(B,E), as E belongs to the segment BC.

Now we will show that d(A,E)+d(E,C) > d(A,D)+d(D,C) using a triangle inequality applied to CED.
d(E,D)+d(C,E)>d(D,C) => d(C,E)>d(D,C)-d(E,D)
As before,
d(A,E)+d(E,C)>d(A,E)+d(D,C)-d(E,D)=d(A,D)+d(D,C),
because d(A,D)=d(A,E)-d(E,D)

Linking both inequalities, we get
d(A,B)+d(B,C)>d(A,E)+d(E,C)>d(A,D)+d(D,C)
Q.E.D.

Punty50, i was indeed trying to get a rigorous prove. This was one of those problems wich were obvious if you looked at a piece of paper. You would say: this is easy. The problem was to give the so called mathematical prove. This was a little bit more hard to get. But i didn''t even had considered the triangle inequality. Once you''ve said it i thought that was the way. (the way i was trying always got sin and cos in the middle)

Well, there''s the rigorous prove. Thanks Alvaro. That was what i wanted.