About integer triangle - It's hard !

[nakamoto2305]

Find all integer triangle ABC ( triangle all of whose sides have lengths that are positive integers ) so that side AC and bisector AI are equal.

 

[Deleted member 4993]

Find all integer triangle ABC ( triangle all of whose sides have lengths that are positive integers ) so that side AC and bisector AI are equal.

If I were to do your assignment, I would:

• Draw a sketch of ABC and AI
  
  
• Name all the relevant points.(A, B, C and I)
  
  
• Does AI bisect the side BC or angle BAC?
  
  
• Apply theorems about AI of a triangle........continue

 

[nakamoto2305]

Here is an image, AI = AC, and $a,b,c$ is a positive integer, I tried to apply theorems but it leads to nowhere.

 

[Dr.Peterson]

View attachment 27988
Here is an image, AI = AC, and $a,b,c$ is a positive integer, I tried to apply theorems but it leads to nowhere.

Please show the theorems you have tried (and any others you have learned that might be relevant). We need to see what you have tried. Also, what methods have you previously used for problems about triangles with integer sides?

 

[nakamoto2305]

I finished it! Let [MATH]AB=c,AC=b,BC=a => AI = b[/MATH], we have [MATH]b=AI=\frac{2bc.cos(\frac{A}{2})}{b+c} => cos(\frac{A}{2})=\frac{b+c}{2c}[/MATH] we also have [MATH]cos(A)=\frac{b^2+c^2-a^2}{2bc}[/MATH] and [MATH]cos(A)=2cos^2(\frac{A}{2})-1[/MATH], so from the above, [MATH]b^3-c^2b+b^2c-c^3+a^2c=0 <=> (c-b)(b+c)^2=a^2c <=> a= (\frac{b}{c}+1)\sqrt{(c^2-cb)}[/MATH], [MATH]a[/MATH] is integer so [MATH]c^2-cb[/MATH] is a square number, and also [MATH]\frac{b}{c}\sqrt{c^2-cb}[/MATH] is an integer. Just do some case of c, we'll have the answer is: [MATH]c=m^3k,b=mk(m^2-n^2),a=(2m^2-n^2)kn[/MATH] , [MATH]\forall m,k \in \mathbb{Z^+}[/MATH] and [MATH]\forall n \in {1,2,...m}[/MATH]