I ABC triangle with integer sides

[littlemathquark]

TL;DR Summary

In the ABC triangle with integer sides, point D on [AB] and point E on [AC] are taken such that DE is tangent to the incircle. If ∣AD∣=5, ∣AE∣=6, ∣DE∣=7, how many different values can ∣BC∣ take?

I can find radius of incircle of ADE and incircle of ABC triangle but I don't know how them help me.

 

[anuttarasammyak]

TL;DR Summary: In the ABC triangle with integer sides, point D on [AB] and point E on [AC] are taken such that DE is tangent to the incircle. If ∣AD∣=5, ∣AE∣=6, ∣DE∣=7, how many different values can ∣BC∣ take?

I can find radius of incircle of ADE and incircle of ABC triangle but I don't know how them help me.

We know cos A by cosine theorem. Does it help you ?

 

[littlemathquark]

We know cos A by cosine theorem. Does it help you ?

Cos A=1/5 but I can't use it.

 

[littlemathquark]

If BC length would not integer, perhaps there would be infinite value of BC I think. But it's integer.

 

[anuttarasammyak]

How much is radius of incircle of triangle ABC you get ? I am interested in how you get it.

 

[littlemathquark]

How much is radius of incircle of triangle ABC you get ? I am interested in how you get it.

Let AKL be isosceles triangle with AK=AL=9
We easly find KL using cosine theorem in AKL triangle. (cosA=1/5)
Let O be incenter of ABC triangle and R ve radius. Cos O=-cosA=-1/5
Using cosine theorem in OKL we can find R but I have'nt got paper and pencil now, I'm out.

 

[anuttarasammyak]

Let AKL be isosceles triangle with AK=AL=9

Thanks. From where 9 comes ?

 

[littlemathquark]

Using tangent properties
DE=4+3=7
AK=5+4=9
AL=6+3=9

 

[martinbn]

You also get a linear relation for the sides of the triangle from the fact that the circle is inscribed the the BCED quadrilateral.

 

[littlemathquark]

You also get a linear relation for the sides of the triangle from the fact that the circle is inscribed the the BCED quadrilateral.

İt's tangential quadrilateral.

 

[anuttarasammyak]

Using tangent properties
DE=4+3=7
AK=5+4=9
AL=6+3=9

I see. Thanks. So
R=9 \tan \frac{A}{2}=\frac{9\sqrt{2}}{\sqrt{3}}
Is it same as yours ?

m=R cot \frac{B}{2}, n=R cot\frac{C}{2}
Cosine theorem
(9+m)^2+(9+n)^2-2(9+m)(9+n)cos A=(m+n)^2

 

[martinbn]

İt's tangential quadrilateral.

Yes. That's what I am saying. But the point is that this gives you a linear equation and the cosine theorem gives you a quadratic equation. The system probably doesn't have many integral solutions.

 

[littlemathquark]

I see. Thanks. So
R=9 \tan \frac{A}{2}=\frac{9\sqrt{2}}{\sqrt{3}}
Is it same as yours ?

m=R cot \frac{B}{2}, n=R cot\frac{C}{2}
Cosine theorem
(9+m)^2+(9+n)^2-2(9+m)(9+n)cos A=(m+n)^2

Yes R is the same as mine. By simplified your cosine theorem I found (if I'm not mistake) m+n=mn/6 - 9

 

[littlemathquark]

Yes R is the same as mine. By simplified your cosine theorem I found (if I'm not mistake) m+n=mn/6 - 9

From this equation (m-6)(n-6)=90 and it can be found m, n integer and values of m+n. Is it ok?

 

[anuttarasammyak]

Factorization of 90 will tell you how many cases.

 

[littlemathquark]

Factorization of 90 will tell you how many cases.

Suitible $(m,n)$ is $(7,96),(8,51),(9,36),(11,24),(12,21),(15,16)$ and $m+n=|BC|=\{103,59,45,35,33,31\}$

 

[anuttarasammyak]

(m-6)(n-6)=2*3^2*5

so m-6 ={1,2,3,5,...} : 2*3*2=12 ways

 

[littlemathquark]

Thank you, but m,n is symmetric and we're interested in m+n