Isosceles Triangle With Integer Sides

**roninpro** · May 12th 2010, 07:27 AM

Hello. I thought up the following problem this morning:

Is it possible to construct an isosceles triangle with integer side length and area?

Mathematica was unable to find any solutions, but it did not say that there were none. I tried to work with the associated Diophantine equation, but I couldn't see immediately how to show that there weren't any solutions.

I'd appreciate any thoughts on the matter.

**undefined** · May 12th 2010, 08:44 AM

Originally Posted by roninpro

Hello. I thought up the following problem this morning:

Is it possible to construct an isosceles triangle with integer side length and area?

Mathematica was unable to find any solutions, but it did not say that there were none. I tried to work with the associated Diophantine equation, but I couldn't see immediately how to show that there weren't any solutions.

I'd appreciate any thoughts on the matter.

Maybe I don't understand the problem correctly, but it seems easy to come up with examples using Pythagorean triples.

Take a 3-4-5 right triangle. Put two of them together to make an isosceles triangle and you're done.

**Failure** · May 12th 2010, 08:46 AM

Originally Posted by undefined

Maybe I don't understand the problem correctly, but it seems easy to come up with examples using Pythagorean triples.

Take a 3-4-5 right triangle. Put two of them together to make an isosceles triangle and you're done.

Problem is: he required the area of the triangle to be an integer (an integral multiple of the unit area) as well.

**undefined** · May 12th 2010, 08:49 AM

Originally Posted by Failure

Problem is: he required the area of the triangle to be an integer (an integral multiple of the unit area) as well.

Say we place the triangles together so that the sides of length four are touching.

Then the area is $\left(\frac{1}{2}\right)(6)(4) = 12 \in \mathbb{Z}$ .

(Fixed typo.)

**Failure** · May 12th 2010, 09:01 AM

Originally Posted by undefined

Say we place the triangles together so that the sides of length four are touching.

Then the area is $\left(\frac{1}{2}\right)(6)(4) = 12 \in \mathbb{Z}$ .

(Fixed typo.)

Right, how foolish of me: I just didn't really do the math but thought that you had not even considered the question of the area in your reply.

**roninpro** · May 12th 2010, 09:40 AM

I guess I should have tried to construct an example explicitly.

Now I'm wondering why Mathematica failed to find a solution. Maybe I typed in the formula incorrectly.

Thanks.

**undefined** · May 12th 2010, 10:30 AM

Originally Posted by roninpro

I guess I should have tried to construct an example explicitly.

Now I'm wondering why Mathematica failed to find a solution. Maybe I typed in the formula incorrectly.

Thanks.

Regarding Mathematica.

Heron's formula:

$A=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{16}}$

Let $a=b$ . Then

$A=\sqrt{\frac{(2a+c)(2a-c)(c)(c)}{16}}$

$=\sqrt{\frac{((2a)^2-c^2)(c^2)}{16}}$

$16A^2=(4a^2-c^2)(c^2)$

$16A^2-(4a^2-c^2)(c^2)=0$

Screenshot from Mathematica:

Edit: Of course it's also possible to avoid Heron's formula by dividing the isosceles triangle into two right triangles to begin with.

Label the two congruent sides $a$ and the other side $b$ . Treat $b$ as the base and draw an altitude from the base to the opposite vertex.

$\left(\frac{b}{2}\right)^2+h^2=a^2$

and

$A=\left(\frac{1}{2}\right)bh$

$h = \frac{2A}{b}$

Substitute

$\left(\frac{b}{2}\right)^2+\left(\frac{2A}{b}\righ t)^2=a^2$

etc.

**roninpro** · May 12th 2010, 11:27 AM

I actually used that formula, originally. I now see that I screwed up the syntax.

Thanks for the computation.

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