integer rectangle $ isosceles, same per. & area (1997 AMQ Concours College p5)

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integer rectangle $ isosceles, same per. & area (1997 AMQ Concours College p5)

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#1 h Mar 7, 2021, 6:07 PM • 3 Y

Y by Mango247, Mango247, Mango247

Find a rectangle and an isosceles triangle with the same perimeter and same area and with all sides of integer lengths.

This post has been edited 3 times. Last edited by parmenides51, Mar 7, 2021, 6:08 PM

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#2 h Mar 7, 2021, 6:19 PM

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A 5,5,6 triangle and 6x2 rectangle both have area=12 and perimeter=16.

@below it's a lot easier to guess the answer, especially since it only asks you to find one and 5,5,6 is the isosceles triangle with smallest integer area (along with 5,5,8)

This post has been edited 1 time. Last edited by asbodke, Mar 7, 2021, 6:26 PM

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#3 h Mar 7, 2021, 6:25 PM

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Notation: Length of rectangle $=l$
Breadth of rectangle $=b$

Congruent Sides of isosceles triangle $=a$
Unequal side of triangle $=k=2p$ (Yup!, its even, so I will show $p$ is also integer.)

We have same perimeter:
$2(l+b)=2a+k$
$\implies k$ is even. So, $k=2p$ .
$\Rightarrow l+b=a+p$

Now equating area:
$lb=p \times \sqrt{a^2-p^2}$
$\Rightarrow l^2b^2=p^2(a^2-p^2)$
$\Rightarrow l^2b^2=p^2(l+b)(a-p)$
What to do?
Idea: I am trying to use inequalities to bound the equation so that I can show $\textbf{no large solutions}$ exists...
With the above statements, following inequalities are obvious:
$l^2b^2>lb>l+b=a+p>a-p$
$lb>a+p>p$
$l^2b^2>a^2+2ap+p^2>p^2$
And I'm not able to bind $p$ very well. So, any ideas?

This post has been edited 11 times. Last edited by SatisfiedMagma, Mar 7, 2021, 7:04 PM

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#5 h Mar 7, 2021, 7:10 PM

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Anyone is here? Little help...

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#6 h Mar 14, 2021, 12:25 PM • 1 Y

Y by Mango247

HI?? Is anyone here?

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#7 h Mar 14, 2021, 2:55 PM

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First, I don't know if you noticed this, but there are infinitely many such triangles/rectangle pairs because we can always multiply the sides of a known pair by a natural number and get a new pair.

Secondly, given that the area of the rectangle is an integer and $b=2p$ , we know the height of the isosceles triangle is an integer. This means we have a right triangle with integer side lengths, making me think of the formula for principle Pythagorean triples. Could you work with this to get something?

Also, I wouldn't try to prove that there are no more principle solutions unless you actually believe that that is true because there could very well be infinitely many principle solutions. Try playing around and see if you can get another pair.

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