Triangles and numbers

Number Theory Level pending

I have 2 right triangles with integer sides and they share the same area. The perimeters of these triangles are 456 and 840.

Find all the sides of these 2 triangles.
Let $a$ denote the smallest side of the triangle with perimeter 456, and
let $b$ denote the smallest side of the triangle with perimeter 840.

Submit your answer as $100a + b$ .

The answer is 9540.

1 solution

Michael Mendrin
Dec 24, 2016

Any integer Pythagorean triplet can be represented by the following, given any positive integers $m,n$

$a=2mn$
$b={m}^{2}-{n}^{2}$
$c={m}^{2}+{n}^{2}$

so that the perimeter length works out to

$2m(m+n)=456$ or $840$

A little extra work gets us

$m=12, n=7$ in the case of $456$ , and
$m=20, n=1$ or $m=15, n=13$ in the case of $840$

Then if we add the condition that the areas are the same, we quickly determine that $m=20, n=1$ in the case of $840$ , and so the triangles have sides, for case $456$ and $840$ respectively

$a=168$
$b=95$
$c=193$

$a=40$
$b=399$
$c=401$

so that the answer to this problem would then be $9540$

I'm guessing it probably took more work to find such an instance of a pair of right triangles.

As always it's nice and construtive to read you.

I did this problem without what you called "little extra work".

There is a solution that does not need any "try and error" part.

Mr X - 4 years, 5 months ago

Well, $228$ factors into $2 \cdot 2 \cdot 3 \cdot 19$ , so $m+n$ had to be $19$ , which means $m=12$ , and then $n=7$ . The other one requires just a bit more ado, but, strictly speaking, can be solve through logic.

Always share your solutions with the readers, even though I have to keep telling myself that.

Michael Mendrin - 4 years, 5 months ago

Sure I will do later. but for now I leave this task to readers like you who by their contributions add more value to the problem.

Mr X - 4 years, 5 months ago

here is how the solution becomes completely algebraic after finding the sides of the first triangle 95, 168 and 193 we calculate the area $a=7980$ . in the second triangle we use this equation which is true in right triangles $x^2-(s+\frac{a}{s})x+2a=0$ where s is half the perimeter and a is the area. the two roots of this equation are the the two right sides of the triangle. $x^2-(420+\frac{7980}{420})x+2*7980=0$

Mr X - 4 years, 5 months ago

@Mr X – It doesn't require much insight to solve for the second triangle's side lengths once you've found the side lengths of the first triangle; you can just let $xy=15960$ , and $x+y+\sqrt{x+y}=840$ , and solve simultaneously for $x$ and $y$ in much the same way.

Miles Koumouris - 3 years, 8 months ago

Triangles and numbers

The answer is 9540.

1 solution

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