right=isosceles

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April 18th 2012, 06:51 AM
Wilmer

right=isosceles

Find a right triangle and an isosceles triangle that have same perimeter and area.
All sides are integers.
April 18th 2012, 08:03 AM
biffboy

Re: right=isosceles

Any right angled isosceles triangle!
April 18th 2012, 08:35 AM
Plato

Re: right=isosceles

Quote:

Originally Posted by biffboy

Any right angled isosceles triangle!

Can you find a right angled isosceles triangle having all side lengths integers?
April 18th 2012, 08:39 AM
biffboy

Re: right=isosceles

Sorry, I'll have start reading the whole question!
November 26th 2012, 08:04 PM
PaddyMac

Re: right=isosceles

Quote:

Originally Posted by Plato

Can you find a right angled isosceles triangle having all side lengths integers?

No.
In a right isosceles triangle, the ratio of hypotenuse to leg length is sqrt(2). Since this irrational, the side lengths can't be all integers.
The closest you can get is a right triangle where the leg lengths differ by 1, for example (3,4,5) or (20,21,29).
November 26th 2012, 08:11 PM
PaddyMac

Re: right=isosceles

Quote:

Originally Posted by Wilmer

Find a right triangle and an isosceles triangle that have same perimeter and area.
All sides are integers.

(8,15,17) and (10,13,13) both have an area of 60 square units.
November 26th 2012, 08:13 PM
PaddyMac

Re: right=isosceles

Oops, I need to read to whole question. The perimeters are different.
November 26th 2012, 09:42 PM
MacstersUndead

Re: right=isosceles

only a start to a possible solution...

Spoiler:

If we allow for the degenerate case, then the triangle of side lengths 0 satisfies all the conditions mentioned in the problem, but I'm assuming that's not the solution you are looking for / I need to prove only the degenerate case works.

Let x,y,z be the lengths of the right angle triangle with z as the hypotenuse
Let a,a,b be the lengths of the isosceles triangle.

We have that $x+y+z=2a+b$ if we assume perimeters must be equal
$x^2+y^2=z^2$ by Pythagoras
$p(p-x)(p-y)(p-z) = p(p-a)(p-a)(p-b)$ by Heron's formula (squaring both sides) since areas are equal. We can use p here on both sides since perimeters are equal and hence the semi-perimeter is equal as well.

If p=0, then we have the degenerate case, so assume p is not equal to 0 and divide both sides by p

so,
$(p-x)(p-y)(p-z) = (p-a)(p-a)(p-b)$
$p^3 -(x+y+z)p^2 + (xy + xz + yz)p - xyz = p^3 - (2a+b)p^2 + (a^2 + 2ab)p - a^2b$

Hence by comparison we have

$x+y+z = 2a+b$ (i)
$xy+xz+yz = a^2+2ab$ (ii)
$xyz = a^2b$ (iii)

By multiplying (ii) by x and using (iii), we get
$x^2y + x^2z + xyz = a^2x + 2abx$
$x^2 [y+z] = a^2(x-b) + 2abx$ (1)

Similarily

$y^2 [x+z] = a^2(y-b) + 2aby$ (2)
$z^2 [x+y] = a^2(z-b) + 2abz$ (3)

I have to use Pythagoras eventually, but it feels like I'm going in circles.

Maybe I'm going about this question incorrectly. Initially I wanted to use the concept for height as an argument. In the xyz right triangle, if I choose x as the base y is the height. I'm thinking about what happens if I draw an isosceles triangle with the base on the same level as the base of the xyz right triangle. Areas are equal if base lengths are equal and height in the same. hmm...
November 27th 2012, 12:20 AM
Wilmer

Re: right=isosceles

Hint: the isosceles triangle is made up of 2 pythagorean congruent right triangles "stuck together"
along one of the legs; so you need to find 2 right triangles: one has area = twice the other's area.

If isosceles triangle made up of 2 right triangles ABC (BC=a, AC=b, AB=c), sharing leg AC,
and other right triangle is DEF (EF=d, DF=e, DE=f), then:
2a + 2c = d + e + f (perimeter)
ab = de / 2 (area)
November 27th 2012, 01:00 AM
MarkFL

Re: right=isosceles

Quote:

Originally Posted by PaddyMac

No.
In a right isosceles triangle, the ratio of hypotenuse to leg length is sqrt(2). Since this irrational, the side lengths can't be all integers.
The closest you can get is a right triangle where the leg lengths differ by 1, for example (3,4,5) or (20,21,29).

You may generate as many such right triangles as you like by taking two successive members of the Pell sequence as "seeds" for Euclid's method for generating Pythagorean triples.

Here are the first few examples (after the two already given):

(119,120,169), (696,697,985), (4059,4060,5741), (23660,23661,33461).
November 27th 2012, 03:48 AM
Wilmer

Re: right=isosceles

Or looping c from 5 increment by 2: a = [SQRT(2c^2 - 1) - 1] / 2
Next 4:
137903, 137904, 195025
803760, 804761, 1136689
4684659, 4684660, 6625109
27304196, 27304197, 38613965