# Just check my solution to this problem

## Homework Statement

Find the number of of all integer sided isosceles obtuse angle triangles with perimeter 2008.

## The Attempt at a Solution

I assumed the obtuse angle as 90+2a
Let the length of the equal sides be L an that of the base = 2X
2(L+x)=1004

Draw the perpendicular to the side opposite to the obtuse angle from the opposite vertex.
It will divide the triangle in to congruent parts. Using sine rule in one of the to triangles formed.
I get
sin (45+a)/X=1/L
sin (45+a)/X=1/(1004-X)

X=1004/ (cosec (45+a)+1)
0<a<45

2<cosec (45+a)+1<2.4142

as 2X is an integer

cosec (45+a)+1 should be an integer and a factor of 1004 which is not possible. Therefore, no such triangle is possible!

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## Homework Statement

Find the number of of all integer sided isosceles obtuse angle triangles with perimeter 2008.

## The Attempt at a Solution

I assumed the obtuse angle as 90+2a
Let the length of the equal sides be L an that of the base = 2X
2(L+x)=1004
You're given that the perimeter is 2008, not 1004, so the equation should be
2L + 2X = 2008, or 2(L + X) = 2008, from which you get L + X = 1004
Draw the perpendicular to the side opposite to the obtuse angle from the opposite vertex.
It will divide the triangle in to congruent parts. Using sine rule in one of the to triangles formed.
I get
sin (45+a)/X=1/L
It's much simpler as sin(45 + a) = X/L
Replacing L by 1004 - X, this becomes
sin(45 + a) = X/(1004 - X)
sin (45+a)/X=1/(1004-X)

X=1004/ (cosec (45+a)+1)
0<a<45

2<cosec (45+a)+1<2.4142
I don't follow your logic here. It looks like it comes from 1 < csc(45 + a) < sqrt(2), but you'll have to explain why that is true. (It might be, but I don't see it.)

Also, you can't replace 1 + sqrt(2) with 2.4142; they are only approximately equal.
as 2X is an integer

cosec (45+a)+1 should be an integer and a factor of 1004 which is not possible. Therefore, no such triangle is possible!

The obtuse angle alpha must indeed be the angle between the equal sides L. Let the side opposite to alpha ideed be 2x. Use of the cosine rule gives
4*x^2=L^2+L^2 -2L^2*cos(alpha).
So cos(alpha)=1-2*x^2/L^2. Because the angle is obtuse cos(alpha)<0 so to solve is:
1-2*x^2/L^2<0, multiply by L^2, since L^2 >0 this gives L^2-2*x^2<0, or L^2<2*x^2,
L<xSqrt(2)=(1004-L)sqrt(2), L(1+sqrt(2))<1004, L<1004/(1+sqrt(2).
The number of solutions are: int(1004/(1+sqrt(2).
greetings

Made an error L has to be greater than x!

Just so you know, there is another way that I might prefer to get to the point where L^2<2*x^2. If you remember our Pythagorean Theorem for right triangles:
$$a^{2} + b^{2} = c^{2}$$
You might also know that for obtuse triangles:
$$a^{2} + b^{2} < c^{2}$$
Because you chose L for the legs, a and b, and 2x for the largest side, the one opposite alpha or c.
$$L^{2} + L^{2} < (2x)^{2}$$
or
$$2L^{2} < 4x^{2}$$
therefore
$$L^{2} < 2x^{2}$$
I like this way, and showing it two ways means it's probably right.

therefore
$$L^{2} < 2x^{2}$$
I like this way, and showing it two ways means it's probably right.
Ok so x<L<sqrt(2)*x, and how many solutions are there?