BuilderBoi

Let's split this problem into cases: 

Case 1 (1 x 1 isosceles triangle) - There are 36 1 x 1 squares, and each square can have a total of 4 triangles which makes for $36 \times 4 = 144$ triangles. 

Case 2 (2 x 2 isosceles triangle) - There are 25 2 x 2 squares, and each square can have a total of 4 triangles, which makes for $25 \times 4 = 100$ triangles.

Case 3 (3 x 3 isosceles triangle) - There are 16 2 x 2 squares, and each square can have a total of 4 triangles, which makes for $16\times 4 = 64$ triangles.

Case 4 (4 x 4 isosceles triangle) - There are 9 4 x 4 squares, and each square can have a total of 4 triangles, which makes for $9 \times 4 = 36$ triangles.

Case 5 (5 x 5 isosceles triangle) - There are 4 5 x 5 squares, and each square can have a total of 4 triangles, which makes for $4 \times 4 = 16$ triangles.

Case 6 (6 x 6 isosceles triangle) - There is 1 6 x 6 square, and each square can have a total of 4 triangles, which makes for $1 \times 4 = 4$ triangles.

So, there are $144 + 100 + 64 + 36 + 16 + 9 + 1 = \color{brown}\boxed{370}$ isosceles right triangles. 

${x \over x+1} + {3 \over x} = {5 \over 2x}$

${x \over x+1} + {6 \over 2x} = {5 \over 2x}$

${x \over x+1} = {5 \over 2x}- {6 \over 2x}$

${x \over x+1} = {-1 \over 2x}$

$2x^2 = -1(x+1)$

$2x^2 = -x - 1$

$2x^2 + x + 1 = 0$

$x = {-1 \pm \sqrt{(1)^2-4(2)(1)} \over 2(2)}$

$x = {-1 \pm \sqrt{-7} \over 4}$

$x = \color{brown}\boxed{{-1 \pm \sqrt{7}i \over 4}}$

Rewrite the second equation to $3a + 2b - mb = k$.

Now, factor out b to get $3a + (2-m)b = k$. 

For the system of equations to have an infinite number of solutions, the 2 equations must be identical. 

This means that $2 -m = 2$, so $\color{brown}\boxed{m = 0}$ and $\color{brown}\boxed{k = 2}$

Oh... I get it now. Thanks so much, Melody, Asinus, and Guest!

Why don't you solve it then?

Thanks for the explanation Melody, but why did you reflect AB over x = 2 in particular?

For the quadratic to be factorable, there must be 2 numbers that sum to $k$ and multiply to $23 \times -5 =-115$. 

There are 4 pairs of numbers that multiply to $-115$: $(115, -1)$, $(23, -5)$, $(5, -23)$, $(1, -115)$. 

This means that k can be the sum of these values or $\color{brown}\boxed{k = -114, -18, 18, 114}$

What if x is 0? or negative?

Then the distance from $(0, 0)$ to $(1.8, -0.4)$ is $\sqrt{(1.8)^2 + (0.4)^2} = {\sqrt{85} \over 5}$ and the distance from $(0, 1)$ to $(1.8, -0.4)$ is $\sqrt{(1.8)^2 + (1.4)^2} = {\sqrt{130} \over 5}$. 

Adding the two gives us ${\sqrt {85 }+ \sqrt {130} \over 5} \approx 4.12426$ but the correct answer was $\sqrt{17} \approx 4.12311$. 

Sorry, but that's incorrect. The correct answer is actually $\sqrt{17}$, even though I don't know how to get it. 

Thanks so much for your attempt, though!