Spiraling Triangles

First, we find the hypotenuse of the smallest triangle.

Let $h_1$ represent the length of the hypotenuse. Applying the Pythagorean theorem ,

$x^2+x^2 = h_1^2,$ so $h_1=(\sqrt{2})x.$

(We chose the positive solution since $h_1$ represents a length.) Next, let $h_2$ represent the hypotenuse of the next triangle (the triangles whose legs have length $h_1$ ). The Pythagorean Theorem gives $(\sqrt{2} x)^2+(\sqrt{2} x)^2=(h_2)^2,$

which simplifies to $2x^2+2x^2=(h_2)^2.$ Solving for the positive value of $h_2$ , we get $h_2=2x.$

Next, let $h_3$ represent the hypotenuse of the triangle whose legs have length $h_2$ and apply the Pythagorean Theorem to this triangle to get $((2x))^2+((2x))^2=(h_3)^2,$ or $h_3 = 2\sqrt{2}x.$ Continuing in this manner, a pattern emerges – each hypotenuse is $\sqrt2$ times the previous hypotenuse. In general, the hypotenuse of a right isosceles triangle is $\sqrt{2}$ times the length of one of its legs – our work finding $h_1$ demonstrates this. Thus, since the hypotenuse of a smaller triangle in the diagram is also a leg of the next larger triangle, the entries in the table can be found by multiplying the preceding entry by $\sqrt{2}$ .

Since $y$ in the diagram represents the hypotenuse of the $6^\text{th}$ smallest triangle, we have $y = 8x$ .

First, lets take notice that in an isosceles triangle the hypotenuse will always equal $\sqrt { 2\times (hypotenuse)^{ 2 } }$ .

Now, bearing that in mind, we can glide through that problem with that simple knowledge.

Let's calculate the first hypotenuse:

$\sqrt { { 2x }^{ 2 } } =x\sqrt { 2 }$ .

Now, this being the leg of our next triangle, we'd use it to calculate the next hypotenuse. Then the next, and so on.

Hypotenuse of second triangle: $\sqrt { 2(x\sqrt { 2 } )^{ 2 } } =(x\sqrt { 2 } )\sqrt { 2 }$ .

Hypotenuse of third triangle: $\sqrt { 2((x\sqrt { 2 } )\sqrt { 2 } )^{ 2 } } =((x\sqrt { 2 } )\sqrt { 2 } )\sqrt { 2 }$ .

Have you noticed the pattern yet? We're just multiplying by $\sqrt { 2 }$ every time. In fact, we can even write up an equation to the find the hypotenuse of the nth triangle: ${ h }_{ n }=x\sqrt { 2 } ^{ n }$ .

So, for the problem above, we would just evalute ${ h }_{ 6 }$ .

${ h }_{ 6 }=x\sqrt { 2 } ^{ 6 }=x\sqrt { 2^{ 6 } } =x\ast 2^{ 3 }=8x$ .

As for the bonus:

${ h }_{ 14 }=x\sqrt { 2 } ^{ 14 }=x\sqrt { 2^{ 14 } } =x\ast 2^{ 7 }=128x$ .

I believe the answer to the bonus question can be expressed as 128x

Each inner edge has a ratio of $\sqrt{2}$ to the two shorter edges. There are 3 inner edges, so the answer is $\sqrt{2}^6=8$ .