# Regular octagon inscribed in a square

Problem: The corners of a 2 meter square are cut off to form a regular octagon. What is the length of the sides of the resulting octagon?

From the picture below, the octagon would form a right isosceles, specifically a right isosceles triangle on the corners. The sides of the octagon were set to "x" and the legs of the triangle were set to $\frac{x}{\sqrt{2}}$. Then add the following cuts of a side of the square: $\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 m, which results to x = 0.828 m.

My inquiry is that, from what I know or learned, a right isosceles triangle has an angle ratio of $45-45-90$ and a side ratio of $1-1-\sqrt{2}$ or in algebra: $x-x-x{\sqrt{2}}$. In the problem he set the hypotenuse as $x$ instead and the legs of the triangle as $\frac{x}{\sqrt{2}}$, which I think is fine. But shouldn't setting the hypotenuse as $x\sqrt{2}$ and the sides as $x$ should equal the first equation?

$\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 should also equal $x + x\sqrt{2} + x = 2$ where 2 is the length of a side of a square. I don't think multiplying or dividing both sides by $\sqrt{2}$ is the answer as that would not satisfy both equations.

This sounds like an easy problem, but it it's confusing me. Sorry. • If you are actually building one of these (like a poker table), then you'll want to use the right-hand diagram and measure your cuts from the corners. (Assuming you actually have a square piece to start.) – John May 21 '18 at 20:27

Let $x$ be the length of your octagon (as in the left picture), and $c$ the length cut from one side of the square edge (which is the $x$ in the right picture).

Then you've correctly stated that $x = \sqrt{2}c$. Now you solve $$c + x + c = 2.$$ This is rewritten as $$2c + x = 2c + \sqrt{2}c = (2 + \sqrt{2})c = 2.$$ Thus $$c = \frac{2}{2+\sqrt{2}}$$ so that $$x = \frac{2\sqrt{2}}{2 + \sqrt{2}}.$$ This final fraction is the length of the sides of the octagon.

In the right picture, everything has been scaled up by $\sqrt{2}$ so that the length of the sides of the octagon will be $$\sqrt{2} \cdot \frac{2\sqrt{2}}{2 + \sqrt{2}} = \frac{4}{2 + \sqrt{2}}.$$ This is why they are unequal.

If you were to solve it with $x$ as in the right picture, then you have $$2x + \sqrt{2}x = 2$$ so that $$x = \frac{2}{2 + \sqrt{2}},$$ and then the length of the octagon is $$\sqrt{2}x = \frac{2\sqrt{2}}{2 + \sqrt{2}}$$ exactly as we had calculated above.

Your confusion stems from using $x$ as a label for two different lengths in either diagram.

The two equations are not equivalent, and they give different results. In the first case, $x$ is the side of the octagon; in the second one, the side of the octagon is $x\sqrt{2}$.

The first one has the following solution: $x=2(\sqrt{2}-1)$.

The second one has solution $x=2-\sqrt{2}$, and to obtain the side of the octagon, you have to multiply it by $\sqrt{2}$, obtaining the same result.

• There were problems wherein I didn't need to multiply or divide back an expression to get the original value of x.. So even though you have explained it clearly, I feel more confused lol. But thanks.~ – Jayce May 21 '18 at 14:46
• Just replace $x$ in the right diagram with $y$ and then equate $x=y\sqrt{2}$. – dan post May 21 '18 at 15:08