From a square of unit length, pieces from the corners are removed to form a regular octagon. Find the value of the removed area.

I got a weird answer, as I considered each side of the removed area will be $\dfrac{1}{3}$ and got $\dfrac{1}{18}$ for area of one triangle. Then I multiplied it by 4, to obtain $\dfrac{2}{9}$ which I thought to be the answer but I was wrong.

  • $\begingroup$ Try it with the algebra. Set the side length of the folded diagonal to be $x$ and see where it goes. $\endgroup$ – Christopher Liu May 6 '14 at 3:57

Say we remove an isosceles right triangle with legs $x$ from each corner.

The hypotenuse is then $x\sqrt{2}$. The amount of material left on say the top side is $1-2x$. The octagon is regular, so we want $1-2x=\sqrt{2}x$. That gives $x=\frac{1}{2+\sqrt{2}}$.

Compute the area of the triangle, and multiply by $4$. We get $\frac{4}{2(2+\sqrt{2})^2}$.

Remark: Things will look nicer if we note that $2+\sqrt{2}=\sqrt{2}(\sqrt{2}+1)$. We have $x=\frac{1}{\sqrt{2}(\sqrt{2}+1)}=\frac{\sqrt{2}-1}{\sqrt{2}}$. (We multiplied top and bottom by $\sqrt{2}-1$.) Square, divide by $2$, and multiply by $4$. We end up with total area removed equal to $(\sqrt{2}-1)^2$, that is, $3-2\sqrt{2}$.


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