From unit circle to regular octagon and back

Geometry Level 4

In a unit circle , square $ABCD$ is inscribed. A new square $A'B'C'D'$ is constructed when the square $ABCD$ is translated by a vector $\vec{AB}$ . Finally, an equilateral $\triangle XYZ$ is drawn, so that rectangle $AB'C'D$ is inscribed in it and the following is fulfilled:

Points $B' \wedge C' \in XY$ , $A \in XZ$ and $D \in YZ$ .

Calculate the area of regular octagon whose side length is equal to the distance between points $O$ and $Z$ . Give answer to 2 decimal places.

The image below represents how everything should look like when the drawing is done. Click here to enlarge the image.

Basically: Calculate the green region

If there is a question regarding the problem, it can be asked here .

The answer is 18.02.

2 solutions

Niranjan Khanderia
Jul 5, 2016

Diameter of the circle = diagonal of the square. So AD = $\sqrt2$ .
Let M be the midpoint of AD. So OM = $\dfrac 1 {\sqrt2}$ .
ADZ is an equilateral triangle with sides $\sqrt2$ .
So altitude ZM = $\dfrac{\sqrt3} 2 *\sqrt2$ .
So a = ZO = ZM + MO = $\dfrac 1{\sqrt2}* ( \sqrt3 +1)$ .
Area of octagonal side a, = $2a^2(\sqrt2 + 1) = 18.0199$

Drex Beckman
Jul 3, 2016

Using the definition of the unit circle and a square, $OA = \frac{\sqrt{2}}{2}$ . This is also equal to the distance from $O$ to the midpoint of $AD$ . Using the definition of an equilateral triangle, $AZO = 30$ . We have a 30 60 90 triangle, so to find the length of the midpoint of $AD$ to $Z$ : $tan(60) = \frac{2x}{\sqrt{2}}$ . To find $OZ$ : $\frac{\sqrt{2} \cdot tan(60)}{2}+ \frac{\sqrt{2}}{2} \approx 1.93185$ . The area of an octagon formula: $2 (1+\sqrt{2})a^{2}$ . For $a$ , the area is approximately $18.02$ .

From unit circle to regular octagon and back

The answer is 18.02.

2 solutions

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