Polygon Bowl

1. Naming :

Let, the given square be $ABCD$ . The hexagon which is attached to the side $AB$ be $ABA_1A'_1A'_2A_2$ , where $A_1A_2$ is a diameter of the circumcircle, and, the naming has been done in the anticlockwise direction. Similarly name the other hexagons.

2.. Observation :

While folding, the point pairs $(A_1, B_2), (B_1, C_2),...(A'_1, B'_2), (B'_1, C'_2)...,$ etc. will coincide. We name these points of co-incidences $H_1, H_2,...H'_1, H'_2,...$ etc. in the order in which they've been stated. Since, the $BH_1, CH_2,$ etc. have the same inclination from the plane of $ABCD$ , and, all the sides of $H_1H_2H_3H_4$ are equal, $H_1H_2H_3H_4$ is a square. In the same way $H'_1H'_2H'_3H'_4$ is also a square.

Now, the entire figure is symmetric about the square $H_1H_2H_3H_4$ , with $ABCD$ and $H'_1H'_2H'_3H'_4$ lying on opposite sides. So, we only consider the part which contains the square $ABCD$ .

3. Compilation : Refer to
figure

$BH_1, CH_2,$ etc. meet at $V$ . The centres of the squares $ABCD$ and $H_1H_2H_3H_4$ be $P$ and $Q$ . Then, the height of the bowl will be twice the length of $PQ$ . $A, B, C, D$ are the mid-points of the sides $VH_4$ , $VH_1,$ etc. If the side-length of the square $ABCD$ is $s$ , then, $H_2H_3=2CD=2s$ . (Properties of a Regular Hexagon).

Consider $\triangle H_1VH_3$ . $VH_1=VH_3=2DH_3=2CD=2s$ . Since, $H_1H_3$ is a diagonal of the square $H_1H_2H_3H_4, H_1H_3= \sqrt{2} H_1H_2=2 \sqrt{2} s.$ So, $(VH_1)^2 + (VH_3)^2 = (H_1H_3)^2$ $\therefore \triangle H_1VH_3$ is an isosceles right-triangle.

Since, $P$ , $Q$ are the mid-points of $BD$ , $H_1H_3$ in the isosceles right triangle $H_1VH_3$ , it can be easily shown that, $PQ=\frac{1}{2}VQ=\frac{1}{2}QH_3=\frac{1}{4}H_1H_3=\displaystyle \frac{2\sqrt{2}}{4}s=\frac{s}{\sqrt{2}}$ .

$\therefore$ Height of the bowl, $h=2.\displaystyle \frac{s}{\sqrt{2}}=s \sqrt{2}= \sqrt{2s^2}.$

For the given problem, $s=22. \therefore a=2\times22^2=\boxed{968}$

Hello. If you draw the diagram (which I have no idea how to put it on here), you notice that if you draw it 2d, it looks 3d so it's easy to imagine. To find the height, you must do the Pythagorean Theorem to find it. The equation for the Pythagorean Theorem would be $\text{Length of Square}^{2} + \text{Side of Hexagon}^{2} = \text{Height}^{2}$ . We know that the length of a hexagon is $22$ , and the length of the square is $22$ also. Therefore, our answer is $22^{2}+22^{2}=22^{2}\cdot 2=\boxed{968}$ .

Vectorial Analysis

First, let a vertex of the square two sides of the squares be $\hat{i}$ and $\hat{j}$ .

Now, after we fold the hexagons, the side which passes through the origin corresponds to a vector. Lets find that out. This vector makes an angle of $120$ degrees with both the x and the y axes. From the arguments of symmetry, we can take the vector to be of the form $a\hat{i} + a\hat{j} +\hat{k}$ . Taking the dot product with either of the two , we'll get $a= \frac { -1 }{ \sqrt { 2 } }$ (for a unit vector). Lets name this vector $\xrightarrow { C }$

Now, we look at the angle made by the (any) hexagonal plane with the square (x-y) plane. For this we must take the cross-product of this vector with $\hat{i}$ . To get the vector $\hat{j} + \frac { 1 }{ \sqrt { 2 } }\hat{k}$ as the perpendicular vector of this plane. Also, $\hat{k}$ is perpendicular to the square. Thus angle between these two vectors would give us the required angle. $\cos\theta = \sqrt { \frac { 1 }{ 3 } }$ and thus $\sin \theta = \sqrt{\frac{2}{3} }$ .

Lastly the height that we are talking about is nothing but projection of a 2 diagonal of hexagon on x-z plane. Thus the answer would be $22\sqrt{3} \times \sin\theta = 22\sqrt{3} \times \sqrt{\frac{2}{3} } = 22\sqrt{2}$

Hence, the answer $h^{2}= 22^{2}\times 2= \boxed{968}$

alt text

So, the height is $\fbox{968}$