Rooster Zen Origami Challenge

Happy Chinese New Year! 新年快樂!

I recommend watching this video before reading the solution in order to understand the geometry. The polygons after unfolding the rooster reveals the triangles and squares.

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First Step

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The first step is to jot down the properties of the rooster. From the given, we can assume that the vertices of the whole rooster form $45^{\circ}$ and $90^{\circ}$ angles, which show that

• $\angle C_4 = \angle C_3C_5C_4 = \angle C_8C_6C_{11} = \angle C_{12} = \angle C_{10}C_{13}C_{12} = 45^{\circ}$
• $\angle C_1 = \angle C_1C_5C_7 = \angle C_4C_3C_5 = \angle C_7 = \angle C_{10} = 90^{\circ}$

Since the creases are equidistant and $|BC| = |CD|$ , the midpoints in the square are formed, which gives sets of congruent segments. Then, the intersection points of the diagonals form congruent segments as well.

However, the left-sided flap is formed. After folding it to create $\angle C_{12}C_{11}C_{13} = 90^{\circ}$ , due to dashed folding line $C_{13}$ , $|C_{12}C_{13}| = |C_1C_{13}|$ .

With all the necessary information gathered, we can answer the problem.

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Second Step

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Given the information, the next step is to determine the total area. Let $x = |C_1C_{14}|$ . Then, $|C_1C_7| = \dfrac{4}{3}x$ , which implies that $|C_1C_5| = |C_7C_{10}| = \dfrac{4}{3\sqrt{2}}x$ . Because $C_1C_5C_7C_{10}$ is a square, $A_{\square}= \dfrac{8}{9}x^2$ . Since the points on the square segments divide into four congruent segments, and there are $32$ of $\Delta C_3C_4C_5$ triangles of the same areas, then $A_{\Delta C_3C_4C_5} = \dfrac{1}{36}x^2$ .

For $\Delta C_{11}C_{12}C_{13}$ , let $y = |C_1C_{13}|$ . Due to symmetry, $y = |C_{12}C_{13}|$ , so $|C_{11}C_{12}| = |C_{11}C_{13}| = \dfrac{1}{\sqrt{2}}y$ . Then, $y + \dfrac{1}{\sqrt{2}}y = \dfrac{2}{3\sqrt{2}}x$ where $y = \dfrac{2}{3\sqrt{2}\left(1 + \frac{1}{\sqrt{2}}\right)}x$ So the area of $\Delta C_{11}C_{12}C_{13}$ is $A_{\Delta C_{11}C_{12}C_{13}} = \dfrac{1}{18\left(1 + \frac{1}{\sqrt{2}} \right)^2}x^2$

For shaded regions, $|C_6C_7| = |C_7C_8| = |C_{10}C_{11}| = \dfrac{2}{3\sqrt{2}}x$ and $|C_9C_{10}| = \dfrac{1}{3\sqrt{2}}x$ . Then, $A_{\Delta C_9C_{10}C_{11}} = \dfrac{1}{18}x^2$ and $A_{\Delta C_6C_7C_8} = \dfrac{1}{9}x^2$ . In this case, the area of the shaded regions to remove is $\dfrac{1}{6}x^2$ .

Therefore, for $x = 1$ , the total area is $\begin{array}{rl} A_{\text{total}} &= A_{\text{white}} + A_{\text{shaded}}\\ &= \dfrac{1}{36}\left(39 - 8\sqrt{2}\right) \end{array}$ So for $a = 1$ , $b = 36$ , $c = 39$ , $d = 8$ and $e = 2$ , $a^c\left(d^e - b\right) = \boxed{28}$ .