New Year's Countdown Day 11: Undecagon

Relevant wiki: Roots of Unity

We can use complex numbers to model this problem. First, let $\omega = e^{2\pi i/11}.$ Place the circle on the complex grid with its center at 0, and let $A$ be located at the complex number 1. Then, the points on the undecagon are the 11th roots of unity, so they are of the form $\omega^n,$ where $1 \leq n \leq 11.$ We can calculate the distance between $A$ and each of the other points by finding the difference between the points and taking the absolute value:

$\begin{aligned} AB &= |1 - \omega| \\ AC &= |1 - \omega^2| \\ AD &= |1 - \omega^3|, \end{aligned}$

and so on. To calculate the desired product, we first note that

$AB \times AC \times AD \times \cdots \times AK = (AB \times AC \times AD \times AE \times AF)^2$

by symmetry i.e. $AB = AK,$ etc. To calculate the left hand side product, we use the fact that the polynomial with roots that are the 11th roots of unity excluding 1 is

$f(x) = \dfrac{x^{11} - 1}{x - 1} = x^{10} + x^9 + x^8 + \cdots + x + 1,$

since $x^{11} - 1$ has all 11 11th roots of unity as roots, and we must exclude 1 as that represents point $A.$ We can factor $f$ as

$f(x) = (x - \omega)(x - \omega^2)(x - \omega^3)\cdots(x-\omega^{10}).$

Finally, since

$\begin{aligned} AB \times AC \times AD \times \cdots \times AK &= |1 - \omega||1 - \omega^2||1 - \omega^3|\cdots|1 - \omega^{10}| \\ &= |(1 - \omega)(1 - \omega^2)(1 - \omega^3)\cdots(1 - \omega^{10})| \\ &= |f(1)| \\ &= |1^{10} + 1^9 + 1^8 + \cdots + 1 + 1| \\ &= 11, \end{aligned}$

we conclude $AB \times AC \times AD \times AE \times AF = \sqrt{11},$ so $k = 11$ and the sum of its digits is $1 + 1 = \boxed{2}.$

$AB\times AC\times AD\times AE\times AF=32r^5\sin(\frac{\pi}{11})\sin(\frac{2\pi}{11})\sin(\frac{3\pi}{11})\sin(\frac{4\pi}{11})\sin(\frac{5\pi}{11})$ $=16r^5(\cos(\frac{\pi}{11})-\cos(\frac{3\pi}{11}))\sin(\frac{3\pi}{11})\sin(\frac{4\pi}{11})\sin(\frac{5\pi}{11})$ $=8r^5(\sin(\frac{4\pi}{11})+\sin(\frac{2\pi}{11})-\sin(\frac{6\pi}{11}))\sin(\frac{4\pi}{11})\sin(\frac{5\pi}{11})$ $=4r^5(1-\cos(\frac{8\pi}{11})+\cos(\frac{2\pi}{11})-\cos(\frac{6\pi}{11})-\cos(\frac{2\pi}{11})+\cos(\frac{10\pi}{11}))\sin(\frac{5\pi}{11})$ $=2r^5(2\sin(\frac{5\pi}{11})-\sin(\frac{13\pi}{11})+\sin(\frac{3\pi}{11})+\sin(\frac{\pi}{11})+\sin(\frac{15\pi}{11})-\sin(\frac{5\pi}{11}))$ $=2r^5(\sin(\frac{\pi}{11})+\sin(\frac{2\pi}{11})+\sin(\frac{3\pi}{11})-\sin(\frac{4\pi}{11})+\sin(\frac{5\pi}{11}))$ So, $k=4r^{10}(\sin(\frac{\pi}{11})+\sin(\frac{2\pi}{11})+\sin(\frac{3\pi}{11})-\sin(\frac{4\pi}{11})+\sin(\frac{5\pi}{11}))^2$ $2\sin(\frac{i\pi}{11})\sin(\frac{j\pi}{11})=\cos(\frac{(i-j)\pi}{11})-\cos(\frac{(i+j)\pi}{11})=c_{i-j}-c_{i+j}$ Therefore $k=\sum_{i=1}^5\sum_{j=1}^5(-)^{\delta_{i4}+\delta_{j4}}(c_{i-j}-c_{i+j})=\sum_{i=1}^5(1-c_{2i})+\sum_{i=1}^4\sum_{j=i+1}^5(-)^{\delta_{i4}+\delta_{j4}}(c_{i-j}-c_{i+j})$ where $\delta_{xy}$ stands for dirac delta function, which equals to 1 when x=y, and equals to 0 otherwise. $=2r^{10}(5-c_2-c_4-c_6-c_8-c_{10}+2(c_1-c_3+c_2-c_4-c_3+c_5+c_4-c_6+c_1-c_5-c_2+c_6+c_3-c_7-c_1+c_7+c_2-c_8-c_1+c_9))$ $=2r^{10}(5+c_2-2c_3-c_4-c_6-3c_8+2c_9-c_{10})$ Where $c_n=\cos(\frac{n\pi}{11})$ . So, $k/r^{10}=12+2(-1+\cos(\frac{\pi}{11})-\cos(\frac{2\pi}{11})+\cos(\frac{3\pi}{11})-\cos(\frac{4\pi}{11})+\cos(\frac{5\pi}{11}))$ $=12-4(\sin(\frac{\pi}{22})\sin(\frac{\pi}{22})+\sin(\frac{\pi}{22})\sin(\frac{5\pi}{22})+\sin(\frac{\pi}{22})\sin(\frac{9\pi}{22}))$ $=12-4\frac{\sin(\frac{\pi}{11})}{\sin(\frac{\pi}{11})}\sin(\frac{\pi}{22})(\sin(\frac{\pi}{22})+\sin(\frac{5\pi}{22})+\sin(\frac{9\pi}{22}))$ $=12-2\frac{\sin(\frac{\pi}{22})}{\sin(\frac{\pi}{11})}(\cos(-\frac{\pi}{22})-\cos(\frac{3\pi}{22})+\cos(\frac{3\pi}{22})-\cos(\frac{7\pi}{22})+\cos(\frac{7\pi}{22})-\cos(\frac{11\pi}{22}))$ $=12-\frac{1}{\cos(\frac{\pi}{22})}(\cos(\frac{\pi}{22})-\cos(\frac{\pi}{2}))=11$ We got k=11, then the sum of its digits is 2.

Mathematica:

s=CirclePoints@11; RootReduce[Times@@EuclideanDistance@@@Table[s[[{1,i}]],{i,2,6}]]

In general, for any regular $n$ -gon inscribed in a circle with radius $r$ we can write a program using the algorithm below:

Let $k \leq n.$

Let $u_{0} = 2r \sin(\dfrac{\pi}{n})$ and for $(1 \leq j \leq k - 1)$ $u_{j} = \sqrt{u_{0}^2 + u_{j - 1}^2 - 2 u_{0} u_{j - 1}\cos(\dfrac{(n - 2)\pi}{n} - (j - 1)\dfrac{\pi}{n})}$

and the product $P = \prod_{j = 0}^{k - 1} u_{j}$ and this problem wants $P^2 = (\prod_{j = 0}^{k - 1} u_{j})^2.$

Note: After doing the problem I wrote a program for the general case based on the algorithm above.

We can apply the cosine theorem to find $AB^2,AC^2,AD^2,AE^2,AF^2$ . So $AB^2=2(1-\cos\frac{2\pi}{11}),\,AC^2=2(1-\cos\frac{4\pi}{11}),\,AD^2=2(1-\cos\frac{6\pi}{11}),\,AE^2=2(1-\cos\frac{8\pi}{11}),\,AF^2=2(1-\cos\frac{10\pi}{11})$ . Multiplication and application of theorems for the transformation a product of trigonometric functions into a sum, give rise to the equality $AB^2AC^2AD^2AE^2AF^2=2(5+\sin\frac{\pi}{22}-\sin\frac{3\pi}{22}+\sin\frac{5\pi}{22}-\sin\frac{7\pi}{22}+\sin\frac{9\pi}{22})$ . To find the last trigonometric sum, we denote $z=e^{i\frac{\pi}{22}}-e^{i\frac{3\pi}{22}}+e^{i\frac{5\pi}{22}}-e^{i\frac{7\pi}{22}}+e^{i\frac{9\pi}{22}}$ . So $z=\dfrac{e^{i\frac{\pi}{22}}(1+e^{i\frac{5\pi}{11}})}{1+e^{i\frac\pi{11}}}$ . The imaginary part of $z$ is the sum $\sin\frac{\pi}{22}-\sin\frac{3\pi}{22}+\sin\frac{5\pi}{22}-\sin\frac{7\pi}{22}+\sin\frac{9\pi}{22}$ , and the imaginary part of the right-side expression is $\frac12$ , so that $AB^2AC^2AD^2AE^2AF^2=11=k$ , meaning the sum of digits of $k$ is $2$ .

Let the centre of the circle be O. Angle AOB = 360/11 = 32.73 deg. AB = 2 x Sin(AOB/2) AC = 2 x Sin(2 x AOB/2) . . AF = 2 x Sin(5 x AOB/2) Substituting the value of angle AOB, AB x AC x AD x AE x AF = 3.3166 Hence k = 3.3166^2 = 11

Given $AB \times AC \times AD \times AE \times AF = \sqrt{k}$ , $AB^2 \times AC^2 \times AD^2 \times AE^2 \times AF^2 = k$ .

To compute the length of $AX$ for $X \in \left\{B, C, D, E, F\right\}$ , we use the formula $c^2 = a^2 + b^2 - 2ab \cos \theta$ where $a$ and $b$ are both the radius. Let $P$ be the centre of the circle. Then, the angle $\theta$ is equal to the angle $APX$ , which is $\frac{n}{11}2\pi rad$ , where $n$ is 1 for $X=B$ , 2 for $X=C$ and so on around the circle. This gives us the equation:

$\prod_{n=1}^{5}{2\left(1-cos\left(\frac{n}{11}2\pi\right)\right)} = 11$

The sum of the digits of 11 is $\boxed{2}$ .

For people bad at geometry:

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from math import *

def varAndPrint(variable):
    try:
        for k,v in globals().items():
            if(v == variable):
                print(k,v)
    except:
        for k,v in locals().items():
            if(v == variable):
                print(k,v)

# Begin:
# We know it's a unit circle, so it's easy to derive the coordinates of the 10 points that are not A:
pointList = []
for i in range(1,6):
    angle = 2*i*pi/11
    pointList.append({'x': cos(angle), 'y':sin(angle)})

# Basic distance formula
def dist(a,b):
    return sqrt((a['x']-b['x'])**2 + (a['y']-b['y'])**2)

# Now get the length from A to each of the other coords
accumulator = 1
for point in pointList:
    accumulator *= dist({'x':1,'y':0}, point)

print(accumulator,"squared is:",accumulator**2)
Rounded = round(accumulator**2)
varAndPrint(Rounded)
accumulator = 0
for dig in str(Rounded):
    accumulator += int(dig)
print("Sum of digits:", accumulator)

Output:

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3.3166247903553994 squared is: 10.999999999999996
Rounded 11
Sum of digits: 2