Equidistant points on a unit circle

Let $\omega = e^{2\pi i/29}.$ Then $x_k = \left|1 - \omega^k\right|$ and $\begin{aligned} \prod_{k=1}^{28} x_k &= \left|\prod_{k=1}^{28}(1-\omega^k)\right| \\ &= \left|\lim_{t\to 1} \prod_{k=1}^{28}(t - \omega^k)\right| \\ &= \left|\lim_{t\to 1} \frac{\displaystyle \prod_{k=0}^{28}(t - \omega^k)}{t-1}\right| \\ &= \left|\lim_{t\to 1} \frac{t^{29}-1}{t-1}\right| \\ &= \left|\lim_{t\to 1} 29t^{28}\right| \\ &= 29 \end{aligned}$

Therefore, $\left\lfloor 100 \prod_{k=1}^{28} x_k \right\rfloor = \left\lfloor 100(29) \right\rfloor = \boxed{2900}$

Let $\zeta$ be a primitive $29$ th root of unity. Then our product is $100 \prod_{k=1}^{28} |\zeta^k-1|.$ Now the $\zeta^k-1$ are the $28$ distinct roots of the degree- $28$ polynomial $\frac{(z+1)^{29}-1}{z}.$ The product of the absolute values of the roots of that polynomial is the absolute value of its constant term, which is $29.$ So the answer is $100 \cdot 29 = \fbox{2900}.$

Thanks to 3Blue1Brown

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Write the distance between two points (x) in terms of their arc distance (a).

$x=2sin(\frac {a}{2R})$ .

In this question, $x_k=2sin(\frac {kπ}{29})$ . Product of these x till k=28 ≈ 29. Thus the answer is 2900.

First, find the equation of a line that has an angle of $360/29\ degrees$ using the tangent function and find the general form of the line which is $x(tan(2a\pi /29))$ . Next, set the general equation equal to a point on the unit circle (or $\sqrt { 1-x^{ 2 } }$ ).

So, $x(tan(2a\pi /29))=\sqrt { 1-x^{ 2 } } \\ x^{ 2 }tan^{ 2 }(2a\pi /29)=1-x^{ 2 }\\ tan^{ 2 }(2a\pi /29)=1/x^{ 2 }-1\\ tan^{ 2 }((2a\pi )/29))+1=1/x^{ 2 }\\ sec^{ 2 }(2a\pi /29)=1/x^{ 2 }\\ sec(2a\pi /29)=\pm 1/x\\ x=\pm cos(2a\pi /29)$

So each individual distance is ${ x }_{ k }=\sqrt { a_{ 2 }-{ a }_{ 1 })^{ 2 }+({ b }_{ 2 }-{ b }_{ 1 })^{ 2 } }$ by the distance formula.

Since the lines intersect at $x=\pm cos(2a\pi /29)$ , the y-value of the intersection is $y=\pm \sqrt { 1-(\pm cos(2a\pi /29)^{ 2 } } =\pm sin(2a\pi /29)$

So the distance between the two points $(1,0)\quad and\quad (\pm cos(2a\pi /29),\pm sin(2a\pi /29))\quad \\$

= $\sqrt { (\pm cos(2a\pi /29)-1)^{ 2 }+(\pm sin(2a\pi /29))^{ 2 } } \\$ = $\sqrt { ( cos(2a\pi /29)-1)^{ 2 }+( sin(2a\pi /29))^{ 2 } } \\$ = $\sqrt { cos^{ 2 }(2a\pi /29)+sin^{ 2 }(2a\pi /29)-2cos(2a\pi /29)+1 }$ = $\sqrt { 2-2cos(2a\pi /29) }$

Now we will simplify this using the identity $cos(2\theta )=1-2sin^{ 2 }\theta$ and obtain $\sqrt { 2-2(1-2sin^{ 2 }(a\pi /29)) } =\sqrt { 4sin^{ 2 }(a\pi /29) }=2sin(a\pi /29)$

So we will have $100\prod _{ k=1 }^{ 28 }{ 2sin(a\pi /29) }$ which will end up being 100*29= 2900

Lacking the mathematical sophistication to solve this as Brian and Patrick did, I just approached it as a trig problem. The length of the first 14 chords can be computed as $2sin(\frac{i\pi}{58}$ ), where i runs from 1 to 14. Since the 14th chord equals the 15th, the 13th equals the 16th, and so on, we just square each of the first 14 terms and multiply them together. This was easy in Excel; 14 rows x three columns did it. The answer is 100(29) = 2900.

The cosine rule $c=\sqrt{a^2+b^2-2ab\cos(\gamma)}$ gives the length c of the opposite side in any triangle with sides of length a and b adjacent to angle $\gamma$ .

In our case a and b are both radii of the unit circle, so the segment lengths are $\sqrt{2-2\cos(\gamma)}$ . The segments are pairwise equal so that taking their product cancels the square roots. The product of the 28 lengths is now given by $(2-2\cos\frac{2\pi}{29})\cdot(2-2\cos\frac{4\pi}{29})\cdot...\cdot(2-2\cos\frac{28\pi}{29})$ I got stuck with multiple angle trig formulas, but after trying $(2-2\cos(\frac{2\pi}{3}))=3$ $(2-2\cos(\frac{2\pi}{5})) (2-2\cos(\frac{4\pi}{5}))=5$ $(2-2\cos(\frac{2\pi}{7})) (2-2\cos(\frac{4\pi}{7}))(2-2\cos(\frac{6\pi}{7}))=7$ I was confident that the product is equal to 29, so the requested answer is 2900.

If you take a regular N-sided polygon, which is inscribed in the unit circle and find the product of all its diagonals (including two sides) carried out from one corner you will get N exactly:

Answer=29×100=2900

The cosine rule puts $x_k$ in terms of k in the relation

${(x_k)}^2 = 1+1 - 2\times1\times1\times cos({\frac{2 \pi}{29}\times k}) \Longrightarrow {(x_k)}^2 = 2(1-cos({\frac{2 \pi}{29}\times k}))$ .

Since the set of intervals has horizontal symmetry, $\prod_{k=1}^{28}x_k$ can be written as $\prod_{k=1}^{14}{(x_k)}^2$ .

Computing $\prod_{k=1}^{14}2(1-cos({\frac{2 \pi}{29}\times k}))$ using a calculator gives 29. So the answer is $\lfloor 100\cdot29 \rfloor = \boxed{2900}$

Firstly , we need to find a way to get the length of the different segments

By using the cosinus law we get $x_k^2 = 1^2 + 1^2 - 2*\cos(k * \frac {2\pi}{29})$ , thus $x_k = \sqrt{2-2*\cos(k * \frac {2\pi}{29})}$

Then with a bit of programming using Python we get :

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import math
angle, s = 2*math.pi/29.0, 1
for a in range(1,15): s*=(2-2*math.cos(angle*a))**0.5
print(math.floor((s**2)*100))

output: 2900

Therefore the answer is $\boxed{2900}$

Hint :- The points lie on a regular icosikaienneagon......Now use roots of unity..........!!!

If one defines the product of the null set as 1, then the resultant product is the number of points, in this case 29, around the unit circle. Therefore, $\lfloor 29\times 100\rfloor=2900$ .