MHB Adam's Circles: Splitting & Connecting Segments

[maxkor]

Adam has a circle of radius $1$ centered at the origin.

- First, he draws $6$ segments from the origin to the boundary of the circle, which splits the upper (positive $y$) semicircle into $7$ equal pieces.

- Next, starting from each point where a segment hit the circle, he draws an altitude to the $x$-axis.

- Finally, starting from each point where an altitude hit the $x$-axis, he draws a segment directly away from the bottommost point of the circle $(0,-1)$, stopping when he reaches the boundary of the circle.

What is the product of the lengths of all $18$ segments Adam drew?

 

[jonah1]

Beer induced query follows.

... What is the product of the lengths of all $18$ segments Adam drew?

Product or sum?

 

[maxkor]

Beer induced query follows.

Product or sum?

Product.

 

[Opalg]

Just to check whether I'm thinking along the right lines, should the answer be

Spoiler

$\dfrac{7^3}{169\cdot2^{12}}$

?

 

[jonah1]

Beer induced reaction follows.

Just to check whether I'm thinking along the right lines, should the answer be

Spoiler

$\dfrac{7^3}{169\cdot2^{12}}$

?

I get the same; although mine's just an approximation, 0.000495504345414
Curious as to how you got an exact expression.
Did you use the math god Wolframalpha?
I was under the impression that while the endpoints of the green lines can be epressed exactly, I settled for an approximation. I guess I didn't took it far enough.
The product of the red and blue lines are of course
$1^6*\bigg[{\displaystyle \prod_{n=1}^{6} \sin\left(\frac{180n}{7}\right)}\bigg]$

 

[Opalg]

Adam has a circle of radius $1$ centered at the origin.

- First, he draws $6$ segments from the origin to the boundary of the circle, which splits the upper (positive $y$) semicircle into $7$ equal pieces.

- Next, starting from each point where a segment hit the circle, he draws an altitude to the $x$-axis.

- Finally, starting from each point where an altitude hit the $x$-axis, he draws a segment directly away from the bottommost point of the circle $(0,-1)$, stopping when he reaches the boundary of the circle.

What is the product of the lengths of all $18$ segments Adam drew?

Spoiler: My solution

The six points on the semicircle have coordinates $\bigl(\cos\frac{k\pi}7,\sin\frac{k\pi}7\bigr)\ (1\leqslant k\leqslant6)$. The red segments all have length $1$ and the $k$th blue segment has length $\sin\frac{k\pi}7$.

The $k$th green segment lies on the line joining $(0,-1)$ and $\bigl(\cos\frac{k\pi}7,0\bigr)$, which has equation $x = (y+1)\cos\frac{k\pi}7$. That meets the semicircle when $ (y+1)^2\cos^2\frac{k\pi}7 + y^2 = 1$, which leads after a bit of simplification to the point $$(x,y) = \Bigl(\frac{2\cos\frac{k\pi}7}{1+\cos^2\frac{k\pi}7},\frac{1-\cos^2\frac{k\pi}7}{1+\cos^2\frac{k\pi}7}\Bigr).$$ So if $d$ is the length of the $k$th green segment then $$d^2 = \Bigl(\frac{2\cos\frac{k\pi}7}{1+\cos^2\frac{k\pi}7} - \cos\tfrac{k\pi}7\Bigr)^2 + \frac{\bigl(1-\cos^2\frac{k\pi}7\bigr)^2}{\bigl(1+\cos^2\frac{k\pi}7\bigr)^2}.$$ Again after some simplification, this becomes $d^2 = \frac{\bigl(1-\cos^2\frac{k\pi}7\bigr)^2}{1+\cos^2\frac{k\pi}7}$, which I prefer to write as $d^2 = \frac{\sin^4\frac{k\pi}7}{2 - \sin^2\frac{k\pi}7}$.

Putting together everything done so far, the product of the lengths of the 18 segments is $$\prod_{k=1}^6 \frac{\sin^3\frac{k\pi}7}{\sqrt{2 - \sin^2\frac{k\pi}7}}.$$ To evaluate that product, notice that the numbers $\sin\frac{k\pi}7\ (1\leqslant k\leqslant6)$, together with $0$, are the solutions of the equation $\sin(7\theta) = 0$. But (either by working with trig. identities or by using de Moivre's theorem) $$\sin(7\theta) = 7\sin\theta - 56\sin^3\theta + 112\sin^5\theta - 64\sin^7\theta.$$ After discarding the solution $\sin\theta=0$, you see that $\sin\frac{k\pi}7\ (1\leqslant k\leqslant6)$ are the solutions of $7 - 56s^2 + 112s^4 - 64s^6 = 0$. The product of the roots of that equation is $\frac{7}{64}$. Therefore $$\prod_{k=1}^6 \sin^3\frac{k\pi}7 = \frac{7^3}{2^{18}}.$$ Next, putting $x = 2 - s^2$ you see that $2 - \sin^2\frac{k\pi}7\ (1\leqslant k\leqslant6)$ are the solutions of $7 - 56(2-x) + 112(2-x)^2 - 64(2-x)^3 = 0$. That simplifies to $64x^3 - 272x^2 + 376x - 169 = 0$, and the product of the roots is $\frac{169}{64}$. Each value of $x$ corresponds to two (equal) values of $s$, so we should square that answer. But then we want to take the square root (getting back to where we started from), for the formula $$\prod_{k=1}^6 \frac{1}{\sqrt{2 - \sin^2\frac{k\pi}7}} = \frac{169}{64}.$$ Finally, $$\prod_{k=1}^6 \frac{\sin^3\frac{k\pi}7}{\sqrt{2 - \sin^2\frac{k\pi}7}} = \frac{7^3/2^{18}}{169/2^6} = \frac{7^3}{169\cdot2^{12}}.$$

 

[maxkor]

It' correct.