Adam's Circles: Splitting & Connecting Segments

In summary: The product of the lengths of all $18$ segments Adam drew is $\displaystyle 1^7*\prod_{n=1}^{6} \sin\left(\frac{180n}{7}\right)$. In summary, Adam drew a circle of radius $1$ centered at the origin and split the upper semicircle into $7$ equal pieces. He then drew $6$ segments from the origin to the boundary of the circle and from each point where a segment hit the circle, he drew an altitude to the $x$-axis. Finally, he drew a segment directly away from the bottommost point of the circle, stopping when he reached the boundary of the circle. The product of the lengths of all $
  • #1
maxkor
84
0
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?
 

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  • #2
Beer induced query follows.
maxkor said:
... What is the product of the lengths of all $18$ segments Adam drew?
Product or sum?
 
  • #3
jonah said:
Beer induced query follows.

Product or sum?
Product.
 
  • #4
Just to check whether I'm thinking along the right lines, should the answer be
$\dfrac{7^3}{169\cdot2^{12}}$
?
 
  • #5
Beer induced reaction follows.
Opalg said:
Just to check whether I'm thinking along the right lines, should the answer be
$\dfrac{7^3}{169\cdot2^{12}}$
?
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
$1^6*\bigg[{\displaystyle \prod_{n=1}^{6} \sin\left(\frac{180n}{7}\right)}\bigg]$
 
  • #6
maxkor said:
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?
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, $$\prod_{k=1}^6 \frac{\sin^3\frac{k\pi}7}{\sqrt{2 - \sin^2\frac{k\pi}7}} = \frac{7^3/2^{18}}{169/2^6} = \frac{7^3}{169\cdot2^{12}}.$$
 
  • #7
It' correct.
 

Related to Adam's Circles: Splitting & Connecting Segments

1. What is the concept behind Adam's Circles?

Adam's Circles is a mathematical concept that involves splitting and connecting segments on a circle to create new shapes and patterns.

2. How is Adam's Circles related to geometry?

Adam's Circles is a geometric concept that explores the relationships between angles, segments, and circles.

3. What are some real-world applications of Adam's Circles?

Adam's Circles can be used to create unique designs in art and architecture, as well as in engineering and computer graphics.

4. Can Adam's Circles be used to solve problems in mathematics?

Yes, Adam's Circles can be used to solve problems involving angles, segments, and circles in geometry and trigonometry.

5. Are there any limitations to using Adam's Circles?

While Adam's Circles can be a useful tool in geometry and design, it is important to note that it is just one concept and may not apply to all situations. It is also important to use caution when applying it to real-world problems, as it may not always provide an accurate solution.

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