geometrypolygons
I am trying to find the vertices of a regular polygon using just the number of sides and 2 vertices. After the second vertex, I will make left turns to find each subsequent vertex that follows.
For example, If I have 4 sides, and 2 points, (0,0) & (0,10), how would I go about find the next to point of the square? I know the points would be (10,10) and then (10,0), but I can't think of a formula to accomplish this.
Thank you for any help, and I hope I provided enough information.
D)
But how did I get that answer? Well, firstly have you drawn a picture of the diagram, with angles labelled? If so, then you’ll notice that the arg (angle from the x axis) of any point on the polygon must be 15 mod 30. The last option doesn’t satisfy this property (the angle it forms is 120 degrees).
However, another approach is to notice — again from a well drawn diagram — that the figure is symmetric in the x axis, the y axis, and the x=y line. Since the first three options are just reflections of each other in these lines, the answer must be D. Meta analysis is useful in multiple choice!
I believe this becomes easier to think about if we bound the circles inside of a larger circle.
Labeled diagram of an example setup with four circles:
With five circles:
Below is some trigonometry that relates the size of the square and the number of circles to the radius of the circles, their distance from the center of the square, and the angle of displacement between the circles. Let $n$ be the number of circles. The other labels are described in the diagram(s).
\begin{align*} e &= \frac{2\pi}{2n} \ \text{rad}=\frac{\pi}{n} \ \text{rad} \\ d &= \pi-\frac{\pi}{2}-e=\frac{\pi}{2}-e \\ b &= a \cdot \sec{d} \\ c &= b+a \\ f &= 2c \end{align*}
Now, to solve for $a$ in terms of $f$ and $n$:
\begin{align*} f &= 2(b+a) \\ f &= 2(a \cdot \sec(d) + a) \\ f &= 2a(\sec(d) + 1) \\ a &= \frac{f}{2(\sec(d) + 1)} \\ a &= \frac{f}{2(\sec(\frac{\pi}{2}-e) + 1)} \\ a &= \frac{f}{2(\sec(\frac{\pi}{2}-\frac{\pi}{n}) + 1)} \\ \end{align*}
And for $b$ in terms of $f$ and $n$:
\begin{align*} b &= a \cdot \sec(d) \\ b &= \frac{f \cdot \sec(d)}{2(\sec(\frac{\pi}{2}-\frac{\pi}{n}) + 1)} \\ b &= \frac{f \cdot \sec(\frac{\pi}{2}-\frac{\pi}{n})}{2(\sec(\frac{\pi}{2}-\frac{\pi}{n}) + 1)} \\ \end{align*}
This works for any number of circles greater than $n=1$.
Bonus eight circle example
Best Answer
For simplicity I am assuming the centre of the polygon is origin.
If two consecutive vertices are given then the angle they make at origin can be found. Call it $\theta=2\pi/n$. Now all points are obtainable from any single point by rotating repeatedly by this angle (centred at origin). As rotation is a linear transformation, this can be achieved by matrix multiplication $$ R_\theta=\pmatrix{\cos\theta & -\sin\theta\cr \sin\theta &\cos\theta\cr}$$ Write the co-ordinates of a vertex of the polygon as a column vector $v_1=\displaystyle {x_1\choose y_1}$. Matrix multiplication $v_2=R_\theta v_1, v_3=R_\theta v_2$ etc will give the co-ordinates of all the vertices.