Round and round they go

Let the angular velocities of $A$ , $B$ and $C$ be $\omega_A= 2\pi$ $\text{ rads}^{-1}$ , $\omega_B= \pi$ $\text{ rads}^{-1}$ and $\omega_C=\dfrac {2\pi}3$ $\text{ rads}^{-1}$ respectively. Instead of all three points are moving we can fixed one. say $C$ , and find the relative position of $A$ and $B$ to $C$ using relative angular velocities. Let the relative angular velocities to $C$ of $A$ and $B$ be $\omega_a = 2\pi - \dfrac {2\pi}3 = \dfrac {4\pi}3$ $\text{ rads}^{-1}$ and $\omega_b = \pi - \dfrac {2\pi}3 = \dfrac \pi 3$ $\text{ rads}^{-1}$ respectively.

Let the centre of the unit circle be $O$ , $\angle AOC = \theta_a$ and $\angle BOC= \theta_b$ . We note that the area of $\triangle ABC$ is given by:

$\begin{aligned} [ABC] & = [ABO] + [BCO] - [ACO] & \small \color{#3D99F6} \text{By formula }A_\triangle = \frac 12 ab \sin \theta \\ & = \frac 12 \sin (\theta_a - \theta_b) + \frac 12 \sin \theta_b - \frac 12 \sin \theta_a \\ & = \frac 12 \left(\sin (\omega_at - \omega_bt) + \sin (\omega_bt) - \sin (\omega_a t) \right) \\ & = \frac 12 \left(\sin (\pi t) + \sin \left(\frac {\pi t}3 \right) - \sin \left(\frac {4\pi t}3\right) \right) & \small \color{#3D99F6} \text{Let }\phi = \frac {\pi t}3 \\ & = \frac 12 \left(\sin 3\phi + \sin \phi - \sin 4\phi \right) \end{aligned}$

The extrema of $[ABC]$ occurs when

$\begin{aligned} \frac d{d\phi} \left(\sin 3\phi + \sin \phi - \sin 4\phi\right) & = 0 \\ 3\cos 3 \phi + \cos \phi - 4 \cos 4\phi & = 0 \\ 3(4\cos^3 \phi - 3\cos \phi) + \cos \phi - 4(8 \cos^4 \phi - 8 \cos^2 \phi + 1) & = 0 & \small \color{#3D99F6} \text{Let }x = \cos \phi \text{ and rearrange.} \\ 8x^4-3x^3-8x^2+2x+1 & = 0 \end{aligned}$

Solving for $x$ by numerical method, we have:

$\begin{cases} x = 1 & \implies \phi = 0 & \implies [ABC] = 0 \\ x = -0.265694309 & \implies \phi = 1.839720373 & \implies [ABC] = -0.303848271 \\ x = 0.5293892 & \implies \phi = 1.012915884 & \implies [ABC] = 0.870295063 \\ x = -0.888694891 & \implies \phi = 2.665287115 & \implies [ABC] = 1.196524967 \end{cases}$

Therefore, the maximum area of $\triangle ABC$ is $\boxed{1.196524967}$ .

既然周期之比为1:2:3，速度比为6:3:2。

现在换参考系，假设速度为3的点为静止，那么其他两个点的速度为3和-1，相等时间内划过的角度为3:-1。

我们知道，三角形面积公式S=ab sinC/2，那么既然三角形内接于单位圆，不妨将三个顶点与圆心相连，将三角形表示为三个三角形面积之和，即：

S=(sinx+sin3x+sin(2π-4x))/2 =(sinx+sin3x-sin4x)/2

分析这个函数，找到在2π内的极大值：1.196525

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

import numpy as np

s_max = 0.0
ta = np.arange(0.0,6.0,0.001) # 0 to 6 sec in steps of 0.001
pi2 = 2*np.pi 

for t in ta: 
    a = (np.cos(t/1*pi2), np.sin(t/1*pi2))
    b = (np.cos(t/2*pi2), np.sin(t/2*pi2))
    c = (np.cos(t/3*pi2), np.sin(t/3*pi2))
    s = abs(0.5*(a[0]*(b[1]-c[1])+b[0]*(c[1]-a[1])+c[0]*(a[1]-b[1])))
    if(s > s_max):
        s_max = s

print ("max area: ",s_max)

> max area: 1.196524791219177

We begin by finding expressions to describe the location of each point at any given time.

We find that the coordinates of point $\textcolor{#3D99F6}{A}$ are described by $(\cos2\pi t,\sin2\pi t)$ , point $\textcolor{#D61F06}{B}$ by $(\cos\pi t,\sin\pi t)$ , and point $\textcolor{#20A900}{C}$ by $(\cos\frac{2\pi}{3}t,\sin\frac{2\pi}{3}t)$ , where $t$ is equal to time, in seconds, after the initial moment where all three points were at the same location. (Note: This is working in radians, not degrees.)

Using this information, we can now find functions to describe the distance between each pair of points (or the length of each side of the triangle) at any given time.

The distance between two points on a graph is described by $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Plugging in our expressions for the locations of each point, we find that the distance between points $\textcolor{#3D99F6}{A}$ and $\textcolor{#D61F06}{B}$ is described by $a(t)=\sqrt{(\cos t-\cos2\pi t)^{2}+(\sin t-\sin2\pi t)^{2}}$ , points $\textcolor{#3D99F6}{A}$ and $\textcolor{#20A900}{C}$ by $b(t)=\sqrt{(\cos\frac{2\pi}{3}t-\cos2\pi t)^{2}+(\sin\frac{2\pi}{3}t-\sin2\pi t)^{2}}$ , and points $\textcolor{#D61F06}{B}$ and $\textcolor{#20A900}{C}$ by $c(t)=\sqrt{(\cos\frac{2\pi}{3}t-\cos t)^{2}+(\sin\frac{2\pi}{3}t-\sin t)^{2}}$ .

Now that we have have described the length of each side of the triangle, we can use Heron's formula to find the area of the triangle at any given time.

Heron's formula states that the area of a triangle is described by $\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are the lengths of the sides of the triangle, and $s$ is the semiperimeter of the triangle, which is described by $s=\frac{a+b+c}{2}$ .

First we create a function to describe the semiperimeter by plugging in our functions for the length of each side of the triangle into the formula, creating $s(t)=\frac{a(t)+b(t)+c(t)}{2}$ .

Now we plug all of our functions into Heron's formula, resulting in $f(t)=\sqrt{s(t)(s(t)-a(t))(s(t)-b(t))(s(t)-c(t))}$

This function models the area of our triangle $\textcolor{#3D99F6}{A}\textcolor{#D61F06}{B}\textcolor{#20A900}{C}$ over time.

Now we simply find the maximum of the function.

The maximum area of the triangle is about $1.1965$ , and occurs for the first time at about $2.545$ seconds after the initial point, and for the second time at at about $3.455$ seconds after the initial point.

Here is an outline of the approach I used: Let $A = (\cos(2\pi t), \sin(2\pi t))$ , $B = (\cos(\pi t), \sin(\pi t))$ and $C = (\cos(\frac{2\pi t}{3}), \sin(\frac{2\pi t}{3}))$ and let $Area(t)$ be the area of triangle $ABC$ at time $t$ .

The triangle $ABC$ is one half of the parallelogram spanned by the vectors $B-A$ and $C-A$ , so:

$\begin{aligned} Area(t) &= \frac{1}{2} \det (B-A, C-A) \\ &= \frac{1}{2} \begin{vmatrix} \cos(\pi t) - \cos(2\pi t) & \cos(\frac{2\pi t}{3}) - \cos(2\pi t) \\ \sin(\pi t) - \sin(2\pi t) & \sin(\frac{2\pi t}{3}) - \sin(2\pi t) \end{vmatrix} \end{aligned}$ After some calculation we get $Area(t) = \frac{1}{2}(-\sin(\frac{\pi t}{3}) - \sin(\pi t) + \sin(\frac{4 \pi t}{3}))$ . From here, we can proceed to find the maximum in the usual way, by setting $\frac{d}{dt}Area(t) = 0$ , and so on.

I originally used the exact same method as David Wilbanks, but I ended up finding a way that is easier than the distance formula to find the distance between any pair of the three points. Lets call the measure of the central angle that intercepts the starting point and point ${\color{#20A900}C}$ 's current position $\theta$ . The central angle that intercepts the starting point and angle ${\color{#3D99F6}A}$ 's current position can be written as $3\theta$ , and similarly this angle for point ${\color{#D61F06}B}$ is $\frac{3\theta}{2}$ .

Image not drawn to scale

If we think about the triangle that is formed with any two points and the center of the unit circle, we can use the law of cosines to find the distance between the points. In the case of point ${\color{#3D99F6}A}$ and point ${\color{#20A900}C}$ , we can first find that central angle that falls in the triangle is $2 \theta$ , and using the law of cosines, we can now say that,

$(d_{A, C})^2 = 2 - 2 \cos 2 \theta$ .

This formula isn't wonderful to look at, so we can use some trigonometric identities to rewrite this as,

$(d_{A, C})^2 = 4(\frac{1 - \cos 2 \theta}{2})$ ,

$d_{A, C} = 2\sqrt{(\frac{1 - \cos 2 \theta}{2}})$ ,

$d_{A, C} = 2 \sin \theta$ .

You can try using the same method for the other two selections of points, but the distances between the points end up being,

$d_{A, B} = 2 \sin \frac{\theta}{4}$ , and

$d_{B, C} = 2 \sin \frac{3 \theta}{4}$ .

Now we can use the Heron's formula to find the equation for the area of the triangle formed by the three points in any possible orientation, which is,

$A = \sqrt{(\sin \theta + \sin \frac{\theta}{4} + \sin \frac{3 \theta}{4})(-\sin \theta + \sin \frac{\theta}{4} + \sin \frac{3 \theta}{4})(\sin \theta - \sin \frac{\theta}{4} + \sin \frac{3 \theta}{4})(\sin \theta + \sin \frac{\theta}{4} - \sin \frac{3 \theta}{4})}$

Finally, we can graph this equation on a graphing calculator to find the maximum value for the area of the triangle formed by the three points, which is approximately $\boxed{1.1965249}$ .

Maximum area is for equilateral triangle when the first point rotated by 120 degrees, second by 240 and 3rd by 360 degrees.

This gives the area = ( 3√3 )/4 = 1.2 approx.

You need the radius of the circle to solve this. Since $A=\frac {abc}{4R}$ and any such triangle has the same $R$, we need to maximize the product of the sides. This is attained when we have an equilateral triangle, which can indeed be made (as when the first particle has gone $\frac {1}{3} $ the way around, the second has gone $\frac {2}{3} $ the way, and the third has gone all the way around). Then we can use $A=\frac{1}{2}ab\sin {\theta} $ to calculate the area of one of the three identical triangles formed when you connect the center of the equilateral triangle to the vertices. This yields $3\cdot\frac {1}{2}(r)(r)(\frac{\sqrt{3}}{2})=\boxed {\frac {3r^2\sqrt {3}} {4}}$

I got the area as $A=(\sin 4x - \sin 3x -\sin x)/2$ . The extrema are at $4 \cos 4x -3 \cos 3x- \cos x=0$ , solved it with Wolfram Alpha. The solutions are coming in the form of $x=n\pi$ , $x=2(n\pi\pm 0.5064)$ and $x=2(n\pi\pm0.09199)$ . Once we calculate the area for a few of these solutions, we can identify the maximum.