3 coins

Relevant wiki: Extrema - Problem Solving - Medium

The white sectors together make a whole circle regardless of the angle. We can ignore them.

The orange rectangles also do not depend on the angle we can ignore them too.

Let the coins have radius 1 and the angle be $2\theta$ then the triangle has area $2\sin{2\theta}$ and the blue rectangle has area $4\sin{\theta}$ so the area as a function of $\theta$ is

$A(\theta)=2\sin{2\theta}+4\sin{\theta}$

$A'(\theta)=4\cos{2\theta}+4\cos{\theta}$

Solving $A'(\theta)=0$

$4\cos{2\theta}+4\cos{\theta}=0$

$2\cos^{2}\theta+\cos{\theta}-1=0$

$\cos\theta = \frac{1}{2} or \cos\theta=-1$

$\theta=60 or 180$

$2\theta=120 or 360$

The solution that makes sense is $2\theta=\boxed{120 degrees}$

Note this picture is not the exact solution but it is close. The middle circle would have to be tangent to the line joining the other circles' centers.

The maximum and minimum angle measure of this are 180 and 60 respectively. So for the optimum area, it will be the median of these measures. Thus, (180+60)/2 = 120.

I did no math for this, but I already knew from rock climbing that a 120-degree angle in your anchor creates maximum tension (that's very bad). So 120 degrees was likely to be the most 'stretched' angle for the blue band.

Relevant wiki: Optimization

The light and dark gray areas are constant, so we must maximize the blue area.

The blue area is the area of this pentagon minus the three green sectors. If we set the circle radii to be 1, the area of a sector is half the angle, so we have:

$A_{green} = \frac{\theta + 2(\pi-\frac{\theta}{2})}{2} = \pi$

Therefore the problem reduces to maximizing the pentagon area, which is equivalent to maximizing one half as follows:

By basic trigonometry:

$\begin{aligned} b &= 2 \cos\frac{\theta}{2} \\ a^2 + b^2 &= 4 \\ b &= \sqrt{4-a^2} \\ \dfrac{\text{d}b}{\text{d}a} &= \frac{1}{2}\frac{1}{\sqrt{4-a^2}}(-2a) = -\frac{a}{b} \\ \end{aligned}$

Now the area to maximize is the area of the rectangle plus the area of the triangle and the solution is where the gradient with respect to $a$ is zero:

$\begin{aligned} A &= a + \frac{ab}{2} \\ &= a(1 + \frac{b}{2}) \\ \dfrac{\text{d}A}{\text{d}a} = 0 &= 1 + \frac{b}{2} + a(\frac{1}{2}\dfrac{\text{d}b}{\text{d}a}) \\ &= 1 + \frac{b}{2} - \frac{a^2}{2b} \\ &= 1 + \frac{b}{2} - \frac{4-b^2}{2b} \\ &= b + \frac{b^2}{2} - 2 + \frac{b^2}{2} \\ &= b^2 + b - 2 \\ &= (b-1)(b+2) \end{aligned}$

and since $0 \leq b \leq 2$ we have $b = 1$ . To confirm this is a maximum, we can easily show that the second derivative at $b=1$ is $-3\sqrt{3}$ .

We could now calculate $\theta$ using an $\arccos$ , however it's clear that the triangle above is the $1,2,\sqrt{3}$ triangle making $\frac{\theta}{2} = 60^\circ$ and $\boxed{\theta = 120^\circ}$ .

Instead of using geometry like Jeremy Galvagni , I used processing and plotted all the points for the three circles, then ran a convex hull calculator to calculate the convex hull around all three circles, filled in that shape with red, and then counted up all the red pixels. I did this for every angle, and then found the angle that gave the greatest number of red pixels.

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// http://leebyron.com/mesh/
import megamu.mesh.*;

// points to hull up
float[][] points = new float[360*3][2];
int currentPoint = 0;

// radius of the cirlce
int r=100;

// the angle in the middle
float ang=120;

int inc=-1;

void setup() {
size(600,600);

}

int tmax=0;
float rmax=0;

void drawCircle(float x, float y, float r, int start, int stop) {
  for (int i=0; i < 360; i++) {
      float x2 = r * cos(radians(i));
      float y2 = r * sin(radians(i));

      //ellipse(x+x2, y+y2, 3,3);
      points[currentPoint][0] = x + x2;
      points[currentPoint][1] = y + y2;
      currentPoint++;
  }
}

void draw() {
currentPoint = 0;
background(0);
translate(width/2, height/2);

ang += inc;
if (ang <=60) { inc=1;}
if ( ang>=180) {inc=-1;}


// 180 degreees total, subtract the angle we have set, then divide by two to find
// the side angles 
float angSide = -1 * (180-ang)/2;

// Circle 1
//ellipse(0,0, r*2,r*2);
drawCircle(0,0,r, 45,135);

// Circle 2
float c2x = 2 * r * cos(radians(angSide));
float c2y = 2 * r * sin(radians(angSide));

//println( c2x,c2y);
//ellipse(c2x, c2y, r*2,r*2);
drawCircle(c2x,c2y,r,270,360);
line(0,0, c2x,c2y);

// Circle 3
float c3x = 2 * r * cos( radians( 180-angSide) );
float c3y = 2 * r * sin( radians( 180-angSide) );
//ellipse(c3x, c3y, r*2,r*2);
drawCircle(c3x,c3y,r, 180,270);
line(0,0, c3x,c3y);

Hull myHull = new Hull( points );

MPolygon myRegion = myHull.getRegion();
fill(255,0,0);
myRegion.draw(this);

int total=0;
for (int x=0; x< width; x++) {
  for (int y=0; y< height; y++) {
    //println(red(get(x,y)));
    if (  red(get(x,y)) == 255 ) {
       total = total + 1;
    }
  }
}

//println( ang, total);

if (total > tmax) { 
  tmax=total; 
  rmax=ang; 
  println("new max:", total, ang);
}
}

Intuitively, I guessed that the area would be max when all the three sides are the same which would be when angle between the known vectors is $120^0$ .

Huh. I just figured that if the lowest areas correspond to 60 degrees and 180 degrees, then the largest area would be halfway between the two at 120.

Observe when more than 120 degree, area is decreasing.