The minimum rectangle area that can contain both triangle and circle.

Question

I got this problem during my interview and I would love to find a good answer to this.

Problem: Given a circle radius equal to r and an equilateral triangle each side has a length equal to a (as in the image below) image. Calculate the minimum rectangle area that can contain both a triangle and a circle.

• A circle can also contain a triangle and a triangle can also contain a circle inside.
• The lines of the triangle and the circle cannot overlap each other (They can touch each other).
• The triangle can rotate freely.

To solve this, I was thinking about 3 scenarios that could happen:

1. The circle is smaller than the inscribed circle of the triangle.
2. The circle is bigger than the circumscribed circle of the triangle.
3. The circle radius is bound between these values.

Here is my code to solve this

def my_function(a, r):
    
    diam = r*2
    h_tri = a*math.sqrt(3)/2
    inner_r = a*math.sqrt(3)/6
    outer_r = a*math.sqrt(3)/3

    # Scenario 1: circle is inside the triangle
    if r <= inner_r:
        return a*h_tri

    # Scenario 2: triangle is inside the circle
    elif r >= outer_r:
        return diam**2

    # Scenario 3: The circle radius is bound between these values.
    else:
        return "I don't know"

The last scenario is a very tricky one and probably my logic is wrong as well. Do you guys have a different approach to this problem?

Answer 1

In the absence of specific values for $r$ and $a$, or even of their size relationship, I think we can at least say the following.

I. If the triangle is within the circle, then the smallest rectangle that contains both is the square with area$$2r\times 2r=4r^2$$On the other hand, if the circle is within the triangle, then since $AB=AC=a$, and $CN=\frac{a\sqrt 3}{2}$, the smallest rectangle that contains both is$$AB\times CN=a\cdot\frac{a\sqrt 3}{2}=\frac{a^2\sqrt 3}{2}$$ II. But even when circle and triangle are external to one another, they can be contained by these same smallest rectangles, as in the two cases below, where in the lefthand figure the altitude of the triangle $$\frac{a\sqrt 3}{2}\le OJ-OP=r\sqrt2-r=r(\sqrt 2-1)$$and in the righthand figure where the radius of the circle$$r\le CH-CD=\frac{a}{2}-r\sqrt 3=\frac{a}{2+2\sqrt 3}$$

III. But generally, if triangle and circle are external to one another, and placed so that side $AC$ of the triangle, and side $AB$ extended, are tangent to the circle at $J$ and $F$, as in the next figure, then III,1): if $GC\le EH$, i.e. if $a\frac{\sqrt 3}{2}\le2r$, the height $EH$ of the smallest rectangle to contain them will be $2r$. And since $DFA$ is a $30^o-60^o-90^o$ triangle, and $DF=r$ and $DA=\frac{2r}{\sqrt 3}$, then$$FA=\sqrt{DA^2-DF^2}=\sqrt{\frac{4r^2}{3}-\frac{3r^2}{3}}=\sqrt{\frac{r^2}{3}}=\frac{r}{\sqrt 3}$$and the base of the rectangle$$EB=EF+FA+AB=r+\frac{r}{\sqrt 3}+a$$and hence the area of the smallest rectangle is$$EH\times EB=2r\left(r+\frac{r}{\sqrt 3}+a\right)=2r^2+\frac{2r^2}{\sqrt 3}+2ar$$

III,2: But if the height of the triangle exceeds the diameter of the circle, i.e. if $GC>2r$, as in this last figure, where $AC$ and $CH\parallel AB$ are tangent to the circle at $J$ and $F$, then the height of the smallest rectangle is$$EH=GC=\frac{a\sqrt3}{2}$$the base is$$HF+FC+CL=r+r\sqrt 3+\frac{a}{2}$$and the area$$EH\times HL=\frac{a\sqrt3}{2}\left(r+r\sqrt 3+\frac{a}{2}\right)=\frac{ar\sqrt3}{2}+\frac{3ar}{2}+\frac{a^2\sqrt 3}{4}$$