Consider the following shape:

enter image description here

Three identical circles are tanged inside a circle.

Which is greater:

  1. The area of the shaded regions.
  2. Twice the area of the unshaded regions.

My attempt: I tried to solve it approximately. Let $R=2$ be the radius of the big circle, and its area is $4\pi$. The diameter of the smaller circles is about $R$ and their radius is $R/2=1$, the area of each small circle is $\pi$. So, the area of the shaded region is $3\pi$. The area of the unshaded region is about $4\pi - 3\pi =\pi$. Twice of it is about $2\pi$ which is smaller than $3\pi$, the area of the shaded region.

However, the correct answer is 2!

  • $\begingroup$ Let use en.m.wikipedia.org/wiki/Descartes%27_theorem $\endgroup$
    – user
    Feb 15 '18 at 8:55
  • $\begingroup$ How can the radius of the big circle be $\;R=2\;$ and the diameter of the smaller circles be "about" (?) $\;R(=2)\;$ ? That'd mean the three circles would intersect at the big circle's center...but they don't. $\endgroup$
    – DonAntonio
    Feb 15 '18 at 9:04
  • 3
    $\begingroup$ Take a peek at math.stackexchange.com/questions/666491/… . It is even poetic... $\endgroup$
    – DonAntonio
    Feb 15 '18 at 9:08
  • $\begingroup$ @DonAntonio As I said, I tried to make an approximation! $\endgroup$
    – Ahmad
    Feb 15 '18 at 9:46
  • 1
    $\begingroup$ @Ahmad Well, that seems like a crude approximation...why not $\;1.5\;$ , too? Or $\;1.7\;$ ? $\endgroup$
    – DonAntonio
    Feb 15 '18 at 9:54

Consider the image below:

enter image description here

Let $R$ be the radius of the big circle and $r$ the radius of the small circles.

If we connect the center of the small circles we get an equilateral triangle with side = $2r$. The center of the triangle is the center of the big circle as well.

The distance between the center of the triangle to the midpoint of one of its sides (aka apothem) is $h/3$, where $h$ is the height of the triangle. In our case, $h = r\sqrt{3}$ (we can easily calculate this with the pythagorean theorem).

As we can see in the image, we can now say that $R = r + \frac{2}{3}r\sqrt{3} = r \left(1 + \frac{2\sqrt{3}}{3}\right) \approx 2.1547r$.

$Area_{shaded} = 3\pi r^2 $

$Area_{bigcircle} = \pi R^2 = \pi (2.1547r)^2 \approx 4.6427\pi r^2$

$Area_{unshaded} = Area_{bigcircle} - Area_{shaded} \approx (4.6427 - 3) \pi r^2 \approx 1.6427 \pi r^2 $

Finally, we can check that $2 (Area_{unshaded}) \approx 3.2854 \pi r^2$ is bigger than $Area_{shaded} = 3 \pi r^2$


Let $r$ be a radius of the little circle and $R$ be a radius of the big circle.

Thus, $$\frac{2}{3}\cdot\frac{2r\sqrt3}{2}+r=R $$ and $$r=(2\sqrt3-3)R.$$ Now, we can see that $$3(2\sqrt3-3)^2<2\left(1-3(2\sqrt3-3)^2\right).$$


Let the distance between the centers of two small circles be $2\sqrt3$. The radius of the circle circumscribed to the equilateral triangle is thus $2$ and the radius of the large circle $2+\sqrt3$.

Hence, compare

$$3{\sqrt3}^2=9$$ and $$2\left(\left(2+\sqrt3\right)^2-3{\sqrt3}^2\right)=8\sqrt3-4.$$

We have





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