Kissing Circles! (Valentines Day)

Join the centers of the three pink, radius $r$ circles. This will form an equilateral triangle with side lengths $2r$ . The incenter of this triangle coincides with the center of the larger circle.

Now draw a radius from the center of the larger circle through the center of one of the pink circles to the point where these two circles are tangent. Since the distance from the incenter of an equilateral triangle to one of its vertices is $\frac{2}{3}$ the length of an altitude of the triangle, this radius will have length

$\dfrac{2}{3}*2r\sin(60^{\circ}) + r = 1 \Longrightarrow r(\dfrac{2}{\sqrt{3}} + 1) = 1 \Longrightarrow r = \dfrac{\sqrt{3}}{2 + \sqrt{3}} = \boxed{2\sqrt{3} - 3}$ .

Let O be the center of the bigger circle

Let r be the radius of one of the small circles

Let A, B and C be the centers of the three small circles

In triangle OAB

OA = OB = 1 - r

AB = 2 r

angle AOB = 120

Applying cosine law in triangle OAB

(AB)^2 = (OA)^2 + (OB)^2 - 2(OA)(OB) cos 120

4 r^2 = 2(1 - r)^2 - 2(1 - r)^2 (-1/2)

4 r^2 = 3(1 - r)^2

2 r = (square root of 3)(1 - r)

Then

r = 2 (square root of 3) - 3

$\frac {2} {\sqrt 3}r+r=1$

$r=2\sqrt 3-3$

$\text{THE SIMPLEST SOLUTION}$ (ONLY WORKING BECAUSE OF THE OPTIONS)

Since the radius of the larger circle is $1$ the options $C$ and $D$ will not work as they will give a value greater than $1$ . If we join the diameter of any 1 of the smaller circles and the center, we would see that the line will not be divided into a ratio of $1:1:1$ by the center of the smaller circle and the point at the circumference which touches this line. Thus option $A$ also won't satisfy. Hence we are only left with one option, that is, $B$

Kissing circles!!! That's quite a good interpretation William .

I see that you haven't entered your question into the Valentine's Day Themed Content Contest .

Do it quickly else you won't get enough votes, and Best of Luck !!

This sum consists of mixed trigonometry, equations & geometric assumptions. Enjoyed a lot solving it.

This solution is a bit funky. I got the idea for this problem after seeing Descartes Theorem for Kissing Circles (I wish I made that up but, Rene beat me to it). So here is the wiki: Click here

Using Curvatures: The Curvature of the larger external circle is $-1$ and the curvatures of the smaller congruent externally tangent circles will all be denoted by $x$ .

Plugging into truncated equation:

$k_{4}=k_{1}+k_{2}+k_{3}\pm 2\sqrt{k_{1}k_{2}+k_{2}k_{3}+k_{3}k_{1}}$

We get

$-1= x+x+x \pm 2\sqrt{ x^{2}+x^{2}+x^{2}}$

Simplifying the equation to solve for x we get

$-1= 3x \pm 2\sqrt{3x^2} \Longrightarrow -1 = 3x \pm 2x\sqrt{3} \Longrightarrow -1 = x(3 \pm 2\sqrt{3}) \Longrightarrow \frac{-1}{3 \pm 2\sqrt{3}} =x$

Just to clean it up a bit, I’ll multiply the top and bottom by $-1$ to get

$x=\frac{1}{-3 \pm 2\sqrt{3}}$

Finally, since x is the curvature, otherwise known as the reciprocal of the radius, we flip the fraction. We reject the negative case because we can’t have a negative radius and get the answer to be $\boxed{x=-3 + 2\sqrt{3}}$ .