# Circles and triangles

Put 4 identical circles inside an equilateral triangle of side length 2, such that a circle touches 2 others and only one side of the triangle. What are the radii?

(ignore the vertical line)

• As stated, the configuration is not unique. Maybe a picture would help in understanding which configuration you are referring to. – Jack D'Aurizio Sep 21 '15 at 9:58
• It would also help if you would tell us where this problem comes from. – Gerry Myerson Sep 21 '15 at 10:00
• It came from an 1st year undergrad worksheet. The question is phrased exactly like that, im guessing the answer will still be the same no matter which config. – minusatwelfth Sep 21 '15 at 10:04
• It should only require high school level theorems – minusatwelfth Sep 21 '15 at 10:31
• The radius $r$ does depends on the configuration. In fact, we have $$\frac{\sqrt{3}}{5+\sqrt{3}} \le r < \frac{\sqrt{3}}{3+2\sqrt{3}}$$ The minimal value of $r$ is achieved when the centers of the circles are stacking up to form a square. The upper bound is not achievable. As $r$ approaches the upper bound, the centers of the circles approaches a rhombus consisting of two equilateral triangles. – achille hui Sep 21 '15 at 13:42

For simplicity of description, let us rescale everything so that the $4$ circles have radius $1$ and the equilateral triangle has side $R = \frac{2}{r}$.

Since each of the circles are touching two other circles, their centers need to form a rhombus with side $2$ with angles $\frac{\pi}{2} \pm \theta$ where $0 \le \theta < \frac{\pi}{6}$.

Since a triangle has $3$ sides and each of the circles are touching one of the sides, two of the circles need to touch the same edge. By a suitable choice of coordinate system, we can assume that edge lying on the $y$-axis and the mid-point between the centers of that two circles lies on the $x$-axis.

Flipping the direction of $y$-axis if necessary, we can position the $4$ centers at

$$(1, 1), (1, -1), (1 + 2\cos\theta, 1 + 2\sin\theta) \;\;\text{ and }\;\; (1 + 2\cos\theta, -1 + 2\sin\theta)$$ Following picture illustrate where the circles go inside the triangle.

To compute $R$, one first look for points on the union of $4$ circles which maximize following two expression

$$\cos\frac{\pi}{3} x +\sin\frac{\pi}{3} y \quad\text{ and }\quad \cos\frac{\pi}{3} x - \sin\frac{\pi}{3} y$$ For the first expression, it is maximized at the perimeter of the upper right circle

$$(x_1,y_1) = ( 1 + 2\cos\theta + \cos\frac{\pi}{3}, 1 + 2\sin\theta + \sin\frac{\pi}{3} ) = ( \frac{3}{2} + 2\cos\theta, \frac{2 + \sqrt{3}}{2} + 2\sin\theta )$$ The tangent line passing that point has the parametrization:

$$\mathbb{R} \ni t \mapsto (x,y) = (x_1 - t, y_1 + \frac{t}{\sqrt{3}})$$

and it hit the $y$-axis at $(0,y_1')$ where $y_1' = y_1 + \frac{1}{\sqrt{3}} x_1$.

For the second expression, it is maximized at the perimeter of the lower right circle.

$$(x_2,y_2) = ( 1 + 2\cos\theta + \cos\frac{\pi}{3}, -1 + 2\sin\theta + -\sin\frac{\pi}{3} ) = ( \frac{3}{2} + 2\cos\theta, -\frac{2 + \sqrt{3}}{2} + 2\sin\theta )$$ The tangent line passing that point has the parametrization:

$$\mathbb{R} \ni t \mapsto (x,y) = (x_2 - t, y_2 - \frac{t}{\sqrt{3}})$$

and it hit the $y$-axis at $(0,y_2')$ where $y_2' = y_2 - \frac{1}{\sqrt{3}} x_2$.

Together with the $y$-axis, these two tangent lines will bound the circles in an equilateral triangle with side length

$$\frac{2}{r} = R = y_1' - y_2' = 2 + \sqrt{3} + \frac{2}{\sqrt{3}}\left(\frac{3}{2}+2\cos\theta\right) = \frac{2}{\sqrt{3}}\left( 3 + \sqrt{3} + 2\cos\theta \right)\\ \implies r = \frac{\sqrt{3}}{ 3 + \sqrt{3} + 2\cos\theta }$$

Since $0 \le \theta < \frac{\pi}{6}$, this leads to

$$\color{blue}{\text{ corr. to } \theta = 0} \rightarrow\quad \frac{\sqrt{3}}{5 + \sqrt{3}} \le r < \frac{\sqrt{3}}{3 + 2\sqrt{3}} \quad\leftarrow \color{blue}{\text{ corr. to } \theta \approx \frac{\pi}{6}}.$$ Please note that when $\theta = 0$, the centers of the 4 circles forms a square and this is probably the configuration your question ask about.

• why must that expression be maximised? – minusatwelfth Sep 21 '15 at 17:14
• @minusatwelfth you already know one edge of the triangle, the $y$-axis. So the outward normals of the other two edges are pointing in the directions $(\cos\frac{\pi}{3},\sin\frac{\pi}{3})$. The points where the circles touch the triangle are the points where the expressions are maximized. One way to visualize this is consider a orthogonal projection onto a line perpendicular to the edge. The point which maximize that expression corresponds to the boundary of the "shadow".... – achille hui Sep 21 '15 at 17:20
• I'm not quite sure I understand. it's a dot product? – minusatwelfth Sep 21 '15 at 17:47
• @minusatwelfth yup, it is a dot product. – achille hui Sep 21 '15 at 18:06
• $$(x_1 - t, y_1 + \frac{t}{\sqrt{3}})$$ i.e. $$(- t, \frac{t}{\sqrt{3}})$$ is not orthogonal to $$t(\cos\frac{\pi}{3}, \sin\frac{\pi}{3})$$ by my calculation. Their dot product is not zero. Should it be $$(x_1 - t, y_1 + \frac{\sqrt{2}t}{\sqrt{3}})$$ instead? – minusatwelfth Sep 21 '15 at 18:21