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In the diagram below is an equilateral triangle with side length of 1 unit. $C_1$, $C_2$ and $C_3$ are circles inside the triangle tangent to each other and the sides of the triangle. Find the radius of each circle.

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Firstly, I'm inclined to think that the circles must all be equal in size, but I'm not sure how to prove that. And I also tried making a smaller triangle inside the outer triangle by connecting the radii of the circles, but I'm not sure how to proceed after that (or if I'm even on the right track for that matter).

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By construction you have $$1=\sqrt 3r+r+r+\sqrt 3r\Rightarrow r=\frac{\sqrt3-1}{4}\approx 0.183012701$$ Later add an explanatory figure.

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Make a right triangle by drawing the segment connecting a vertex of your triangle to the center of the nearest circle, and dropping the perpendicular from that circle. As this is a $30-60-90$ triangle we see that the leg along the triangle side has length $\sqrt 3 \,r$. Inspection quickly shows that $$1=2\sqrt 3\, r+2r\implies r = \frac {\sqrt3 -1}4$$

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  • $\begingroup$ by r I'm guessing you mean the radius of the circles? But could you explain how the length would be √3 r? I thought it would be 1/2 instead because what you did was dropping a perpendicular bisector to the base. $\endgroup$ – space Sep 19 '16 at 15:37
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    $\begingroup$ Yes, $r$ is the radius. The hypotenuse, the segment from vertex to center, is the angle bisector (clearly) so the right triangle is $30-60-90$. Now, $\tan(30)=\frac 1{\sqrt 3}=\frac rx$ where $x$ is the leg along the triangle base. It follows that $x=\sqrt 3 r$ $\endgroup$ – lulu Sep 19 '16 at 15:40
  • $\begingroup$ Just to say, the other posted solution (from @Piquito) is the same as mine, but comes with a very helpful illustration. $\endgroup$ – lulu Sep 19 '16 at 15:44
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Nobody says the circles have the same radii. One can prove it and that's the point of this problem. The condition is that the three circles are in the equilateral triangle (because there are configurations in which one circle can be outside and it's bigger than the other two, but still touches both and each circle touches a pair of "extended" edges of the triangle!).

Proof 1. One way is to use the uniqueness of solution to Apollonius' problem. Indeed, let us assume that the three circles $c_1, c_2, c_3$ are in the equilateral triangle $A_1A_2A_3$ and each circle $c_i$ is simultaneously tangent to the edges $A_iA_j$ and $A_iA_k$ for $i\neq j\neq k \in \{1,2,3\}$. Draw the angle bisector $A_3H_3\,\,$ ($H_3 \in A_1A_2$), which is also a line of reflection symmetry of triangle $A_1A_2A_3$ ($A_3H_3$ is also the median, the altitude and the orthogonal edge bisector to the edge $A_1A_2$). Then the center $O_3$ of circle $c_3$ lies on $A_3H_3$ and if we perform a reflection with respect to $A_3H_3$ the triangle $A_1A_2A_3$ is mapped to itself and $c_3$ is mapped to itself. Now since $c_1$ is tangent to circle $c_3$ and edges $A_1A_2$ and $A_1A_3$, which after reflection are mapped to $c_3, $ $A_2A_1$ and $A_2A_3$, the reflective image of $c_1$ is a circle $c_1'$ which is tangent to circle $c_3$ and edges $A_2A_1$ and $A_2A_3$ and is inside the triangle. Now, by the uniqueness of the solution to Apollonius' problem applied to circle $c_3$ and edges $A_2A_1$ and $A_2A_3$, there is only one circle tangent to $c_3$, $A_2A_1$ and $A_2A_3$ and located inside the triangle and since both $c'_1$ and $c_2$ are solutions to Apollonius' problem with these conditions, $c'_1 = c_2$, so $c_2$ is the reflective image of $c_1$ under the reflection with respect to $A_3H_3$. Consequently, the radius $r_1$ of $c_1$ and the radius $r_2$ of $c_2$ are equal, $r_1 = r_2$. Now, carry out the same argument but with respect to vertex $A_2$ and axis of symmetry $A_2H_2$. That would lead to the conclusion that the radius $r_1$ of $c_1$ and the radius $r_3$ of $c_3$ are equal, and consequently $r_1 = r_2=r_3$.

Proof 2. Assume the radii of the three circles are $r_1, r_2, r_3 > 0$. Since the circles are inside the triangle, $r_1, r_2, r_3 < 1$. Take for instance the two tangent circles $c_1$ and $c_2$ of radii $r_1$ and $r_2$ and centers $O_1$ and $O_2$ respectively. Let $T_1$ and $T_2$ be the points of tangency of $c_1$ and $c_2$ twith the triangle edge $A_1A_2$ tangent to both of them simultaneously. Then $O_1T_1$ and $O_2T_2$ are perpendicular to $A_1A_2$ and are therefore parallel. We have a trapezoid $T_1T_2O_2O_1$ with $T_1T_2$ orthogonal to the parallel $O_1T_1$ and $O_2T_2$. Moreover, $|O_1T_1| = r_1, \,\, |O_2T_2| = r_2$ and $|O_1O_2| = r_1 + r_2$. Then by drawing a line through say $O_2$ parallel to $T_1T_2$, we obtain a right angled triangle with hypotenuse $|O_1O_2| = r_1+r_2$ and another edge of length $|r_1-r_2|$. By Pythagoras' theorem, $$|T_1T_2| = \sqrt{(r_1+r_2)^2 - (r_1-r_2)^2} = 2 \sqrt{r_1r_2}.$$ Analogously, $$|T_2T_3| = \sqrt{(r_2+r_3)^2 - (r_2-r_3)^2} = 2 \sqrt{r_2r_3}.$$ $$|T_3T_1| = \sqrt{(r_3+r_1)^2 - (r_3-r_1)^2} = 2 \sqrt{r_3r_1}.$$ The two tangents from $A_1$ to the circle $c_1$ (on of them is $AT_1$) have equal lengths $\sqrt{3}\, r_1$ because triangle $AT_1O_1$ is right angled with angle $\angle \, T_1A_1O_1 = 30^{\circ}$, and since $|A_1O_1| = 2 \, r_1, \, |O_1T_1| = r_1$ and hence $|A_1T_1| = \sqrt{3}\, r_1$. Similarly, for the other vertices of the triangle, the common tangents from the given vertex and the corresponding closest circle have lengths respectively $ \sqrt{3}\, r_2$ and $ \sqrt{3}\, r_3$. Thus the length of the three edges of the equilateral triangle (they are all equal and of length $1$) can be expressed as \begin{align} \sqrt{3}(r_1 + r_2) + 2 \sqrt{r_1\, r_2} &= 1\\ \sqrt{3}(r_2 + r_3) + 2 \sqrt{r_2\, r_3} &= 1\\ \sqrt{3}(r_3 + r_1) + 2 \sqrt{r_3\, r_1} &= 1 \end{align} To solve this system, set $r_2 = x_2^2 \, r_1$ and $r_3 = x_3^2 \, r_1$ and plug in the equations these two expression to arrive at \begin{align} \sqrt{3}(1 + x^2_2) + 2 \, x_2 &= \frac{1}{r_1}\\ \sqrt{3}(x^2_2 + x^2_3) + 2 \, {x_2\, x_3} &= \frac{1}{r_1}\\ \sqrt{3}(1 + x^2_3) + 2 \, x_3 &= \frac{1}{r_1} \end{align} This is a system that simultaneously finds $r_1$ and $x_2, \, x_3$. Look at the polynomial $$f(x) = \sqrt{3}\, x^2 + 2 \, x + \sqrt{3} - \frac{1}{r_1}$$ Then the first and the third equations from above are actually $$f(x_2) = 0 \,\,\, \text{ and } \,\,\, f(x_3) = 0$$ i.e. $x_2$ and $x_3$ are two roots (different or the same) of the polynomial $f$. Consequently \begin{align} f(x_2) = \sqrt{3}\, x^2_2 + 2 \, x_2 + \sqrt{3} - \frac{1}{r_1} &=0\\ \sqrt{3}\big((x_2 + x_3)^2 - 2\, x_2x_3\big) + 2 \, {x_2\, x_3} &= \frac{1}{r_1}\\ f(x_3) = \sqrt{3}\, x^2_3 + 2 \, x_3 + \sqrt{3} - \frac{1}{r_1} &=0 \end{align} or in another form \begin{align} f(x_2) = \sqrt{3}\, x^2_2 + 2 \, x_2 + \sqrt{3} - \frac{1}{r_1} &=0\\ \sqrt{3}(x_2 + x_3)^2 + (2- 2\sqrt{3})\, x_2 \, x_3 &= \frac{1}{r_1}\\ f(x_3) = \sqrt{3}\, x^2_3 + 2 \, x_3 + \sqrt{3} - \frac{1}{r_1} &=0 \end{align} Case1: Assume $x_2$ and $x_3$ are the two roots of $f$ (could be that $x_2\neq x_3$ or could be that $x_2=x_3$ is a double root). By Vieta's formulas $$x_2+x_3 = - \, \frac{2}{\sqrt{3}} \,\,\,\,\, x_2\, x_3 = 1- \frac{1}{\sqrt{3} \, r_1}$$ so if we plug them in the second equaiton $$\frac{4 \sqrt{3}}{3} + (2- 2\sqrt{3}) - (2- 2\sqrt{3})\frac{1}{\sqrt{3} \, r_1} = \frac{1}{r_1}$$ $$2 - \frac{2 \sqrt{3}}{3} + \left(2 - \frac{2 \sqrt{3}}{3}\right)\frac{1}{ r_1} = \frac{1}{r_1}$$ $$1 + \frac{1}{r_1}=\frac{3}{6-2\sqrt{3}} \, \frac{1}{r_1}$$ $$1 = \frac{3}{6-2\sqrt{3}} \, \frac{1}{r_1} - \frac{1}{r_1} = \left( \frac{3}{6-2\sqrt{3}} - 1\right) \, \frac{1}{r_1}$$ when you solve for $r_1$ one gets $$r_1 = \frac{\sqrt{3}-1}{4}.$$ The solutions $x_2$ and $x_3$ are different, one of them is, let's say $x_2$ is $1$ and the other, $x_3$, is greater than $1$ in absolute value, which leads to $r_3>1$. This is not a solution to our configuration.

Case 2: $x_2 = x_3 = x$. Then the equations reduce to \begin{align} f(x) = \sqrt{3}\, x^2 + 2 \, x + \sqrt{3} - \frac{1}{r_1} &=0\\ (2 \sqrt{3} + 2 )\, x^2 - \frac{1}{r_1} &= 0 \end{align} which leads to the equation $$(\sqrt{3} + 2) \, x^2 - 2\, x - \sqrt{3} = 0$$ and it is easy to see that $x=1$ is a solution. Then the other solution is $x^* = - \,\frac{\sqrt{3}}{\sqrt{3} + 2}$.

If $x=1$ then plug it in the second equation of the system above. This leads to $$r_3 = r_2 = r_1 = \frac{\sqrt{3}-1}{4}$$ If $x=x^*$, then one sees that $r_1 > 1$ which is not a solution to our problem, as then circle $c_1$ should not be inside the triangle $A_1A_2A_3$.

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Here is a simple proof that three circles, each of them touching two sides of the triangle $\triangle$ from the inside, necessarily have the same radius:

Each circle has its center on a symmetry axis of the triangle. Given such a circle $\gamma_c$ with its center on $m_c$, and at least so large that $\gamma_c$ intersects the incircle of $\triangle$, the radius $r_a$ of a circle having its center on $m_a$ that touches $\gamma_c$ from the outside is uniquely determined, and due to symmetry with respect to the line $m_c$ it is equal to the radius $r_b$ of the circle having its center on $m_b$ that touches $\gamma_c$. From $r_a=r_b$ it follows that all three radii are equal.

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  • $\begingroup$ Your proof is a bit of a mess. Only because a circle touches another one and its center lies on a symmetry axis doesn't make it unique. Even the stronger statement, which is more crucial, that a circle touches another one from the configuration and two of the edges of the triangle doesn't make it unique. Apollonius' problem has several solutions and definitely not only one in this case. Only the fact that the circle is inside the triangle makes it unique. $\endgroup$ – Futurologist Sep 23 '16 at 20:41
  • $\begingroup$ @Futurologist: This has nothing to do with Apollonius' problem. It's about three circles with disjoint interiors, contained in an equilateral triangle. – I suggest you leave out the m-word. $\endgroup$ – Christian Blatter Sep 24 '16 at 7:54
  • $\begingroup$ It has to do with a circle tangent to one circle and two lines. That is part of Apollonius' problem. When you reflect one of the circles in another symmetry axis how do you justify its image is one of the other circle? What is the argument? $\endgroup$ – Futurologist Sep 24 '16 at 12:36

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