# Equilateral triangle with vertices on 3 concentric circles

Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?

My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ I will manipulate the formula afterwards,,,

• What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle? Commented Mar 15, 2019 at 15:50
• I have changed your singular title (pun intended) "Equilateral triangle on a concentric circle" into its plural version. No objection ? Commented May 1, 2020 at 16:54

Using the construction that @Michael Rozenberg suggested

I will leave the following exercise for you (which isn't that hard)

Prove that the quadrilateral $$ABCD$$ is cyclic.

Thus $$\angle BDC=180°-\angle CAB=120°$$. In virtue of the law of Cosines $$\begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·\cos(\angle BDC)\\ &=1+4-2·1·2·(-0.5)\\ &=5+2=7 \end{array}$$

• Nice! I thought about algebraic solution only. Commented Mar 15, 2019 at 17:51

The hint.

Take $$A$$ on the biggest circle and rotate the smallest circle by $$60^{\circ}$$ around $$A$$.

Now, take an intersection point $$B$$ with the middle circle.

Thus, $$AB$$ is a side of the needed triangle.

I took $$A(-3,0)$$ and got $$AB=\sqrt7.$$

I'll use the process of Dr. Mathva in a different way.

We'll first prove $$ABCD$$ is cyclic.

Let $$AB=AC=BC=s$$

$$DC=1, DB=2, DA=3$$

We see that $$AB×DC+BD×AC=1s+2s=3s=AD×BC$$

By Converse of Ptolemy's theorem, we conclude that $$ABCD$$ is cyclic.

After this you can find out a through pure trigonometric means. I got $$s = \sqrt{7}$$

Following the notations of the figure by @Dr. Mathva, knowing that $$AD=3, BD=2, CD=1,$$ and setting $$x=AB=AC=BC$$, let us express that the Cayley-Menger determinant of the four points $$A,B,C,D$$ is zero (see here):

$$\begin{vmatrix}0&x^2&x^2&3^2&1\\ x^2&0&x^2&2^2&1\\ x^2&x^2&0&1^2&1\\ 3^2&2^2&1^2&0&1\\ 1&1 &1 &1& 0\end{vmatrix}=0\tag{1}$$

Expanding and factoring, one obtains :

$$-2x^2(x^2 - 7)^2=0\tag{2}$$

therefore with unique "geometrically acceptable" answer $$x=\sqrt{7}$$.

Generalization, with radii $$a, b, c$$.

$$\begin{vmatrix}0&x^2&x^2&c^2&1\\ x^2&0&x^2&b^2&1\\ x^2&x^2&0&a^2&1\\ c^2&b^2&a^2&0&1\\ 1&1 &1 &1& 0\end{vmatrix}=0\tag{2}$$

giving, by setting $$X=x^2$$, the equation :

$$2X(X^2 - sX+ \underbrace{(s^2-3r)}_q)=0 \ \text{where} \ r:=a^2b^2+a^2c^2+b^2c^2, \ s:=a^2+b^2+c^2\tag{3}$$

which, setting apart solution $$X=0$$, is a quadratic equation. Therefore, conditionned by the following constraint on the sign of its discriminant:

$$\Delta=3(4r-s^2)\ge 0\tag{4}$$

warranting real solutions.

$$\Delta$$ possesses a remarkable factorization:

$$\Delta=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\tag{5}$$

As a consequence, when $$\Delta > 0$$, $$\sqrt{\Delta}$$ is 4 times the area of the triangle with sides $$a,b,c$$, due to Heron formula.

The signs of the roots are governed by the sign of $$q$$ which is in fact always positive, meaning that there are two possible sidelengths for the equilateral triangle if $$\Delta >0$$,

$$x=\sqrt{\frac12 (s \pm \sqrt{\Delta})}\tag{6}$$

Graphical representation: We can, without loss of generality, work in a plane with coordinates $$(a/c,b/c)$$ or said otherwise in a plane with coordinates $$(a,b)$$ with $$c=1$$. Therefore, equation $$\Delta=0$$ becomes the equation of a curve... which is plainly the union of four straight lines, 3 of them appearing in the first quadrant. Here is the result of a simulation : in green, resp. red: values of $$(a/c, b/c)$$ giving a (double) solution, resp. not yielding a solution.

Different remarks:

1. 3 lines are materialized in this simulation : $$a+b-1=0, a-b+1=0, and -a+b+1=0$$

2. the particular case $$a=3, b=2, c=1$$ considered in the first part corresponds to a limit case between the green and the red zone.

3. Here is a case where we have two solutions $$a=1.8, \ b=1.5, \ c=1$$ with two different values of sidelength $$x$$.

Prompted by an additional point in a duplicate question asking whether the construction is always possible for arbitrary concentric circles, the following is about the general case of radii $$\,A,B,C\,$$ (where OP's question here is about $$\,A=1,B=2,C=3\,$$).

Let $$\,a,b,c\,$$ be points in the complex plane on the respective circles, and assume WLOG that $$\,a=A \in \mathbb R\,$$, since the problem is invariant to rotations. Let $$\,\omega\,$$ be a complex cube root of unity, then $$\,\omega^2+\omega+1=0\,$$, and the condition for the triangle $$\,a,b,c,\,$$ to be equilateral is:

$$c-a + \omega(b-a)=0 \;\;\iff\;\; c = - \omega b + (1+\omega)a \tag{1} \;\;\iff\;\; c = - \omega b - \omega^2a$$

Multiplying by the conjugate, and using that $$\,\omega\overline\omega=|\omega|^2=1\,$$, $$\,\overline\omega=\omega^2\,$$:

\begin{align} |c|^2 &= |b|^2+|a|^2 + a \left(\omega\overline\omega^2b+\overline\omega\omega^2\overline b\right) = |b|^2+|a|^2 + 2a \,\text{Re} \left(\omega^2 b\right) \end{align}

It follows that:

$$\text{Re} \left(\omega^2 b\right) \,=\,\frac{|c|^2-|a|^2-|b|^2}{2a}=\frac{C^2-A^2-B^2}{2A} \tag{2}$$

Since $$\,\left|\omega^2 b\right|=|b|\,$$, the condition for existence of a solution is:

$$\left|\text{Re} \left(\omega^2 b\right)\right| \le |b| \quad\iff\quad \left(C^2-A^2-B^2\right)^2 \le 4 A^2 B^2 \tag{3}$$

In that case:

$$\text{Im} \left(\omega^2 b\right) \,=\,\pm \sqrt{|b|^2 - \left(\text{Re} \left(\omega^2 b\right)\right)^2}=\pm\sqrt{B^2-\left(\frac{C^2-A^2-B^2}{2A}\right)^2} \tag{4}$$

When a solution exists according to $$\,(3)\,$$, equations $$(2),(4)$$ give $$\,b\,$$ (up to a reflection about the real axis), then $$\,(1)\,$$ gives $$\,c\,$$.

Quick verification for OP's case $$\,a=A=1, B=2, C=3\,$$, which satisfies $$\,(3)\,$$ so solutions exist.

• From $$\,(2)\,$$: $$\;\text{Re}\left(\omega^2b\right)=\frac{9-1-4}{2}=2\,$$.

• From $$\,(4)\,$$: $$\;\text{Im}\left(\omega^2b\right)=\pm\sqrt{4-2^2}=0\,$$.

• Then $$\,\omega^2 b=\text{Re}(\omega^2b)+i\,\text{Im}(\omega^2b)=2\,$$, so $$\,b = 2\omega\,$$.

• From $$\,(1)\,$$: $$\;c=-\omega\cdot 2 \omega - \omega^2 \cdot 1 =-3\omega^2\,$$.

Therefore the solution is $$\,\big\{\,1, \,2 \omega, \,-3\omega^2 \,\big\}\,$$ up to rotations and reflections, and it can be easily verified that the side length of the equilateral triangle is $$\,|b-1|=\sqrt{7}\,$$.

[ EDIT ] $$\;$$ Condition $$\,(3)\,$$ can be written as $$\,\left(2AB\right)^2 - \left(A^2+B^2-C^2\right)^2 \ge 0\,$$ and, after factoring the differences of squares twice, is equivalent to:

$$(A+B+C)(A+B-C)(A-B+C)(-A+B+C) \ge 0 \tag{3'}$$

This matches $$\,(5)\,$$ in Jean Marie's answer and, when $$\,A \le B \le C\,$$, is equivalent to $$\,C \le A + B\,$$ which matches the answer to another related question linked in Jean Marie's comment.

• An old reference coping very clearly with the general case here to which I have just added a new graphics. Commented Apr 12, 2022 at 22:39
• @JeanMarie Nice find, thanks for the pointer. I edited my answer and added a short note at the end.
– dxiv
Commented Apr 12, 2022 at 23:39

While the posted geometric solutions are much simpler, it is possible to do this with algebra and coordinate geometry.

Centering the circles at the origin, we get the equations that you provided: $$x^2+y^2=1$$ $$x^2+y^2=4$$ $$x^2+y^2=9$$ Let's choose an arbitrary point on the smallest circle, say $$(0, 1)$$ for simplicity. Let $$l$$ be the length of each side of the equilateral triangle. So the vertices on the other two circles must be a distance of $$l$$ from our chosen point $$(0, 1)$$. Equivalently, the two vertices must be on the circle with radius $$l$$ centered at $$(0, 1)$$ We can set up an equation to represent this: $$x^2+(y-1)^2=l^2$$

Red is the circle of radius 1, Blue is the circle of radius 2, Green is the circle of radius 3, Dotted Black is the circle centered at $$(0, 1)$$ with radius $$l$$.

Finding the intersection of this circle with the other two circles, we get the following two equations to represent the vertices: $$x^2+y^2-4=x^2+(y-1)^2-l^2$$ $$x^2+y^2-9=x^2+(y-1)^2-l^2$$ Solving the equations for $$y$$, we get the following. $$y_1$$ is the y-coordinate of the vertex on the circle of radius 2, and $$y_2$$ is the y-coordinate of the vertex on the circle of radius 3: $$y_1=\frac{5-l^2}{2}$$ $$y_2=\frac{10-l^2}{2}$$ We can plug this into their respective equations to find the x-coordinates: $$x_1=\sqrt{4-\left(\frac{5-l^2}{2}\right)^2}$$ $$x_2=\sqrt{9-\left(\frac{10-l^2}{2}\right)^2}$$ These coordinates are a distance of $$l$$ from the point on the smallest circle. It now remains to make these two points a distance of $$l$$ from each other: $$l=\sqrt{\left(\sqrt{9-\left(\frac{10-l^2}{2}\right)^2}-\sqrt{4-\left(\frac{5-l^2}{2}\right)^2}\right)^2+\left(\frac{10-l^2}{2}-\frac{5-l^2}{2}\right)^2}$$ Solving this equation for $$l$$ yields the answer of $$l=\sqrt{7}$$

One can prove that the quadrilateral $$ABDC$$ is cyclic using Ptolemy’s Theorem. But how to prove it if one does not know that theorem? Well, here is one way to go.

Let $$E$$ be a point at the same side of $$AD$$ as $$C$$, such that $$\triangle ADE$$ is equilateral. Then $$\triangle ACE$$ and $$\triangle ABD$$ are congruent because of SAS: $$|AC| = |AB|$$, $$\angle BAD = 60° - \angle DAC = \angle CAE$$, and $$|AD| = |AE|$$. Therefore, $$|CE|=|BD|=2$$ and $$\angle BDA = \angle CEA$$. Because $$|DE|=3$$, $$|CE|=2$$ and $$|CD|=1$$, $$C$$ lies on $$DE$$. Hence, $$\angle CEA = \angle DEA = 60°$$ and therefore also $$\angle BDA = 60°$$. Because $$\angle BDA = 60° = \angle BCA$$, the quadrilateral $$ABDC$$ is cyclic.