# How to construct an equilateral triangle on 2 concentric circles

Construct an equilateral triangle with the given vertex so that the other vertices lie on the concentric circles respectively.

I constructed the triangle, but I don't know how it works. How does this construction work? Is there any proof?

My construction.

Let the smaller circle be $$a$$, the larger circle $$b$$, and the point $$c$$.

• Step 1: Construct a circle with radius of $$b$$ at the point $$c$$.

• Step 2: The circle will intersect circle $$a$$ at $$2$$ points. Let the two points be $$x$$ and $$y$$. Construct a perpendicular bisector of the line connecting $$x$$ and the common centre of circle $$a$$ and $$b$$.

• Step 3: The bisector intersect the circle $$a$$ at a point which is another vertex of the equilateral triangle.

For more context, this is from the game "Euclidea" level 13.3. Video solutions can be found here.

• How did you construct the one in the diagram? Did you use GeoGebra?
– user754135
Commented May 1, 2020 at 15:35
• How did you construct it?
– user754135
Commented May 1, 2020 at 15:37
• Why don't you answer to the questions ? Commented May 1, 2020 at 16:35
• Why don't you want to have a dialog with us ? Why do you want solutions "served on a tray" ? I have been looking at your 5 previous questions. You always give the raw text of your homwork, then write a single sentence, ending by something like "i don't know how to do it.how it works.". This is not the way you will progress in mathematics... Commented May 1, 2020 at 16:45
• I recognize this game...is it Euclidea Commented May 3, 2020 at 9:16

Task. Given a point $$P$$ on the plane and two (not necessarily distinct and not necessarily concentric) circles $$c$$ and $$k$$, construct an equilateral $$PAB$$ such that $$A$$ is a point of $$c$$ and $$B$$ is a point of $$k$$.

Construction. Denote by $$c'$$ and $$k'$$ the images of $$c$$ and $$k$$, respectively, under the counterclockwise rotation about $$P$$ by $$\dfrac{\pi}{3}$$. Suppose that $$c$$ meets $$k'$$ at $$A$$ and $$A'$$, and that $$c'$$ meets $$k$$ at $$B''$$ and $$B'''$$. Let $$B$$, $$B'$$, $$A''$$, and $$A'''$$ be the images of $$A$$, $$A'$$, $$B''$$, and $$B'''$$ under the clockwise rotation about $$P$$ by $$\dfrac{\pi}{3}$$. Then, $$PAB$$, $$PA'B'$$, $$PA''B''$$, and $$PA'''B'''$$ are equilateral triangles. The number of such triangles can be $$0$$, $$1$$, $$2$$, $$3$$, and $$4$$, depending how $$c$$ and $$k$$ intersect $$c'$$ and $$k'$$.

Explaination. If $$PAB$$ is a desired triangle, then $$A$$ is the image of counterclockwise rotation about $$P$$ by $$\theta\in\left\{-\dfrac{\pi}{3},+\dfrac{\pi}{3}\right\}$$. If $$\theta=+\dfrac{\pi}{3}$$, then clearly, $$A$$ is a point of intersection between $$c$$ and $$k'$$. If $$\theta=-\dfrac{\pi}{3}$$, then $$B$$ is the point of intersection between $$c'$$ and $$k$$.

The OP's construction works when the two circles are concentric. I have not yet found out why. If I know the answer, I will come back to give a proof. For now, I attach a figure showing that the OP's steps do lead to a correct construction.

• @endgameendgame I am not sure how to help you with "getting the idea." I have seen a similar solution before (it was a different problem), so the use of rotation was the first thing that came into my mind. Commented May 3, 2020 at 10:31
• Thanks for the reply Commented May 3, 2020 at 10:36
• @Batominovski I explained in my answer why the OP's constructions works. Commented Jun 19, 2020 at 9:19

As shown in one of the other answers, to find another vertex of the equilateral triangle one can rotate the outer circle $$b$$ by $$60°$$ about the given vertex $$P$$: each intersection between the rotated circle and the inner circle $$a$$ is then a possible second vertex of the equilateral triangle to be constructed.

The construction you found works because, instead of rotating $$b$$ about $$P$$ by $$60°$$ counterclockwise, we can obtain the same result by rotating a circle equal to $$b$$ but centred at $$P$$ by $$60°$$ clockwise about the common center $$O$$ of $$a$$ and $$b$$.

In figure below, the red circle is obtained by rotating circle $$b$$ about $$P$$ by $$60°$$ counterclockwise; its intersection $$A'$$ with circle $$a$$ is the second vertex of the equilateral triangle to be constructed.

But we can also find the red circle by constructing first a circle equal to $$b$$ centred at $$P$$ (blue circle in the figure) and then rotating it about $$O$$ by $$60°$$ clockwise. Point $$A'$$ can then be quickly obtained by rotating point $$A$$ (the intersection between $$a$$ and blue circle) by $$60°$$ clockwise about $$O$$: as triangle $$AOA'$$ is equilateral, $$A'$$ is thus the intersection between $$b$$ and the perpendicular bisector of $$OA$$.

COMMENT.-An easily verifiable fact is that given two points $$Q, R$$ one in each circle, there is always a point $$P$$ such that the triangle $$\triangle PQR$$ is equilateral, but another problem is the one in which the point $$P$$ occurs and the points to be determined are $$Q$$ and $$R$$. Look just at first glance an analytical solution.

Being $$R$$ and $$r$$ the radius and the point $$P=(a,0)$$ we have for the two circles and the three equal distances the four equations determining the points $$Q=(x,y)$$ and $$R=(w,z)$$ in both circles

$$x^2+y^2=r^2\\z^2+w^2=R^2\\(x-a)^2+y^2=(z-a)^2+w^2=(x-z)^2+(y-w)^2$$

We have $$z=B+x$$ where $$B=\dfrac{R^2-r^2}{2a}$$ and $$w=\dfrac{C+2Dx-2x^2}{2y}$$ where $$C=R^2-a^2$$ and $$D=a-B$$. Then $$x^2+y^2=r^2\\(x+B)^2+\left(\dfrac{C+2Dx-2x^2}{2y}\right)^2=R^2$$ so the resultant, where the coefficients $$c_i$$ are constant $$c_1x^3+c_2x^2+c_3x+c_4=0$$.

This equation has always a real root but this does not guarantee that there is always a solution to the problem posed. In fact, the smallest and greatest possible distance between two points, one in each circle, is $$R-r$$ and $$R + r$$ respectively. Consequently, for every point outside the interior of the circle of radius $$2R+r$$ we can assure that there is no solution.

• In fact the question of the OP isn't quite different from this other issue : math.stackexchange.com/q/3149465 Commented May 2, 2020 at 15:20
• @Jean Marie: Thank you for your comment. But something prevents me from fully agreeing with your point of view. It happens that the analytical solution in this problem is given with a cubic equation while the other problem is solved without odd powers. And we all know that you cannot build with a ruler and compass cube roots. Commented May 2, 2020 at 15:48
• @Piquito I don't know about the cubic equation you got. It is highly possible that this cubic polynomial can be factorized. It is important to check whether the cubic is irreducible before concluding that the roots of the cubic are not constructible. The triangles for this problem are constructible. Commented May 3, 2020 at 8:09
• @Batominovski: What do you think I am, please? If I had thought what you accuse me of, then I would have put it in my answer. Obviously the cubic of the case could have a linear factor in some cases but that this is "highly possible", as you say, that is obviously false, the radius $R$ and $r$ being arbitraries. There are many points of the plane from which there is no possible construction responding to the problem which I have highlighted in my answer. Nowhere have I said that the problem is impossible in all cases.(Another thing, my answer to Jean Marie is absolutely pertinent). Commented May 3, 2020 at 15:17
• @Piquito Excuse me? What did I accuse you of? I found a geometric construction. Therefore, I could say that the cubic you got couldn't be irreducible (otherwise a geometric construction would exist). I don't know what your cubic looks like, you never explicitly wrote the coefficients. I can't say for sure about its property. There is no need for such an impolite response. Commented May 3, 2020 at 15:21