# Conditions required to allow an equilateral triangle to be constructed with a vertex on each of three concentric circles

I have come across this problem proposed for IMO 26, and I really don't know how to construct this. Any ideas would be very much appreciated.

Given three concentric circles, construct an equilateral triangle with a vertex on each of the three circles. Give a necessary and sufficient condition for this to be possible.

Picture isn't so clear, do excuse me for that.

(new figure done by JeanMarie ; please note the pink and purple triangles $$AQB$$ and $$AOC$$)

Well, the construction steps:

Given three circles ($$C_1,C_2,C_3$$), with resp. radii satisfying $$r_1>r_2>r_3$$ and $$r_2+r_3 \ge r_1$$

$$1$$. Take an arbitrary point on the circle with radius $$r_1$$, name it $$A$$, draw the circle with center $$A$$ and radius $$r_1$$.

$$2$$. Let $$Q$$ be the point where the drawn circle intersects circle $$C_1$$.

$$3$$. Now, draw the circle with centre $$Q$$ and radius $$r_3$$.

$$4$$. Let $$B$$ be one of the points of intersection of the drawn circle and circle $$C_2$$.

$$5$$. Now rotate clockwise $$60^0$$ $$AB$$ onto $$AC$$, completing the triangle $$ABC$$.

We need to prove that the point $$C$$ lies on the circle $$C_3$$.

In $$\triangle OAQ$$,

$$OQ=OA=AQ$$ (According to the construction). Therefore, $$\triangle OAQ$$ is equilateral.

Observe $$\triangle OCB$$ and $$\triangle BQA$$,

Let $$\angle OAC=x$$

$$\angle OAB=60^0-x$$

Therefore, $$\angle OAQ=x$$.

$$AC=AB$$(Construction)

$$OA=QA$$ (Equilateral triangle).

Therefore, by CPCT , $$BQ=CO$$

$$BQ=r_1$$

Therefore, $$C$$ lies on circle $$C_3$$.

There's no triangle formed if $$r_1 > r_2+r_3$$

Note: There is one more triangle which can formed from an arbitrary point(As the circle cuts $$C_1$$ in two points. :)

• Indication: CPCT stands for “Corresponding parts of Congruent triangles”, Commented Apr 12, 2022 at 12:44
• I have taken the liberty to realize a new figure that I think clearer. I have left the former one. Commented Apr 13, 2022 at 10:07