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. :)
