Denote by $O$ the centre and by $R$ the radius of the circle circumscribed to $\triangle{ABC}$. Since $B=30^\cdot$ we get that $AC=R$, hence $\triangle{OAC}$ is equilateral. Now, it is enough to use the following Lemma (from a Romanian TST 2002 ): Lemma. If $X$ is a variable point on the disk centred at $A$ (with radius $R_1$) and $Y$ is a variable point on the disk centered at $C$ (with radius $R_2$), then the locus of points $Z$ such that $\triangle{XYZ}$ is equilateral, is the disk centred at $O$ and with radius $R_1+R_2$ (and, of course, its reflection in $AC$, but that's redundant here). and the conclusion follows.

Quote: Now, it is enough to use the following Lemma (from a Romanian TST 2002 ): Lemma. If is a variable point on the disk centered at (with radius ) and is a variable point on the disk centered at (with radius ), then the locus of points such that is equilateral, is the disk centered at and with radius (and, of course, its reflection in , but that's redundant here). Sorry, Sailor, could you give a demonstration to your lemma, please: I can't find the "Romanian TST 2002" on the Web , and I'm very intrested in this problem.

Sorry for bumping such an old topic but can somebody please prove the stated lemma?? Thanks in advance..

Note: In my diagram, the triangle is $ABC$ in counterclockwise order. Proof of the lemma: Let the discs centered at $A,C$ be called $D_1,D_2$ respectively. Let $Y$ be fixed and let $X$ vary on $D_2$. For each $X$, $Z$ is simply the $60^{\circ}$ counterclockwise rotation of $X$ with respect to $Y$ so, the locus of $Z$ as $X$ varies is simply the circle that results from rotating $D_1$ $60^{\circ}$ degrees counterclockwise about $Y$. Let this circle have center $P$, so $YAP$ is equilateral. Now if we let $Y$ vary as well, the locus of $P$ is simply the $60^{\circ}$ clockwise rotation of $D_2$ with respect to $A$, which is easily seen to be the circle centered at $O$ with radius $R_2$. Then the locus of $Z$ is just the union of all possible circles centered at $P$, which is clearly a circle centered at $O$ with radius $R_2+R_1$.

Without interfering with the above solutions, but the problem states 'with one vertex in each of the discs...'. Best regards, sunken rock

Consider the circumcentre of $ABC$, $O$. Then $OAC$ is an equilateral triangle. Suppose in 3 the circles centered around $A,B,C$ there are points $J,K,L$ forming an equilateral triangle. Consider the spiral similarity sending $JKL$ to $JMC$, then that of $JMC$ to $AOC$. Then it is easy to see that $KM=LC<\frac{AC}{3}$, $OM=AJ<\frac{AC}{3}$, but $OK>\frac{2AC}{3}$, contradiction.

Since you revived this ... sunken rock wrote: Without interfering with the above solutions, but the problem states 'with one vertex in each of the discs...'. ... but it also states "closed disks". There does exist a (unique) such equilateral triangle. For a detailed analysis, see RMC 2006.

Ahhh. If we look at the inequalities established above we see that such a triangle is only possible when $AJ=BK=CL=\frac{AC}{3}$, and that $O,M,K$ collinear. Then using the same spiral similarities it is easy to determine that the triangle $JKL$ is determined by $AJ$ as reflection of $AC$ across $AB$ and similarly $CL$ reflection of $CA$ across $CB$, and by doing so we have a triangle that works.