let $ \alpha=\angle PAB$ and $ \theta=\frac{1}{2}\angle AO_1B$... note that $ ADC$ is isosceles, so $ AL$ is perpendicular to $ CD$... also note that $ \angle EDC=\angle ECB=\angle EAB=90-\alpha-\theta$, and since $ \angle ABD=\alpha+\theta$ it follows that $ DE$ is perpendicular to $ AB$, or $ DE$ is parallel to $ O{}_1O{}_2$... this implies that $ DEO{}_1O{}_2$ is a parallelogram, therefore, $ FPO{}_2{}_O{}_1$ is a parallelogram too, and this implies that $ F$ lies in $ C_1$... note that symmetry arguments imply that $ \angle PFB=\angle PAB=\alpha$, son this implies that $ AC$ is perpendicular to $ BF$... since $ \angle CAL+\angle ALB=(90-\theta)+\theta=90$ it follows that $ AC$ is also perpendicular to $ BL$, which implies that $ F,B,L$ are collinear... note that $ \angle PFB=\angle CFB=\alpha$ and $ AC$ is perpendicular to $ BF$, therefore $ FP=FC$... also since $ AFL$ is isosceles we have that $ FP=PL$, so $ FPLC$ is a kite (rhombus)... this implies that $ CL=CF$... let $ l$ be a line through $ F$ such that $ l$ and $ FL$ form an angle of $ 30$, and let $ X$ be another point on $ l$ such that $ CF=CX$... (in at least one of the two possible lines $ l$ $ X$ and $ F$ are different)... then, $ C$ is the circumcenter of triangle $ FLX$ and since $ \angle XFL=30$ we have that $ \angle XCL=60$, which implies that $ XLC$ is equilateral... we can prove that this point $ X$ satisfies the other two conditions as well ($ CL=CF=O{}_1O{}_2$ and the other one is similar), and we're done

my solution is like campos's above the key point lies on the proof of PF=DE=O1O2, F,L are symmetry w.r.t AC => FC=CL=LP=PF=ED=EC=r => XDC is an equilateral triangle => see my figure X is the intersection of the two red cirlces with the same radius r and centered at C,L respectively

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Hey this is my problem

can someone please tell me what equal circles are ?? equal radius?

[asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(20.7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.77, xmax = 8.93, ymin = -5.69, ymax = 10.98; /* image dimensions */ /* draw figures */ draw(circle((-4.15,4.53), 4.89)); draw(circle((4.49,5.01), 4.9)); draw((0.04,7.05)--(-2.27,0.02)); draw((7.16,9.11)--(-2.27,0.02)); draw((0.04,7.05)--(7.16,9.11)); draw((-1.48,8.63)--(6.37,0.49)); draw((-1.48,8.63)--(-0.31,6)); /* dots and labels */ dot((-4.15,4.53),dotstyle); label("$O_1$", (-4.02,4.73), NE * labelscalefactor); dot((0.29,2.49),dotstyle); label("$B$", (0.44,2.7), NE * labelscalefactor); dot((4.49,5.01),dotstyle); label("$O_2$", (4.61,5.22), NE * labelscalefactor); dot((0.04,7.05),dotstyle); label("$A$", (0.19,7.26), NE * labelscalefactor); dot((-0.31,6),dotstyle); label("$P$", (-0.16,6.21), NE * labelscalefactor); dot((-2.27,0.02),dotstyle); label("$C$", (-2.12,0.24), NE * labelscalefactor); dot((7.16,9.11),dotstyle); label("$D$", (7.32,9.33), NE * labelscalefactor); dot((-1.48,8.63),dotstyle); label("$E$", (-1.35,8.84), NE * labelscalefactor); dot((6.37,0.49),dotstyle); label("$L$", (6.51,0.7), NE * labelscalefactor); dot((-0.9,7.31),dotstyle); label("$M$", (-0.76,7.54), NE * labelscalefactor); dot((-8.95,5.51),dotstyle); label("$F$", (-8.83,5.71), NE * labelscalefactor); dot((-0.83,6.45),dotstyle); label("$X$", (-0.69,6.66), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $ C_1$ and $ C_2$ be two congruent circles centered at $ O_1$ and $ O_2$, which intersect at $ A$ and $ B$. Take a point $ P$ on the arc $ AB$ of $ C_2$ which is contained in $ C_1$. $ AP$ meets $ C_1$ at $ C$, $ CB$ meets $ C_2$ at $ D$ and the bisector of $ \angle CAD$ intersects $ C_1$ and $ C_2$ at $ E$ and $ L$, respectively. Let $ F$ be the symmetric point of $ D$ with respect to the midpoint of $ PE$. Prove that there exists a point $ X$ satisfying $ \angle XFL = \angle XDC =30^\circ$ and $ CX =O_1O_2$. Author: Arnoldo Aguilar (El Salvador)