in which 
and 
.
and 
such that 
.
.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
Prove that
Let 
Prove that
be an acute, scalene triangle, and let 
,
,
be the midpoints of 
,
,
,
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
,
,
.
,
,
intersects diagonal 
at points 
and 
with 
,
,
,
.
,
and 
are positive integers, and 
.
and 
intersect at points 
and 
.
is tangent to 
closer to point 
passes through 
and intersecting 
.
,
are collinear, 
,
,
.
.
and 
intersect at 
intersect at 
intersect at 
.
intersects line 
intersects line 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
,
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
,
,
,
are collinear if and only if 
.
Y by samrocksnature, Adventure10, Mango247, Rounak_iitr
Let 
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
such that 
is an obtuse angle. Let the circumcircle of 
intersect and respectively in points 
and 
. Let the line 
intersect in 
; let the line 
intersect in 
. Prove that, the points 
are concyclic.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
such that 
is an obtuse angle. Let the circumcircle of 
intersect and respectively in points 
and 
. Let the line 
intersect in 
; let the line 
intersect in 
. Prove that, the points 
are concyclic.This post has been edited 1 time. Last edited by integrated_JRC, Jan 22, 2018, 2:10 AM
is a diameter of 
.
is a diameter of 
.
is a straight line parrallel to 
.
meet 
,
.
is cyclic implying 
giving 
are collinear. Similarly, 
are collinear.
.
.
and 
.
is not isosceles, 
and 
gives 
is cyclic. So, 
.
.
and 
are co-linear.
we get,
intersects 
at 
are co-linear. (Proved)
are co-linear.
are co-linear.
are concyclic [Q.E.D.]
,
and 
,
and 
.
is obtuse, 
and 
,
,
.
is the reflection of 
is the reflection of 
because there is only one point other than 
whose reflections in 
(why?) 
and the orthocenter clearly satisfies this condition. Hence, 
is the orthocenter of 
and 
are the feet of altitudes through 
is cyclic.
again at 
.
.
)
and 
are collinar in that order .
and since 
and 
are the radiuses of their respective circles , we get that 
wich completes the proof.
which can never be equal to 
so your claim is wrong
I think my solution works fine.
are collinear. Similarly 
are collinear.
(Angle in a semicircle). By inscribed angles theorem, 
and 
which implies the conclusion
are coliner. Similarly 
and 
.Note that 
since it's radial axis of 
.
is isosceles .And hence 
and by similar argument we have 
and hence 
.
.
.
are concyclic . 
.
are Coliner and 
parallel to 
which gives 
.
parallel to 
parallel to 
are concyclic so 
.
hence 
.
we get 
so we are done .
is cyclic..
are collinear, so is 
and 
and we are done.
are collinear, hece 
,
are concyclic
are collinear.
so 
so 
is cyclic.
and 
are the reflection of 
and 
respectively, then note that since both 
,
is the orthocenter of 
and therefore, 
are in fact collinear, the conclusion now follows immediately.
Let 
be the perpendicular bisector of segment 
be the reflection of 
.
.
such that 
.
be the perpendicular bisector of segment 
.
be the mid-point of 
and let 
be the mid-point of 
.
This gives
Now, if we let 
be the mid-point of 
,
.
.
be the perpendicular bisector of 
.
are parallel, it follows by Equal Intercept Theorem that 
Now, 
Also, since 
Combining, we get 
This means that 
are concyclic.
and 
,
(exterior angle of cyclic quadrilateral equals interior opposite angle). This proves that 
is the perpendicular bisector of 
,
.
which has been left as an exercise to the reader (you can send your proof as a reply to this message).
Please note that this proof may seem overly extended, however I have made use of many properties so that the reader recalls them and gains benefit from reading this proof.
,
,
.
since 
,
.
,
.
,
.
as desired.
.
.
?
).
a
and 
and so you get 
which finishes.
and 
become two lines that intersect at 
becomes the line 
,
and 
,
are collinear as 
;
is cyclic by a quick angle chase.
.
are collinear, 
are collinear, 
are collinear, 
are concyclic, and 
are concyclic.
.![\[\angle AF^*C^* = \angle C^*F^*O_1^* = \angle C^*D^*B^*\]](http://latex.artofproblemsolving.com/9/1/b/91b6741d9e16784dd56e24881e4c35ee658cc428.png)
![\[\measuredangle AO_1D = \measuredangle AO_2D = 2\measuredangle ADO_2 = 2\measuredangle AO_1O_2 = \measuredangle AO_1B\]](http://latex.artofproblemsolving.com/a/0/9/a09b3d926c5041c685b4efc06635f8b14a6b233e.png)
from which it is clear that ![\[\measuredangle FCE = \measuredangle FCB = 90^\circ = \measuredangle BDE = \measuredangle FDE\]](http://latex.artofproblemsolving.com/4/7/e/47ee4a61c9c7358236097e5c38bf28ba788daacc.png)
from which it is clear that the points 
As 
,
,
,
,
are collinear,
and 
,
is a cyclic quadrilateral.
is collinear.
Then, we have that 
goes to a line 
goes to a line 
goes to the reflection of 
goes to the reflection of 
going to the intersections of 
with 
We also have that 
go to the intersections of 
and 
since 
Thus, 
so 
is cyclic. However, since 
we have that 
is cyclic. Inverting back, we get our desired claim.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(13cm);
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 = -25.51570034117379, xmax = 40.82314408852699, ymin = -16.69519592790025, ymax = 21.93029312646862; /* image dimensions */
pen qqccqq = rgb(0,0.8,0); pen ttffqq = rgb(0.2,1,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffqqff = rgb(1,0,1); pen ffwwqq = rgb(1,0.4,0); pen xfqqff = rgb(0.4980392156862745,0,1);
/* draw figures */
draw(circle((-3,-1), 9.501880502737109), linewidth(1) + blue);
draw(circle((12.983545658162647,-1.0859342715935283), 10.371101277912132), linewidth(1) + red);
draw(circle((4.978910084569212,-3.4354033592755004), 8.342313567590866), linewidth(1) + qqccqq);
draw((-3,-1)--(12.983545658162647,-1.0859342715935283), linewidth(1) + red);
draw((4.455727850687627,4.890488545509315)--(12.983545658162647,-1.0859342715935283), linewidth(1) + ttffqq);
draw((4.455727850687627,4.890488545509315)--(-3,-1), linewidth(1) + ttffqq);
draw((-0.10455216140722312,-10.049978723858084)--(21.50932605200054,4.81921690282699), linewidth(1) + blue);
draw((9.085859717659737,-10.69675036732216)--(-10.445019518138734,4.904017061529285), linewidth(1) + red);
draw((-10.445019518138734,4.904017061529285)--(21.50932605200054,4.81921690282699), linewidth(1) + ffdxqq);
draw((-10.445019518138734,4.904017061529285)--(-0.10455216140722312,-10.049978723858084), linewidth(1) + ffdxqq);
draw((-0.10455216140722312,-10.049978723858084)--(9.085859717659737,-10.69675036732216), linewidth(1) + ffdxqq);
draw((9.085859717659737,-10.69675036732216)--(21.50932605200054,4.81921690282699), linewidth(1) + ffdxqq);
draw((4.455727850687627,4.890488545509315)--(4.419722115642268,-6.935777692513815), linewidth(1) + ffqqff);
draw((4.455727850687627,4.890488545509315)--(9.085859717659737,-10.69675036732216), linewidth(1) + ffwwqq);
draw((4.455727850687627,4.890488545509315)--(-0.10455216140722312,-10.049978723858084), linewidth(1) + ffwwqq);
draw(circle((5.532254937295013,4.899928348557101), 15.977274978600997), linewidth(1) + linetype("4 4") + ffqqff);
draw((-3,-1)--(-0.10455216140722312,-10.049978723858084), linewidth(1) + xfqqff);
draw((12.983545658162647,-1.0859342715935283)--(9.085859717659737,-10.69675036732216), linewidth(1) + xfqqff);
/* dots and labels */
dot((-3,-1),dotstyle);
label("$O_1$", (-4.514173220805075,-1.4528504714676924), NE * labelscalefactor);
dot((4.455727850687627,4.890488545509315),dotstyle);
label("$A$", (4.362760922855928,5.995113785652763), NE * labelscalefactor);
dot((12.983545658162647,-1.0859342715935283),dotstyle);
label("$O_2$", (13.889226833126274,-1.7126631781114292), NE * labelscalefactor);
dot((4.419722115642268,-6.935777692513815),dotstyle);
label("$B$", (4.2328545695340605,-9.030721081910016), NE * labelscalefactor);
dot((-0.10455216140722312,-10.049978723858084),dotstyle);
label("$C$", (-0.6169826211490247,-11.67215026612134), NE * labelscalefactor);
dot((9.085859717659737,-10.69675036732216),dotstyle);
label("$D$", (9.125993877991101,-12.105171443860902), NE * labelscalefactor);
dot((21.50932605200054,4.81921690282699),dotstyle);
label("$E$", (22.72285885901332,4.782654487981991), NE * labelscalefactor);
dot((-10.445019518138734,4.904017061529285),dotstyle);
label("$F$", (-11.745626889055746,4.955862959077816), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/2/2/0/220f188c4b86850fc0d6a363922ee1beeaed194d.png)
By Angle Chasing we get, 
Also, 
Since, 
and 
are collinear and 
are collinear and parallel to 
Now By Angle Chasing we get, 
And 
Therefore 
are concyclic points.
is the perpendicular bisector of 
.
.
which gives 
is a circle so 
collinear, finishes by PoP.
