AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least 
of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly 
leaders present. The chairperson must seat them in a square-shaped conference hall of size 
, where each leader will be seated in a designated 
cell. It is known that exactly 
of these leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly 
neighbors and will vote "FOR" only if both of their neighbors voted "FOR." supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least of all leaders?
be a quadrilateral with 
. Diagonals 
and 
meet at 
. Let 
and 
denote the circumcircle and the circumcenter of triangle 
. Let 
and 
denote the circumcircle and circumcenter of triangle 
. Segment 
meets and again at 
and 
(other than 
and 
), respectively. Let 
and 
be the midpoints of minor arcs 
(not including ) and 
(not including ). Prove that 
.
such that 
. Find the minimum value of:
be positive real numbers such that 
and 
, prove that
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-regular graph , there exists a coloring using 
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
be positive real numbers. Prove the inequality![\[
\frac{x_1 + 3x_2}{x_2 + x_3}
\;+\;
\frac{x_2 + 3x_3}{x_3 + x_4}
\;+\;
\frac{x_3 + 3x_4}{x_4 + x_1}
\;+\;
\frac{x_4 + 3x_1}{x_1 + x_2}
\;\ge\;8.
\]](http://latex.artofproblemsolving.com/5/c/0/5c062a37f6ff120232ae87eb0093a4ce73ee0eb1.png)
, where 
is a positive integer. with exactly 
elements such that the sum of its elements is a prime number.
with at most four natural divisors, which have the property that for any two distinct proper divisors 
and 
of 
, the positive integer 
divides .
satisfying the equation 
with complex coefficients such that 
for any complex number 
and be two fixed points in the plane. For each point 
of the plane, outside of the line , let 
be the barycenter of the triangle 
. Determine the locus of points such that 
.
lies inside triangle such that 
and 
. Point 
is the midpoint of segment . Point 
lies on segment with 
. Prove that 
.
be, the circumcircle of the triangle 
with 
with 
and 
it is easy to show that 
and so, we have that 
and let 
be, the midpoint of the segment 
we can say that the points 
are collinear, because of 
is the center of 
from 
we have that 
is a diameter of and from 
taken as its transversal the line segment 
applying again the Menelaos theorem,
From 
and because of 
we conclude that 
and 
we conclude that 
and the proof is completed.
is midpoint of 
and 
reflection of 
in 
then 
Since 
bisects 
is therefore equilateral and 
is on circumcircle 
of the isosceles 
Let 
cut again at 
is equilateral with 
Let 
be circumcircle with diameter of the right 
Since 
the circles 
are similar with center 
and coefficient 2. Since 
and 
it follows that 
with diameter and 
on so that 
. It is easy to see that 
. Let 
be the reflection of across . 
, and 
, so 
is equilateral, so 
is cyclic. Since 
, 
, so 
is cyclic. 
and 
, so 
, so 
must be a diameter of 
, so 
. Since 
, a homothety centered at with factor 1/2 gives 
, as desired.
meets 
at 
. Let 
be the mid-point of 
and is that of . Now apply cosine rules to show 
is an equilateral triangle. So 
and 
is cyclic.
, we must have 
![\[\angle EFD=\angle EFA-60^{\circ}=\angle BF'A-60^{\circ} =\angle BDA-60^{\circ}=\angle BDW=60^{\circ}-\angle DAB\]](http://latex.artofproblemsolving.com/d/1/7/d1709c4fa84c5c84fd05aabb48c335569b841098.png)
At the same time 
and 
is cyclic. So 
. Let 
be the circumcenter of 
and let 
. We have 
since 
, and we can compute 
. So 
is a parallelogram and 
is the midpoint of 
. Then we can compute that 
, implying .
and the conclusion follows.
and 
. Let 
intersect the circumcircle 
again at . Also, let the circle with diameter 
intersect again at 
. Let be the reflection of over 
. We see that lies on 
and 
is a 30, 60, 90 triangle with right angle at . Thus there is a homothety centered at that takes 
. is the midpoint of , then the intersection of 
with closer to is the midpoint of . So point lies on , and the perpendicular condition is prove.
is a homogenized coordinate while 
is not.
be the point on for which 
is an equilateral triangle. We will use as our reference triangle instead of 
, with 
, 
and 
. Then it is easy to see 
, since is the midpoint of 
.
implies that lies on the circumcircle 
. Thus, if 
, then 
. Now, we can calculate the points and , giving
Now, we can calculate the displacement vectors 
and 
. We get
Finally, it remains to show that the vectors are perpendicular, which happens iff![\[
\big((3v-2)w+(3w-2)v\big)+\big((3u+1)w+(3w-2)(u-1)\big)+\big((3u+1)v+(3v-2)(u-1)\big)=0
\]](http://latex.artofproblemsolving.com/c/3/6/c360ea105ffd16a0a69bc2775d38fb38ba849fd5.png)
by the perpendicularity criterion. But the LHS is equal to![\[ 6(vw+uw+uv)-4(u+v+w)+4 = 0-4+4=0, \]](http://latex.artofproblemsolving.com/3/c/d/3cdd2c6748bd819c11f70c38ffe4cceaacddc1ee.png)
and we are done.
lies on the perpendicular bisector of . be the midpoint of , 
.
intersects at . with scale . 
, 
, 
, 
and 
. It now suffices to prove that lies on on 
.
and 
. lies on . Since![\[CD\cdot CD' = 2CD^2 = \tfrac{2}{3} AC^2 = AC\cdot AF'\]](http://latex.artofproblemsolving.com/c/f/a/cfafa60b9a99ea7c47039b70f47083549882260c.png)
Now, note that 
is an equilateral triangle, since, 
and 
. lies on since 
.
. Then 
and 
are tangent to each other and 
is their common tangent.
was the circumcenter of 
are tangent to the 
, thus 
is symmedian of 
and subsequently median of 
, thus, if 
was the midpoint of , then 
(actually 
is a parallelogram, thus 
). But from 
and 
we get 
, hence 
and 
is 
right-angled.
be the midpoint of and let be the point on 
such that 
. Since 
, 
is tangent to 
. Since 
, quadrilateral is cyclic and 
. Taking a homothety at by scale factor 
, we find that 
. Thus, quadrilateral is cyclic and 
, as desired.
be the midpoint of and let 
be the foot of perpendicular from to , then,
Hence, 
, Let be the reflection of in , then, 
is the angle bisector of 
, hence now, 
, hence, is cyclic 
, since, 
, hence, 
and 
, now, 
, therefore, 
is cyclic, 
![[asy]
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draw((5.3383604845092725,-1.492379251671836)--(9.48,2.12), linewidth(1) + dbwrru);
draw((xmin, -0.5644774419752964*xmin + 1.5210048189659622)--(xmax, -0.5644774419752964*xmax + 1.5210048189659622), linewidth(1) + white); /* line */
draw((5.3383604845092725,-1.492379251671836)--(13.630832927947854,-6.173292884194681), linewidth(1) + wrwrwr);
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draw((14.3,-2)--(13.630832927947854,-6.173292884194681), linewidth(1) + dtsfsf);
/* dots and labels */
dot((9.48,2.12),dotstyle);
label("B", (9.56,2.32), NE * labelscalefactor);
dot((2.64,-6.28),dotstyle);
label("A", (2.72,-6.08), NE * labelscalefactor);
dot((19.12,-6.12),dotstyle);
label("$C$", (19.2,-5.92), NE * labelscalefactor);
dot((10.833811978464832,-1.4426337818774861),linewidth(4pt) + dotstyle);
label("$D$", (10.92,-1.28), NE * labelscalefactor);
dot((5.3383604845092725,-1.492379251671836),linewidth(4pt) + dotstyle);
label("$O$", (5.42,-1.34), NE * labelscalefactor);
dot((13.630832927947854,-6.173292884194681),linewidth(4pt) + dotstyle);
label("$F$", (13.72,-6.02), NE * labelscalefactor);
dot((14.3,-2),linewidth(4pt) + dotstyle);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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[/asy]](http://latex.artofproblemsolving.com/5/4/f/54fb3e0da95d45d090376e6b8517748ceeac625d.png)
and 
. We know that:
This means that 
. Note that:
Hence, 
. Since 
and 
with 
, 
. Now, assume WLOG 
(if this is not the case, we can always scale the diagram). Then, 
. Given that 
, we can deduce that 
. Hence, 
, so is a parallelogram. By Stewart’s theorem:
Note that the intersection of and is the midpoint of ( is a parallelogram). The homothety that takes to (from ) also takes the midpoint of to , so the distance from the midpoint of to is 
. It follows that 
is a right triangle.
over be 
, the midpoint of be and the point at which the circumcircle of 
intersects be .
; since lies on , 
and we have that 
, so 
.
, lies on the circumcircle of . Now, consider the homothety sending the circumcircle of 
to the circumcircle of 
centered at . It has scale factor , so it maps to , and hence 
is cyclic. This gives 
as desired.
was not on segment . Then I misread that is outside . What am I doing?![[asy]
size(250);
pair A = origin, C = (3, 0), D = (3/2, sqrt(3)/2);
pair X = extension(C, D, (0, 0), (0, 50));
pair B = circumcenter(A, D, X)+abs(circumcenter(A, D, X)-A)*dir(87);
draw(A--C--D--B--cycle);
pair EE = (B+C)/2;
pair F = C*2/3;
draw(B--C);
draw(D--EE--F--cycle, dashed);
pair M = (D+F)/2;
draw(EE--M);
draw(EE--circumcenter(A, B, D)--F);
draw(circumcenter(A, B, D)--B);
draw(circumcenter(A, B, D)--C, dashed);
draw(circumcenter(A, B, D)--D--A);
dot("$O$", circumcenter(A, B, D), SW);
dot("$A$", A, SW);
dot("$C$", C, SE);
dot("$B$", B, N);
dot("$F$", F, S);
dot("$M$", M, SW);
dot("$E$", EE, NE);
dot(D);
label("$D$", D+(0, 0.05), N);
[/asy]](http://latex.artofproblemsolving.com/8/9/3/8932564ef0e570bbdd036744e74282df905f03b6.png)
be the midpoint of . It suffices to show that 
.
. From the Law of Cosines, we obtain 
so it suffices to show that 
.
is 1 by the Extended Law of Sines, so compute 
. Furthermore, since 
and 
, is a parallelogram. It follows that there is a homothety at taking to 
with ratio 
, and hence is the midpoint of 
. with ratio 
taking 
to 
. It follows that 
for all choices of , and as desired. 
be the reflection of over and be the reflection of over and be the midpoint of . It is trivial that 
. Now if 
then 
and 
and since 
, we have 
is a 30-60-90 triangle.
is cyclic and that 
. In addition we get 
. This means 
but since 
, we have that lies on 
.
from sends 
to 
respectively, so they are all cyclic. And thus ![\[\angle DEF=180-\angle DMF=90\]](http://latex.artofproblemsolving.com/a/b/c/abcdc9443899373a5b61bad6d6faa48f4e97f868.png)
so we are done.
outside triangle 
such that 
and 
. Consider the circle centered at with radius 
, which we let equal 1. Denote this circle as 
. We then consider the configuration on the complex plane, letting be 
(or the center/origin), making the unit circle. Since , we know that lies on .
, 
, 
, and 
. We will keep these as reference, as these will be useful later on., it suffices to show that 
. We substitute the values above that were kept as reference to obtain that
We know that 
, 
, 
, and 
from substitution. Therefore, we get the following:
Combining denominators and simplifying, the numerator becomes
It turns out that this expression is nice AF and everything CANCELS OUT, so therefore 
, as desired.
. Note that the midpoint of AC=M lies on since ADC is isosceles, 
is the midpoint of CD since CFD is isosceles (proof: DMC is 30-60-90, MF/FC=(1/6)/(1/3)=1/2=MD/DC implies DF bisects angle MDC which is 60 deg), and F is the midpoint of 
with C, since APD=120 means DPF=60, whence FPD is equilateral implies FP=FD=FC; in particular, (ADP) goes to (MNF), with A going to E by homothety at C; in particular, since B lies on (ADP), E lies on (DNFM), as desired. 
, 
, 
, 
, and 
where 
. We see that 
so we must have that 
. Since , 
, as desired.
. Therefore, the circle with diameter passes through the midpoints of and . Thus, taking a homothety with scale factor from will take this circle to the circle with , , and the reflection of over . Therefore, it suffices to show that is on this circle as well. But this is just because 
. 
. Moreover, let be the circumcenter of , and let intersect again at point 
. It is clear that 
, which implies that 
. is a parallelogram, so denoting the midpoint of 
as , we find that is also the midpoint of 
. Thus, 
, with scale factor 
:![\[ME = \frac{1}{2} OB = 1 = MD = MF, \]](http://latex.artofproblemsolving.com/e/5/a/e5a77be2d7c43a1947c02228e90521d46dd5fc71.png)
is a right triangle.
. So 
also passes through the midpoints of and . So a homothety at with scale factor finishes. was literally only used once.
be the point on the line parallel to through and which lies on the circle centered at with radius 
and so we will consider 
as our reference triangle and will be its center. Furthermore, let 
.
is a parallelogram and quadrilateral 
is an isosceles trapezoid with radius 
, which sends to and as 
, it sends to , hence 
and is a parallelogram.
, hence is an isosceles trapezoid. be the reflection of across , therefore 
.
is a cyclic quadrilateral centered at 
and we claim is tangent to it.
, consequently 
, hence is the center of .
is cyclic, which follows from 
, proving our claim.
, which combined with yields that 
is the side bisector of , hence 
, as desired. is almost the center of (we just have to get rid of the angle 
), which is the motivation to add . Besides that, the condition of 
is only used when proving 
, that is the parallelogram and isosceles trapezoid are also true when 
is arbitrary. Moreover, if is the point satisfying , will always be a cyclic quadrilateral with center , however won't satisfy 
.
P![$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$](http://latex.artofproblemsolving.com/6/b/6/6b6011a32b103fe94652974a521eeae479cba1d8.png)
lies inside triangle 
such that 
and 
.
is the midpoint of segment 
.
.
.
is the midpoint of segment 
.
,
.
.
and 
.
is inscribed, which equivalent to prove 
is inscribed :
.
is cyclic, which is equivalent to proving that 
.
homothety centered at 
.
implies 
is fixed with only 
.
are concyclic 
are concyclic 
at 
and WLOG 
We know 
is cyclic since 
It is also a trapezoid because 
Thus, 
and 
with scale factor 
Thus, 
meaning that 
which implies that 
![[asy]
unitsize(1cm);
pair A, B, C, D, E, F;
D=(0,0);
A=(-3sqrt(3),-3);
C=(3sqrt(3),-3);
B=(-2sqrt(3),2sqrt(3));
E=(B+C)/2;
F=(2C+A)/3;
draw(circle((-2sqrt(3),0),2sqrt(3)));
draw(circle((D+F)/2,distance(D,F)/2));
draw(F--E--D--A--C--D--B--A--B--C);
dot("$O$", (-2sqrt(3),0));
label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$D$", D, W);
label("$E$", E, N);
label("$F$", F, SE);
[/asy]](http://latex.artofproblemsolving.com/9/f/3/9f329ffad335791c51a5dd6b72c02e4edb4e1725.png)
,
.
and 
,
is the circle with diameter 
.
and 
.
is equilateral.
is a 30-30-120 triangle, so 
and 
.
,![\[\angle DEF = \angle XBY = 180 - \angle YAX = 180-60-30 = 90. \quad \blacksquare\]](http://latex.artofproblemsolving.com/8/b/5/8b589864bb85beac2918cfb1a5443bea8f5acc92.png)
,
,
is free.
,
and 
,
and 
.
,
,
,