[/quote]
,
and 
then prove 
and 
.
are positive real numbers such that 
,![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](http://latex.artofproblemsolving.com/0/c/e/0ce097b0b507403c959c2992cae1217ecdd7e581.png)
be a finite sequence of positive integers. Prove that there exist nonnegative integers 
and 
such that 
holds for all integers 
be a positive integer. Dominoes are placed on a 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each 
or 
.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
and circumcenter 
.
,
,
.
with diameter 
,
at a point 
.
meets line 
at 
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
and incenter 
and let 
are selected on sides 
,
such that 
,
,
.
intersects 
other than 
and 
meet on 
,
,
,
,
,
initially contains one coin. The following operations are allowed
,
,
;
,
,
and 
.
coins.
Prove that
Y by Davi-8191, RAMUGAUSS, alex_g, Groots, HamstPan38825, megarnie, Lamboreghini, Adventure10, Mango247, Rounak_iitr, and 1 other user
In acute triangle 
, 
and 
. Let 
be a variable point on side 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Prove that as moves along side , the circumcircle of triangle 
passes through a fixed point other than 
.
, 
and 
. Let 
be a variable point on side 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Prove that as moves along side , the circumcircle of triangle 
passes through a fixed point other than 
.
,
as 
,![$\angle DPE=180-[360-2(\angle B+\angle C)]=180-2\angle A$](http://latex.artofproblemsolving.com/c/b/8/cb8851036450b2e820843cb0155cf9afdde720f2.png)
.
,
are concyclic, furthurmore, 
bisects 
since 
.
at 
.
,
(
and 
(
)
which is a fixed value.
,
stays the same. which means that the locus of 
is the intersection of the 
be midpoints of 
respectively and let 
intersects the circumcircle of 
and let 
are concurrent at a point other than 
be the intersection of 
lies on the circumcircle of 
regardless of the location of the point ![[asy]
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.300000000000005,xmax=7.960000000000012,ymin=-5.500000000000006,ymax=6.300000000000007;
pair A=(-0.6800000000000008,4.980000000000006), B=(3.160000000000004,-0.7400000000000008), C=(-1.760000000000002,-1.060000000000001), P=(1.723475284670574,-0.8334325018100450), D=(2.353907271759469,0.4607422931082958), H=(-0.3487737647763798,-0.1126033665631657), M=(1.779593460306762,1.316230574751391);
draw(A--C); draw(C--B); draw(B--A); draw((-1.465637979375667,0.5862468560842407)--P); draw(P--D); draw((-1.465637979375667,0.5862468560842407)--D); draw(circle((0.5090884022823820,2.500265814908376),2.750093245174520)); draw((-1.465637979375667,0.5862468560842407)--H,linetype("2 2")); draw(H--D,linetype("2 2")); draw((-1.465637979375667,0.5862468560842407)--H,linetype("2 2")); draw((-1.465637979375667,0.5862468560842407)--H,linetype("2 2")); draw(H--(-1.552124096897579,0.1025652358691018),linetype("2 2")); draw(M--H,linetype("2 2"));
dot(A,ds); label("$A$",(-0.6000000000000003,5.100000000000006),NE*lsf); dot(B,ds); label("$B$",(3.480000000000006,-1.160000000000001),NE*lsf); dot(C,ds); label("$C$",(-2.220000000000003,-1.440000000000002),NE*lsf); dot(P,ds); label("$P$",(1.720000000000003,-1.320000000000002),NE*lsf); dot((-1.465637979375667,0.5862468560842407),ds); label("$E$",(-1.380000000000001,0.7000000000000008),NE*lsf); dot(D,ds); label("$D$",(2.440000000000004,0.5800000000000006),E*lsf); dot(H,ds); label("$H$",(-0.4600000000000001,-0.5600000000000007),NE*lsf); dot((-1.552124096897579,0.1025652358691018),ds); label("$N$",(-2.380000000000003,0.02000000000000002),NE*lsf); dot(M,ds); label("$M$",(1.860000000000003,1.440000000000002),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/1/7/5/17598af92edcb46b3c6b7e1d459be9be59a8220d.png)
or 
,
be ![\[ \frac{MH}{NH} = \frac{\cos B } {\cos C} \]](http://latex.artofproblemsolving.com/0/4/d/04d20818ff9cdf558e4ec9492effa6ea7e7dbed8.png)
with some trigonometry on 
and 
.
,![\[ \frac{ BC-2BP } { 2CP - BC } =1 .\]](http://latex.artofproblemsolving.com/2/1/9/219b54613758ad2ac8e595bec8dae9dc54939665.png)
Noting that 
(because 
are isosceles), we can further find that![\[ \frac{MD}{NE} = \frac{ BC \cos B - 2BP \cos B } { 2CP \cos C - BC \cos C } \]](http://latex.artofproblemsolving.com/7/1/f/71fc02fc6deb3d98a0ca19e753022b7be170c785.png)
![\[ = \frac{ \cos B }{ \cos C} \cdot \frac{BC-2BP}{2CP-BC} = \frac{ \cos B }{ \cos C}.\]](http://latex.artofproblemsolving.com/6/1/2/612ca825ee6b26d6909b1cf4b4db2c80223bfd46.png)
and 
are similar. Since they share a common vertex and are directly similar, 
is similar to 
,
the foots of altitudes from 
.Observe that 
and 
is cyclic Done!!!.. 
into lines and 
,
,
,
,
,
where 
and 
depend on 
(we will find them later). Then ![\[\overrightarrow{BP} = (0,p - 1,1 - p) \\ \overrightarrow{DP} = (-q,-1 + q + p,1 - p) \\ \overrightarrow{CP} = (0,p,-p) \\ \overrightarrow{EP} = (r - 1,p,1 - p - r)\]](http://latex.artofproblemsolving.com/f/8/3/f838302ed645090d2e9a1c4c9fe97d433ff1de9b.png)
Thus by the distance formula we get ![\[a^2(1 - p)^2 = -a^2(-1 + q + p)(1 - p) - b^2(-q)(1 - p) - c^2(-q)(-1 + q + p)\]](http://latex.artofproblemsolving.com/5/d/d/5ddfcb1c7d0912e76700d66f49b3fa31b9eeea82.png)
i.e. ![\[q = \frac{(1 - p)(a^2 - b^2 + c^2)}{c^2}\]](http://latex.artofproblemsolving.com/b/6/a/b6a66220033e666c0288bd9b4a414a074f0ceff5.png)
and ![\[a^2p^2 = -a^2(p)(1 - p - r) - b^2(r - 1)(1 - p - r) - c^2(r - 1)(p)\]](http://latex.artofproblemsolving.com/3/9/f/39f7b98f14633817079cfb900bf5eb551af908db.png)
i.e. ![\[r = \frac{b^2 - p(a^2 + b^2 - c^2)}{b^2}\]](http://latex.artofproblemsolving.com/9/0/b/90bae983e6459e9aeac544bf9c0d536bd45db483.png)
The general equation of circle is ![\[-a^2yz - b^2zx - c^2xy + (ux + vy + wz)(x + y + z) = 0\]](http://latex.artofproblemsolving.com/0/8/b/08b4dab92998e16d548d007add15c7982bec6a9e.png)
Plugging in points 
we get that ![\[-a^2yz - b^2zx - c^2xy + \big(c^2qy + b^2(1 - r)z\big)(x + y + z) = 0\]](http://latex.artofproblemsolving.com/8/7/f/87fd635151f6bc25ea6aae1101cc693147afae01.png)
Plugging in values of 
Note that ![\[H = \left(\frac{1}{-a^2 + b^2 + c^2}:\frac{1}{a^2 - b^2 + c^2}:\frac{1}{a^2 + b^2 - c^2}\right)\]](http://latex.artofproblemsolving.com/f/6/6/f663bbffb0f5c6067bdcef558d2545a1d76d0d61.png)
Plugging this into equation of 
)
,
)
and 
.
and 
at the beginning (instead of 
)
be 
respectively. Then by considering projections on 
we see that the midpoint of 
is 
to 
,
and so 
,
are concyclic, so 
intersects 
.
,
is a cyclic quadrilateral.
.
.
is cyclic 
.
,
and 
such that 
.
goes through circumcircle of 
is cyclic,so again angle chase we get that for all 
is fixed and 
is fixed
such that 
,
to 
,
is similar to 
and from this is obvious that 
and 
this parabola is tangent to 
so done. Note we can define this, by taking 
,
.
,
is similar to the orthic triangle of 
,
is the orthocenter 
of 
pass through a fixed point 
be points on 
be their corresponding points. Because 
,
then 
,
.
be 
at 
respectively.
,
which means 
is cyclic.
,
and 
of 
and 
determined as 
.
to have 
and for 
to hold 
the corresponding circumcircles pass through 
and two points 
each of them moving at a constant speed along 
,
(the speeds may be different). Consider another two points 
doing the same along the lines 
then 
and 
along the entire motion. Saying it in other words, consider a linear mapping 
,
and 
of the points 
satisfy 
for some constants 
.
is another linear mapping and 
for some points 
then 
.
.
coincide one needs three points they are equal at.
lie on a unit circle with orthocenter 
![\[\frac{e-d}{h-d} \div \frac{e-a}{h-a} \in\mathbb{R}\]](http://latex.artofproblemsolving.com/9/8/e/98e961b8701f081224bea9488c5ba03e26a9b002.png)
Since 
Easy to find, that 
![\[\frac{e-d}{h-d} \div \frac{e-a}{h-a} = \frac{ba\overset{-}{p}-ac\overset{-}{p}}{ba\overset{-}{p}+b+c-p}\cdot \frac{b+c}{p-ac\overset{-}{p}}\]](http://latex.artofproblemsolving.com/f/0/c/f0c022b2f45922d7726873da85bda9d2ebf3c41c.png)
![\[\overset{-------------}{(\frac{e-d}{h-d} \div \frac{e-a}{h-a})}=\frac{cp-bp}{cp+ab+ac-abc\overset{-}{p}}\cdot \frac{a(b+c)}{abc\overset{-}{p}-bp}\]](http://latex.artofproblemsolving.com/a/b/5/ab54f6368432d2992d0fe6673d27cf0dc19dcea4.png)
It remains to prove, that![\[\frac{p}{cp+ab+ac-abc\overset{-}{p}}= \frac{b\overset{-}{p}}{ba\overset{-}{p}+b+c-p}\]](http://latex.artofproblemsolving.com/7/f/e/7fe1a0a618676522451c07f71b6f0bf344a62b72.png)
or![\[abp\overset{-}{p}+bp+cp-p^2=bcp\overset{-}{p}+ab^2\overset{-}{p}+abc\overset{-}{p}-ab^2c\overset{-}{p}^2 \implies\]](http://latex.artofproblemsolving.com/0/f/5/0f5cbbd71496790d84185759d36dd43bd2c822bf.png)
![\[(p-b-c+bc\overset{-}{p})(ab\overset{-}{p}-p)=0\]](http://latex.artofproblemsolving.com/9/5/3/95372294e79178ee362dd46b96db489420624162.png)
which is true, since 
We are done!
denote the feet of altitude from 
is a consequence of homothety at 
.
is projective because 
.
on 
.
are the feet of altitudes from 
,
)
,
is a altitude and that 
over 
.
is concyclic as desired.
with 
Since 
by LoC, 
Thus, 
Similarly, 
The equation of 
where 
and 
are constants. From 
we know 
from 
from 
Since 
the orthocenter of 
is our desired point. 
be the midpoints of 
respectively, 
meet 
at 
respectively. Trivially, we know 
and 
.
.
Now, observe 
so 
Now, notice 
and 
so 
where the last equality follows from 
.
which finishes. 
again at 
denote the reflection of 
,
to 
,
,
is also cyclic by the Spiral Center Lemma.
so 
is actually a cyclic hexagon.
and 
so 
.
to 
.
are the reflections of 
so 
which finishes. 
Let 
for some real 
Furthermore, let 
and 
for reals 
Notice that we have 
is isosceles, notice that we have ![\[\dfrac{EC/2}{PC} = \dfrac{bu}{2at} = \cos(C) \implies u = \dfrac{2at\cos(C)}{b}.\]](http://latex.artofproblemsolving.com/7/d/f/7df055082566b4a4aa5bf496247d5d13f57b6ae1.png)
Similarly, we can compute that ![\[v = \dfrac{2a(1 - t)\cos(B)}{c}.\]](http://latex.artofproblemsolving.com/4/9/3/493358a12162ff93e68941856de9fe046c42d17e.png)
Now, we know that the equation of ![\[-a^yz - b^2xz - c^2xy + (i_1x + i_2y + i_3z)(x + y + z) = 0\]](http://latex.artofproblemsolving.com/c/a/1/ca1e4d68fe4df8c3736a9c7d81fb91054d967ca8.png)
for some constants 
and 
Plugging the point 
gives 
putting in the point 
and 
Thus, after substituting the values of 
and 
the full equation is ![\[-a^2yz - b^2xz - c^2xy + (2ac(1 - t)\cos(B)y + 2ab(1 - t)z)(x + y + z).\]](http://latex.artofproblemsolving.com/5/8/b/58b469b61b67d7e631ad1e364bf1a79d4e214f90.png)
We now claim that the fixed point is the orthocenter 
Indeed, it suffices to prove that ![\[-a^2\tan(B)\tan(C) - b^2\tan(C)\tan(A) - c^2\tan(A)\tan(B) + (2act\sin(B) + 2ab(1 - t)\sin(C))(\tan(A) + \tan(B) + \tan(C)) = 0.\]](http://latex.artofproblemsolving.com/2/b/b/2bbdefe2f309b2d8f197fd073c7b9f057c89ced9.png)
Since 
where ![\[-a^2\tan(B)\tan(C) - b^2\tan(C)\tan(A) - c^2\tan(A)\tan(B) + 4S\tan(A)\tan(B)\tan(C) = 0\]](http://latex.artofproblemsolving.com/7/3/d/73dfeba03f64323f62fec43eba668473bf683a44.png)
since 
for any acute triangle. Dividing by 
we see that it only suffices to show that ![\[4S = \dfrac{a^2}{\tan(A)} + \dfrac{b^2}{\tan(B)} + \dfrac{c^2}{\tan(C)}.\]](http://latex.artofproblemsolving.com/e/8/0/e80d1da8df9b210fec34a7a7adf6dfc613493eff.png)
Now, notice that 
Thus, it suffices to prove that ![\[a^2(b^2 + c^2 - a^2) + b^2(c^2 + a^2 - b^2) + c^2(a^2 + b^2 - c^2) = 16S^2,\]](http://latex.artofproblemsolving.com/1/5/e/15e40e184c2f8427e154279fd7c57cff512c1bdc.png)
but the left-hand side factors as 
which can be seen to be 
by Heron's Formula. Thus, 
so we're done!
passes is the orthocentre
.
also as 
,
are concyclic.Similarly 
and 
are concyclic.Finally we do an angle chase,![$$\angle GDF=180-\angle PDF=180-(90-\frac{1}{2}\angle DPF)=180-[90-(90-\angle BAC)]=180-\angle BAC=\angle GHF$$](http://latex.artofproblemsolving.com/c/1/0/c10bfe82e04463253d22c6fdd56c1be13c2a51f3.png)
Which implies 
are concyclic which inturn implies 
all lie on a circle and thus 
and 
before going to bed and so found that the fixed point was the circumcenter of 
helped a lot and i was able to finish with miquel. Today morning when I visited this thread to type the sol I realised I misread the problem and tried to solve the correct problem. Somehow the same method worked and here I am with this solution.
and let 
and similarly 
,
,
which is a fixed point, the end 
and 
to find that 
This means that 
This shows that 
is a fixed point. However, the correct fixed point should be the orthocenter, and 
.
and 
.
Simplifying further, we get 
And then just plugging in the orthocenter which is 
,
where 
.
and 
.
,
,
.
and 
and then 
and 
and 
.
and 
.
-
.
is fixed (in fact it is 
,
.
remains fixed. Let 
.
.
and 
.
while 
![\[\frac{AD'}{D'C}\div\frac{AE'}{E'B}=\frac{1-\frac{c(b+c-a-x)}{r^2}}{\frac{bx}{r^2}-1}=\frac cb,\]](http://latex.artofproblemsolving.com/d/1/0/d10f4d0ce8eef39739757cbb2a28b8063bd0d52b.png)
So, we want 
and hence 
.![\[AI=\sqrt{\frac{(s-a)bc}{s}} \text{ and } AL=\frac2{b+c}\sqrt{s(s-a)bc} \Longrightarrow AI\cdot AL=\frac{2bc(s-a)}{b+c}=\frac{bc(b+c-a)}{b+c}=r^2\]](http://latex.artofproblemsolving.com/a/8/a/a8abeb2503b0cac6195e79f9b91bb70548d6e558.png)
We are done. 
be circle with radius 
be circle with radius 
centered at 
and let 
be circle 
r
are fixed points where 
and 
are colinear, and since 
we are done
