and 
Prove that
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
,
is an homoemorphisme too.
,
,
,
be non-negative real numbers such that![\[2a+2b+2c+2d+ab+bc+cd+da+3=abcd.\]](http://latex.artofproblemsolving.com/d/e/8/de8ccd816cfe032a97d940982192b1abea21dc18.png)
prove that : ![\[\sqrt[4]{abc}+\sqrt[4]{bcd}+\sqrt[4]{cda}+\sqrt[4]{dab}\le\sqrt[4]{27(1+a)(1+b)(1+c)(1+d)}.\]](http://latex.artofproblemsolving.com/a/f/c/afc556cd8f5c7f4dedf153ffb92fb1f978c56823.png)
such that 
.
May everyone not to upload solutions on this problem anymore until May 15/2025, this is an active problem on Mathematical Reflection! (Thank you Victoria_Discalceata1)
Y by tenplusten, Pluto1708, nguyendangkhoa17112003, AlastorMoody, Muaaz.SY, megarnie, aaaa_27, Adventure10, Mango247, and 2 other users
Let 
be a line parallel to the side 
of a triangle 
, with 
on the side 
and 
on the side 
. The lines 
and 
meet at point 
. The circumcircles of triangles 
and 
meet at two distinct points and 
. Prove that 
.
Liubomir Chiriac, Moldova
be a line parallel to the side 
of a triangle 
, with 
on the side 
and 
on the side 
. The lines 
and 
meet at point 
. The circumcircles of triangles 
and 
meet at two distinct points and 
. Prove that 
.Liubomir Chiriac, Moldova
This post has been edited 3 times. Last edited by v_Enhance, May 22, 2015, 10:55 PM
Reason: \equal{} -> =
Reason: \equal{} -> =
is a complete quadrilateral, therefore, 
,
are the sets of concyclic points. Hence, 
.
.
,
triangles have the same direction, hence, through the spiral similarity with center 
,
,
,
,
,
,
.
,
.
,
,
,
and since, 
.
is the symmedian of 
,
post 
, i.e. the point 
-
- the distance of the point 
to the line 
. I"ll present below two simple and well-known properties :
The geometrical locus of an interior point 
of the angle 
for which 
is the 
The geometrical locus of an interior point 
is the 
two mobile points point 
(constant)
and the second intersection 
, i.e. the geometrical locus of the point 
be the midpoint of 
are collinear then 
is the median of triangle 
be the projection of 
.
QED
meet 
at 
.
,
and 
.
is the Miquel Point of the compelte quadrilateral with vertices 
.
and 
so 
(1).
and 
so 
(2).
,
is a symmedian of 
.
as 
.
is a median, so 
,
.
,
.![\[\measuredangle ABQ=\measuredangle MBQ=\measuredangle MPQ =-\measuredangle QPM =-\measuredangle QPC =\]](http://latex.artofproblemsolving.com/d/d/9/dd9ea7ab8cbeb24e7acc169a889f50529eef0b81.png)
![\[=-\measuredangle QNC=-\measuredangle QNA =\measuredangle ANQ.\]](http://latex.artofproblemsolving.com/1/3/a/13a81cbc36712362adae528d412d8ba45ad12f9d.png)
Hence, 
is cyclic. Similarly, 
is cyclic.
.
.
to 
.![\[\measuredangle QMB =\measuredangle QCN =\measuredangle QFN =-\measuredangle NFQ =-\measuredangle NFM=\measuredangle MFN.\]](http://latex.artofproblemsolving.com/3/a/a/3aa6278e2b6a479723fd52b730313fe84e8e8402.png)
So 
,
.
be the point such that 
is a parallelogram. Then 
.
Similarly, 
So 
is the center of the unique spiral similarity sending 
to 
,
and hence 
are cyclic.
and![\[\measuredangle NPA=\measuredangle NPH =\measuredangle NGH=\measuredangle NGM=-\measuredangle MGN =-\measuredangle MGQ=\]](http://latex.artofproblemsolving.com/7/8/b/78baa9683467fd16c79b3a875cece78ad314cc7a.png)
![\[=-\measuredangle MBQ=\measuredangle QBM=\measuredangle QBA.\]](http://latex.artofproblemsolving.com/6/3/2/632290a7eba3f3af452dffb4af7ed5495f9d5a4a.png)
So 
,
,
are good. Now we sense it suffices to show that 
are cyclic quads. Apply 
inversion with centre 
.
goes to 
,
go to 
and 
go to 
and 
is the intersection of 
and 
.
is a median in triangle 
.
pass through the isogonal conjugate of the complement of 
,
is A-median of 
pass through the complement of 
.
and 
are cyclic.
and 
.
,
and 
as desired. 
(equal since 
)
(as 
)
.
-
so there is a spiral similarity mapping 
to 
Hence, the ratio of the distance from 
and 
is 
Since the ratio of the distance from the midpoint of 
to 
we see that 
is the 
Since 
.
and 
so 
and 
are similar.
so 
is symmedian.
By Ceva's Theorem, 
but because 
so 
Since 
and 
implying that it is in fact the midpoint of 
s
First note that 
and the second intersection of 
and 
is 
so there is a spiral similarity centered at 
As a result, 
denote the distance from 
denote the distance from 
Because 
if and only if 
where 
denotes the distance between 
Since 
as desired.
![[asy]
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -1.1, xmax = 0.6, ymin = -0.8, ymax = 0.8; /* image dimensions */
pen wwccqq = rgb(0.4,0.8,0);
/* draw figures */
draw((-0.5649135165662259,0.536594333380709)--(-0.8852842704686563,-0.32441412262196817), linewidth(0.5));
draw((-0.8852842704686563,-0.32441412262196817)--(0.34335010783265624,-0.3302004697725238), linewidth(0.5));
draw((0.34335010783265624,-0.3302004697725238)--(-0.5649135165662259,0.536594333380709), linewidth(0.5));
draw((-0.752919505397859,0.03132116411857255)--(-0.03190918098033823,0.02792551109933775), linewidth(0.5));
draw((-0.8852842704686563,-0.32441412262196817)--(-0.03190918098033823,0.02792551109933775), linewidth(0.5));
draw((-0.752919505397859,0.03132116411857255)--(0.34335010783265624,-0.3302004697725238), linewidth(0.5));
draw((xmin, -2.938976377952756*xmin-1.1236731473936517)--(xmax, -2.938976377952756*xmax-1.1236731473936517), linewidth(0.5) + wwccqq); /* line */
draw(circle((-0.6122085315419402,-0.22352897588386156), 0.29111539296770506), linewidth(0.5) + blue);
draw(circle((-0.052563305306497866,-0.3693858698488498), 0.3978478682645369), linewidth(0.5) + blue);
draw((xmin, -7.985780076992614*xmin-3.9746807724376945)--(xmax, -7.985780076992614*xmax-3.9746807724376945), linewidth(0.5) + wwccqq); /* line */
draw(circle((-0.534292834968941,0.03509349622859525), 0.502434787615023), linewidth(0.5) + linetype("4 4") + red);
draw(circle((-0.1271159426332545,0.08608123799082845), 0.6281948462534274), linewidth(0.5) + linetype("4 4") + red);
/* dots and labels */
dot((-0.5649135165662259,0.536594333380709),linewidth(3pt) + dotstyle);
label("$A$", (-0.6821395092458079,0.5800735245046977), NE * labelscalefactor);
dot((-0.8852842704686563,-0.32441412262196817),linewidth(3pt) + dotstyle);
label("$B$", (-0.9701803985921769,-0.2699294920958601), NE * labelscalefactor);
dot((0.34335010783265624,-0.3302004697725238),linewidth(3pt) + dotstyle);
label("$C$", (0.37118648806983545,-0.3122884464115025), NE * labelscalefactor);
dot((-0.752919505397859,0.03132116411857255),linewidth(3pt) + dotstyle);
label("$M$", (-0.8600471173715064,0.05764642127844126), NE * labelscalefactor);
dot((-0.03190918098033823,0.02792551109933775),linewidth(3pt) + dotstyle);
label("$N$", (-0.03263554307262299,0.06611821214156974), NE * labelscalefactor);
dot((-0.3475011702611344,-0.10237541668523897),linewidth(3pt) + dotstyle);
label("$P$", (-0.35173966558379643,-0.05813472051764801), NE * labelscalefactor);
dot((-0.44030864113663953,-0.4584727983210283),linewidth(3pt) + dotstyle);
label("$Q$", (-0.4477532953659194,-0.4167738670567538), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/3/9/b/39b3818bbab5063771eb7584791db96f514fceaa.png)
is cyclic and similarly, we also get that 
is also cyclic.
followed by a reflection along the angle-bisector of 
.
.
from which we get that 
.
and 
and thus 
swaps with 
and we are done.
.
and thus as 
,
,
and 
and proceed similarly as the rest above.
and 
;
Now the 
inversion swaps 
,
and 
,
,
,
Furthermore, by Ceva, 
and 
and 
,
gives 
Thus, the equation of 
Intersecting this with the A-symmedian, plugging in 
gives 
The important thing about this is that it is symmetric in 
so 
are cyclic.
and 
![\[\angle CBS=\angle LKS=\angle NKB=\angle NAB=\angle A\]](http://latex.artofproblemsolving.com/f/8/e/f8efe3f948fdc0f248efc4cf7a1d6631fffb6fde.png)
![\[\angle SCB=\angle SLK=\angle CLM=\angle CAM=\angle A\]](http://latex.artofproblemsolving.com/d/7/f/d7fe845f2c743749654730b0d2d7a2bcc82e3958.png)
So 
and 
are tangents to 
at 
are concurrent which gives that 
are collinear. 
is symedian so 
as desired
and 
where 
.
.
,
and 
for the circumcircle 
.
.
.
from ![\[a^2yz + b^2xz + c^2xy = \left((xc^2n + z\frac{a^2n+b^2m-c^2mn}{n+1}\right)(x+y+z)\]](http://latex.artofproblemsolving.com/5/1/b/51b506e4ba1e08e5bf487c2067f28b54c66831ff.png)
Thus symmetrically the equation for 
is ![\[a^2yz + b^2xz + c^2xy = \left((xb^2n + y\frac{a^2n+c^2m-b^2mn}{n+1}\right)(x+y+z)\]](http://latex.artofproblemsolving.com/0/b/a/0ba6a4563001cb9b60651e39375d05b5b3e062f8.png)
We can see that the point 
satisfies both equations. Obviously this point lies on the cevian of the isongonal conjugate of 
)![\[\frac{\text{d}(X, AB)}{\text{d}(X, AC)} = \frac{AB}{AC}, \]](http://latex.artofproblemsolving.com/0/f/7/0f77aa6f99490fe89d4bfcade07532a8976525ef.png)
where 
denotes the distance from point 
is where the 
The opposite direction is proved similarly. 
,![\[\frac{BM}{BQ} = \frac{NC}{NQ} \implies \frac{AB}{AC} = \frac{BM}{NC} = \frac{BQ}{NQ}, \]](http://latex.artofproblemsolving.com/5/5/2/552d8ef682554e589e810a4b6d91c9830dcfa6cf.png)
which by the above lemma is enough to show that 
,
,
.
,
stays unchanged
yields to the fact that 
is the 
in 
yields to the result 
,
and 
