and 
Prove that
Let 
and 
Prove that
prove that![\[ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{d^2} + \frac{d^2}{a^2}\geq\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\]](http://latex.artofproblemsolving.com/0/0/4/004e7f148bbb0c797688f5195aa4002dbc316470.png)
If you know same tasks, link them please
Prove that
Let 
Prove that
Let 
and 
Prove that
Let 
Prove that
.
,
,
are concurrent and the point of concurrence is 
,
(
,
,
,
,
.
.
be reals such that 
Prove that
Where 
and 
Prove that 
.
Prove that 
,
,
Point 
is chosen on the circle with radius 
is chosen on the circle with radius 
is chosen on the circle with radius 
,
in terms of 
.
of moving 
unit in the positive direction and probability 
of moving 
.
Y by buratinogigle, phantranhuongth, Ramanujan_1729, Davi-8191, Tawan, HolyMath, thczarif, SSaad, centslordm, Adventure10, Mango247
Let 
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray at 
and ray at 
. Let 
be a point on segment 
. The line through parallel to intersects line at 
. Let 
be the intersection of lines 
and , and let 
be the intersection of line and the line through parallel to . Prove that line 
is tangent to .
Warut Suksompong, Thailand
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray at 
and ray at 
. Let 
be a point on segment 
. The line through parallel to intersects line at 
. Let 
be the intersection of lines 
and , and let 
be the intersection of line and the line through parallel to . Prove that line 
is tangent to .Warut Suksompong, Thailand
This post has been edited 2 times. Last edited by MellowMelon, May 18, 2017, 3:30 AM
Reason: add proposer
Reason: add proposer
be the infinity point on 
From 
we get the second tangent from 
to 
is parallel to 
so from the converse of Brianchon's theorem for the degenerate hexagon 
we conclude that 
is tangent to 
be a diameter of 
now (at infinity).![[asy]size(5cm);
defaultpen(fontsize(9pt));
pair E = dir(140);
pair F = dir(40);
pair A = 2*E*F/(E+F);
pair O = origin;
pair X = dir(100);
pair N = 2*E*X/(E+X);
pair M = extension(A, F, N, X);
pair G = -E;
pair P = 2*F*G/(F+G);
pair R = extension(N, P, E, F);
filldraw(unitcircle, palecyan, blue);
draw(E--A--P, blue);
draw(O--P, mediumcyan);
draw(E--F, mediumcyan);
draw(M--R, red);
draw(A--E, red);
draw(G--P, dashed+red);
draw(N--P, blue);
draw(E--G, dotted+red);
draw(N--X, blue);
draw(X--M, dotted+blue);
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$A$", A, dir(A));
dot("$O$", O, dir(225));
dot("$X$", X, dir(X));
dot("$N$", N, dir(N));
dot("$M$", M, dir(M));
dot("$G$", G, dir(G));
dot("$P$", P, dir(P));
dot("$R$", R, dir(250));
[/asy]](http://latex.artofproblemsolving.com/6/3/c/63cc0d2f11a0c7e4622ce3d9fa65fcb6068cc0db.png)
to a rectangle centered at 
.
collinear, done.
.
,
.
But 
,
Let us view 
,
-
at 
,
then prove 
.
is tangent to 
at 
,
then prove 
.
Next, let's compute 
.
Then likewise 
.
.
Clearly 
is the diameter of 
moreover 
be the intersection of 
and 
It's easy to see that 
lies on the polar of 
hence we conclude that 
(here 
follows by parallel condition), which amounts to 
is tangent to 
![[asy]
size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt);
void b(){
pair O=D("O",origin,S);
pair E=D("E",dir(150),dir(150));
pair F=D("F",dir(30),dir(-100)*1.5);
pair A=D("A",IP(L(-E*dir(90)+E,E,5,5),L(-F*dir(-90)+F,F,5,5)),N);
pair P=D("P",IP(L(F-E,O,5,5),L(A,F,5,5)),dir(45));
pair R=D("R",6*F/11+5*E/11,dir(-130));
pair N=D("N",extension(P,R,A,E),dir(135));
pair X=D("X",IP(CP(N,E),unitcircle),dir(60));
pair M=D("M",extension(N,X,A,F),dir(45));
pair Y=D("Y",extension(X,R,E,O),dir(-60));
D(unitcircle);
D(A--E--F--cycle);
D(N--M,blue+dashed);
D(X--Y--E,red);
D(O--P--F);
D(M--R);
D(CP(M,R),dotted);
D(X--O--F);
D(Y--P,dashed);
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/2/c/a/2ca5a977dada55ae57fafcd0631bdc301616f2ff.png)
be a diameter of 
now.
.
meet 
at 
.
,
it suffices to prove that
,
,
,
and 
.
,
.
,
,
ane 
.
.
or 
is the point at infinity? They are not parallel, are they?![[asy] pair E = dir(220); pair F = -E; pair X = dir(140); pair G = -X; pair O = origin; pair N = 2*E*X/(E+X); pair P = 2*F*G/(F+G); pair M = extension(P, F, N, X); pair R = extension(N, P, E, F); pair T = extension(N, E, P, G);
filldraw(unitcircle, invisible, blue); draw(M--P, blue); draw(O--P, heavycyan); draw(E--F, heavycyan);
draw(M--T, red); draw(N--T--P, red); draw(N--P, blue); draw(E--G, dotted+red); draw(N--X, blue); draw(X--M, dotted+blue);
dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$X$", X, dir(X)); dot("$G$", G, dir(G)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$R$", R, dir(315)); dot("$T$", T, dir(T));
/* TSQ Source:
E = dir 220 F = -E X = dir 140 G = -X O := origin N = 2*E*X/(E+X) P = 2*F*G/(F+G) M = extension P F N X R = extension N P E F R315 T = extension N E P G
unitcircle 0.1 lightcyan / blue M--P blue O--P heavycyan E--F heavycyan
M--T red N--T--P red N--P blue E--G dotted red N--X blue X--M dotted blue
*/ [/asy]](http://latex.artofproblemsolving.com/8/a/3/8a33d6f17aa64182bda0c17dc50ffd90278faf46.png)
be the second tangent through 
by 
and 
and 
Since OP is parallel to FE it follows that 
Also 
Now it needs to be proven that 
Using 
and 
we get the desired result.
and 
is tangent to 
be the point at infinity along 
and 
.
are collinear, which follows from Newton's Theorem in 
.
Hence degree of the line 
is at most 
So the degree of the pole of 
which hence moves on a conic, and so by the projective dual of this 
.
we need to check this for 
positions of 
since a conic is uniquely determined by 
and the result is trivial here. Take 
such that 
and the result is again true here. For the last position, take 
.
concur at the point at infinity 
.
are collinear.
is a tangent to 
be a point on 
is a tangent to 
and 
we get that 
are collinear. So, 
.
its orthocenter 
the foot from 
Let 
to 
and 
to 
are collinear
a point on 
, 
prove that 
and let 
the orthocenter of 
and from the lemma above 
are collinear thus 
the circle with center 
are tangent to it and since 
we have 
is the midpoint of 
then 
,
be a triangle with 
as 
e
touches 
at 
.
.
and 
is center of 
.
.
and 
Thus, the tangents at 
to 
be a point on ray 
is tangent to 
is parallel to 
.
is tangent to 
.
be the point at infinity on line 
,
,
,
being the intouch triangle, and 
be the 
respectively. Then the intersection of the line through 
which is parallel to 
is the reflection of 
and lies on the 
inversion at 
is mapped to 
are parallel. Taking a reflection in the midpoint of 
at 
as the reference triangle, we see that 
(since 
coincides with the 
,
at 
resp. 
meet line through 
meet 
.
we see that tangent from 
.
we have that 
.
is a diameter.
the point at infinity defined by 
.
are still collinear and 
'
.
.
at 
is tangent to 
,
and 
.
,
and 
.
.
and 
.
.
are collinear. Let calculate the degrees of the points (here degrees are at most).
are fixed. Therefore 
.
has 
cases that 
then easily 
.
then same as previous.
degrees we have to show only 
then collinear and when 
then also collinear.
gives us the desired. 
Let distinct from 
since angle 
is fixed, mapping 
is composition of rotation and central projection, so it's a homography. Also composition of central projections 
is a homography, so to prove 
it's suffice to check it for three cases; trivial when 
When 
we see 
and 
and let the tangent from 
) intersect line 
all are projective
is projective
be the tangency point of 
with 
all are projective hence 
is also projective
,
so 
so 
then 
so to sho
but note that :
and 
are tangent to 
Points 
lies on 
,
.
.![\[FL \perp PO \parallel EF \implies EL \text{ is a diameter of } \omega.\]](http://latex.artofproblemsolving.com/3/3/4/334df3f61856e90f40bc9c546315e0c8b75c06a3.png)
by the given parallel condition, we know 
and ![\[\angle FKR = \angle FEL = \frac{180 - \angle FOE}{2} = \frac{\angle A}{2} = \frac{\angle FAE}{2} = \frac{\angle FMR}{2}.~{\color{blue} \Box}\]](http://latex.artofproblemsolving.com/0/2/3/023e5de4975c66ff67e50597920b566d71b222b7.png)
,
is tangent to ![[asy]
size(270);
pair A, B, C, O, E, F, P, R, N, M, K, L;
A = dir(95);
B = dir(210);
C = dir(320);
O = incenter(A, B, C);
E = foot(O, A, C);
F = foot(O, A, B);
P = extension(O, O+F-E, A, B);
R = .2E + .8F;
N = extension(A, C, P, R);
M = extension(A, B, R, A+R-E);
K = foot(O, M, N);
L = 2*foot(F, P, O)-F;
draw(C--A--B^^E--F^^K--L);
draw(M--R^^P--N--K^^O--P--L);
draw(M--K^^L--extension(L, F, E, K)--E, dashed);
draw(incircle(A, B, C));
dot("$A$", A, dir(90));
dot("$B$", B, dir(270));
dot("$C$", C, dir(270));
dot("$E$", E, dir(0));
dot("$F$", F, dir(180));
dot("$K$", K, dir(90));
dot("$L$", L, dir(270));
dot("$O$", O, dir(270));
dot("$P$", P, dir(180));
dot("$M$", M, .6dir(140));
dot("$N$", N, dir(0));
dot("$R$", R, dir(305));
dot(extension(L, F, E, K));
[/asy]](http://latex.artofproblemsolving.com/9/f/d/9fd21d8452b493e277a84c500f94d0c6bd500130.png)
denote the line through 
.
denote the antipode of 
is the polar of 
be the point such that 
is tangent to 
,
.
.
we deduce that 
.
and 
is the line through 
is tangent to 
is projective. When 
the result is obvious. To show 
works, it suffices to prove that 
.
,
has degree 
over 
has degree 
are collinear, which is now a degree 
and the collinearity is obvious
we have 
and the reflected line is the other line parallel to 
,
and the collinearity follows.
such that 
.
,
,
concur.
,
.
and 
,
,
we need to solve
hence the intersection 
.
.
;
and 
,
so we're done. 
;
are collinear. In what proceeds all ratios are unsigned.
.
,
.
;
,
Clearly none of 
we conclude. 
(say they are concurrent at 
,
to a rectangle. Thus: by converse of Brianchon's theorem on 
,
is tangent to 
.![\[ \tfrac{ER}{RF} \cdot \tfrac{AN}{NE} = \tfrac{AM}{MF} \cdot \tfrac{AN}{NE} = \tfrac{2c}{a+b-c} \tfrac{2b}{a-b+c} = \tfrac{2}{1 - \tfrac{b^2 - c^2-a^2}{2bc}} = \tfrac{2}{1 - \cos \angle MAN } = \tfrac{1}{\sin^2 \angle MAN} = \tfrac{AP}{PO} \cdot \tfrac{PO}{PF} = \tfrac{AP}{PF},\]](http://latex.artofproblemsolving.com/5/d/5/5d524dc087a15746f48ea5476681a90bc5ac06d9.png)
implying the collinearity by Menelaus 
and 
)
Observe that 
and so they are concurrent at the point of infinite along line 
to a rectangle
becomes the center of the circle.
obviously becomes a rhombus, where 
is the point of infinity along 
