[/quote]
,
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
such that 
for all 
.
,
,
be positive real numbers such that 
.![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
and 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
such that ![\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]](http://latex.artofproblemsolving.com/2/b/a/2baee1d8464992fc38a16561b48b18f4b35807bd.png)
for all 
.
and in 
steps and has to reach 
is valid but 
is not valid.
,
lie on a line in this order and satisfy 
.
be the midpoints of 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
such for any 
.
satisfy,
Where 
}
be an integer. We are given a balance and 
weights of weight 
.
to satisfy that the equation 
has at least 1 roots 
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
.
of the equation![\[n^3=p^2-p-1\]](http://latex.artofproblemsolving.com/f/f/8/ff88093bb0aa93afcfdff3ae25149ec5fbacb61f.png)
is a prime number and 
?
Y by Ifhml, mathematicsy, Adventure10, Mango247
In the isosceles triangle 
with 
, the center of 
is the midpoint of the side 
, and 
are tangent to the circle at points 
respectively. Point 
is on with 
. A tangent line of passes through , and meets 
at 
. Prove that line 
bisects 
.
with 
, the center of 
is the midpoint of the side 
, and 
are tangent to the circle at points 
respectively. Point 
is on with 
. A tangent line of passes through , and meets 
at 
. Prove that line 
bisects 
.This post has been edited 1 time. Last edited by MarkBcc168, Jul 31, 2018, 7:29 AM
is a harmonic pencil, which is equivalent to 
being harmonic.
intersect the circle 
a second time at 
,
,
are symmetric wrt 
is tangent to 
is harmonic, which is equivalent to 
being a harmonic quadrilateral, if 
.
is the polar of 
:
be the reflection of 
are collinear.
and 
are cyclic quadrilaterals, so 
.
will also tangent to 
,
must be collinear. So, 
which implies that 
.
and 
are tangent to 
is the polar of 
and 
is the polar of 
.
which is 
is the projection of 
is the midpoint of 
is cyclic."
are collinear. Q.E.D.
are colinear so quadrilateral 
is harmonic. Taking pencil at 
,
,
and project it to 
Moreover, by taking pencil at 
hence 
where 
at 
meet 
at 
,
denotes the point at infinity on 
is tangent to 
so 
so 
and 
.
.
,
.
,
,
are all harmonic. By Ptolemy's, 
,
,
Therefore, 
is harmonic. Now, 
as desired.
and 
.
Since 
,
and
By computation, 
which is true. 
and 
are tangent to a circle 
at 
and 
and 
the following points are collinear: 
Similarly, by Pascal's Theorem on "hexagon" 
the following points are collinear: 
Therefore, 
and 
be the second intersection point of 
and 
and 
and 
By the above lemma, it suffices to show that 
is tangent to 
quadrilateral 
is cyclic, so 
.
is tangent to 
at 
so 
This implies that 
and thus 
,
and 
,
meet 
.
is harmonic, so 
.
meet 
,
![[asy]
import markers;
unitsize(2inch);
pair A, B, C, O, E, F, G, J, K, L, X, Y, M, N;
A = dir(90); B = dir(205); C = dir(335);
draw(B--A--C); draw(B--C, lightblue);
O = .5B + .5C;
E = foot(O, A, B); F = foot(O, A, C); draw(E--F, lightblue); X = .5E + .5F;
J = 2*foot(F, B, C) - F; L = 2*foot(E, B, C) - E; G = extension(A, J, X, L);
draw(circumcircle(A, G, E), dashed+lightmagenta); draw(circumcircle(E, F, J), dashed+lightmagenta);
draw(A--O); draw(E--G); Y = extension(E, G, A, X);
K = extension(.5A+.5E, G, A, F);
draw(K--B, dotted); draw(G--L, lightred); draw(G--F); draw(L--O);
M = .5A+.5E; draw(G--M--X, green); draw(G--K); draw(K--Y, lightblue); draw(K--O, lightred);
N = extension(K, O, F, X);
draw(G--O--F, green);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(200)); dot("$F$", F, dir(F)); dot("$O$", O, dir(O)); dot("$X$", X, dir(315)); dot("$G$", G, dir(40)); dot("$Y$", Y, dir(135)); dot("$K$", K, dir(K)); dot("$M$", M, dir(M)); dot("$N$", N, dir(310));
[/asy]](http://latex.artofproblemsolving.com/c/d/b/cdb19f83b83ed2e068148348fdaf68bd02127257.png)
.
and 
are tangents to 
,
and 
are collinear. From 
,
.
and 
concur on 
shows that 
bisects 
.
,
is parallel to 
,
.
;
.
,
.
are reflections over 
tangent to 
is harmonic and we're done.
,
and 
.
is the polar of 
so 
are collinear and the tangent at 
are collinear.
.
and 
.
by similar triangles.
and we are done.
are collinear
are cocyclic
and immediately 
is the midpoint of 
,
be 
.![\[\angle F'GE = \angle AGE = 90^{\circ},\]](http://latex.artofproblemsolving.com/c/9/e/c9edab9093d09e711689b921e73b48ada9f03852.png)
is a diameter, which implies that 
.
,
is tangent.
,
is a harmonic quadrilateral since the intersection of the tangents at ![\[-1 = (G, F; T, F') \stackrel{F'}{=} (A, F; K, B) \stackrel{C}{=} (E, F; M', \infty) \]](http://latex.artofproblemsolving.com/4/c/0/4c02e9636a64edc1c97ea17fc9b273f49e4dae80.png)
be an intersection of 
with 
,
,
is tangent to 
and 
intersect at 
be the point at infinity on 
so 
Let 
and 
.
is harmonic. Taking perspectivity at 
gives,![\[(G,F;X,Z) \stackrel{Z}{=} (A,F;K,B) = -1\]](http://latex.artofproblemsolving.com/5/e/8/5e8cdc0843d213ed474b05805d08a71f122d81f4.png)
. Now to finish we take perspectivity at ![\[(A,F;K,B) \stackrel{C}{=} (E,F;D,P_{\infty})=-1\]](http://latex.artofproblemsolving.com/2/3/9/239d622a0b900a192dc698eec3fc6415a0e92ad7.png)
which implies that 
is isosceles therefore 
L
,
Polar of 
are collinear.
and therefore 
therefore
is the midpoint of 
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draw((0.027980405959922017,0.6017273866500511)--(0.14421777207501246,0.23396495613663015)--(0.5119802025884335,0.3502023222517206)--(0.395742836473343,0.7179647527651416)--cycle, linewidth(1) + qqwuqq);
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draw(shift((-2.7968531467322975,2.520250947223354))*xscale(3.666183934108524)*yscale(3.666183934108524)*arc((0,0),1,64.52561847876126,244.52561847876123), linewidth(1) + zzffff);
draw(shift((-2.7968531467322975,2.520250947223354))*xscale(3.666183934108524)*yscale(3.666183934108524)*arc((0,0),1,-115.47438152123875,64.52561847876125), linewidth(1) + zzffff);
draw((-1.22,5.83)--(1.8021821645099834,-3.731856874613141), linewidth(1));
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draw((-4.373706293464595,-0.7894981055532924)--(0.395742836473343,0.7179647527651416), linewidth(1));
draw((-4.373706293464595,-0.7894981055532924)--(1.8256884618951665,-0.8398875813570109), linewidth(1) + qqccqq);
draw((-5.06,-2.23)--(1.3638897013666347,0.17142590538703653), linewidth(1));
draw((0.395742836473343,0.7179647527651416)--(1.3638897013666347,0.17142590538703653), linewidth(1));
draw((-4.373706293464595,-0.7894981055532924)--(1.8021821645099834,-3.731856874613141), linewidth(1));
/* dots and labels */
dot((-1.22,5.83),dotstyle);
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dot((-5.06,-2.23),dotstyle);
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dot((2.488475871045388,-2.2913549801664335),dotstyle);
label("$C$", (2.554545454545448,-2.1027272727272694), NE * labelscalefactor);
dot((-1.2857620644773058,-2.260677490083217),linewidth(4pt) + dotstyle);
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dot((-4.373706293464595,-0.7894981055532924),linewidth(4pt) + dotstyle);
label("$E$", (-4.3,-0.6481818181818154), NE * labelscalefactor);
dot((1.8256884618951665,-0.8398875813570109),linewidth(4pt) + dotstyle);
label("$F$", (1.9,-0.70272727272727), NE * labelscalefactor);
dot((0.395742836473343,0.7179647527651416),linewidth(4pt) + dotstyle);
label("$G$", (0.4636363636363574,0.8609090909090931), NE * labelscalefactor);
dot((1.3638897013666347,0.17142590538703653),linewidth(4pt) + dotstyle);
label("$K$", (1.445454545454539,0.3154545454545478), NE * labelscalefactor);
dot((1.8021821645099834,-3.731856874613141),linewidth(4pt) + dotstyle);
label("$Z$", (1.8818181818181756,-3.5936363636363597), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/b/3/4/b3423b03934b764a1f04885d3a286d6bc1511344.png)
and 
by some easy angle chasing, and by symmetry (or just angle chasing) we get that 
is tangent to 
So, it suffices to show that 
and 
it suffices to show that 
Another interesting fact: This midpoint lies on 
as 
Let 
be the reflection of 
Since 
and 
we know 
Hence, 
is tangent to 
Therefore, 
where 
Noticing 
finishes. 
![[asy]
size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6;
pen greenfill,greendraw,turquoisedraw,lightbluedraw,bluedraw,purpledraw,pinkdraw;
greenfill = RGB(204,255,204);
greendraw = RGB(0,187,0);
turquoisedraw = RGB(0,170,153);
lightbluedraw = RGB(68,204,255);
bluedraw = RGB(0,102,255);
purpledraw = RGB(170,34,255);
pinkdraw = RGB(255,17,255);
pair A,B,C,M,E,F,EE,G,K,X,L;
A = (0,3);
B = (-2,0);
C = (2,0);
M = (B+C)/2;
E = foot(M,A,B);
F = foot(M,A,C);
path c = circle(M,distance(M,E));
EE = 2M-E;
G = intersectionpoints(c,A--EE)[0];
K = extension(M,foot(M,F,G),A,C);
X = extension(B,K,E,F);
L = intersectionpoints(c,EE--K)[0];
filldraw(c,greenfill,greendraw);
draw(A--B--C--cycle,bluedraw);
draw(A--EE,pinkdraw);
draw(E--EE,turquoisedraw);
draw(E--F,purpledraw);
draw(B--K,lightbluedraw);
draw(G--K,lightbluedraw);
draw(EE--K,lightbluedraw);
draw(C--EE,bluedraw);
dot("$A$",A,dir(90));
dot("$B$",B,dir(180));
dot("$C$",C,dir(0));
dot("$M$",M,dir(240));
dot("$E$",E,dir(140));
dot("$F$",F,dir(40));
dot("$E'$",EE,dir(320));
dot("$G$",G,dir(50));
dot("$K$",K,dir(50));
dot("$X$",X,dir(110));
dot("$L$",L,dir(215));
[/asy]](http://latex.artofproblemsolving.com/3/5/7/357cac761c6337f1aa66e23b8077a38ad59460a7.png)
and 
,
intersect 
and let 
be the point at infinity on ![\[-1=(G,F;L,E')\overset{E'}{=}(A,F;K,C)\overset{B}{=}(E,F;X,\infty_{BC}),\]](http://latex.artofproblemsolving.com/b/1/8/b1814e88194d40b8ad8cc2ecdc97835dcb11a67f.png)
so 
and 
.
is the center of 
and as 
is tangent to 
is also tangent to 
are collinear.
,
is tangent to 
are collinear. Finally, 
.
,
is a parallelogram, which further gives that 
are collinear.
is the midpoint of 
,
is tangent to 
which implies that 
be a convex cyclic quadrilateral. Then 
,
,
are collinear.
,
,
in cyclic fashion. By Brianchon on 
,
are concurrent. By Brianchon on 
,
are concurrent. Hence 
as desired. 
,
intersect 
hence 
.
passes through 
to get that 
.
such that 
is tangent to 
and hence solves the problem.
,
and 
respectively, so we can easily calculate 
.
.
.
,
(although this is not hard either): instead, notice that 
should bisect 
for homothety reasons, hence its slope is twice that of 
,
.
,
passing through 
.
,
.
are collinear, we calculate the slope of 
We will now calculate the slope of 
,
beforehand this is just equal to
hence the desired collinearity holds and we are done. 
![\[ \measuredangle FE'M = \measuredangle FE'E = \measuredangle FEE' + \measuredangle E'FE = \measuredangle BFE' + \measuredangle E'FE = \measuredangle BFE = \measuredangle FBM. \]](http://latex.artofproblemsolving.com/2/b/d/2bd7de52fe2682f80f0ad8be36c223a496e14cc3.png)
![\[ (G,F;\overline{KE'} \cap \Gamma,E') \overset{E'}= (AF;KB) \overset{C}= (EF;I\infty) = -1 \]](http://latex.artofproblemsolving.com/8/8/3/8833362dda5a7b05310eab1dacb06ccd0053fa0e.png)
and the result follows.
be the intersection of 
,
is a diameter of 
,
as 
is tangent to 
is tangent to 
be the intersection of 
and 
and 
are tangents to 
,
is harmonic, so that 
.
This forces 
This problem was copied from APMO 2016 P3 
.
and 
are homothetic about 
and 
are homothetic about 
bisects 
.![\[-1 = (H, HK \cap \Gamma; FG) \overset{H}{=} (BK, FA) \overset{M}{=} (\infty, KM \cap FN; FN). \quad \blacksquare\]](http://latex.artofproblemsolving.com/0/6/f/06ff4d14c7711e015380657797ef4dd01a953b1a.png)
.
.
is a harmonic quadrilateral, so 
.
so 
.
.
.
.
.
.
if L is the midpoint of FE, then 
and since 
its obvious that 
and 
Since 
,
Finally taking perspectivity at 
which implies 
,
,
.![\[-1 = (F,G;D,E') \overset{E'}{=} (F,A;K,B) \overset{C}{=} (F,E;N, \infty). \ \square\]](http://latex.artofproblemsolving.com/d/f/0/df0ebc456819b75752147a6cc6271409a2872a0e.png)
