[/quote]
,
,
denote the number of positive divisors of n.![\[
n, d(n), d(d(n)), \ldots
\]](http://latex.artofproblemsolving.com/b/1/e/b1e33fd2ac00e4ee128feb59ee66f7536c7016fd.png)
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
be a positiv integer. Ana chooses the positive integers 
and for any non-empty subset 
she computes the sum ![\[s_A=\sum_{k
\in A}a_k.\]](http://latex.artofproblemsolving.com/7/8/a/78afe6d1036a006b2252023d8f9619285fe0c6d5.png)
She orders these sums 
Prove that there exists a subset 
with 
elements such that, regardless of the integers 
with 
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
,
to be the largest value in row 
,
.
to be the largest value in column 
.
?
,
be all the positive integers smaller than 
such that
for all 
.
is placed in
r
t
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
be a sequence of perfect squares such that 
can be obtained by concatenating a digit to the right of 
.
intersects the circumcircle of the acute-angled 
at point 
and 
intersect sides 
is the circumcenter of 
,
students each student has at least 
acquaintances among the remaining members of the group. What is more, there exists a student that has at least 
acquaintances in the group. Prove that for every pair of students 
,
in the group such that for every index 
,
and 
are acquaintances.
,
be positive real numbers. Prove that![\[\frac{a^4 + 3}{b} + \frac{b^4 + 3}{c} + \frac{c^4 + 3}{a} \ge 12.\]](http://latex.artofproblemsolving.com/1/f/d/1fdfd64c9aafa8741ad1594db46cc16a403b1a99.png)
When does equality hold?
dividing 
,
.
that satisfy![\[f(xf(y) + f(x)) = f(x)f(y) + 2f(x) + f(y) - 1,\]](http://latex.artofproblemsolving.com/4/7/f/47f094ce57fe35d339ec3b93152ec3459a22ec48.png)
for every 
,
for every 
and 
,
.
Y by DrMath, anantmudgal09, Catoptrics, itslumi, 554183, pog, MehmetBurak, Newmaths, Lcz, Adventure10, Mango247, Rounak_iitr, Funcshun840, X.Luser, and 1 other user
In triangle 
, 
, 
and 
are the points of tangency of incircle with the center of 
to 
, 
and 
respectively. Let 
be the foot of the perpendicular from to 
. 
is on 
such that 
. If 
is the orthocenter of 
, prove that 
bisects 
.
, 
, 
and 
are the points of tangency of incircle with the center of 
to 
, 
and 
respectively. Let 
be the foot of the perpendicular from to 
. 
is on 
such that 
. If 
is the orthocenter of 
, prove that 
bisects 
.This post has been edited 1 time. Last edited by khashi70, May 16, 2009, 6:07 PM
, 
, 
and 
are the points of tangency of incircle with the center of 
to 
, 
and 
respectively . Let 
be the feet of perpendicular from 
is on 
such that : 
. If 
be the orthocenter of 
bisects 
cut 
at 
and 
cut 
.
.
is cyclic and
.
and 
.
be midpoint of 
and points 
it's enough to prove that
.
we get 
,
.
.
as
is harmonical.
is orthic triangle of 
.
is bisector of 
.
we get that 
is exterior bisector of 
,
.
and 
(this result can be proved easily by harmonic division). Hence 
.
and 
be the projection of 
,
,
.
,
.
.
,
is the midpoint of 
Here is the proof without calculation.
to BC
are collinear.
midpoint of 
radius of incircle of 
.
at 
so 
)
)
(1)
so 
(2)
(3).
,
so
(4).
)
and 
be the feet of the perpendicular from 
and 
,
.
and 
of triangle 
respectively, so 
and therefore 
.
,![[asy]import graph; size(12.73cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-7.08,xmax=5.65,ymin=-3.59,ymax=3.72;
pen qqfftt=rgb(0,1,0.2), ttwwqq=rgb(0.2,0.4,0), zzttqq=rgb(0.6,0.2,0), ccqqqq=rgb(0.8,0,0);
pair A=(-3,2), B=(-4,-3), C=(3,-3), I=(-1.86,-1.24), D=(-1.86,-3), F=(-3.58,-0.9), H=(-1.86,2.92), B_1=(0.19,0.43), C_1=(-3.19,-0.76), M=(-2.71,-0.59), P=(-2.28,-1.79), X_C=(-4.46,1.37), X_B=(0.75,1.37);
D(A--B--C--cycle,linewidth(1.2)+dotted+qqfftt); D(D--(-0.73,0.11)--F--cycle,linewidth(1.2)+dotted+ttwwqq); D(B--C--H--cycle,zzttqq); D(B_1--C_1--D--cycle,blue);
D(A--B,linewidth(1.2)+dotted+qqfftt); D(B--C,linewidth(1.2)+dotted+qqfftt); D(C--A,linewidth(1.2)+dotted+qqfftt); D(CR(I,1.76),linetype("2 2")+ttwwqq); D(D--(-0.73,0.11),linewidth(1.2)+dotted+ttwwqq); D((-0.73,0.11)--F,linewidth(1.2)+dotted+ttwwqq); D(F--D,linewidth(1.2)+dotted+ttwwqq); D(B--C,zzttqq); D(C--H,zzttqq); D(H--B,zzttqq); D(B_1--C_1,blue); D(C_1--D,blue); D(D--B_1,blue); D(D--M); D(CR(I,0.9),linewidth(1.6)+linetype("2 2")+blue); D(P--H,linewidth(1.6)+linetype("2 2")+ccqqqq); D(C_1--X_C,linetype("4 4")+blue); D(B_1--X_B,linetype("4 4")+blue); D(CR(H,3.03),linetype("4 4")+blue);
D(A); MP("A",(-2.96,2.07),NE*lsf); D(B); MP("B",(-3.96,-2.93),NE*lsf); D(C); MP("C",(3.04,-2.93),NE*lsf); D(I); MP("I",(-1.82,-1.18),NE*lsf); D(D); MP("D",(-1.82,-2.93),NE*lsf); D((-0.73,0.11)); MP("E",(-0.68,0.17),NE*lsf); D(F); MP("F",(-3.54,-0.84),NE*lsf); D(H); MP("H",(-1.82,2.99),NE*lsf); D(B_1); MP("B_1",(0.23,0.49),NE*lsf); D(C_1); MP("C_1",(-3.15,-0.7),NE*lsf); D((-2.15,-0.39)); MP("N",(-2.11,-0.33),NE*lsf); D(M); MP("M",(-2.67,-0.52),NE*lsf); D(P); MP("P",(-2.24,-1.73),NE*lsf); D(X_C); D(X_B);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/c/f/6/cf68e9012a71459db08c5c91d589b4623ea78d1e.png)
,
.
is cyclic, from which it follows that 
.
,
is the orthic triangle of 
.
and 
now so equation of circle 
be 
.
point be 
,
in terms of 
.
is 
, now basically we need to show now slope of the line is 
and that’s quite easy to show. Note here we’re saying , solution is easy by CG because no stuffs involving any power is coming, and that makes our calculation easier. (I got a trig solution too, but there calculation is damn hard to do.)
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -1.535709806738513, xmax = 7.699401969463958, ymin = 1.537544229253762, ymax = 7.025735038013770; /* image dimensions */
pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000);
draw((4.120000000000004,6.740000000000007)--(0.6000000000000006,2.120000000000002)--(4.046266301600477,2.770413386903411)--cycle, zzttqq);
/* draw figures */
draw((4.120000000000004,6.740000000000007)--(0.6000000000000006,2.120000000000002), zzttqq);
draw((0.6000000000000006,2.120000000000002)--(4.046266301600477,2.770413386903411), zzttqq);
draw((4.046266301600477,2.770413386903411)--(4.120000000000004,6.740000000000007), zzttqq);
draw((2.219650741964543,4.245791598828465)--(4.061766128857575,3.604874537240598));
draw((3.226139448965370,2.615630953617537)--(3.623387848646997,3.757397078019321));
draw((0.6000000000000006,2.120000000000002)--(2.823005108246210,4.751669850437923));
draw((3.424763648806184,3.186514015818429)--(2.823005108246210,4.751669850437923));
draw((3.623387848646997,3.757397078019321)--(4.046266301600477,2.770413386903411));
draw((2.823005108246210,4.751669850437923)--(4.046266301600477,2.770413386903411));
draw((3.623387848646997,3.757397078019321)--(4.046266301600477,2.770413386903411));
draw((4.046266301600477,2.770413386903411)--(2.355699169206497,4.198457018789681));
draw((0.6000000000000006,2.120000000000002)--(3.385861897857764,3.840038173964708));
draw((3.035843169874220,3.623930701373472)--(2.823005108246210,4.751669850437923));
draw((3.226139448965370,2.615630953617537)--(3.035843169874220,3.623930701373472));
/* dots and labels */
dot((4.120000000000004,6.740000000000007),dotstyle);
label("$A$", (4.153046114633986,6.797820137317756), NE * labelscalefactor);
dot((0.6000000000000006,2.120000000000002),dotstyle);
label("$B$", (0.6340400478875363,2.175705951202600), NE * labelscalefactor);
dot((4.046266301600477,2.770413386903411),dotstyle);
label("$C$", (4.080113346411262,2.822984269179279), NE * labelscalefactor);
dot((3.035843169874220,3.623930701373472),dotstyle);
label("$I$", (3.068171187320962,3.679944295796290), NE * labelscalefactor);
dot((2.219650741964543,4.245791598828465),dotstyle);
label("$F$", (2.256794140843153,4.299872825689447), NE * labelscalefactor);
dot((3.226139448965370,2.615630953617537),dotstyle);
label("$D$", (3.259619703905613,2.668002136705990), NE * labelscalefactor);
dot((4.061766128857575,3.604874537240598),dotstyle);
label("$E$", (4.098346538466943,3.661711103740609), NE * labelscalefactor);
dot((3.623387848646997,3.757397078019321),dotstyle);
label("$M$", (3.660749929130597,3.807576640186058), NE * labelscalefactor);
dot((3.424763648806184,3.186514015818429),dotstyle);
label("$P$", (3.460184816518106,3.242347686459944), NE * labelscalefactor);
dot((2.823005108246210,4.751669850437923),dotstyle);
label("$H$", (2.858489478680629,4.810402203248517), NE * labelscalefactor);
dot((3.140708435411058,3.925333068034533),dotstyle);
label("$N$", (3.177570339655048,3.980791964715028), NE * labelscalefactor);
dot((2.355699169206497,4.198457018789681),dotstyle);
label("$J_c$", (2.393543081260761,4.254289845550244), NE * labelscalefactor);
dot((3.385861897857764,3.840038173964708),dotstyle);
label("$J_b$", (3.423718432406743,3.898742600464463), NE * labelscalefactor);
dot((2.967592072231618,3.985564490881980),dotstyle);
label("$X$", (3.004355015126078,4.044608136909912), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/1/3/b/13bb9b268912000c3fd2779f9bbac94e28b291c2.png)
and 
be the feet of perpendiculars from 
respectively. Since 
and 
and 
are collinear. Furthermore, since 
is cyclic and since 
is also cyclic. Now, note that 
and 
so 
implying 
are collinear. Similarly, 
are collinear. Combining them, we see that 
are collinear. Let this line intersect 
Using cyclic quadrilaterals 
and 
we deduce that![\[\angle J_cJ_bI = 90^{\circ} - \angle HJ_bJ_c = 90^{\circ} - \angle CJ_bE = \angle IJ_bC,\]](http://latex.artofproblemsolving.com/c/a/b/cabeb187952837e752997606337f505af928bb9e.png)
internally bisects 
Since 
externally bisects 
so by the angle bisector theorem we have 
and 
so 
implying 
is harmonic, or 
![\[\dfrac{PM}{PN} \cdot \dfrac{DH}{DX} \cdot \dfrac{XN}{MN} = 1,\]](http://latex.artofproblemsolving.com/2/3/4/2342f29f772fb8bd1a3c6b768eb4183c5672d27b.png)
we get
be the midpoint of 
and 
we get
.
.
,
are collinear as desired.
note that 
is cyclic so 
thus 
are collinear similarly 
are collinear now let the perpendicular lines from 
intersect 
note that 
(because both of them are perpendicular to 
)
thus 
so 
is perpendicular bisector of 
similarly 
is perpendicular bisector of 
so 
now using 
are collinear.
,
,
.
and 
the pentagon 
is cyclic with 
and hence 
.
the homothety centered at 
.
from which the assertion follows since 
since 
,
.
so 
which means the pentagon 
is cyclic since 
.
is the perpendicular bisector of 
,
which yields 
,
so we are done.
and 
,
as 
and 
are the feet of the 
,
is the orthic triangle of acute 
,
,
,
,
,
are on 
and 
are homothetic, and if 
is the reflection of 
is the same cevian in 
in 
(isotomic to altitude), we have 
as desired.
and 
denote the feet from 
and the feet from 
respectively. Note that 
is an orthocentric system, so 
are collinear as well as 
By that egmo lemma, 
Thus, 
,
are 
in 
.
Since 
is just the midpoint of 
,
.
Thus, 
.
is cyclic and thus 
which means that indeed 
is the foot of the perpendicular from 
the feet of the perpendiculars from 
to the opposite sides. Let 
be the 
W
)
.
and 
maps to 
and thus the 
maps to the 
under this inversion. Further, since 
remains fixed under this inversion 
)
be the reflection of 
be the centroid of 
the intersection of 
and 
(with 
)![\[GM \cdot GW = \frac{AG \cdot GW}{2} = \frac{A'G\cdot GY_A}{GD \cdot GY_A} = GD \cdot GY_A\]](http://latex.artofproblemsolving.com/3/e/8/3e87e56bf62f0e0252be7d04be612b480a9df243.png)
Thus, 
is cyclic. Now, let 
.![\[X_A'D\cdot X_A'M = X_A'Y_A\cdot X_A'W = X_A'B \cdot X_A'C\]](http://latex.artofproblemsolving.com/3/4/b/34b33627a06e3a7e2d4a2956bfa9371f4210b4e8.png)
Thus, 
and we have that 
.
be the 
and 
is cyclic. Now, note that from this it follows that,![\[\measuredangle X_AY_AD = \measuredangle WY_AD = \measuredangle AMD=\measuredangle H_AMD = \measuredangle H_AHA = \measuredangle X_AHD\]](http://latex.artofproblemsolving.com/5/f/6/5f68a355b87b448551eaeeefc680e9b315f2dfd1.png)
and thus, 
is cyclic and 
.
.
is cyclic. Further, since 
and 
and we also have,![\[\measuredangle H_AY_AD = \measuredangle H_AA' =90^\circ\]](http://latex.artofproblemsolving.com/7/9/b/79b887b58c94a718e5ad2b1ababae7a3ece2cc6a.png)
H
.
is the 
.
.
by Midpoint Theorem, which implies that 
must indeed be cyclic.![\[\measuredangle DY_AM = \measuredangle DNM = \measuredangle DX_AZ = \measuredangle DY_AZ\]](http://latex.artofproblemsolving.com/f/5/b/f5b26a89df4f18b11aab72088234f0adf489d2d0.png)
This clearly implies that 
and we put these two together and conclude that indeed 
are collinear as claimed.
and the foot of the perpendicular from 
is the perpendicular bisector of 
bisects 
![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
,
a
.
with incenter 
a
and 
.
and from 
,
is the orthic triangle of 
satisfies 
,
respectively. By the Iran Lemma, we see that 
are collinear. The condition that 
Since 
is the orthic triangle of 
we see that 
so the foot from 
is the 
Additionally, it is clear that 
.
,
.
trivially and cyclic variants. Now we write 
. Now we find 
. The circumcenter of 
,
.
.
.
,
,
.
is real, which is equivalent to showing 
which is obviously self conjugating, so we are done.
,
and 
be the midpoint of ![\[-1=(EF\cap BC,D;C,B)\stackrel{K}{=}(RD;HI)\stackrel{\infty_{\perp EF}}{=}(RM;SQ)\stackrel{H}{=}(D,M;\infty_{DM},HQ\cap DM),\]](http://latex.artofproblemsolving.com/f/1/4/f14d21a14f82237ebfaf055a0f65ee7d071de314.png)
so 
bisects ![\[(MD;P\infty) \overset{N}{=} (MN \cap HD, D; HI) = -1. \quad \blacksquare\]](http://latex.artofproblemsolving.com/c/f/5/cf5626d5f1492aa78b2ca5077f82c8b8c94ca28d.png)
and 
be the feet of the altitudes from 
and 
are cyclic.![\[\measuredangle H_BFI=\measuredangle H_BBI=90^\circ-\measuredangle BIH_B=90^\circ-\measuredangle BIC=-\tfrac{\measuredangle BAC}{2}=\measuredangle EFI\]](http://latex.artofproblemsolving.com/4/7/2/4720a84e40c3b1a10adce3b1505b9eb086fd380d.png)
with 
with 
and 
and 
.
be 
.
is linear. It suffices to show that 
.
.
traversing 
,![\[\frac{MN}{TN}\cdot \frac{TH}{DH}\cdot \frac{DP}{MP}=1\implies \frac{MN}{TN}=\frac{DH}{TH}\implies \frac{NT}{MT}=\frac{TH}{DH+TH}\]](http://latex.artofproblemsolving.com/f/7/2/f724e3b447b2087c44d611b11fcbe96833160a09.png)
Therefore,![\[f(N) = \frac{TH}{DH+TH}f(M)\]](http://latex.artofproblemsolving.com/f/4/1/f41bf130da346d3ba2f6f045d53855ac7a05fd94.png)
Since 
and 
,
.![\[\frac{TH}{DH+TH}=\frac{d(H,H_BH_C)}{2d(H,H_BH_C)+d(D,EF)}=\frac{H_BH_C}{2H_BH_C+EF}\]](http://latex.artofproblemsolving.com/6/5/3/6533ba724e44840f342667d06a5b03c625c6c0aa.png)
Let 
,
,
,
.
.
Now, we relate 
with 
and 
with 
.
;
.
implying 
while 
so![\[y=\cot{\angle C}=\frac{1-\tan^2\left(\frac{\angle C}{2}\right)}{2\tan\left(\frac{\angle C}{2}\right)} = \frac{1-u^2}{2u}\]](http://latex.artofproblemsolving.com/3/b/1/3b15b3181ef6abbf8b3c261d6d562b468ebebb29.png)
and the result follows.
and we are done.
Let 
.
be the medial triangle of 
.
and 
.
and 
;
and so 
and also 
.
.
which is trivial just because 
is a parallelogram and gliding principle or whatever.
