[/quote]
,
such that for any real value of 
we have:
with circumcenter 
at 
respectively. 
is another intersection point of external bisector of 
with 
is 
incircle touch point to 
be points lie on 
intersect 
at 
. Assume that 
.
and 
are concyclic, and 
are concurrent.
are concurrent.
,
are concurrent.
be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
,
and 
are two (not necessarily distinct) numbers on his napkin.
,
,
are four (not necessarily distinct) numbers from the first line.
be a positive integer. Show that there exist a polynomial 
with integer coefficient that satisfy the following
for each 
such that 
coprime, 
and 
.
,such that 
P
be the number of positive divisors of 
be the number of positive divisors of 
when divided by 
.
.
be a scalene triangle with the orthocenter 
be the reflection of 
and 
be the reflection of 
.
and 
is perpendicular to 
.
be three mutually tangent circles such that 
are externally tangent at 
,
are internally tangent at 
are internally tangent at 
be the centers of 
.
meets 
intersects the 
,
,
.
and 
be the midpoints of 
and 
,
is perpendicular to 
.
.
.![$(a+b+c)^4(ab+bc+ca) - 9\sum{\frac{a}{c}} \ge 54[(a+b)(b+c)(c+a)+abc-1]$](http://latex.artofproblemsolving.com/d/3/c/d3cf439c79cb84ca23083e51b6190ab5118f34e8.png)
Y by HWenslawski, Adventure10, jhu08, Miku_, ehuseyinyigit
Given a triangle 
. Let 
be the circumcenter of this triangle . Let 
, 
, 
be the feet of the altitudes of triangle from the vertices 
, 
, 
, respectively. Denote by 
, 
, 
the midpoints of these altitudes 
, 
, 
, respectively. The incircle of triangle has center 
and touches the sides 
, 
, 
at the points 
, 
, 
, respectively. Prove that the four lines 
, 
, 
and 
are concurrent. (When the point concides with , we consider the line as an arbitrary line passing through .)
. Let 
be the circumcenter of this triangle . Let 
, 
, 
be the feet of the altitudes of triangle from the vertices 
, 
, 
, respectively. Denote by 
, 
, 
the midpoints of these altitudes 
, 
, 
, respectively. The incircle of triangle has center 
and touches the sides 
, 
, 
at the points 
, 
, 
, respectively. Prove that the four lines 
, 
, 
and 
are concurrent. (When the point concides with , we consider the line as an arbitrary line passing through .)
is the tangency point of the circle passing through B,C tangent to the incircle.
can be proved by radical axis.
inscribed in a circle with center 
.
,
,
be the feet of the altitudes of triangle 
,
,
,
,
,
the midpoints of these altitudes 
,
,
,
and touches the sides 
,
,
at the points 
,
,
,
,
,
and 
are concurrent. (When the point 
,
,
of triangle ABC (proof: see below). Hence, these three lines coincide with the lines 
,
,
,
are homothetic (since their corresponding sidelines are parallel, what can be easily seen); hence, the lines 
its semiprojection on a given line is an affine mapping).
and 
are homothetic...
of the 
in the points 
respectively. Then 
of the 
touch the sides 
in the points 
respectively. Then 
:
.
,
,
be the projections of 
,
,
are the midpoints of 
,
,
,
,
,
,
be the intersection of 
with 
.
the intersection of 
with 
.
is perpendicular to 
is similar to triangle 
.
is the midpoint of 
,
is also perpendicular to 
,
.
,
are equal. With the same argument for 
is the radical axis wrt 
be the intersection of these lines, we get that, 
is perpendicular to 
are collinear. Which leads to the result that ![[asy]import graph; size(12cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-10.98,xmax=12.96,ymin=-6.14,ymax=7.52;
pen zzttqq=rgb(0.6,0.2,0), qqttff=rgb(0,0.2,1), zzzzzz=rgb(0.6,0.6,0.6), zzzzqq=rgb(0.6,0.6,0), qqzzzz=rgb(0,0.6,0.6), aqgreen=rgb(0,0.4,0.4);
pair A=(-1.88,-3), B=(9.12,-3), C=(0.12,6), L=(0.12,-3), O=(3.62,0.5), I=(1.86,0), L_C=(-8.78,-3), C_0=(0.12,1.5), F=(1.86,-3), P=(0.81,-0.29), P_C=(1.11,-3), K_C=(2.74,-3);
D(A--B--C--cycle,zzttqq);
D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,3),linetype("4 4")+qqttff); D(C--L,zzzzzz); D(C_0--F,zzzzqq); D(O--L_C,linewidth(1.2)+green); D(C--P_C,qqzzzz); D(L_C--A); D(C--K_C,qqzzzz); D(F--I,linetype("0 3 4 3")+zzzzzz); D(CR(O,6.52),linewidth(1.3)+aqgreen);
D(A); MP("A",(-1.81,-2.89),SW*lsf); D(B); MP("B",(9.2,-2.89),SE*lsf); D(C); MP("C",(0.19,6.12),NW*lsf); D(L); MP("L",(0.07,-3.52),NE*lsf); D(O); MP("O",(3.69,0.62),NE*lsf); D(I); MP("I",(1.95,0.13),NE*lsf); D(L_C); MP("L_C",(-8.98,-2.79),NE*lsf); D(C_0); MP("C_0",(0.19,1.62),NE*lsf); D(F); MP("F",(1.91,-3.48),NE*lsf); D(P); MP("P",(0.89,-0.17),NE*lsf); D(P_C); MP("P_C",(0.99,-3.42),NE*lsf); D(K_C); MP("K_C",(2.73,-3.44),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/7/9/f/79f0974d7d9506d2adfc4e97da3f20e80887a4de.png)
,
and 
.
.
be a point at infinity on line 
.
with respect to 
and 
.
can be computed and simplified using the identities 
,
and 
as ![\begin{align*}
L_C &= \left( \det \left[ \begin{array}{cc} a^2S_A & c^2S_C \\ a & c \end{array}\right] :
\det \left[ \begin{array}{cc} b^2S_B & c^2S_C \\ b & c \end{array} \right] :
0\right) \\
&= \left( a(aS_A - cS_C) : b(bS_B - cS_C) : 0 \right) \\
&= \left( a(a-c)(s-b) : b(b-c)(s-a) : 0 \right)
\end{align*}](http://latex.artofproblemsolving.com/b/2/c/b2cf7134db7b2b1866039decbf044d694e242480.png)
is a harmonic bundle. Taking a perspective at 
,
is a harmonic bundle. Taking perspective at 
.
when adding componentwise, then 
.
such that ![\[ \frac{r(s-b) + a(a-c)(s-b)}{r(s-a) + b(b-c)(s-a)} = \frac{a}{b} \]](http://latex.artofproblemsolving.com/8/c/a/8ca2e11ac9314f79002aaf5cd17f2f90d602a55d.png)
because this will give us the coordinates of 
![\[
\frac{-r(s-b) + a(a-c)(s-b)}{-r(s-a) + b(b-c)(s-a) } = \frac{ab(s-b) + a(a-c)(s-b)}{ab(s-a) + b(b-c)(s-a) } = \frac{a(s-b)}{b(s-a)}. \]](http://latex.artofproblemsolving.com/5/5/4/554781f1de3acda769007d7d666d8769074dbdbf.png)
Hence, 
.
and 
analogously, we see that they concur at the point ![\[ P = (a(s-b)(s-c) : b(s-c)(s-a) : c(s-a)(s-b) ). \]](http://latex.artofproblemsolving.com/3/f/c/3fc75af1e9c252cef81b0b174a2582e052a9c0e7.png)
It remains to show that this point actually lies on ![\[
\det \left[ \begin{array}{ccc} a^2S_A & b^2S_B & c^2S_C \\ a & b & c \\ a(s-b)(s-c) & b(s-c)(s-a) & c(s-a)(s-b) \end{array} \right] \]](http://latex.artofproblemsolving.com/4/d/7/4d74b60629dd60f332b936315c65573c35d59359.png)
which simplifies as ![\[ \sum_{\text{cyc}} a(s-b)(s-c) \left( bc(bS_B - cS_C) \right)
= abc(s-a)(s-b)(s-c) \sum_{\text{cyc}} (b-c)
= 0.
\]](http://latex.artofproblemsolving.com/a/9/7/a97da381bf2904025705f89581ea0b46dbda2a2e.png)
This implies the points are collinear, so we're done.
O.K.
be the excentral triangle of triangle 
,
,
,
,
is the Bevan point of triangle 
is the midpoint of the altitudes 
,and we can get 
is 
similarily we can get line 
is 
and line 
is 
and circumcentre 
then line 
the rest part is to calculate the Det which is easy only takes 5minutes
.
concur at 
are the excenters. But just note that 
and similarly to other two sides, so 
and 
,
colinear and we are done.
,
,
,
are collinear. Observing that the sides of 
are parallel to those of 
,
respectively. Let the incenter of 
,
,
be the 
be the 
be the incircle of 
be the circumcircle of 
be the exsimilicenter of 
.
is the midpoint of arc 
.
.
be the foot of 
be the tangency point of the 
be the antipode of 
to 
.
and 
,
.
and 
also, so 
denote the 
-
and 
and 
are concurrent since it is well known that 
and 
and 
are concurrent at the incenter w.r.t. 
and 
are concurrent at the Gergonne point w.r.t. 
so by Cevian Nest, we know that the three lines meet at some point 
as well. To do so, we consider the homothety centered at 
since 
it is mapped to the circumcenter of 
)
and 
and 
,
,
with 
;
is reflection of 
then: 
is harmonic quadrilateral
is harmonic quadrilateral so: 
,
,
or 
is bisector of 
tangents 
tangents 
tangents 
be midpoint of 
which not containing 
,
,
,
be external homothetic center of 
then: 
,
,
concurrent at a point 
be the feet of the altitudes from 
respectively. Denote by 
.
be the feet from 
.
concur.
concur.)
concur
,
and 
are collinear 
.
and 
are concurrent. Note that in 
and 
,
and 
are collinear (by 
and 
.
we can conclude that there exists a homothety 
that maps 
)
are collinear 
.
.
and 
we can conclude that 
and 
is the Nine-Point Circle of 
we can say that 
are collinear. Note that ![[asy]
import graph; size(31.668149578472285cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.4012066955558575,xmax=28.266942882916428,ymin=-7.3203210542584065,ymax=14.233001568944603;
pen xdxdff=rgb(0.49019607843137253,0.49019607843137253,1.), qqccqq=rgb(0.,0.8,0.), ffttcc=rgb(1.,0.2,0.8), ffqqtt=rgb(1.,0.,0.2);
pair A=(9.011309345848499,7.215059903683832), B=(7.060803027120685,-0.9462221037893297), C=(14.294050373775246,-1.2914678077732946), I=(9.97820131745357,1.2452595365401127), O=(10.848355810206165,2.462293101426641), F=(7.713890745265604,1.7864186429857236), D=(9.867207656966674,-1.0801728862885847), I_A=(11.232649004007717,-6.499969895476381), I_B=(20.375805631921757,9.05569770470921), I_C=(1.8067677366971093,6.0481844618646425), P=(11.71851037050289,3.6793262492811234);
draw(A--B,linewidth(2.)); draw(B--C,linewidth(2.)); draw(C--A,linewidth(2.)); draw(circle((10.848355810206167,2.4622931014266416),5.095442276078402),linewidth(2.)+xdxdff); draw(circle(I,2.3280797979904406),linewidth(2.)+qqccqq); draw(D--(11.955936837608213,2.4734768455411915),linewidth(2.)); draw((11.955936837608213,2.4734768455411915)--F,linewidth(2.)); draw(F--D,linewidth(2.)); draw((xmin,1.3986407873809124*xmin-12.710659810748405)--(xmax,1.3986407873809124*xmax-12.710659810748405),linewidth(2.)+linetype("4 4")+ffttcc); draw(circle(P,10.190884714789368),linewidth(2.)+ffqqtt); draw(I_C--I_B,linewidth(2.)); draw(I_B--I_A,linewidth(2.)); draw(I_A--I_C,linewidth(2.));
dot(A,ds); label("$A$",(9.11120887133264,7.464808717394179),NE*lsf); dot(B,ds); label("$B$",(6.488846327373972,-1.2763997624679915),NE*lsf); dot(C,ds); label("$C$",(14.630657654331356,-1.3762992879521305),NE*lsf); dot(I,linewidth(4.pt)+ds); label("$I$",(10.085229244803001,1.445862306974799),NE*lsf); dot(O,linewidth(4.pt)+ds); label("$O$",(10.5847268722237,2.6696314941555026),NE*lsf); dot((11.955936837608213,2.4734768455411915),linewidth(4.pt)+ds); label("$E$",(12.05824487311476,2.6696314941555026),NE*lsf); dot(F,linewidth(4.pt)+ds); label("$F$",(7.23809276850502,1.420887425603764),NE*lsf); dot(D,linewidth(4.pt)+ds); label("$D$",(9.885430193834722,-1.5760983389204086),NE*lsf); dot(I_A,linewidth(4.pt)+ds); label("$I_A$",(11.45884772020992,-6.9456978336928845),NE*lsf); dot(I_B,linewidth(4.pt)+ds); label("$I_B$",(20.474779895153528,9.263000176108683),NE*lsf); dot(I_C,linewidth(4.pt)+ds); label("$I_C$",(1.3440207649407778,6.191089767471405),NE*lsf); dot(P,linewidth(4.pt)+ds); label("$P$",(11.583722127065096,3.9183755627072414),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]](http://latex.artofproblemsolving.com/0/3/c/03c1f8341e4e8b7cdab3ee7ba67781237b4e0108.png)
be the reflection of 
on 
such that 
,
.
be the antipedal triangle of 
,
.
lie on 
respectively. 
concur at some point 
through 
,
.
and 
.
,
concur ar 
,
be the 
as the circumcenter of 
,
,
.
.
and 
,
,
.
and 
,
,
,
concur at a homothetic center of 
is the Nine-Point Circle of 
.
.
passes through any of the homothetic centers between the two triangles. 
,
concur. In fact, this was how I initially proved the concurrency.
lie on 
respectively. Hence if suffices to show that 
,
.
,
,
.
,
passes through 
be the excenters corresponding to vertices 
, 
, 
are collinear.
are concurrent at the homothety center say 
.
are collinear. But, since 
are collinear. Thus, 
as 
respectively.. It is well-known that
collinear,
collinear,
collinear,
concur.
are concurrent at the orthocenter of 
are concurrent at the Gergonne point of 
concur, say at point 
and 
are parallel. It is well-known that 
and 
are collinear, so it suffices to show that 
.
and 
,
is a kite.
and 
similarly, with the previous concurrency we find that 
are collinear as desired. 
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(5); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -14.725320829842607, xmax = 19.245388209431944, ymin = -11.567096893730422, ymax = 12.654170986290742; /* image dimensions */
pen wwwwww = rgb(0.4,0.4,0.4); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0);
/* draw figures */
draw((-2.4807267297757383,5.812209977015955)--(-4,-3), linewidth(0.5) + blue);
draw((-4,-3)--(6,-3), linewidth(0.5) + blue);
draw((-2.4807267297757383,5.812209977015955)--(6,-3), linewidth(0.5) + blue);
draw((-2.4807267297757383,5.812209977015955)--(-2.4807267297757387,-3), linewidth(0.5) + wwwwww);
draw((-4,-3)--(1.1916166684739693,1.996326956111153), linewidth(0.5) + wwwwww);
draw((6,-3)--(-3.7113434389898203,-1.3257111957945942), linewidth(0.5) + wwwwww);
draw(circle((1,0.6750431171822711), 6.205315617528958), linewidth(0.5) + fuqqzz);
draw(circle((-0.6439913101582455,-0.17307439890085252), 2.8269256010991475), linewidth(0.5) + ffvvqq);
draw((-9.586207811039753,3.6317139544889536)--(11.586207811039749,10.129003514933505), linewidth(0.5));
draw((11.586207811039749,10.129003514933505)--(2.6439913101582455,-10.887470601792524), linewidth(0.5));
draw((2.6439913101582455,-10.887470601792524)--(-9.586207811039753,3.6317139544889536), linewidth(0.5));
draw((-3.429817618862806,0.30721731649646655)--(1.39288983457389,1.7871866950314028), linewidth(0.5));
draw((1.39288983457389,1.7871866950314028)--(-0.643991310158246,-3), linewidth(0.5));
draw((-0.643991310158246,-3)--(-3.429817618862806,0.30721731649646655), linewidth(0.5));
draw((xmin, -2.39887843477509*xmin-4.544856866121172)--(xmax, -2.39887843477509*xmax-4.544856866121172), linewidth(0.5) + qqwuqq); /* line */
draw((xmin, 0.8183612871845586*xmin + 0.6472995771032272)--(xmax, 0.8183612871845586*xmax + 0.6472995771032272), linewidth(0.5) + qqwuqq); /* line */
draw((xmin, -0.540007461225735*xmin-1.5449097883329328)--(xmax, -0.540007461225735*xmax-1.5449097883329328), linewidth(0.5) + qqwuqq); /* line */
draw((xmin, 0.5158892938439469*xmin + 0.1591538233383234)--(xmax, 0.5158892938439469*xmax + 0.1591538233383234), linewidth(0.5) + dotted + ccqqqq); /* line */
/* dots and labels */
dot((-2.4807267297757383,5.812209977015955),linewidth(2pt) + dotstyle);
label("$A$", (-2.3100168536054966,5.951430189303753), NNE * labelscalefactor);
dot((-4,-3),linewidth(2pt) + dotstyle);
label("$B$", (-3.8333670347388846,-2.8459171067416698), NE * labelscalefactor);
dot((6,-3),linewidth(2pt) + dotstyle);
label("$C$", (6.144576651684806,-2.8459171067416698), NE * labelscalefactor);
dot((-0.643991310158246,-0.1730743989008523),linewidth(2pt) + dotstyle);
label("$I$", (-0.29157786360375754,-0.7513106076832358), NE * labelscalefactor);
dot((-2.4807267297757387,-3),linewidth(2pt) + dotstyle);
label("$H$", (-2.3100168536054966,-2.8459171067416698), NE * labelscalefactor);
dot((1.1916166684739693,1.996326956111153),linewidth(2pt) + dotstyle);
label("$K$", (1.3460235811146346,2.1430547364702366), NE * labelscalefactor);
dot((-3.7113434389898203,-1.3257111957945942),linewidth(2pt) + dotstyle);
label("$L$", (-3.5667807530405415,-1.1702319074949226), NE * labelscalefactor);
dot((1,0.6750431171822711),linewidth(2pt) + dotstyle);
label("$O$", (1.155604808472961,0.810123327978506), NE * labelscalefactor);
dot((2.6439913101582455,-10.887470601792524),linewidth(2pt) + dotstyle);
label("$I_A$", (2.7932062531913533,-10.729254294107049), NE * labelscalefactor);
dot((-9.586207811039753,3.6317139544889536),linewidth(2pt) + dotstyle);
label("$I_C$", (-9.431678950404086,3.780656181188649), NE * labelscalefactor);
dot((11.586207811039749,10.129003514933505),linewidth(2pt) + dotstyle);
label("$I_B$", (11.742888567350008,10.292978205533961), NE * labelscalefactor);
dot((-0.643991310158246,-3),linewidth(2pt) + dotstyle);
label("$D$", (-0.48199663624543104,-2.8459171067416698), NE * labelscalefactor);
dot((1.39288983457389,1.7871866950314028),linewidth(2pt) + dotstyle);
label("$E$", (1.6887773718696468,1.6860496821302147), NE * labelscalefactor);
dot((-3.429817618862806,0.30721731649646655),linewidth(2pt) + dotstyle);
label("$F$", (-3.262110716813864,0.46736953722348945), NE * labelscalefactor);
dot((-2.4807267297757383,1.4061049885079777),linewidth(2pt) + dotstyle);
label("$A_0$", (-2.3100168536054966,1.5717984185452092), NE * labelscalefactor);
dot((1.1443282805050898,-2.162855597897297),linewidth(2pt) + dotstyle);
label("$C_0$", (1.3079398265862998,-2.0080745071182964), NE * labelscalefactor);
dot((-1.4041916657630154,-0.5018365219444235),linewidth(2pt) + dotstyle);
label("$B_0$", (-1.3198392358687945,-0.9036456257965765), NE * labelscalefactor);
dot((2.6439913101582446,1.5231606332653922),linewidth(2pt) + dotstyle);
label("$O_2$", (2.7932062531913533,1.6860496821302147), NE * labelscalefactor);
dot((-1.6138543882150664,-0.6734163773649031),linewidth(2pt) + dotstyle);
label("$N$", (-1.9672630628504844,-1.474901943721604), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/e/d/1/ed148af71cb04a98dbee53128aba4462752cb675.png)
It is well known that 
so there is a homothety that maps 
and 
concur. But it is well known that 
are collinear, 
concur at 
Note that the center of our homothety is also 
respectively
we see that the circumcenter of 
maps to the circumcenter of 
Thus, 
are collinear. But 
are collinear because they are on the euler line, so line 
which are the feet of the altitudes of 
.
since both are perpendicular to ![\[\angle EDF = 180 - \angle FDB - \angle EDC = 180 - (90 - \frac12 B) - (90-\frac12 C) = 180 - (90 + \frac12 A) = 180-\angle BIC = \angle BI_A C\]](http://latex.artofproblemsolving.com/d/9/5/d95fd6a2ed36a73d1323322e05c52e395eb0e409.png)
,
are the orthocenter and nine point center of 
are also collinear and we're done. 
the 
Now,
Similarly, we find that
Thus,
Thus, by the converse of Ceva's Theorem on triangle 
and 
lies on 
and 
lies on 
this means that 
Thanks in advance!
(and 
and 
)
concur on 
.
,
.
and 
.
.
.
gives 
.
which is true upon expansion.
is given by 
We can find the equtaions of lines 
by permuting variables.
however this is quite clear since 
which we can solve with Cramer's Rule to be 
,
which (after factoring out 
)
:
.
are collinear.
are collinear. Similarly 
and 
are collinear. We then have that the homothety sending 
sends 
.
is the Euler line of 
)
point of 
are collinear. (Further we actually have 
)
to 
are collinear. (We have the map sends 
,
,
,
,
are collinear. But 
as they're both perpendicular to 
,
,
are homothetic, and the circumcenters of these two triangles are 
,
.
are collinear.
,
be the intersection of the line parallel to 
,
and 
to 
are collinear. Now, consider the homothety centered at 
,
and the midpoint 
becomes the reference triangle 
,
,
,
.
,
.
and 
.
and 
-
,
be the orthic triangle.
be the midpoint of 
Define 
similarly.
be the intouch triangle.
denote the second intersection of 
with the incircle. 2002 ISLG 7 implies 
with the incircle.
similarly.
be the second intersection of 
with 
By curvilinear properties, 
Define 
similarly.
denote the midpoint of arc 
Let 
are collinear.
have circumcenter 
This moves 
Thus, 
which is fixed. Thus, 
Thus, 
are concurrent, and also the line connecting the circumcenters, which is 
