[/quote]
,
be nonnegative real numbers such that 
.
and 
Prove that
Where 
,
,
cut three common chords concurrent at 
,
which are internally tangent to all 
.
Prove that
Proposed by Phan Ngoc Chau
intersects the circumcircle 
distinct from point 
.
and 
intersect 
and 
respectively.
grid has had the central unit square removed. For which positive integers 
is it possible to cut the remaining area into 
and 
rectangles?
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
and 
be positive real numbers such that 
.![\[
\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)}
\geq \frac{3}{4}.
\]](http://latex.artofproblemsolving.com/5/6/5/56568d27a296f1c900fabbb1042549cd2e552e29.png)
![\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1. \]](http://latex.artofproblemsolving.com/b/2/f/b2ff456632d4dd3b9ed93e4cba437bdd34d54cd8.png)
Y by Mathuzb, Adventure10, Mango247
Let 
be the foot of perpendicular from 
to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle 
. A circle 
with centre 
passes through and , and it intersects sides 
and 
at 
and 
respectively. Let 
be the foot of altitude from to 
, and let 
be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
be the foot of perpendicular from 
to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle 
. A circle 
with centre 
passes through and , and it intersects sides 
and 
at 
and 
respectively. Let 
be the foot of altitude from to 
, and let 
be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
we have the relations 
We now determine the coordinates of the circumcenter 
.
and 
lie on a circle centered at 
,
(the midpoint of 
)![\[ w = s + \frac{(x-s)(y-s)}{x+y-2s}. \]](http://latex.artofproblemsolving.com/d/4/c/d4c218f7d5b4b07f183ad09e6b3b58ff2daa0e7a.png)
Then 
Thus ![\[ \frac{w - \frac{a+b+c}{2}}{b-c} = (k-1) \cdot \frac{b^2+bc+c^2}{b^2-c^2} \]](http://latex.artofproblemsolving.com/6/3/3/633215ba56ed9b00b7c39b9848304bc78d6a72ef.png)
is pure imaginary. So the line through 
and when 
.
.
move uniformly on lines 
has a fixed shape 
,
.
is the circumcenter of 
,
or 
.
has a fixed shape. So as 
,
has a fixed shape. Then by similarity using determinants, we have that if the displacement vector 
is 
,
cuts the Euler line at 
is the diameter
be the projections of 
on 
; 
are the midpoints of 
are concurrent 
be the centers of these circles, then 
because it's parrallel with 
be the centers of 
, then we have 
so the rotation - homothetic centre 
transform 
to 
then 
perpendicular bisector of 
is the NPC so 
perpendicular bisector of 
, it followed that the circumcenter of 
(all in the unit circle), and apply the formula for circumcenter of 
is 
.
,
(after shifting), so a lot of terms cancel and the bash ends very nicely.
and 
the statement is true, hence it is also true for 
from continuity.
case was problematic as well, but it is not. The circumcenter 
of 
![[asy]
size(10cm);
pair A=dir(70); pair B=dir(220); pair C=dir(-40);
pair O=(0, 0);
pair H=A+B+C;
pair D = foot(A, O, H);
pair D1=2*D-A;
pair O1 = B+C;
pair H1=2*foot(A, B, C)-H;
pair E=circumcenter(B, O, C);
draw(unitcircle);
draw(A--B--C--cycle);
draw(A--D1, orange+dashed);
draw(circumcircle(O, H, D1), red);
draw(D1--H1, dotted);
draw(A--H1);
draw(O--O1);
draw(D1--O1, heavygreen);
draw(D--2*H-O, heavygreen);
string[] names = {"$A$", "$B$", "$C$", "$O$", "$H$", "$D$", "$D'$", "$O'$", "$H'$", "$E$"};
pair[] pts = {A, B, C, O, H, D, D1, O1, H1, E};
pair[] labels = {A, B, C, dir(90), dir(45), D, D1, O1, H1, dir(225)};
for(int i=0; i<names.length; ++i){
dot(names[i], pts[i], dir(labels[i]));
}
[/asy]](http://latex.artofproblemsolving.com/c/7/4/c74887f9d0c24ce022b5049ecc9edd419808be4e.png)
be the circumcenter and orthocenter of 
.
be the circumcenter of 
,
again at 
,
be the reflection of 
be the reflection of 
are collinear. Since 
is an isosceles trapezoid, it is cyclic. 
is a parallelogram, so 
is an isosceles trapezoid as well. Hence 
all lie on a circle. The inverse of ![[asy]
size(10cm);
pair A=dir(70); pair B=dir(220); pair C=dir(-40);
pair O=(0, 0);
pair H=A+B+C;
pair Q=1.5*H-0.5*O;
pair X1=2*foot(Q, A, B)-A;
pair Y1=2*foot(Q, A, C)-A;
pair D = foot(A, O, H);
pair D1=2*D-A;
pair P=foot(A, B, C);
pair M=midpoint(B--C);
pair H1=2*foot(A, B, C)-H;
pair E=circumcenter(B, O, C);
pair T=circumcenter(Q, X1, Y1);
draw(unitcircle);
draw(X1--A--Y1);
draw(B--C);
draw(D1--H1, dotted);
draw(A--H1);
draw(O--T, red);
draw(circumcircle(Q, X1, Y1), orange);
draw(circumcircle(A, X1, Y1), blue+dashed);
string[] names = {"$A$", "$B$", "$C$", "$O$", "$D'$", "$H'$", "$E$", "$Q$", "$X_1$", "$Y_1$", "$T$", "$P$", "$M$"};
pair[] pts = {A, B, C, O, D1, H1, E, Q, X1, Y1, T, P, M};
pair[] labels = {A, B, C, dir(90), D1, H1, dir(225), dir(90), X1, Y1, T, P, dir(135)};
for(int i=0; i<names.length; ++i){
dot(names[i], pts[i], dir(labels[i]));
}
[/asy]](http://latex.artofproblemsolving.com/5/9/1/591754469db6f5f0b7828af08330c9946666b080.png)
at 
to 
.
.
is cyclic, 
to 
,
to ![\[\measuredangle D'ET = \measuredangle D'BX_1 = \measuredangle D'BA = \measuredangle D'H'A. \]](http://latex.artofproblemsolving.com/f/f/0/ff065c6e8c65f91f4c6b425761301701d4b8c19f.png)
Therefore 
,
.
,
.
as well and the statement still holds.)
.
.
are inverses of each other with respect to 
where 
by sine rule in 
and with this we are done .
let 
and let 
.![[asy]
size(10cm);
pair A=(8.227184338605278,8.917180830035385);
pair B=(5.836037486558229,-5.602600636257744);
pair C=(22.852265578227577,-5.038029693332168);
pair O=(14.151548373898125,0.48476445751051883);
pair H=(8.61239065559483,-2.6929784145755735);
pair D=(13.333865353835646,0.015670324291781175);
pair S=(14.5510849593405,6.629546963587926);
pair X=(7.827568156749471,6.490588073520484);
pair Y=(17.132485773397857,0.41976777226100737);
pair P=(8.705768203458195,-5.507387609724802);
pair M=(14.344151532392903,-5.320315164794956);
pair T=(11.291157898982467,1.632964751029067);
pair N9=(11.381969514746476,-1.1041069785325273);
pair E=(8.419787497100053,3.1121012077299053);
pair Q=(10.408968970806821,0.28080888219357325);
pair G=(8.39272486321513,3.927771862972893);
pair L=(12.480026965073659,3.455177922890749);
draw(A--B--C--A,purple);
draw(A--P,orange);
draw(circumcircle(A,X,Y),red);
draw(circumcircle(A,B,C),orange);
draw(circumcircle(S,G,D),blue+dashed);
draw(circumcircle(T,Q,D),cyan+dashed);
draw(E--M,magenta);
draw(O--H,green);
draw(T--Q--N9--D--T,lightblue);
draw(T--N9,lightblue);
draw(Q--D,magenta);
draw(S--A,green);
draw(S--Y,green);
draw(S--X,green);
draw(X--Y,green);
draw(G--D,lightblue);
draw(S--T,red);
draw(E--D,magenta+dotted);
draw(A--O,orange);
draw(A--D,cyan);
draw(S--G,green);
draw(S--D,green);
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,SE);
dot("$M$",M,SE);
dot("$H$",H,W);
dot("$P$",P, SE);
dot("$S$",S,NE);
dot("$X$",X,NW);
dot("$Y$",Y,SW);
dot("$G$",G,W);
dot("$E$",E,SW);
dot("$T$",T,N);
dot("$N_9$",N9,SE);
dot("$Q$",Q,SW);
dot("$L$",L,N);
dot("$O$",O,NE);
[/asy]](http://latex.artofproblemsolving.com/8/6/e/86e0940367d7248d6ba60ddded3e9d9bbeb8bc38.png)
as the intersection of 
with 
.
is cyclic where 
and hence 
is cyclic. Now notice that 
and with this we are done 
and notice that this is exactly twice the distance from 
is cyclic where 
is the nine point center.
are inverses with respect to 
.
and this finishes the proof 
and with this we are done 
meet 
and 
and 
and 
respectively.
are collinear.
Hence 
.
be the reflection of 
be the midpoint of 
.
.
Since 
,
by shooting lemma. 
be the reflection of 
be the second intersection of 
and 
Together with 
and 
,
.
are collinear, while 
to 
.
Hence
which implies 
and 
,![\[o = \frac{1}{2} \left(a+e+\frac{bc(1-a\overline{e})}{b+c}\right) = \frac{1}{2} \cdot \frac{ab+bc+ca+k(b^2+bc+c^2)}{b+c}\]](http://latex.artofproblemsolving.com/3/6/8/368660e2bba99a3ffb864d811ffc2534b94bdd53.png)
Notice that 
is real, so 
is constant. Since 
is imaginary, we have 
is a constant not depending on 
.
is a real number in other words we have that 
.
as the following 
,
later on.
.
,![\[\frac{a-d}{\overline{a-d}}=-\frac{h}{h}=-1\]](http://latex.artofproblemsolving.com/4/c/e/4ce5f0592b85e321f1bc19bae0cd759795a0a46c.png)
this implies that 
.
,
where 
,
this implies that 
,
Now we calculate 
.
,
,
,
let 
,
this implies that:
giving us that 
.
.
,![\[ g=\frac{\begin{vmatrix}
x & x\overline{x} & 1 \\
s & s\overline{s} & 1 \\
y & y\overline{y} & 1
\end{vmatrix}}
{\begin{vmatrix}
x & \overline{x} & 1 \\
s & \overline{s} & 1 \\
y & \overline{y} & 1
\end{vmatrix}} \]](http://latex.artofproblemsolving.com/b/8/5/b85c2da95f463fff6f919d41ae19f0be821d0078.png)
Calculating this(calculation done in 30 min), we get that 
,
and 
.
,
be the orthic triangle of 
be the midpoint of 
and 
,
to segment 
.
to segment 
.
.![\[\measuredangle QXY=90^\circ-\measuredangle YSX=90^\circ-\measuredangle ERF=\measuredangle N_9FE.\]](http://latex.artofproblemsolving.com/4/0/0/400724e24e7ce8ecb34bd09e422c1cacbca86b2a.png)
and similarly we have 
.![\[\measuredangle PN_9Q=\measuredangle PFX=\measuredangle PFA=\measuredangle PHA.\]](http://latex.artofproblemsolving.com/6/0/a/60a5696847b9835d4581e7c98c4924649ca12163.png)
and so 
is parallel to 
,
and the result follows.
on the unit circle. But 
?
circumcenter and orthocenter and let 
Move 
points 
are midpoints of 
so 
form isosceles trapezoid. When 
circumcenter of 
so it does for every 
.
,
,
be the circumcircle of 
be the point on 
such that 
.
be the 9-point circle of 
,
and 
.
.
,
is due to 
.
.
,
lie on the same side of 
be the radius of 
The radius of 
,
,
Thus
But we also have
.
![[asy]
size(7cm);
defaultpen(fontsize(10pt));
pair A, B, C;
A = dir(110);
B = dir(-162);
C = dir(-18);
pair O, H, D, P, M;
O = circumcenter(A, B, C);
H = orthocenter(A, B, C);
D = foot(A, H, O);
P = foot(A, B, C);
M = (B + C)/2;
pair K, S, X, Y, N, S1;
K = 0.57 * O + 0.43 * H;
S = (A + K)/2;
X = foot(K, A, B);
Y = foot(K, A, C);
N = (X + Y)/2;
S1 = 2 * circumcenter(S, X, Y) - S;
draw(A--B--C--A--P);
draw(circumcircle(A, B, C));
draw(K--A, deepgreen+dashed);
draw(X--K--Y, deepgreen);
draw(circumcircle(A, D, K), red);
draw(A--D, blue);
draw((4.1*O-3.1*H)--(3.1*H-2.1*O), blue);
draw(X--Y, purple);
draw(S--S1, purple+dotted);
draw(circumcircle(S, X, Y), purple+dotted);
dot("$A$", A, dir(110));
dot("$B$", B, dir(-162));
dot("$C$", C, dir(-18));
dot("$O$", O, dir(80));
dot("$H$", H, dir(190));
dot("$D$", D, dir(200));
dot("$P$", P, dir(260));
dot("$M$", M, dir(260));
dot("$K$", K, dir(-100));
dot("$S$", S, dir(60));
dot("$X$", X, dir(185));
dot("$Y$", Y, dir(5));
dot("$N$", N, dir(-95));
dot("$S'$", S1, dir(-95));
[/asy]](http://latex.artofproblemsolving.com/e/4/3/e439855dc55fd7d6b71336bd7545340c4f4c8a3b.png)
and 
be the perpendicular bisector of 
be the antipode of ![\[\frac{SN}{NS'} = \left(\frac{SN}{NY}\right)^2 = (\tan{\angle NSY})^2 = \tan^2{A},\]](http://latex.artofproblemsolving.com/6/d/9/6d9ce8268b9f68123fb7e8309d2799570c4664a6.png)
which is constant. Since 
,
,
.
.
,
,
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
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 = -4.36471658380817, xmax = 14.498175842200546, ymin = -5.39223726459042, ymax = 4.219781556321238; /* image dimensions */
pen zzttff = rgb(0.6,0.2,1);
draw((3.12,3.61)--(0.46,-3.87)--(11.44,-3.87)--cycle, linewidth(0.4) + zzttff);
/* draw figures */
draw((3.12,3.61)--(0.46,-3.87), linewidth(0.4) + zzttff);
draw((0.46,-3.87)--(11.44,-3.87), linewidth(0.4) + zzttff);
draw((11.44,-3.87)--(3.12,3.61), linewidth(0.4) + zzttff);
draw(circle((4.535,1.0003208556149736), 2.968610826066321), linewidth(0.8) + blue);
draw((3.12,3.61)--(5.95,-1.6093582887700522), linewidth(0.8));
draw((5.95,-1.6093582887700522)--(5.95,-3.87), linewidth(0.8));
draw((3.12,3.61)--(3.12,-3.87), linewidth(0.8));
draw((1.79,-0.13)--(4.535,1.0003208556149739), linewidth(0.8));
draw((4.535,1.0003208556149739)--(7.28,-0.13), linewidth(0.8));
draw((1.79,-0.13)--(7.28,-0.13), linewidth(0.8));
/* dots and labels */
dot((3.12,3.61),dotstyle);
label("$A$", (3.177662346473665,3.7475147356406078), NE * labelscalefactor);
dot((0.46,-3.87),dotstyle);
label("$B$", (0.5107438296889277,-3.725413191599959), NE * labelscalefactor);
dot((11.44,-3.87),dotstyle);
label("$C$", (11.497892510817715,-3.725413191599959), NE * labelscalefactor);
dot((5.95,-1.6093582887700522),linewidth(4pt) + dotstyle);
label("$O$", (6.039043671773957,-1.7668949058361674), NE * labelscalefactor);
dot((3.12,-0.9112834224598962),linewidth(4pt) + dotstyle);
label("$H$", (3.177662346473665,-0.7945808632583985), NE * labelscalefactor);
dot((2.0687040121969074,-0.6519606994429188),linewidth(4pt) + dotstyle);
label("$D$", (2.1220071002463734,-0.5445572523098293), NE * labelscalefactor);
dot((1.79,-0.13),linewidth(4pt) + dotstyle);
label("$X$", (1.8442030880812965,-0.016729629196183367), NE * labelscalefactor);
dot((7.28,-0.13),linewidth(4pt) + dotstyle);
label("$Y$", (7.330832328341564,-0.016729629196183367), NE * labelscalefactor);
dot((4.535,1.0003208556149739),linewidth(4pt) + dotstyle);
label("$S$", (4.594462808515557,1.1083766200723777), NE * labelscalefactor);
dot((3.12,-3.87),linewidth(4pt) + dotstyle);
label("$P$", (3.177662346473665,-3.7531935928164666), NE * labelscalefactor);
dot((5.95,-3.87),linewidth(4pt) + dotstyle);
label("$M$", (6.011263270557449,-3.7531935928164666), NE * labelscalefactor);
dot((4.535,-2.8979750100503),linewidth(4pt) + dotstyle);
label("$O''$", (4.594462808515557,-2.780879550238698), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/3/7/6/376e53ff55491b374e733111eb4fb663190bc2fd.png)
be the circumcenter of 
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
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 = -1.7419188729163546, xmax = 12.001886155190885, ymin = -4.429974238189027, ymax = 2.573496365235269; /* image dimensions */
pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0);
draw((2.78,2.05)--(0.76,-3.09)--(10.56,-3.09)--cycle, linewidth(0.8) + zzttff);
/* draw figures */
draw((2.78,2.05)--(0.76,-3.09), linewidth(0.8) + zzttff);
draw((0.76,-3.09)--(10.56,-3.09), linewidth(0.8) + zzttff);
draw((10.56,-3.09)--(2.78,2.05), linewidth(0.8) + zzttff);
draw(circle((2.78,1.0087548638132295), 1.0412451361867703), linewidth(0.8) + qqwuqq);
draw(circle((4.22,-1.0406225680933903), 2.504705144005615), linewidth(0.8) + ccqqqq);
/* dots and labels */
dot((2.78,2.05),dotstyle); label("$A$", (2.8224817660205854,2.148430230345355), NE * labelscalefactor);
dot((0.76,-3.09),dotstyle);
label("$B$", (0.7983573141638491,-2.9928458773707467), NE * labelscalefactor);
dot((10.56,-3.09),dotstyle);
label("$C$", (10.605240283409735,-2.9928458773707467), NE * labelscalefactor);
dot((5.66,-2.0487548638132296),linewidth(4pt) + dotstyle);
label("$O$", (5.696738487657151,-1.9706630291830964), NE * labelscalefactor);
dot((2.78,-0.03249027237354091),linewidth(4pt) + dotstyle);
label("$H$", (2.8224817660205854,0.05346142267363664), NE * labelscalefactor);
dot((1.8016049241530079,0.6524761678588333),linewidth(4pt) + dotstyle);
label("$D$", (1.8407814068700683,0.7315431140456422), NE * labelscalefactor);
dot((2.78,1.0087548638132295),linewidth(4pt) + dotstyle);
label("$S$", (2.8224817660205854,1.0857648931205706), NE * labelscalefactor);
dot((2.071079344262295,0.24611278688524568),linewidth(4pt) + dotstyle);
label("$X$", (2.1140382078707276,0.3267182236742956), NE * labelscalefactor);
dot((3.7377807425127645,1.4172245480057057),linewidth(4pt) + dotstyle);
label("$Y$", (3.7738202583932514,1.5007104057512006), NE * labelscalefactor);
dot((4.22,-1.0406225680933852),linewidth(4pt) + dotstyle);
label("$N_{9}$", (4.259610126838869,-0.9586008032547298), NE * labelscalefactor);
dot((2.78,-3.09),linewidth(4pt) + dotstyle);
label("$P$", (2.8224817660205854,-3.013087121889314), NE * labelscalefactor);
dot((5.66,-3.09),linewidth(4pt) + dotstyle);
label("$M$", (5.696738487657151,-3.013087121889314), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/b/a/f/baf36d819c6f5e80449ed266d156f78973b19320.png)
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
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 */
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.
to be the 
,
Here, the last line follows from the shifted circumcenter formula, observing that 
lie on a circle centered at 
,
is pure imaginary. The rest is pure computation:
This is a constant multiple of 
,
clearly equals its own conjugate. This finishes the problem.
and on circle 
we have the map of moving points 
is linear. Since 
,
is fixed, which implies that 
is a linear map.
cases.
is the midpoint of 
.
.
