[/quote]
,
,
,
cut three common chords concurrent at 
.
,
which are internally tangent to all 
.
be nonnegative real numbers such that 
.
Prove that
Proposed by Phan Ngoc Chau
at 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle 
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)
be a parallelogram of center 
,
and 
such that 
.
Y by Adventure10, ImSh95, Mango247, Funcshun840, ItsBesi, and 1 other user
Let 
be the circumcenter of an acute-angled triangle 
with 
. The line 
meets the side 
at 
. The circumcenters of the triangles 
and 
are 
and 
, respectively. Extend the sides 
and 
beyond 
, and choose on the respective extensions points 
and 
such that 
and 
. Prove that the quadrilateral 
is a rectangle if and only if 
.
Proposed by Hojoo Lee, Korea
be the circumcenter of an acute-angled triangle 
with 
. The line 
meets the side 
at 
. The circumcenters of the triangles 
and 
are 
and 
, respectively. Extend the sides 
and 
beyond 
, and choose on the respective extensions points 
and 
such that 
and 
. Prove that the quadrilateral 
is a rectangle if and only if 
.Proposed by Hojoo Lee, Korea
This post has been edited 1 time. Last edited by djmathman, Aug 1, 2015, 2:52 AM
Reason: Official version is better than non-official one
Reason: Official version is better than non-official one
.
are isosceles, and have the same angles, so they are similar. This means that 
is similar to 
,
.
share the same altitude starting from 
,
,
,
is equilateral. With the given conditions, this is equivalent to 
,
. Anyway, here is the way the problem was posed on our TST:
.
acute chordal angle of the chord CA in the circumcircle of triangle ACD
(since < BCA = C and < ECA = B - A)
(since B + C = 180° - A by the sum of the angles in triangle ABC)
.
![$\Rightarrow 2sinAcosC[1-2cos(B-C)]=0$](http://latex.artofproblemsolving.com/c/b/3/cb30d21c7f924b9d96f5eccdd52eb774ae78a99b.png)
.
was equilateral - see Grobber's, as I did, then from 
get 
,
getting 
.
intersects 
(second time) at 
respectively, then we know that 
,
is cyclic; from 
we get the same 
.
,
.
and 
is similar to 
.
as 
.
which means 
is equaliteral. So 
.
.
.
.
and 
we easily get other angles are also 
,Since, 
is a Isosceles Trapezium. But, 
.
If 
is Equilateral. Hence, 
& 
.
Hence, 
and since, 
& 
are Isosceles 
is always an isosceles trapezoid, and in particular 
.
always holds. (One clean way to see this is to note that 
and 
are antiparallel through 
.
.![\[ \triangle AEF \overset{+}{\sim} \triangle ABC. \]](http://latex.artofproblemsolving.com/f/4/6/f46430b129a849e07b97cbd0b642f3e18b891949.png)
![[asy] size(6cm); pair B = dir(200); pair C = dir(-20); pair A = dir(30); pair O = origin; pair D = extension(A, O, B, C); pair E = circumcenter(A, B, D); pair F = circumcenter(A, C, D);
filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); filldraw(A--E--F--cycle, invisible, red); pair G = A+(A-B)*abs(C-A)/abs(B-A); pair H = A+(A-C)*abs(B-A)/abs(C-A);
filldraw(A--C--G--cycle, invisible, blue); filldraw(A--B--H--cycle, invisible, blue); filldraw(A--G--H--cycle, invisible, blue);
dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A$", A, dir(330)); dot("$O$", O, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(270)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H));
/* TSQ Source:
!size(10cm); B = dir 200 C = dir -20 A = dir 30 R330 O = origin R270 D = extension A O B C R270 E = circumcenter A B D F = circumcenter A C D R270
A--B--C--cycle 0.1 lightblue / blue A--D blue A--E--F--cycle 0.1 lightred / red G = A+(A-B)*abs(C-A)/abs(B-A) H = A+(A-C)*abs(B-A)/abs(C-A)
A--C--G--cycle 0.1 lightcyan / blue A--B--H--cycle 0.1 lightcyan / blue A--G--H--cycle 0.1 lightblue / blue
*/ [/asy]](http://latex.artofproblemsolving.com/f/8/8/f88d7ebb09a2070263a04214731e43caffc96d47.png)
if and only if 
.
.
always, triangle 
is equilateral if and only if 
.
,
,
we get the earlier congruence. 
by symmetry through the oppositely congruent triangles 
and 
as needed.
![[asy]
unitsize(5cm);
pair A, B, C, D, E, F, G, H;
A=(-sqrt(3)/2,1/2);
B=(-1,0);
C=(1,0);
D=(0,0);
E=(-1/2,1-sqrt(3)/2);
F=(1/2,1+sqrt(3)/2);
G=(1/2-sqrt(3)/2,3/2+sqrt(3)/2);
H=(-1/2-sqrt(3)/2,3/2-sqrt(3)/2);
draw(D--A--B--C--H--A--G--H--E--F--G--A--E--(1/2,1-sqrt(3)/2)--F);
dot(E);
dot(F);
label("$A$", A, NW);
label("$C$", B, SW);
label("$B$", C, SE);
label("$D$", D, S);
label("$F$", E, S);
label("$E$", F, dir(0));
label("$H$", G, N);
label("$G$", H, W);
label("$X$", (1/2,1-sqrt(3)/2), SW);
[/asy]](http://latex.artofproblemsolving.com/c/d/2/cd2becfc841432f34eb71764f2cc58a8e6611aa3.png)
and 
,
.
is the perpendicular bisector of 
,
.
and 
.
,
if and only if 
if and only if 
,
if and only if 
.
.
,
.
is isosceles. Therefore, we have
Therefore, 
and so on. Notice 
since 
is the radical axis of 
and 
Also, 
as 
Finally, 
implies 
is equilateral as 
Hence, 
so 
and 
Similarly, 
so 
Similarly, 
so 
Then, 
so 
We have ![\[\angle GAX=\angle OAB=90^\circ-\angle ACB=90^\circ-\angle AGX.\]](http://latex.artofproblemsolving.com/b/0/f/b0f63d64b15dc3daa96282a5e04266438cf0ab79.png)
Thus, 
We also have 
so 
It is clear why F is inside and E is outside so now, 
so 
is cyclic. Similarly, 
is cyclic. We have ![\[\angle FOE+\angle FAE=360^\circ-\angle FOD-\angle DOE+\angle FDE\]](http://latex.artofproblemsolving.com/e/f/d/efdef04bcc81c514c6f2343d7dfc6d4b1efb6e90.png)
![\[=\angle FCD+\angle EBD+180^\circ-\angle FDC-\angle EDB=180^\circ\]](http://latex.artofproblemsolving.com/5/9/a/59ab66f5a688a99704f9f00c6beac8ce02f80993.png)
which implies that 
is cyclic. We have 
What's important to see is that ![\[\angle OAE=\angle OFE=\angle FOD-90^\circ=90^\circ-\angle FCD=\angle OAC\]](http://latex.artofproblemsolving.com/8/f/3/8f3d9ed33fb8edfb984542ee9ccfd8006c5e7a70.png)
and similarly, 
Thus, 
Therefore, we have ![\[\angle ACB=\angle ABC\iff \triangle ACF \text{ is equilateral}\]](http://latex.artofproblemsolving.com/b/9/b/b9b94d70269f570894a9531288e167d60e700eac.png)
We know that 
and 
so 
is equilateral 
Since 
rectangle, as desired.
.
and 
,
.
.![[asy]
import graph; size(11cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(14); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7, xmax = 6, ymin = -4.184372376555464, ymax = 10.16202343462137; /* image dimensions */
pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffqqff = rgb(1.,0.,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.);
draw((-0.8535898390715775,2.4107656245533)--(4.846554211506446,-2.620429350613437)--(-3.14,-2.7)--cycle, linewidth(1.1) + yqqqyq);
draw((-0.8535898390715775,2.4107656245533)--(-5.051246248590594,6.115799681176306)--(2.251206267261163,9.350852527005179)--cycle, linewidth(1.1) + yqqqyq);
draw((-0.8535898390715775,2.4107656245533)--(-0.895059345715962,-0.6375021281980235)--(3.0810645932710234,1.12395692083598)--cycle, linewidth(1.1) + ffqqff);
draw((-2.6885720014591716,6.552852553843345)--(-2.1789546962989768,6.77861765259111)--(-2.4047197950467414,7.288234957751304)--(-2.9143371002069363,7.0624698590035395)--cycle, linewidth(1.1) + qqwuqq);
draw((0.7780870069641727,0.10371691540088587)--(0.5523219082164079,0.6133342205610806)--(0.04270460305621318,0.3875691218133157)--(0.268469701803978,-0.12204818334687892)--cycle, linewidth(1.1) + qqwuqq);
/* draw figures */
draw((-0.8535898390715775,2.4107656245533)--(4.846554211506446,-2.620429350613437), linewidth(1.1) + yqqqyq);
draw((4.846554211506446,-2.620429350613437)--(-3.14,-2.7), linewidth(1.1) + yqqqyq);
draw((-3.14,-2.7)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + yqqqyq);
draw((-0.8535898390715775,2.4107656245533)--(1.3905292426795337,-2.654861991247058), linewidth(1.1) + blue);
draw((-0.8535898390715775,2.4107656245533)--(-5.051246248590594,6.115799681176306), linewidth(1.1) + yqqqyq);
draw((-5.051246248590594,6.115799681176306)--(2.251206267261163,9.350852527005179), linewidth(1.1) + yqqqyq);
draw((2.251206267261163,9.350852527005179)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + yqqqyq);
draw((-2.9143371002069363,7.0624698590035395)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + blue);
draw((-0.8535898390715775,2.4107656245533)--(-0.895059345715962,-0.6375021281980235), linewidth(1.1) + ffqqff);
draw((-0.895059345715962,-0.6375021281980235)--(3.0810645932710234,1.12395692083598), linewidth(1.1) + ffqqff);
draw((3.0810645932710234,1.12395692083598)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + ffqqff);
draw((-0.895059345715962,-0.6375021281980235)--(-3.14,-2.7), linewidth(1.1) + ffqqff);
draw((-5.051246248590594,6.115799681176306)--(-0.895059345715962,-0.6375021281980235), linewidth(1.1) + blue);
draw((3.0810645932710234,1.12395692083598)--(2.251206267261163,9.350852527005179), linewidth(1.1) + blue);
/* dots and labels */
dot((-0.8535898390715775,2.4107656245533),dotstyle);
label("$A$", (-1.6,2.1217356723134735), NE * labelscalefactor);
dot((4.846554211506446,-2.620429350613437),dotstyle);
label("$B$", (5.163488257557553,-2.712947165152712), NE * labelscalefactor);
dot((-3.14,-2.7),dotstyle);
label("$C$", (-3.796440261877091,-2.844324416170815), NE * labelscalefactor);
dot((-5.051246248590594,6.115799681176306),dotstyle);
label("$G$", (-5.819649927555882,6.06305320285656), NE * labelscalefactor);
dot((2.251206267261163,9.350852527005179),dotstyle);
label("$H$", (2.3520150857701427,9.610238980345336), NE * labelscalefactor);
dot((0.8408620061861096,-1.4141036218163907),linewidth(4.pt) + dotstyle);
label("$O$", (1.1170689261999718,-1.5305519059897863), NE * labelscalefactor);
dot((1.3905292426795337,-2.654861991247058),linewidth(4.pt) + dotstyle);
label("$D$", (1.3798234282361783,-3.2384561692251235), NE * labelscalefactor);
dot((-0.895059345715962,-0.6375021281980235),linewidth(4.pt) + dotstyle);
label("$F$", (-1.0900688909041634,-1.2677974039535806), NE * labelscalefactor);
dot((3.0810645932710234,1.12395692083598),linewidth(4.pt) + dotstyle);
label("$E$", (3.1928294922860037,1.3334721662048563), NE * labelscalefactor);
dot((-2.9143371002069363,7.0624698590035395),linewidth(4.pt) + dotstyle);
label("$D'$", (-3.139554006786575,7.481927513852071), NE * labelscalefactor);
dot((0.268469701803978,-0.12204818334687892),linewidth(4.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/f/7/b/f7bcae943aec07b6e4f683c3052787e94d2da575.png)
and 
.![\[\angle AFE = \frac{1}{2} \angle AFD = \angle ACB.\]](http://latex.artofproblemsolving.com/d/d/a/dda2c2e056d628aabfe14d094db4ff9687cd59e9.png)
follows symmetrically proving the similarity. For the parallel part,
and the claim is proven. ![\[\overline{EF} = \overline{GH} = \overline{BC}.\]](http://latex.artofproblemsolving.com/1/6/a/16ac4f2f7b9c13a4ec78f1decdaac4034eabfb2e.png)
This would give 
.
giving 
.
,
as desired.
and 
are isosceles. These triangles would also imply 
.![\[\angle AFG = \frac{1}{2} A - 30^\circ \qquad \text{and} \qquad \angle AEH = \frac{1}{2} A + 30^\circ.\]](http://latex.artofproblemsolving.com/f/7/3/f730b5267f151daced326828d09827a006667c8a.png)
Finally observe that![\[\angle GFE = C + \frac{1}{2} A - 30^\circ = 90 ^\circ = \angle HEF\]](http://latex.artofproblemsolving.com/e/b/e/ebe4d10c244eac61bf10f31e2989b19174f8a39c.png)
which proves that 
Now note the cyclicislscelestrapezoid since 
and that 
.
where the last step follows from knowing that 
Finally, we note that 
,
It is now evident that EF=BC=HG 
)
whence AFC is equilateral, and 
is acute. First 
.
and thus if 
and then 
is a paralellogram. In particular, 
.
implies that 
in 
.
is just 
(by a simple angle chase), proving one side of the story.
in triangles 
and 
.
and 
it follows that 
is a parallelogram, implying 
as 
.
and 
we have 
and so 
.
be the intersection of the altitude from 
Notice that since 
are the perpendicular bisectors of 
we have 
so 
is cyclic and 
Also, by symmetry, we have 
so 
are similar and similarly oriented. Thus, there exists a spiral similarity at 
Notice that this spiral similarity also sends 
so we have 
and quadrilaterals 
and 
is parallel to the tangent to the circumcircle of 
so 
and 
Also, we have 
so 
so the tangent to the circumcircle of 
and thus also bisects 
by symmetry. Therefore, 
and 
if and only if 
Since 
this holds if and only if 
is equilateral, or equivalently 
Now, let 
be the point such that 
is an isosceles trapezoid with 
and let 
Then we have 
so we are done.
,
.
be the foot of ![\[\measuredangle DAE=\measuredangle LAB\]](http://latex.artofproblemsolving.com/0/1/6/016732016aad0b96b4b9de58b277ec21221265cb.png)
![\[\measuredangle DAF=\measuredangle LAC\]](http://latex.artofproblemsolving.com/c/0/b/c0b89aeb4780e74bfd43cb358610f5ad03f5ae5a.png)
so 
.![\[90^{\circ}-\angle ADC=180^{\circ}-(\angle A+2\angle B)\]](http://latex.artofproblemsolving.com/b/4/0/b40af3fa42aea4977c84aed9a56b4ef8de7a1b08.png)
![\[90^{\circ}-(\angle B+90^{\circ}-\angle C)=180^{\circ}-(\angle A+2\angle B)\]](http://latex.artofproblemsolving.com/9/f/6/9f6d7368ce1509f1c865f81db959adba240f4482.png)
which is true.
in order to have ![\[\angle ADC=\angle B+90^{\circ}-\angle C=30^{\circ}\implies \angle C-\angle B=60^{\circ}\]](http://latex.artofproblemsolving.com/7/6/3/7631cd3d376430d2c932a1bb733f75232bc2f6b2.png)
and we are done.
,
due to 
.
.
sending 
.
,
,
.
which implies that 
.
,
and 
,
,![\[180^{\circ}=30^{\circ}+\angle DAC+\angle C=30^{\circ}+(90^{\circ}-\angle B)+\angle C\Longrightarrow \angle C-\angle B=60^{\circ}.\]](http://latex.artofproblemsolving.com/7/b/f/7bf9c34c29fa757fa57c60d6e78f431b93fb4b34.png)
,
.
,
.
,
,
,
,
through some simple angle chasing. Hence, it suffices to show that 
.![\[\angle AEF = \frac{1}{2} \angle AEB = \angle ABD,\]](http://latex.artofproblemsolving.com/5/9/0/59071f55f1b1be782fa6263cbd214cf6d1de6f6b.png)
.
.
,
is obtained from 
passing through 
.
,
.
,
,
.
,
from isogonality properties. To prove it is equivalent to 
,![\[BC = EF = BC \cdot 2 \cos \angle BAE\]](http://latex.artofproblemsolving.com/c/e/6/ce6ab43e27558144ea944653c22b56dc4449ca1a.png)
![\[\iff 60 = \angle BAE = 90 - \angle ADC = \angle A + 2 \angle C - 180 = \angle C - \angle B. \quad \blacksquare\]](http://latex.artofproblemsolving.com/5/a/e/5aeeb89b61cd9f94924938b22ed09458f514b4e0.png)
