[/quote]
,
and 
. Prove that
Where 
and 
Prove that
Where 
![\[
|MA|^2 + |MB|^2 + |MC|^2 = |GA|^2 + |GB|^2 + |GC|^2 + 3|MG|^2
\]](http://latex.artofproblemsolving.com/0/7/0/070ff15920d19d474defb77ceb606b2811fa85b4.png)
?
with circumcenter 
such that 
.
meets the circumcircle of 
.
meets the circumcircle of 
for the second time at 
.
meets the circumcircle of 
.
,
and 
are concurrent.
be a cyclic quadrilateral, and let 
,
be the midpoints of 
,
,
respectively. Let 
,
,
and 
be the orthocenters of triangles 
,
,
and 
,
have the same area.
,![\[
A = \left\{ r \, \middle| \, r = \frac{a}{3}\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) + b\sqrt[3]{xyz}, \quad x, y, z \in \left[1, \frac{a}{b}\right] \right\},
\]](http://latex.artofproblemsolving.com/e/9/2/e92becb6264b8e5a9753f2288436a48523a25f29.png)
and![\[
B = \left[2\sqrt{ab}, a + b\right].
\]](http://latex.artofproblemsolving.com/2/d/2/2d25aa634389b768dc26408480bb71b316c0bf17.png)
.
,
touch sides 
at 
,
lie on the line through 
be the intersections of 
with 
,
be the projections of 
onto line 
be the second intersections of 
with the incircle 
be the intersection of 
and 
.
bisects segment 
.
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that: 
.
![\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]](http://latex.artofproblemsolving.com/5/6/b/56b8cb8c13616501dd8308989cea2b338e91598b.png)
Y by jam10307, Davi-8191, yojan_sushi, blacksheep2003, Ultroid999OCPN, HsuAn, Toinfinity, OlympusHero, sugar_rush, Adventure10, megarnie, jhu08, suvamkonar, Lamboreghini, mathmax12, Mango247, ESAOPS, Rounak_iitr, and 6 other users
Let 
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
.
.
on 
.
.
,
,
.
,
.
and 
.
.
is the circumcenter of 
then 
so 
is cyclic. Then 
.
.
is a right triangle.
,
is taken to 
is taken to 
and is the midpoint of 
,
all lie on a circle.
to 
.
to 
.
and 
.
is a symmedian.
is a median, use the lemma again to show that 
,
,
.
and similarly 
,
.
so 
.
and 
.
so 
.
,
to the midpoint of 
and 
is cyclic.
.
and 
is 
and 
.
,
.
.
,
.
intercepts the minor arc 
in the circumcircle of 
.
,
is cyclic.
is directly similar to 
![\[ \angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A
\]](http://latex.artofproblemsolving.com/e/1/3/e13a8654a1709a7a02d1fa646bda4c1385cebc6a.png)
.![\[ \angle ENM = 90 - \angle MNC = 90 - A
\]](http://latex.artofproblemsolving.com/2/4/4/244049c0ce519505cd45904aa098efa03c182832.png)
,
is the medial triangle of 
.
.
since 
.
.![\[ \angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM
\]](http://latex.artofproblemsolving.com/1/6/b/16b4796a6589fb88d210d716ae46f00788d4dc96.png)
.
.
so 
by SAS similarity. Hence![\[ \angle EMO = \angle ENF = \angle ONF
\]](http://latex.artofproblemsolving.com/2/e/c/2ecaec60677d316cdf0e8c0f9bb44e2536dfcdc4.png)
is directly similar to 
.
is directly similar to 
.![\[ \angle EMO = \angle DMO = \angle DPF = \angle OPF
\]](http://latex.artofproblemsolving.com/5/e/a/5ea9ca703fd13b16d514858e04d72f2abd8cc4a7.png)
,
is cyclic. In other words, 
.
,
is cyclic. In other words, 
be an isosceles triangle with 
(
and 
on 
such that 
.
denotes the intersection of 
and 
and the gable(an art term) pass through 
.
,
intersect the line through 
meets 
(
.
and 
we want to show that 
is a harmonic range. This follows from![\[ \frac {AF}{FE} = \frac {AA'}{EE'} = \frac {AC}{EC}
\]](http://latex.artofproblemsolving.com/4/f/7/4f7833c91c0644bccae109bfa016346fb8ac6d43.png)
)
.
intersect at 
,
is a harmonic range and by the problem above, we conclude that 
.
,![\[ \frac {AF'}{F'E} = \frac {PA^2}{PE^2} = \frac {AC}{EC}
\]](http://latex.artofproblemsolving.com/5/b/0/5b024deeae4d226b5fbccb9b8c4ec40e4ade8137.png)
are six points on a circle then 
concur iff 
This result come from a slightly stronger version of the 1988 IMO Longlist Problem 88.
where 
is the median.
which is obviously true.
be the symmedian, and 
be a point on the symmedian such that 
is perpendicular to 
,
and similarly 
,
,
as the reference triangle, with A = (1:0:0), B=(0:1:0) and C = (0:0:1). Calculate the midpoints of BC, AC and AB and we get (0:1/2:1/2), (1/2:0:1/2) and (1/2:1/2:0) the points are M, N and P. So let's calculate the perpendicular bissector of AB and AC. By the Evan's Forgotten Trick (Perpendicularity Criterion), two vectors 
with coordinates 
and 
with coordinate
.
and AC :
are in the form (t :1/2 : 1/2), plugging these we can calculate t and ,furthermore, calculate the coordinates o D and E :
Now, let's calculate the intersection of lines BD and CE :
and points on CE are in the form : 
.
and CE by 
,
equation that is :
and find the constants u,v,w.
.
and 
.
.
which is clearly equal to zero, so, ANFP is cyclic.
.
s![[asy]
import olympiad;
size(10cm);
pair A = dir(120);
pair B = dir(210);
pair C = dir(330);
pair M = (B+C)/2;
pair N = (A+C)/2;
pair P = (A+B)/2;
pair O = circumcenter(A,B,C);
pair D = extension(A,M,P,O);
pair E = extension(N,O,A,M);
pair F = extension(B,D,E,C);
pair Q = extension(A,F,B,C);
draw(A--B--C--A, darkmagenta);
draw(P--N, orange);
draw(N--M, orange);
draw(P--M, orange);
draw(A--M, orange);
draw(P--O, orange);
draw(E--N, orange);
draw(B--D, orange);
draw(C--F, orange);
draw(O--F, orange);
draw(A--Q, fuchsia);
draw(circumcircle(A,P,N), purple + dashed);
markscalefactor = 0.02;
draw(anglemark(D,B,A), deepgreen);
draw(anglemark(B,A,D), deepgreen);
draw(anglemark(A,C,E), cyan);
draw(anglemark(E,A,C), cyan);
markscalefactor = 0.006;
draw(rightanglemark(O,F,A), blue);
markscalefactor = 0.008;
draw(anglemark(B,F,Q));
draw(anglemark(Q,F,C));
markscalefactor = 0.01;
draw(anglemark(B,F,Q));
draw(anglemark(Q,F,C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$P$", P, dir(P));
dot("$O$", O, dir(O));
dot("$D$", D, NE);
dot("$E$", E, dir(E));
dot("$F$", F, NW);
dot("$Q$", Q, dir(Q));
[/asy]](http://latex.artofproblemsolving.com/0/3/4/034c594f58b53d14c17014dd4fda7e8950ace0c1.png)
is the angle bisector of 
.
.
and 
.![\[\frac{\overline{FB}}{\sin(\angle BAF)} = \frac{\overline{AF}}{\sin(\angle ABF)} \qquad \text{and} \qquad \frac{\overline{AF}}{\sin(\angle ACF)} = \frac{\overline{CF}}{\sin(\angle FAC)}.\]](http://latex.artofproblemsolving.com/7/8/a/78a7a08804be7595a64a85ebe7c83f03279a3350.png)
Dividing both equations, eliminating 
and putting 
and 
,![\[\frac{\sin(\angle ACF)}{\sin(\angle ABF)} = \frac{\overline{BF}}{\overline{FC}} \cdot \frac{\sin(\angle FAC)}{\sin(\angle FAB)} = \frac{\sin(\angle MAC)}{\sin(\angle MAB)}.\]](http://latex.artofproblemsolving.com/b/7/f/b7f89c57ff7af1cde76dbeaabbbcc6fc0417f145.png)
Applying ``Ratio Lemma'' on 
would yield us![\[\frac{\overline{BF}}{\overline{FC}} \cdot \frac{\overline{QC}\cdot \overline{AB}}{\overline{BQ}\cdot \overline{AC}} = \frac{\overline{AB}}{\overline{AC}} \iff \frac{\overline{FB}}{\overline{FC}} = \frac{\overline{QB}}{\overline{QC}}\]](http://latex.artofproblemsolving.com/0/9/3/0931db781296545a84247bddecda8a35fd795e35.png)
By the converse of ``Angle Bisector Theorem'', we get that 
and the claim is proven. 
e
since 
and 
are exterior angle bisectors by symmetry. Due to the above claim, 
is exterior angle bisector of 
.
.![\[\angle OFN = \angle ONA = \angle OPA = 90^\circ\]](http://latex.artofproblemsolving.com/e/c/e/ece57554de860ccc1a91e3e87834f377b8989d78.png)
we get that 
with diameter 
as desired. 
,
are concyclic, where 
be the point on 
such that 
is a symmedian. Then 
in 
,
,
,
.
-
.
.
.
is the 
.
are collinear. Similarly we also get that 
are collinear and we are done.
,
is the intersection of the A-symmedian with the circumcircle. Note that 
,
to 
ratio and it satisfies the circumcircle equation. Thus, letting 
(we make them the same sum so we can just average them), we have that the midpoint of 
Denote this point by 
intersects 
.
and 
,
The equation of 
,
Note that 
and 
This is done so that 
and 
(strictly speaking this is not exactly 
,
which is clearly true. Hence, by EFFT, 
so 
.
.
by definition (and thus 
)
,
,
and 
,
inversion, then a homothety of scale factor 
centered at 
be the 
be the midpoints of 
and 
respectively. Let 
and 
to 
and 
meet on 
.
of 
and suppose it intersects 
.
,
is the midpoint of arc 
,
.
is right (since 
;
![[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 = -6.56, xmax = 5.32, ymin = -4.42, ymax = 7.74; /* image dimensions */
draw((-1.16,6.34)--(-2.88,-0.96)--(4.18,-0.9)--cycle, linewidth(0.4));
draw(arc((4.18,-0.9),0.6,158.05618236276638,180.48692124781314)--(4.18,-0.9)--cycle, linewidth(2) + blue);
draw(arc((-1.16,6.34),0.6,-76.01940149306334,-53.58866260801658)--(-1.16,6.34)--cycle, linewidth(2) + blue);
draw(arc((-1.16,6.34),0.6,-103.25801748053517,-76.01940149306334)--(-1.16,6.34)--cycle, linewidth(2) + red);
draw(arc((-2.88,-0.96),0.6,0.48692124781311374,27.725537235284936)--(-2.88,-0.96)--cycle, linewidth(2) + red);
draw(arc((-2.88,-0.96),0.6,49.50336653199302,76.74198251946484)--(-2.88,-0.96)--cycle, linewidth(2) + red);
draw(arc((4.18,-0.9),0.6,126.41133739198344,148.8420762770302)--(4.18,-0.9)--cycle, linewidth(2) + blue);
/* draw figures */
draw((-1.16,6.34)--(-2.88,-0.96), linewidth(0.4));
draw((-2.88,-0.96)--(4.18,-0.9), linewidth(0.4));
draw((4.18,-0.9)--(-1.16,6.34), linewidth(0.4));
draw((-1.16,6.34)--(0.65,-0.93), linewidth(0.4));
draw((-2.88,-0.96)--(0.2481425288175455,0.6840905057991408), linewidth(0.4));
draw((0.2481425288175455,0.6840905057991408)--(4.18,-0.9), linewidth(0.4));
draw((-2.88,-0.96)--(-0.14104396110317108,2.2472870702873227), linewidth(0.4));
draw((-0.4421832207949546,1.8946563452047962)--(4.18,-0.9), linewidth(0.4));
/* dots and labels */
dot((-1.16,6.34),linewidth(2pt) + dotstyle);
label("$A$", (-1.08,6.42), NE * labelscalefactor);
dot((-2.88,-0.96),linewidth(2pt) + dotstyle);
label("$B$", (-3.28,-0.9), NE * labelscalefactor);
dot((4.18,-0.9),linewidth(2pt) + dotstyle);
label("$C$", (4.26,-0.82), NE * labelscalefactor);
dot((0.65,-0.93),linewidth(1pt) + dotstyle);
label("$M$", (0.74,-0.9), NE * labelscalefactor);
dot((0.2481425288175455,0.6840905057991408),linewidth(2pt) + dotstyle);
label("$Q$", (0.32,0.76), NE * labelscalefactor);
dot((-0.14104396110317108,2.2472870702873227),linewidth(1pt) + dotstyle);
label("$D$", (0.12,2.3), NE * labelscalefactor);
dot((0.015671121441022638,1.6178292525545666),linewidth(1pt) + dotstyle);
label("$E$", (0.16,1.52), NE * labelscalefactor);
dot((-0.4421832207949546,1.8946563452047962),linewidth(2pt) + dotstyle);
label("$F$", (-0.7,1.82), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/b/d/9bd60ac35e48a1b5c5a42e5a74d222ea83eb6fcc.png)
H
and 
then 
i.e the 
sends 
(where 
with 
finishes. 
.
.
,
is collinear. Similarly 
is collinear, hence 
.
and 
.![\[\frac{AD}{DM}\cdot\frac{MB}{BC}\cdot\frac{CF_B}{F_BA}=-1\]](http://latex.artofproblemsolving.com/5/e/5/5e54170ce2ac7024f3333172ae62de301b4e5615.png)
![\[\frac{AE}{EM}\cdot\frac{MC}{CB}\cdot\frac{BF_C}{F_CA}=-1\]](http://latex.artofproblemsolving.com/7/a/1/7a1664eaa620dd943d59f799bb20b5160a416700.png)
![\[\Longrightarrow\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=\frac{DM}{AD}\cdot\frac{AE}{EM}\]](http://latex.artofproblemsolving.com/c/e/1/ce165eb9c69492f598abd43877c54c1ebe4d3daf.png)
![\[\frac{BF_A}{F_AC}\cdot\frac{CF_B}{F_BA}\cdot\frac{AF_C}{F_CB}=1\]](http://latex.artofproblemsolving.com/a/e/e/aee8d8b39f862aabd18e788fb5e77ad67d0a0b0e.png)
![\[\Longrightarrow\frac{BF_A}{F_AC}=\frac{AD}{DM}\cdot\frac{EM}{AE}=\frac{BD}{DM}\cdot\frac{EM}{CE}=\frac{\sin(\angle FCM)}{\sin(
\angle MBF)}=\frac{BF}{FC}\]](http://latex.artofproblemsolving.com/0/5/5/055bf1dd252f3a766680ede6383b59b84bbee079.png)
Hence, 
.![\[\angle BFC=\angle BAC+\angle FBA+\angle ACF=\angle BAF+\angle FAC+\angle FBA+\angle ACF=2\angle BAC.\]](http://latex.artofproblemsolving.com/f/b/e/fbe29c96f075f66804ea6bc2c775490d29f7ab4d.png)
Hence, 
are concyclic. Moreover, ![\[\angle BAF+\angle FBA=\angle FAC+\angle ACF=\angle BAC=\angle BAF+\angle FAC\Longrightarrow \angle FBA=\angle FAC,\angle BAF=\angle ACF\]](http://latex.artofproblemsolving.com/b/1/5/b1590a555d679dea87082d4811e29284a18bd8f4.png)
This means 
,
are collinear, which would imply by symmetry that 
are collinear, solving the problem.
This sends 
is a parallelogram. Additionally, the perpendicular bisector of 
We wish to show that 
are concyclic. Indeed, 
so we are done.
and 
.
