AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a parallelogram of center 
. Prove that for any point 
, there exist unique points 
and 
such that is the center of mass of 
.
with 
be given . On the extension of the segment 
beyond 
, a point 
is chosen arbitrarily. Let the line through and 
intersect the line 
at the point 
. Let 
be the intersection of the lines 
and 
. Prove that the line 
is tangent to the circumcircle of the triangle 
.
be any point. Let 
be projection point from to 
. Circumcircle 
cuts circumcircle 
at 
. Let 
be antipode of wrt . Prove that 
are cyclic.
, 
be real numbers, such that 
. Prove that 
.
, angle 
. Point lies on 
such that 
. Line parallel to 
passing trough intersects 
at 
. Prove that 
.
Compute -![\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}](http://latex.artofproblemsolving.com/8/6/f/86fb48db976012106bd30f74cb6ada8c6e6d7800.png)
For 
define 
Determine the open interval(s) (if any) where 
is decreasing and all the open interval(s) (if any) where is increasing.
Determine all the local minima of (if any) and all the local maxima of (if any) 
of positive integers such that
, then it must not contain edge 
.
and 
meet at 
. Let 
be the incentres of triangles 
and 
respectively. It's easy to see 
are collinear as 
. excentre of 
and the 
excentre of 
both lie on the line perpendicular to 
through . So it follows is both excentres and 
.
. So 
is the angle bisector of 
, hence the result.
be a convex quadrilateral for which denote 
and 
. Prove that
is a tangential quadrilateral.
be a convex quadrilateral and denote . 
is a tangential quadrilateral.[/color is the incentre of quad. 
so the excenters are same.
be a convex quadrilateral. Let be the intersection of and , and let be the intersection of and . , , lie in that order, and , 
, lie in that order. Then 
has an incircle if and only if 
. be a convex quadrilateral. Let be the intersection of and , and let be the intersection of and . , , lie in that order, and , , lie in that order. Then there exists a circle tangent to 
, , , 
, that is a - excircle of concave quadrilateral , if and only if 
.
with 
and 
with 
. Then, from given relation we get 
.
be the intersection of perpendicular bisectors of 
and 
. Obviously, 
and 
are external angle bisectors of 
respectively and 
. With 
we get 
and 
are concyclic points. Fromk this fact, with 
we easily see that 
, so is the common excenter of the triangles 
, done.
be a triangle and for 
, 
denote 
.
is tangential 
of and 
. Therefore, 
.
. be a convex quadrilateral for which denote and . Suppose that exists what is tangent to the lines , , , . Prove that 
. 
. Therefore, 
.
.
be a convex quadrilateral for which denote and . Prove that is a tangential quadrilateral.
, and points 
on the ellipse such that 
intersect at . We can draw many diagrams and guess the point we want is the intersections of the tangent lines at on the ellipse. By properties of tangent lines to the ellipse, we have that the tangent line at is the external angle bisector of 
, and likewise for the tangent at . So the point we want is the intersection of the external angle bisector of , the external angle bisector 
, and the angle bisector of 
, which we shall call 
. These three angle bisectors intersect because of the tangential quadrilateral formed by 
, which is true by the angle conditions. Let the incenter of 
be 
, and the incenter of 
be 
. Letting the tangent at be called 
, we can let intersect 
at 
, and we want to show that that 
. By the tangent properties and external angle bisectors, we have that 
, so we have that 
is a harmonic bundle. But since lies on the angle bisector of 
, we also have that 
is a right angle because of the harmonic bundle. But this implies that 
is the angle bisector of , so 
. We can do the same for the angle bisector of , so we're done.
denote the intersection of the tangents to the ellipse at points and .![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(200);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.6) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.6499720477319775, xmax = 16.746605276580397, ymin = -5.5310671563144735, ymax = 8.857885711116099; /* image dimensions */
pair A = (1.51395092,2.694633594), B = (5.276470894,3.27487536),
C = (0.83338,-0.87), D = (8.03,-0.87), X = (3.2526330,3.51410666),
P = (2.442608,7.5406758);
/* draw figures */
draw(C--P--D,rgb(0.4,0.1,0.4));
draw(A--X--B,rgb(0.9,0.1,0.1));
draw(A--D^^B--C,rgb(0.1,0.6,0.1));
pair C1 = 0.9*X+0.1*C, D1 = 0.9*X+0.1*D, P1 = 0.8*X+0.2*P;
draw(C--C1,orange+linetype("4 4"), EndArrow);
draw(D--D1,orange+linetype("4 4"), EndArrow);
draw(P--P1,orange+linetype("4 4"), EndArrow);
draw(shift((4.43169,-0.863405))*xscale(5.521706)*yscale(4.18824184)*unitcircle,rgb(0.1,0.1,0.7));
/* dots and labels */
dot("$C$",C,SW);
dot("$D$",D,SE);
dot("$A$",A,NW);
dot("$B$",B,NE);
dot("$X$",X,N);
dot("$P$",P,N);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/1/d/7/1d76b313b5964bdd35ab9a445e5020a915d99412.png)
lies on the angle bisectors of 
and 
.
lies on the angle bisectors of and .
and 
be the reflections of across 
and 
, respectively. By the reflection property of ellipses, 
(as both are equal to the major axis of the ellipse), and furthermore 
. Therefore triangles 
and 
are congruent, implying 
bisects 
. The other part of the claim is symmetrical. is the incenter of both 
and 
. The result follows.
and 
Prove that
