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be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
have 
as the 
-mixtilinear incircle, and let 
be the tangency point of with the circumcircle 
. Let 
be a diameter of . Define 
, 
similarly. Prove that the lines , 
, and 
are concurrent.
such that 
, for all 
.
be a triangle and 
its incenter. The lines 
and 
intersect the circumcircle of again at 
and 
, respectively. Let 
and 
be the circumferences of diameters 
and 
, respectively. The circle intersects 
at 
and 
, and the circle intersects 
at 
and 
. Show that , , and are concyclic.
and 
are given on side 
of convex quadrilateral 
(with closer than to 
). It is known that 
and 
. Prove that 
.
be a positive integer. On an 
board, players and take turns in a game. On each turn, a player selects an edge of the board (not on the board border) and makes a cut along that edge. If after the move the board is split into more than one piece, then that player loses the game. moves first. Depending on the value of , determine whether one of the players has a winning strategy.
be a quadrilateral with 
, 
, and 
. Let be a point in the plane not on 
, 
, 
, and 
be the point on such that 
. If 
and 
, show that 
, 
, and 
are collinear.
be a triangle with altitude 
. The -excircle touches 
at 
, and intersects the circumcircle at two points and 
. Prove that one can select points and on lines 
and such that quadrilateral 
is a rhombus.
be non-negative real numbers, such that 
Prove that![\[52 + 17(\sqrt a + \sqrt b + \sqrt c + \sqrt d)^2\geqslant 9(ab+ bc + ca + da + db + dc)^2.\]](http://latex.artofproblemsolving.com/6/e/0/6e06719b43ee8df511295af3581b487ddfb17f5c.png)
and be two positive five-digit palindrome integers, such that 
and there is no other palindrome number between them. Determine the possible values of 
.
). And indeed, the 
should be 
; mind you, when it works for the latter, it will a fortiori work for the former! of the quadrilateral 
is 
, by the known fact that the area of a convex quadrilateral is half the product of its diagonals with the sine of their angle, followed by the fact that the sine of an angle is at most 
, followed by Ptolemy's inequality.
, so if we prove that 
, i.e. 
, that will be enough. But this is a simple vectorial computation.
and 
, so 
.
, by the triangle inequality.
. Then 
, 
, 
, so the inequality is equivalent to 
, which, as far as I remember, is the triangle inequality, expressed in the Euclidean norm (in any dimension) by Minkovski's inequality. The fact it is not appearing in its usual form
,
(having as last recourse the triangle inequality - what else?); otherwise it is difficult to foresee a change in the chain of formulae.
and 
prove that![\[\sqrt a + \sqrt b + \sqrt c \geqslant ab+bc+ca.\]](http://latex.artofproblemsolving.com/c/e/9/ce9dc64acc63c6bcce137e963ca8f86ab75018bf.png)
[/quote]
such that![\[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\]](http://latex.artofproblemsolving.com/b/4/2/b42cd6305fab848a60c79689f17bdb55e1ba0d31.png)
![\[\tfrac{AB}{AE}=\tfrac{CD}{CF}=n\]](http://latex.artofproblemsolving.com/f/8/2/f82128eb770639038420fce432811dd37da6243d.png)
and 
than we have 
hence a mistake in question?
instead of 
it seems.
such as 
are parallelograms.
collinear 
such as 
.
.
.
and 
be perpendicular lines on 
,
so 
.
inequality is equivalent with 
.
.
and 
.
.