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.
,
and 
. Interior angle bisector of 
intersects 
at 
, and 
at 
.Prove that circumcircle of 
passes from circumcenter of 
.
be inscribed in the circle 
. A line through point 
intersects 
and 
at points 
and 
, respectively. Let 
be the reflection of across the midpoint of , and 
be the reflection of across the midpoint of . Prove that:
of triangle across line 
lies on the circle .
and 
coincide. is a positive integer. Call all positive divisors of which are different from 
and beautiful divisors.We call a special number when it has at least 
beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
and 
are three complex numbers with product . Assume that none of and are real or have absolute value . Define
Given that both 
and 
are real numbers, find all possible values of the ordered pair 
.
be positive integers and let 
be a non-increasing list of positive real numbers. Prove that there exists 
sets 
which partition the set 
such that 
+1 w
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
+1 w
, 
, and 
, are marked on the plane. You have a pencil and a double-edged ruler of width . Using them, you may perform the following operations:
and 
are marked, draw the line connecting and .
and 
are drawn, mark the intersection of and .
is drawn, draw a line parallel to that is at distance away from (note that two such lines may be drawn). using these operations.
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
, define 
to be the set of all triplets 
of natural numbers, with 
odd and 
, such that 
and divides 
. Find all values of for which is finite.
be positive real numbers . Prove that
and 
in the plane, draws the perpendicular bisector of 
. Show that if three lines forming a triangle are drawn, Janabel can mark the orthocenter of the triangle using this device, a pencil, and no other tools.
be a convex equilateral hexagon with sides of length . Let 
be the area of the region contained within both 
and 
, and let 
be the area of the region within the hexagon outside both triangles. Prove that: ![\[ \min \{ [ACE], [BDF] \} + R_2 - R_1 \le \frac{3\sqrt{3}}{4}. \]](http://latex.artofproblemsolving.com/c/c/a/cca874d6f354636f36af4229e45af97f8c153bcd.png)
![\[[ABCDEF] - \max \{[ACE], [BDF]\} \le \frac{3\sqrt{3}}{4}\]](http://latex.artofproblemsolving.com/5/d/1/5d1b29ce1a095617013c158df5ebdf92e97a8c35.png)
If 
are the corresponding vertex angles, this is equivalently ![\[\min \{\frac{1}{2}(1)^2[\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}], \frac{1}{2}(1)^2[\sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F}] \} \le \frac{3\sqrt{3}}{4}\]](http://latex.artofproblemsolving.com/3/c/6/3c6be09305e9ccc097283e35143c63a8f28e9f44.png)
By the convexity of 
on the interval ![$[0, \pi]$](http://latex.artofproblemsolving.com/0/a/1/0a151f5b93ce5735f72cdd79c46051f52b9d820d.png)
(where all angles lie, by the convexity of the figure), this is at most ![\[ \min \{\frac{3}{2} \sin \left({\frac{\theta_A + \theta_C + \theta_E}{3}} \right), \frac{3}{2}\sin \left( {\frac{\theta_B + \theta_D + \theta_F}{3}} \right) \} \]](http://latex.artofproblemsolving.com/e/1/b/e1bc167033d9696e359bc30610eff6b19c0141bf.png)
Since 
,![\[\frac{3}{2} \min \{\sin{\theta}, \sin \left( {\frac{4\pi}{3} - \theta} \right) \}\]](http://latex.artofproblemsolving.com/7/8/f/78f1993f6b47492c5c59591743be347a1d67eb18.png)
where 
If ![$\theta \in [\pi / 3, 2\pi /3]$](http://latex.artofproblemsolving.com/5/d/a/5da25d1641ee4c04a2fe2e1b22829166537f2da9.png)
then ![\[ \sin \left( \frac{4\pi}{3} - \theta \right) \le \sin \left( \frac{4\pi}{3} - \frac{2\pi}{3} \right) = \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/3/e/9/3e90b1fc98ec8bd2c381935f5b142a65bfb2083b.png)
Similarly, if ![$\theta \in [2\pi /3, \pi]$](http://latex.artofproblemsolving.com/d/7/4/d743a6829e6f5cb930a7b673eaa4be1f1a0686d0.png)
,![\[ \sin \left(\theta \right) \le \sin \left(\frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/9/b/2/9b2df59dbf8f609076e7bf48bf69e95a9cf12cbf.png)
so in all cases the expression is at most ![\[\frac{3}{2} \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/8/5/c/85c4144d07507d4614d9e6c2d383dc93516103e1.png)
as desired.![\[\dfrac{1}{2}\min (\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}, \sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F} ) \le \dfrac{3\sqrt{3}}{4}\]](http://latex.artofproblemsolving.com/c/b/5/cb580ef1c8d2b47b1d90fe92b0e6b30e7a5dfd27.png)
or ![\[\min (\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}, \sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F} ) \le \dfrac{3\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/7/7/7/777ecb55aa409155e90a21547fc1c2256ce83737.png)
![\[\min (\sin{\theta_A} + \sin{\theta_C} + \sin{\theta_E}, \sin{\theta_B} + \sin{\theta_D} + \sin{\theta_F} ) >\dfrac{3\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/c/1/e/c1e3c06bd2b8738f44eeca4407a67d6984dd2836.png)
![\[\sum\sin\theta > 3\sqrt{3}\]](http://latex.artofproblemsolving.com/6/5/6/65654411670949364d31d7f513f4c8397cb3cdd4.png)
is concave for 
,
so 
, contradiction.