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.
, where 
, then the number of integers in the range of 
is __
be a triangle and 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
, then which of the following is/are true?
is obtuse
is acute
, where 
is semi perimeter
, where is semi perimeter
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
. Prove that
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
. Prove that
satisfy that: 
and 
. Find the maximum: 
+1 w
, such that 
Prove that :
equality cases : ?
. The point 
is in the interior of . The following ratio equalities hold:![\[\angle PAD:\angle PBA:\angle DPA=1:2:3=\angle CBP:\angle BAP:\angle BPC\]](http://latex.artofproblemsolving.com/9/3/8/9388a91325842f1ec67a4693f0f10af46ed66c6d.png)
Prove that the following three lines meet in a point: the internal bisectors of angles 
and 
and the perpendicular bisector of segment 
.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
be the set of positive real numbers. Determine all functions 
such that for all positive real numbers 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be 
. The tangents to at 
meet at point 
. For a point 
on line 
which is not on the segment , let the midpoint of 
be 
. Lines 
meet again at points 
respectively. Let 
be the midpoint of . Prove that the points 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
=
and 
=
.
= and =.
.. Since 
= 
= , and 
= (analogously), then 
= 
. So, by the reflection: 
= . We can conclude that D' lies on 
. Again, the reflection also gives us that 
= 
= 
. Then, E is the center of 
. By this, we have:
= 
= 
- . Analogously: 
= - . So: 
= and D' must lie on 
. (In fact, D' is the Miquel point of the quadrilateral ). Noting that AD' is the radical axis of and 
, we conclude that 
is perpendicular to 
. Analogously, 
is perpendicular to 
. Then: 
= 
- 
= - = - . As we want to show: O lies on .
be the intersection of and , notice that 
belongs to 
, and 
iff 
. So notice we just need to prove 
, and since we already have a common side 
and a common angle (
, which is true). Then, we'll prove 
to conclude by SAA congruence.
, but since 
, I spent very much time on angle chase, but I couldn't do anything, then I thought: what if I tried the SAS similarity to prove it? Here it is:
, which is true because is the miquel point of 
, and there's a spiral similarity taking 
.
be the unit circle.
lies on the perpendicular from to 
. Also 
is cyclyc 
satisfying 
and 
Prove that
