AoPS Academy
, with 
.Initially, he folds the sheet along a straight line 
, where 
is a point on the side 
, so that vertex 
is located on side 
, as shown in the figure. Then folds the sheet again along a straight line 
, where 
is a point on side , so that vertex 
lies on the line ; and finally folds the sheet along the line 
. Matías observed that the vertices and 
were located on the same point of segment after making the folds. Calculate the measure of the angle 
.
be relatively prime positive integers. Calculate 
are different at 
. If 
are also different at , we call the ordered 
-tuple 
lucky. For which positive integers , one can find a lucky -tuple?
and 
.
.
.
, we have
satisfied
for all 
and all 
.
. Get all the 
such that 
is uniformly continuous on 
.
be a closed map with the property that the inverse image of each point in 
is a compact subset of 
. Prove that 
is compact whenever 
is compact in .
converge?
be real numbers. For any set 
let 
. Assume that the function 
takes on at least 
values where 
runs over all 
subsets of 
. Prove that the number of sets 
for which 
does not exceed 
.
satisfying the following two conditions for all 
:![\[ f(x+y)=f(x)+f(y), \;\;\; f \left( \max \{x, y \} \right) = \max \left\{ f(x),f(y) \right\}. \]](http://latex.artofproblemsolving.com/3/9/1/391164166941b57fe54bde43b1f3ef3b49494180.png)
. By one question, we may take three different indices 
and find out the set of numbers 
(unordered, of course). Find the minimal number of questions, which are necessary to find out all numbers 
.
. By one question, we may take three different indices and find out the set of numbers (unordered, of course). Find the minimal number of questions, which are necessary to find out all numbers .
, actually, and, for general instead of 
, I think it's 
.
questions, where 
is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other 
indices, the ones we haven't touched so far, and the procedure is repeated. questions suffice, assume first that is even (but 
; for 
we need 
questions, 
don't suffice). We ask questions 
for odd 
between 
and 
, and after that we ask the question 
. For odd we do something very similar.
numbers after few questions concerning next group? For example: 123, then 234, then 567, then 514?
is the third one.
If there are more questions using numbers already used, just choose one of them. We do this until we no longer can, i.e. until there are no questions using the numbers previously used.
questions, where is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other indices, the ones we haven't touched so far, and the procedure is repeated.
. Ask 
. Let 
of be asked once. If 
are asked once, then they cannot be asked in the same set hence there are at least sets asked. Consider the bipartite graph whose vertices are 
and the asked sets. If one could find in 
questions, then by looking at degrees, we get 
or 
however, there must be at least sets which contradicts as desired.
and 
![$A = \sum_{r=0}^{24} \frac{d(F(r,x))}{dx} \left[\frac{1}{F(r,x) + \frac{1}{F(r,x)}}\right]$](http://latex.artofproblemsolving.com/4/2/b/42bd0116f69f03884746ba8d06493ed999a567d6.png)
![$B = \sum_{r=0}^{24} \frac{-d(G(r,x))}{dx} \left[\frac{1}{G(r,x) + \frac{1}{G(r,x)}}\right]$](http://latex.artofproblemsolving.com/9/f/5/9f5ff4b5d991249940fa0521b222e5e4201a9fe0.png)
![$R = \frac{A - B}{\tan x\left[PQ - \frac{1}{3^{25}}\right]}$](http://latex.artofproblemsolving.com/b/f/4/bf447429d8f9e71117178abb46c688d7d4a036d6.png)
where 
is divided by 
?
questions 
where the 
'
'
'![$[1, 2005].$](http://latex.artofproblemsolving.com/b/a/0/ba07258b82d281889375fb4235b2b902ca265f90.png)
Call a positive integer 
sketchy if it only appears one time among the 
,
)
is. Now, let's define the function 
for 
,
is the number of times that 
is exactly equal to 
for all 
Indeed, if it was greater than 
are 
However, this means that the values of 
and 
are indistinguishable, impossible. Therefore, as 
contradiction. Hence, our assumption that we finished in 