Difference between revisions of "2013 USAJMO"

 
(15 intermediate revisions by 4 users not shown)
Line 1: Line 1:
==Day 1==
+
'''2013 [[USAMO|USAJMO]]''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.
===Problem 1===
 
  
Are there integers and such that and are both perfect cubes of integers?
+
*[[2013 USAJMO Problems]]
 +
*[[2013 USAJMO Problems/Problem 1]]
 +
*[[2013 USAJMO Problems/Problem 2]]
 +
*[[2013 USAMO Problems/Problem 1|2013 USAJMO Problems/Problem 3]]
 +
*[[2013 USAJMO Problems/Problem 4]]
 +
*[[2013 USAJMO Problems/Problem 5]]
 +
*[[2013 USAMO Problems/Problem 4|2013 USAJMO Problems/Problem 6]]
  
[[2013 USAMO Problems/Problem 1|Solution]]
+
{{USAJMO newbox|year= 2013 |before=[[2012 USAJMO]]|after=[[2014 USAJMO]]}}
 
 
===Problem 2===
 
Each cell of an board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:
 
 
 
(i) The difference between any two adjacent numbers is either or .
 
(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to .
 
 
 
Determine the number of distinct gardens in terms of and .
 
 
 
[[2013 USAMO Problems/Problem 2|Solution]]
 
 
 
===Problem 3===
 
In triangle , points lie on sides respectively. Let , , denote the circumcircles of triangles , , , respectively. Given the fact that segment intersects , , again at respectively, prove that .
 
 
 
[[2013 USAMO Problems/Problem 3|Solution]]
 
 
 
==Day 2==
 
===Problem 4===
 
Let be the number of ways to write as a sum of powers of , where we keep track of the order of the summation. For example, because can be written as , , , , , and . Find the smallest greater than for which is odd.
 
 
 
[[2013 USAMO Problems/Problem 4|Solution]]
 
 
 
===Problem 5===
 
 
 
Quadrilateral is inscribed in the semicircle with diameter . Segments and meet at . Point is the foot of the perpendicular from to line . Point lies on such that line is perpendicular to line . Let be the intersection of segments and . Prove that
 
[[2013 USAMO Problems/Problem 5|Solution]]
 
 
 
===Problem 6===
 
 
 
Find all real numbers satisfying
 
 
 
[[2013 USAMO Problems/Problem 6|Solution]]
 
 
 
== See Also ==
 
{{USAJMO newbox|year= 2013|before=[[2012 USAJMO]]|after=[[2014 USAJMO]]}}
 

Latest revision as of 19:05, 30 April 2014

2013 USAJMO problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.

2013 USAJMO (ProblemsResources)
Preceded by
2012 USAJMO
Followed by
2014 USAJMO
1 2 3 4 5 6
All USAJMO Problems and Solutions