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| − | ==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===
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| − | Are there integers <math>a</math> and <math>b</math> such that <math>a^5b+3</math> and <math>ab^5+3</math> 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]] |
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| − | [[2013 USAJMO Problems/Problem 1|Solution]]
| + | {{USAJMO newbox|year= 2013 |before=[[2012 USAJMO]]|after=[[2014 USAJMO]]}} |
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| − | ===Problem 2===
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| − | Each cell of an <math>m\times n</math> 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:
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| − | (i) The difference between any two adjacent numbers is either <math>0</math> or <math>1</math>.
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| − | (ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to <math>0</math>.
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| − | Determine the number of distinct gardens in terms of <math>m</math> and <math>n</math>.
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| − | [[2013 USAJMO Problems/Problem 2|Solution]]
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| − | ===Problem 3===
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| − | In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively. Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively. Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math>.
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| − | [[2013 USAMO Problems/Problem 1|Solution]]
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| − | ==Day 2==
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| − | ===Problem 4===
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| − | Let <math>f(n)</math> be the number of ways to write <math>n</math> as a sum of powers of <math>2</math>, where we keep track of the order of the summation. For example, <math>f(4)=6</math> because <math>4</math> can be written as <math>4</math>, <math>2+2</math>, <math>2+1+1</math>, <math>1+2+1</math>, <math>1+1+2</math>, and <math>1+1+1+1</math>. Find the smallest <math>n</math> greater than <math>2013</math> for which <math>f(n)</math> is odd.
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| − | [[2013 USAJMO Problems/Problem 4|Solution]]
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| − | ===Problem 5===
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| − | Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>. Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>. Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>. Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>. Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>. Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath>
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| − | [[2013 USAJMO Problems/Problem 5|Solution]]
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| − | ===Problem 6===
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| − | Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
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| − | [[2013 USAMO Problems/Problem 4|Solution]]
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| − | == See Also ==
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| − | {{USAJMO newbox|year= 2013|before=[[2012 USAJMO]]|after=[[2014 USAJMO]]}} | |
