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Difference between revisions of "1997 USAMO Problems"

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== Problem 1 ==
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== Day 1 ==
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=== Problem 1 ===
 
Let <math>p_1,p_2,p_3,...</math> be the prime numbers listed in increasing order, and let <math>x_0</math> be a real number between <math>0</math> and <math>1</math>. For positive integer <math>k</math>, define
 
Let <math>p_1,p_2,p_3,...</math> be the prime numbers listed in increasing order, and let <math>x_0</math> be a real number between <math>0</math> and <math>1</math>. For positive integer <math>k</math>, define
  
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where <math>\{x\}</math> denotes the fractional part of <math>x</math>. (The fractional part of <math>x</math> is given by <math>x-\lfloor{x}\rfloor</math> where <math>\lfloor{x}\rfloor</math> is the greatest integer less than or equal to <math>x</math>.) Find, with proof, all <math>x_0</math> satisfying <math>0<x_0<1</math> for which the sequence <math>x_0,x_1,x_2,...</math> eventually becomes <math>0</math>.
 
where <math>\{x\}</math> denotes the fractional part of <math>x</math>. (The fractional part of <math>x</math> is given by <math>x-\lfloor{x}\rfloor</math> where <math>\lfloor{x}\rfloor</math> is the greatest integer less than or equal to <math>x</math>.) Find, with proof, all <math>x_0</math> satisfying <math>0<x_0<1</math> for which the sequence <math>x_0,x_1,x_2,...</math> eventually becomes <math>0</math>.
  
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[http://www.artofproblemsolving.com/Wiki/index.php/Problem_1 Solution]
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[[1997 USAMO Problems/Problem 1|Solution]]
  
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== Problem 2 ==
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=== Problem 2 ===
 
Let <math>ABC</math> be a triangle, and draw isosceles triangles <math>BCD, CAE, ABF</math> externally to <math>ABC</math>, with <math>BC, CA, AB</math> as their respective bases. Prove that the lines through <math>A,B,C</math> perpendicular to the lines <math>\overleftrightarrow{EF},\overleftrightarrow{FD},\overleftrightarrow{DE}</math>, respectively, are concurrent.
 
Let <math>ABC</math> be a triangle, and draw isosceles triangles <math>BCD, CAE, ABF</math> externally to <math>ABC</math>, with <math>BC, CA, AB</math> as their respective bases. Prove that the lines through <math>A,B,C</math> perpendicular to the lines <math>\overleftrightarrow{EF},\overleftrightarrow{FD},\overleftrightarrow{DE}</math>, respectively, are concurrent.
  
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[http://www.artofproblemsolving.com/Wiki/index.php/Problem_2 Solution]
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[[1997 USAMO Problems/Problem 2|Solution]]
  
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== Problem 3 ==
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=== Problem 3 ===
 
Prove that for any integer <math>n</math>, there exists a unique polynomial <math>Q</math> with coefficients in <math>\{0,1,...,9\}</math> such that <math>Q(-2)=Q(-5)=n</math>.
 
Prove that for any integer <math>n</math>, there exists a unique polynomial <math>Q</math> with coefficients in <math>\{0,1,...,9\}</math> such that <math>Q(-2)=Q(-5)=n</math>.
  
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[http://www.artofproblemsolving.com/Wiki/index.php/Problem_3 Solution]
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[[1997 USAMO Problems/Problem 3|Solution]]
  
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== Problem 4 ==
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== Day 2 ==
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=== Problem 4 ===
 
To ''clip'' a convex <math>n</math>-gon means to choose a pair of consecutive sides <math>AB, BC</math> and to replace them by three segments <math>AM, MN,</math> and <math>NC,</math> where <math>M</math> is the midpoint of <math>AB</math> and <math>N</math> is the midpoint of <math>BC</math>. In other words, one cuts off the triangle <math>MBN</math> to obtain a convex <math>(n+1)</math>-gon. A regular hexagon <math>P_6</math> of area <math>1</math> is clipped to obtain a heptagon <math>P_7</math>. Then <math>P_7</math> is clipped (in one of the seven possible ways) to obtain an octagon <math>P_8</math>, and so on. Prove that no matter how the clippings are done, the area of <math>P_n</math> is greater than <math>\frac{1}{3}</math>, for all <math>n\ge6</math>.
 
To ''clip'' a convex <math>n</math>-gon means to choose a pair of consecutive sides <math>AB, BC</math> and to replace them by three segments <math>AM, MN,</math> and <math>NC,</math> where <math>M</math> is the midpoint of <math>AB</math> and <math>N</math> is the midpoint of <math>BC</math>. In other words, one cuts off the triangle <math>MBN</math> to obtain a convex <math>(n+1)</math>-gon. A regular hexagon <math>P_6</math> of area <math>1</math> is clipped to obtain a heptagon <math>P_7</math>. Then <math>P_7</math> is clipped (in one of the seven possible ways) to obtain an octagon <math>P_8</math>, and so on. Prove that no matter how the clippings are done, the area of <math>P_n</math> is greater than <math>\frac{1}{3}</math>, for all <math>n\ge6</math>.
  
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[http://www.artofproblemsolving.com/Wiki/index.php/Problem_4 Solution]
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[[1997 USAMO Problems/Problem 4|Solution]]
  
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== Problem 5 ==
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=== Problem 5 ===
 
Prove that, for all positive real numbers <math>a, b, c,</math>
 
Prove that, for all positive real numbers <math>a, b, c,</math>
  
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<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.
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<math>\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}</math>.
  
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== Problem 6 ==
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[[1997 USAMO Problems/Problem 5|Solution]]
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=== Problem 6 ===
 
Suppose the sequence of nonnegative integers <math>a_1,a_2,...,a_{1997}</math> satisfies  
 
Suppose the sequence of nonnegative integers <math>a_1,a_2,...,a_{1997}</math> satisfies  
  
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<math>a_i+a_j\lea_{i+j}\lea_i+a_j+1</math>
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<math>a_i+a_j \le a_{i+j} \le a_i+a_j+1</math>
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for all <math>i, j \ge 1</math> with <math>i+j \le 1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\le x</math>) for all <math>1 \le n \le 1997</math>.
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[[1997 USAMO Problems/Problem 6|Solution]]
  
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for all <math>i, j\ge1</math> with <math>i+j\le1997</math>. Show that there exists a real number <math>x</math> such that <math>a_n=\lfloor{nx}\rfloor</math> (the greatest integer <math>\lenx</math>) for all <math>1\len\le1997</math>.
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== See Also ==
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{{USAMO newbox|year= 1997|before=[[1996 USAMO]]|after=[[1998 USAMO]]}}
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{{MAA Notice}}

Latest revision as of 13:31, 12 April 2023

Day 1

Problem 1

Let $p_1,p_2,p_3,...$ be the prime numbers listed in increasing order, and let $x_0$ be a real number between $0$ and $1$. For positive integer $k$, define

$x_{k}=\begin{cases}0&\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&\text{ if }x_{k-1}\ne0\end{cases}$

where $\{x\}$ denotes the fractional part of $x$. (The fractional part of $x$ is given by $x-\lfloor{x}\rfloor$ where $\lfloor{x}\rfloor$ is the greatest integer less than or equal to $x$.) Find, with proof, all $x_0$ satisfying $0<x_0<1$ for which the sequence $x_0,x_1,x_2,...$ eventually becomes $0$.

Solution

Problem 2

Let $ABC$ be a triangle, and draw isosceles triangles $BCD, CAE, ABF$ externally to $ABC$, with $BC, CA, AB$ as their respective bases. Prove that the lines through $A,B,C$ perpendicular to the lines $\overleftrightarrow{EF},\overleftrightarrow{FD},\overleftrightarrow{DE}$, respectively, are concurrent.

Solution

Problem 3

Prove that for any integer $n$, there exists a unique polynomial $Q$ with coefficients in $\{0,1,...,9\}$ such that $Q(-2)=Q(-5)=n$.

Solution

Day 2

Problem 4

To clip a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$. In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon. A regular hexagon $P_6$ of area $1$ is clipped to obtain a heptagon $P_7$. Then $P_7$ is clipped (in one of the seven possible ways) to obtain an octagon $P_8$, and so on. Prove that no matter how the clippings are done, the area of $P_n$ is greater than $\frac{1}{3}$, for all $n\ge6$.

Solution

Problem 5

Prove that, for all positive real numbers $a, b, c,$

$\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}$.

Solution

Problem 6

Suppose the sequence of nonnegative integers $a_1,a_2,...,a_{1997}$ satisfies

$a_i+a_j \le a_{i+j} \le a_i+a_j+1$

for all $i, j \ge 1$ with $i+j \le 1997$. Show that there exists a real number $x$ such that $a_n=\lfloor{nx}\rfloor$ (the greatest integer $\le x$) for all $1 \le n \le 1997$.

Solution

See Also

1997 USAMO (Problems • Resources)
Preceded by
1996 USAMO 		Followed by
1998 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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