Difference between revisions of "1997 USAMO Problems"
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− | == Problem 1 == | + | == Day 1 == |
+ | === 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>. | ||
− | [ | + | [[1997 USAMO Problems/Problem 1|Solution]] |
− | == Problem 2 == | + | === 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. | ||
− | [ | + | [[1997 USAMO Problems/Problem 2|Solution]] |
− | == Problem 3 == | + | === 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>. | ||
− | [ | + | [[1997 USAMO Problems/Problem 3|Solution]] |
− | == Problem 4 == | + | == Day 2 == |
+ | === 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>. | ||
− | [ | + | [[1997 USAMO Problems/Problem 4|Solution]] |
− | == Problem 5 == | + | === 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> | ||
− | <math> | + | <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>. |
− | [ | + | [[1997 USAMO Problems/Problem 5|Solution]] |
− | == Problem 6 == | + | === 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 | ||
− | <math>a_i+a_j\ | + | <math>a_i+a_j \le a_{i+j} \le a_i+a_j+1</math> |
− | for all <math>i, j\ | + | 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>. |
+ | |||
+ | [[1997 USAMO Problems/Problem 6|Solution]] | ||
+ | |||
+ | == See Also == | ||
+ | {{USAMO newbox|year= 1997|before=[[1996 USAMO]]|after=[[1998 USAMO]]}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:31, 12 April 2023
Contents
[hide]Day 1
Problem 1
Let be the prime numbers listed in increasing order, and let be a real number between and . For positive integer , define
where denotes the fractional part of . (The fractional part of is given by where is the greatest integer less than or equal to .) Find, with proof, all satisfying for which the sequence eventually becomes .
Problem 2
Let be a triangle, and draw isosceles triangles externally to , with as their respective bases. Prove that the lines through perpendicular to the lines , respectively, are concurrent.
Problem 3
Prove that for any integer , there exists a unique polynomial with coefficients in such that .
Day 2
Problem 4
To clip a convex -gon means to choose a pair of consecutive sides and to replace them by three segments and where is the midpoint of and is the midpoint of . In other words, one cuts off the triangle to obtain a convex -gon. A regular hexagon of area is clipped to obtain a heptagon . Then is clipped (in one of the seven possible ways) to obtain an octagon , and so on. Prove that no matter how the clippings are done, the area of is greater than , for all .
Problem 5
Prove that, for all positive real numbers
.
Problem 6
Suppose the sequence of nonnegative integers satisfies
for all with . Show that there exists a real number such that (the greatest integer ) for all .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.