Difference between revisions of "1977 USAMO Problems"

(Problem 5: fixed LaTeXed equation)
m (Problem 5: eww tiny parentheses)
 
(9 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 +
Problems from the '''1977 [[United States of America Mathematical Olympiad | USAMO]]'''.
 +
 
==Problem 1==
 
==Problem 1==
 
Determine all pairs of positive integers <math> (m,n)</math> such that
 
Determine all pairs of positive integers <math> (m,n)</math> such that
<math> (1\plus{}x^n\plus{}x^{2n}\plus{}\cdots\plus{}x^{mn})</math> is divisible by <math> (1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{m})</math>.
+
<math> (1+x^n+x^{2n}+\cdots+x^{mn})</math> is divisible by <math> (1+x+x^2+\cdots+x^{m})</math>.
 +
 
 +
[[1977 USAMO Problems/Problem 1 | Solution]]
  
 
==Problem 2==
 
==Problem 2==
<math> ABC</math> and <math> A'B'C'</math> are two triangles in the same plane such that the lines <math> AA',BB',CC'</math> are mutually parallel. Let <math> [ABC]</math> denotes the area of triangle <math> ABC</math> with an appropriate <math> \pm</math> sign, etc.; prove that
+
<math> ABC</math> and <math> A'B'C'</math> are two triangles in the same plane such that the lines <math> AA',BB',CC'</math> are mutually parallel. Let <math> [ABC]</math> denote the area of triangle <math> ABC</math> with an appropriate <math> \pm</math> sign, etc.; prove that
<cmath> 3([ABC] \plus{} [A'B'C']) \equal{} [AB'C'] \plus{} [BC'A'] \plus{} [CA'B'] \plus{} [A'BC] \plus{} [B'CA] \plus{} [C'AB].</cmath>
+
<cmath> 3([ABC] + [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B'] + [A'BC] + [B'CA] + [C'AB].</cmath>
 +
 
 +
[[1977 USAMO Problems/Problem 2 | Solution]]
  
 
==Problem 3==
 
==Problem 3==
If <math> a</math> and <math> b</math> are two of the roots of <math> x^4\plus{}x^3\minus{}1\equal{}0</math>, prove that <math> ab</math> is a root of <math> x^6\plus{}x^4\plus{}x^3\minus{}x^2\minus{}1\equal{}0</math>.
+
If <math> a</math> and <math> b</math> are two of the roots of <math> x^4+x^3-1=0</math>, prove that <math> ab</math> is a root of <math> x^6+x^4+x^3-x^2-1=0</math>.
 +
 
 +
[[1977 USAMO Problems/Problem 3 | Solution]]
  
 
==Problem 4==
 
==Problem 4==
 
Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.
 
Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.
 +
 +
[[1977 USAMO Problems/Problem 4 | Solution]]
  
 
==Problem 5==
 
==Problem 5==
 
If <math> a,b,c,d,e</math> are positive numbers bounded by <math> p</math> and <math> q</math>, i.e, if they lie in <math> [p,q], 0 < p</math>, prove that
 
If <math> a,b,c,d,e</math> are positive numbers bounded by <math> p</math> and <math> q</math>, i.e, if they lie in <math> [p,q], 0 < p</math>, prove that
<cmath> (a + b + c + d + e)(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} + \frac {1}{e}) \le 25 + 6\left(\sqrt {\frac {p}{q}} - \sqrt {\frac {q}{p}}\right)^2</cmath>
+
<cmath> (a + b + c + d + e)\biggl(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} + \frac {1}{e}\biggr) \le 25 + 6\left(\sqrt {\frac {p}{q}} - \sqrt {\frac {q}{p}}\right)^2</cmath>
 
and determine when there is equality.
 
and determine when there is equality.
 +
 +
[[1977 USAMO Problems/Problem 5 | Solution]]
 +
 +
== See Also ==
 +
{{USAMO box|year=1977|before=[[1976 USAMO]]|after=[[1978 USAMO]]}}
 +
{{MAA Notice}}

Latest revision as of 14:46, 29 December 2023

Problems from the 1977 USAMO.

Problem 1

Determine all pairs of positive integers $(m,n)$ such that $(1+x^n+x^{2n}+\cdots+x^{mn})$ is divisible by $(1+x+x^2+\cdots+x^{m})$.

Solution

Problem 2

$ABC$ and $A'B'C'$ are two triangles in the same plane such that the lines $AA',BB',CC'$ are mutually parallel. Let $[ABC]$ denote the area of triangle $ABC$ with an appropriate $\pm$ sign, etc.; prove that \[3([ABC] + [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B'] + [A'BC] + [B'CA] + [C'AB].\]

Solution

Problem 3

If $a$ and $b$ are two of the roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.

Solution

Problem 4

Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.

Solution

Problem 5

If $a,b,c,d,e$ are positive numbers bounded by $p$ and $q$, i.e, if they lie in $[p,q], 0 < p$, prove that \[(a + b + c + d + e)\biggl(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} + \frac {1}{e}\biggr) \le 25 + 6\left(\sqrt {\frac {p}{q}} - \sqrt {\frac {q}{p}}\right)^2\] and determine when there is equality.

Solution

See Also

1977 USAMO (ProblemsResources)
Preceded by
1976 USAMO
Followed by
1978 USAMO
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png