Difference between revisions of "1987 USAMO Problems"
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+ | Problems from the '''1987 [[USAMO]].''' | ||
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==Problem 1== | ==Problem 1== | ||
Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers. | Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers. | ||
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==Problem 2== | ==Problem 2== | ||
− | The feet of the angle bisectors of | + | The feet of the angle bisectors of <math>\Delta ABC</math> form a right-angled triangle. If the right-angle is at <math>X</math>, where <math>AX</math> is the bisector of <math>\angle A</math>, find all possible values for <math>\angle A</math>. |
− | [[ | + | [[1987 USAMO Problems/Problem 2|Solution]] |
==Problem 3== | ==Problem 3== | ||
− | X is the smallest set of polynomials <math>p(x)</math> such that: | + | <math>X</math> is the smallest set of polynomials <math>p(x)</math> such that: |
− | 1. <math>p(x) = x</math> belongs to X | + | : 1. <math>p(x) = x</math> belongs to <math>X</math>. |
− | 2. If <math>r(x)</math> belongs to X, then <math>x\cdot r(x)</math> and <math>(x + (1 - x) \cdot r(x) )</math> both belong to X. | + | : 2. If <math>r(x)</math> belongs to <math>X</math>, then <math>x\cdot r(x)</math> and <math>(x + (1 - x) \cdot r(x) )</math> both belong to <math>X</math>. |
− | Show that if <math>r(x)</math> and <math>s(x)</math> are distinct elements of X, then <math>r(x) \neq s(x)</math> for any <math>0 < x < 1</math>. | + | Show that if <math>r(x)</math> and <math>s(x)</math> are distinct elements of <math>X</math>, then <math>r(x) \neq s(x)</math> for any <math>0 < x < 1</math>. |
− | [[ | + | [[1987 USAMO Problems/Problem 3|Solution]] |
==Problem 4== | ==Problem 4== | ||
M is the midpoint of XY. The points P and Q lie on a line through Y on opposite sides of Y, such that <math>|XQ| = 2|MP|</math> and <math>\frac{|XY|}2 < |MP| < \frac{3|XY|}2</math>. For what value of <math>\frac{|PY|}{|QY|}</math> is <math>|PQ|</math> a minimum? | M is the midpoint of XY. The points P and Q lie on a line through Y on opposite sides of Y, such that <math>|XQ| = 2|MP|</math> and <math>\frac{|XY|}2 < |MP| < \frac{3|XY|}2</math>. For what value of <math>\frac{|PY|}{|QY|}</math> is <math>|PQ|</math> a minimum? | ||
− | [[ | + | [[1987 USAMO Problems/Problem 4|Solution]] |
==Problem 5== | ==Problem 5== | ||
− | <math>a_1, a_2, \cdots, a_n</math> is a sequence of 0's and 1's. T is the number of triples <math>(a_i, a_j, a_k) with < | + | <math>a_1, a_2, \cdots, a_n</math> is a sequence of 0's and 1's. T is the number of triples <math>(a_i, a_j, a_k)</math> with <math>i<j<k</math> which are not equal to (0, 1, 0) or (1, 0, 1). For <math>1\le i\le n</math>, <math>f(i)</math> is the number of <math>j<i</math> with <math>a_j = a_i</math> plus the number of <math>j>i</math> with <math>a_j\neq a_i</math>. Show that <math>T=\sum_{i=1}^n f(i)\cdot\left(\frac{f(i)-1}2\right)</math>. If n is odd, what is the smallest value of T? |
+ | |||
+ | [[1987 USAMO Problems/Problem 5|Solution]] | ||
− | [[ | + | == See Also == |
+ | {{USAMO box|year=1987|before=[[1986 USAMO]]|after=[[1988 USAMO]]}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:42, 18 July 2016
Problems from the 1987 USAMO.
Problem 1
Find all solutions to , where m and n are non-zero integers.
Problem 2
The feet of the angle bisectors of form a right-angled triangle. If the right-angle is at
, where
is the bisector of
, find all possible values for
.
Problem 3
is the smallest set of polynomials
such that:
- 1.
belongs to
.
- 2. If
belongs to
, then
and
both belong to
.
Show that if and
are distinct elements of
, then
for any
.
Problem 4
M is the midpoint of XY. The points P and Q lie on a line through Y on opposite sides of Y, such that and
. For what value of
is
a minimum?
Problem 5
is a sequence of 0's and 1's. T is the number of triples
with
which are not equal to (0, 1, 0) or (1, 0, 1). For
,
is the number of
with
plus the number of
with
. Show that
. If n is odd, what is the smallest value of T?
See Also
1987 USAMO (Problems • Resources) | ||
Preceded by 1986 USAMO |
Followed by 1988 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.