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Difference between revisions of "1987 USAMO Problems"

(→‎Problem 2)
(→‎Problem 3)
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==Problem 3==
 
==Problem 3==
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X is the smallest set of polynomials <math>p(x)</math> such that:  
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<math>X</math> is the smallest set of polynomials <math>p(x)</math> such that:  
  
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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>.  
  
 
[[1987USAMO Problems/Problem 3|Solution]]
 
[[1987USAMO Problems/Problem 3|Solution]]

Revision as of 14:06, 24 July 2011

Problem 1

Find all solutions to $(m^2+n)(m + n^2)= (m - n)^3$, where m and n are non-zero integers.

Solution

Problem 2

The feet of the angle bisectors of $\Delta ABC$ form a right-angled triangle. If the right-angle is at $X$, where $AX$ is the bisector of $\angle A$, find all possible values for $\angle A$.

Solution

Problem 3

$X$ is the smallest set of polynomials $p(x)$ such that:

1. $p(x) = x$ belongs to $X$. 2. If $r(x)$ belongs to $X$, then $x\cdot r(x)$ and $(x + (1 - x) \cdot r(x) )$ both belong to $X$.

Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $r(x) \neq s(x)$ for any $0 < x < 1$.

Solution

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 $|XQ| = 2|MP|$ and $\frac{|XY|}2 < |MP| < \frac{3|XY|}2$. For what value of $\frac{|PY|}{|QY|}$ is $|PQ|$ a minimum?

Solution

Problem 5

$a_1, a_2, \cdots, a_n$ is a sequence of 0's and 1's. T is the number of triples $(a_i, a_j, a_k)$ with $i<j<k$ which are not equal to (0, 1, 0) or (1, 0, 1). For $1\le i\le n$, $f(i)$ is the number of $j<i$ with $a_j = a_i$ plus the number of $j>i$ with $a_j\neq a_i$. Show that $T=\sum_{i=1}^n f(i)\cdot\left(\frac{f(i)-1}2\right)$. If n is odd, what is the smallest value of T?

Solution

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