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Difference between revisions of "1987 USAMO Problems"

(Created page with "==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. Solution ==Problem 2== The ...")
 
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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?  
  
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[[1987USAMO Problems/Problem 3|Solution]]
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[[1987USAMO 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>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>\displaystyle T=\sum_{i=1}^n f(i)\cdot\frac{f(i)-1}2$. If n is odd, what is the smallest value of T?
 
<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>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>\displaystyle T=\sum_{i=1}^n f(i)\cdot\frac{f(i)-1}2$. If n is odd, what is the smallest value of T?
  
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[[1987USAMO Problems/Problem 3|Solution]]
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[[1987USAMO Problems/Problem 5|Solution]]

Revision as of 14:03, 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 the triangle 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$\displaystyle T=\sum_{i=1}^n f(i)\cdot\frac{f(i)-1}2$. If n is odd, what is the smallest value of T?

Solution

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