Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1987 USAMO Problems"

(→‎Problem 5)
(→‎Problem 2)
Line 5: Line 5:
  
 
==Problem 2==
 
==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.
+
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>.
  
 
[[1987USAMO Problems/Problem 2|Solution]]
 
[[1987USAMO Problems/Problem 2|Solution]]

Revision as of 15:05, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_USAMO_Problems&oldid=40639"