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 "2010-2011 Mock USAJMO Problems/Solutions"

(→‎Problem 1)
Line 9: Line 9:
 
<cmath>\frac{3x+1}{y+z}+\frac{3y+1}{z+x}+\frac{3z+1}{x+y}\ge\frac{4}{2x+y+z}+\frac{4}{x+2y+z}+\frac{4}{x+y+2z},</cmath>
 
<cmath>\frac{3x+1}{y+z}+\frac{3y+1}{z+x}+\frac{3z+1}{x+y}\ge\frac{4}{2x+y+z}+\frac{4}{x+2y+z}+\frac{4}{x+y+2z},</cmath>
 
with equality if and only if <math>x=y=z</math>.
 
with equality if and only if <math>x=y=z</math>.
 +
 +
 +
[[2010-2011 Mock USAJMO Problems/Solutions/Problem 2|Solution]]
 +
  
 
==Problem 3==
 
==Problem 3==
Line 16: Line 20:
  
 
b) <math>n=8</math>
 
b) <math>n=8</math>
 +
 +
[[2010-2011 Mock USAJMO Problems/Solutions/Problem 3|Solution]]
  
 
==Problem 4==
 
==Problem 4==
 
The sequence <math>\{a_i\}_{i\ge0}</math> satisfies <math>a_0=1</math> and <math>a_n=\sum_{i=0}^{n-1}(n-i)a_i</math> for <math>n\ge1</math>. Prove that for all nonnegative integers <math>m</math>, we have
 
The sequence <math>\{a_i\}_{i\ge0}</math> satisfies <math>a_0=1</math> and <math>a_n=\sum_{i=0}^{n-1}(n-i)a_i</math> for <math>n\ge1</math>. Prove that for all nonnegative integers <math>m</math>, we have
 
<cmath>\sum_{k=0}^{m}\frac{a_k}{4^k} < \frac{9}{5}.</cmath>
 
<cmath>\sum_{k=0}^{m}\frac{a_k}{4^k} < \frac{9}{5}.</cmath>
 +
 +
[[2010-2011 Mock USAJMO Problems/Solutions/Problem 4|Solution]]
  
 
==Problem 5==
 
==Problem 5==
Line 25: Line 33:
 
<cmath>5\cdot4^{3m+1}-4^{2m+1}-1=15n^{2m},</cmath>
 
<cmath>5\cdot4^{3m+1}-4^{2m+1}-1=15n^{2m},</cmath>
 
where <math>m</math> and <math>n</math> are positive integers.
 
where <math>m</math> and <math>n</math> are positive integers.
 +
 +
[[2010-2011 Mock USAJMO Problems/Solutions/Problem 5|Solution]]
  
 
==Problem 6==
 
==Problem 6==
Line 30: Line 40:
 
<cmath>\frac{[EFGH]}{[ABCD]} \le \frac{1}{2}.</cmath>
 
<cmath>\frac{[EFGH]}{[ABCD]} \le \frac{1}{2}.</cmath>
 
'''Note:''' A tangential quadrilateral has concurrent angle bisectors, and <math>[X]</math> denotes the area of figure <math>X</math>.
 
'''Note:''' A tangential quadrilateral has concurrent angle bisectors, and <math>[X]</math> denotes the area of figure <math>X</math>.
 +
 +
[[2010-2011 Mock USAJMO Problems/Solutions/Problem 6|Solution]]

Revision as of 22:55, 27 September 2010

Problem 1

Given two fixed, distinct points $B$ and $C$ on plane $\mathcal{P}$, find the locus of all points $A$ belonging to $\mathcal{P}$ such that the quadrilateral formed by point $A$, the midpoint of $AB$, the centroid of $\triangle ABC$, and the midpoint of $AC$ (in that order) can be inscribed in a circle.

Solution

Problem 2

Let $x, y, z$ be positive real numbers such that $x+y+z = 1$. Prove that \[\frac{3x+1}{y+z}+\frac{3y+1}{z+x}+\frac{3z+1}{x+y}\ge\frac{4}{2x+y+z}+\frac{4}{x+2y+z}+\frac{4}{x+y+2z},\] with equality if and only if $x=y=z$.


Solution


Problem 3

At a certain (lame) conference with $n$ people, each group of three people wishes to meet exactly once. On each of the $d$ days of the conference, every person participates in at most one meeting with two other people, with no limit on the number of groups that can meet. Determine $\min d$ for

a) $n=6$

b) $n=8$

Solution

Problem 4

The sequence $\{a_i\}_{i\ge0}$ satisfies $a_0=1$ and $a_n=\sum_{i=0}^{n-1}(n-i)a_i$ for $n\ge1$. Prove that for all nonnegative integers $m$, we have \[\sum_{k=0}^{m}\frac{a_k}{4^k} < \frac{9}{5}.\]

Solution

Problem 5

Find all solutions to \[5\cdot4^{3m+1}-4^{2m+1}-1=15n^{2m},\] where $m$ and $n$ are positive integers.

Solution

Problem 6

In convex, tangential quadrilateral $ABCD$, let the incircle touch the four sides $AB, BC, CD, DA$ at points $E, F, G, H$, respectively. Prove that \[\frac{[EFGH]}{[ABCD]} \le \frac{1}{2}.\] Note: A tangential quadrilateral has concurrent angle bisectors, and $[X]$ denotes the area of figure $X$.

Solution

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010-2011_Mock_USAJMO_Problems/Solutions&oldid=35864"