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Difference between revisions of "2010-2011 Mock USAJMO Problems/Solutions"

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==Problem 1==
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== Problems ==
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===Problem 1===
  
 
Given two fixed, distinct points <math>B</math> and <math>C</math> on plane <math>\mathcal{P}</math>, find the locus of all points <math>A</math> belonging to <math>\mathcal{P}</math> such that the quadrilateral formed by point <math>A</math>, the midpoint of <math>AB</math>, the centroid of <math>\triangle ABC</math>, and the midpoint of <math>AC</math> (in that order) can be inscribed in a circle.
 
Given two fixed, distinct points <math>B</math> and <math>C</math> on plane <math>\mathcal{P}</math>, find the locus of all points <math>A</math> belonging to <math>\mathcal{P}</math> such that the quadrilateral formed by point <math>A</math>, the midpoint of <math>AB</math>, the centroid of <math>\triangle ABC</math>, and the midpoint of <math>AC</math> (in that order) can be inscribed in a circle.
  
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==Problem 2==
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[[2010-2011 Mock USAJMO Problems/Solutions/Problem 1|Solution]]
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===Problem 2===
 
Let <math>x, y, z</math> be positive real numbers such that <math>x+y+z = 1</math>. Prove that
 
Let <math>x, y, z</math> be positive real numbers such that <math>x+y+z = 1</math>. Prove that
 
<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>.
  
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==Problem 3==
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[[2010-2011 Mock USAJMO Problems/Solutions/Problem 2|Solution]]
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===Problem 3===
 
At a certain (lame) conference with <math>n</math> people, each group of three people wishes to meet exactly once. On each of the <math>d</math> 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 <math>\min d</math> for
 
At a certain (lame) conference with <math>n</math> people, each group of three people wishes to meet exactly once. On each of the <math>d</math> 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 <math>\min d</math> for
 +
 
a) <math>n=6</math>
 
a) <math>n=6</math>
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b) <math>n=8</math>
 
b) <math>n=8</math>
  
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==Problem 4==
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[[2010-2011 Mock USAJMO Problems/Solutions/Problem 3|Solution]]
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===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>
  
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==Problem 5==
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[[2010-2011 Mock USAJMO Problems/Solutions/Problem 4|Solution]]
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===Problem 5===
 
Find all solutions to
 
Find all solutions to
 
<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.
  
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==Problem 6==
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[[2010-2011 Mock USAJMO Problems/Solutions/Problem 5|Solution]]
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===Problem 6===
 
In convex, tangential quadrilateral <math>ABCD</math>, let the incircle touch the four sides <math>AB, BC, CD, DA</math> at points <math>E, F, G, H</math>, respectively. Prove that
 
In convex, tangential quadrilateral <math>ABCD</math>, let the incircle touch the four sides <math>AB, BC, CD, DA</math> at points <math>E, F, G, H</math>, respectively. Prove that
 
<cmath>\frac{[EFGH]}{[ABCD]} \le \frac{1}{2}.</cmath>
 
<cmath>\frac{[EFGH]}{[ABCD]} \le \frac{1}{2}.</cmath>
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\textbf{Note:} A tangential quadrilateral has concurrent angle bisectors, and <math>[X]</math> denotes the area of figure <math>X</math>.
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'''Note:''' A tangential quadrilateral has concurrent angle bisectors, and <math>[X]</math> denotes the area of figure <math>X</math>.
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[[2010-2011 Mock USAJMO Problems/Solutions/Problem 6|Solution]]
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-----
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== Forum Threads ==
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* [http://www.artofproblemsolving.com/community/c5h345916p1850492 Initial Discussion]
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* [http://www.artofproblemsolving.com/community/c5h345916p1850492 Problems PDF]
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* [http://www.artofproblemsolving.com/community/c5h345916p1858538 Solutions PDF]
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=== See Also: ===
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* [[Mock USAJMO]]
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* [[AoPS Past Contests]]

Latest revision as of 18:36, 28 June 2015

Problems

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

Forum Threads

See Also:

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