Difference between revisions of "2010-2011 Mock USAJMO Problems/Solutions"
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<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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+ | [[2010-2011 Mock USAJMO Problems/Solutions/Problem 2|Solution]] | ||
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==Problem 3== | ==Problem 3== | ||
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b) <math>n=8</math> | b) <math>n=8</math> | ||
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+ | [[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> | ||
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+ | [[2010-2011 Mock USAJMO Problems/Solutions/Problem 4|Solution]] | ||
==Problem 5== | ==Problem 5== | ||
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<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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+ | [[2010-2011 Mock USAJMO Problems/Solutions/Problem 5|Solution]] | ||
==Problem 6== | ==Problem 6== | ||
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<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>. | ||
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+ | [[2010-2011 Mock USAJMO Problems/Solutions/Problem 6|Solution]] |
Revision as of 22:55, 27 September 2010
Problem 1
Given two fixed, distinct points and
on plane
, find the locus of all points
belonging to
such that the quadrilateral formed by point
, the midpoint of
, the centroid of
, and the midpoint of
(in that order) can be inscribed in a circle.
Problem 2
Let be positive real numbers such that
. Prove that
with equality if and only if
.
Problem 3
At a certain (lame) conference with people, each group of three people wishes to meet exactly once. On each of the
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
for
a)
b)
Problem 4
The sequence satisfies
and
for
. Prove that for all nonnegative integers
, we have
Problem 5
Find all solutions to
where
and
are positive integers.
Problem 6
In convex, tangential quadrilateral , let the incircle touch the four sides
at points
, respectively. Prove that
Note: A tangential quadrilateral has concurrent angle bisectors, and
denotes the area of figure
.