Difference between revisions of "2001 IMO Shortlist Problems/N2"
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== Problem == | == Problem == | ||
− | Consider the system <math>x + y = z + u,</math> <math>2xy | + | Consider the system <math>x + y = z + u,</math> <math>2xy = zu.</math> Find the greatest value of the real constant <math>m</math> such that <math>m \leq x/y</math> for any positive integer solution <math>(x,y,z,u)</math> of the system, with <math>x \geq y</math>. |
==Solution== | ==Solution== | ||
First consider the real solutions to the system. We have by AM-GM that <math>\frac{z+u}{2}\ge\sqrt{zu}</math> and substituting we get <math>\frac{x+y}{2}\ge\sqrt{2xy}</math>. Squaring and simplifying and dividing by <math>y^2</math>, we get the inequality <math>r^2-6r+1\ge0</math>, where <math>r=\frac{x}{y}</math>. Then <math>r^2-6r+9\ge8</math>, so <math>r\ge3+2\sqrt2</math> or <math>r\le3-2\sqrt2</math>. Since <math>r\ge1</math>, we discard the second inequality and have that <math>3+2\sqrt2</math> is a lower bound for <math>r</math> | First consider the real solutions to the system. We have by AM-GM that <math>\frac{z+u}{2}\ge\sqrt{zu}</math> and substituting we get <math>\frac{x+y}{2}\ge\sqrt{2xy}</math>. Squaring and simplifying and dividing by <math>y^2</math>, we get the inequality <math>r^2-6r+1\ge0</math>, where <math>r=\frac{x}{y}</math>. Then <math>r^2-6r+9\ge8</math>, so <math>r\ge3+2\sqrt2</math> or <math>r\le3-2\sqrt2</math>. Since <math>r\ge1</math>, we discard the second inequality and have that <math>3+2\sqrt2</math> is a lower bound for <math>r</math> | ||
− | This bound is also attainable for real values when <math>z=u</math>. Since <math>\mathbb{Q+}</math> is dense, it is always possible to assign rational values to <math>x, y, z,</math> and <math>w</math> so that <math>r</math> approaches <math>\ | + | This bound is also attainable for real values when <math>z=u</math>. Since <math>\mathbb{Q+}</math> is dense, it is always possible to assign rational values to <math>x, y, z,</math> and <math>w</math> so that <math>r</math> approaches <math>3+2\sqrt2</math>, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of <math>r</math>. Thus, <math>\boxed{m=3+2\sqrt2}</math>. |
== Resources == | == Resources == |
Latest revision as of 14:38, 1 September 2017
Problem
Consider the system
Find the greatest value of the real constant
such that
for any positive integer solution
of the system, with
.
Solution
First consider the real solutions to the system. We have by AM-GM that and substituting we get
. Squaring and simplifying and dividing by
, we get the inequality
, where
. Then
, so
or
. Since
, we discard the second inequality and have that
is a lower bound for
This bound is also attainable for real values when . Since
is dense, it is always possible to assign rational values to
and
so that
approaches
, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of
. Thus,
.