Difference between revisions of "1974 IMO Problems/Problem 5"
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The above equation can be further simplified to | The above equation can be further simplified to | ||
<cmath>S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.</cmath> | <cmath>S = \frac{1}{2-m}+\frac{k}{2-l}+m+\frac{l}{2-k}.</cmath> | ||
− | Note that <math>S</math> is a continuous function and that <math>f(m) = m + \frac{1}{2-m}</math> is a strictly increasing function. We can now decrease <math> | + | Note that <math>S</math> is a continuous function and that <math>f(m) = m + \frac{1}{2-m}</math> is a strictly increasing function. We can now decrease <math>k</math> and <math>l</math> to make <math>m</math> tend arbitrarily close to <math>1</math>. We see <math>\lim_{m\to1} m + \frac{1}{2-m} = 2</math>, meaning <math>S</math> can be brought arbitrarily close to <math>2</math>. |
+ | Now, set <math>a = d = x</math> and <math>b = c = y</math> for some positive real numbers <math>x, y</math>. Then | ||
+ | <cmath>S = \frac{2x}{2x+y} + \frac{2y}{2y+x} = \frac{2y^2 + 8xy + 2x^2}{2y^2 + 5xy + 2x^2}.</cmath> | ||
+ | Notice that if we treat the numerator and denominator each as a quadratic in <math>y</math>, we will get <math>1 + \frac{g(x)}{2y^2 + 5xy + 2x^2}</math>, where <math>g(x)</math> has a degree lower than <math>2</math>. This means taking <math>\lim_{y\to\infty} 1 + \frac{g(x)}{2y^2 + 5xy + 2x^2} = 1</math>, which means <math>S</math> can be brought arbitrarily close to <math>1</math>. Therefore, we are done. | ||
<cmath> </cmath> | <cmath> </cmath> | ||
~Imajinary | ~Imajinary |
Revision as of 13:12, 7 November 2020
Problem 5
Determine all possible values of where
are arbitrary positive numbers.
Solution
Note that We will now prove that
can reach any range in between
and
.
Choose any positive number . For some variables such that
and
, let
,
, and
. Plugging this back into the original fraction, we get
The above equation can be further simplified to
Note that
is a continuous function and that
is a strictly increasing function. We can now decrease
and
to make
tend arbitrarily close to
. We see
, meaning
can be brought arbitrarily close to
.
Now, set
and
for some positive real numbers
. Then
Notice that if we treat the numerator and denominator each as a quadratic in
, we will get
, where
has a degree lower than
. This means taking
, which means
can be brought arbitrarily close to
. Therefore, we are done.
~Imajinary