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 | ||
+ | |||
+ | == See Also == {{IMO box|year=1974|num-b=4|num-a=6}} |
Latest revision as of 16:00, 29 January 2021
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
See Also
1974 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |