Difference between revisions of "1978 USAMO Problems/Problem 1"
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If <math>\mu=0</math>, then <math>\lambda=0</math> according to <math>(6)</math> and <math>\lambda=1</math> according to <math>(7)</math>, so <math>\mu \neq 0</math>. Setting the right sides of <math>(3)</math> and <math>(4)</math> equal yields <math>\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b</math>. Similar steps yield that <math>a=b=c=d</math>. Thus, <math>(1)</math> becomes <math>4d+e=8</math> and <math>(2)</math> becomes <math>4d^{2}+e^{2}=16</math>. Solving the system yields <math>e=0,\frac{16}{5}</math>, so the maximum possible value of <math>e</math> is <math>\frac{16}{5}</math>. | If <math>\mu=0</math>, then <math>\lambda=0</math> according to <math>(6)</math> and <math>\lambda=1</math> according to <math>(7)</math>, so <math>\mu \neq 0</math>. Setting the right sides of <math>(3)</math> and <math>(4)</math> equal yields <math>\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b</math>. Similar steps yield that <math>a=b=c=d</math>. Thus, <math>(1)</math> becomes <math>4d+e=8</math> and <math>(2)</math> becomes <math>4d^{2}+e^{2}=16</math>. Solving the system yields <math>e=0,\frac{16}{5}</math>, so the maximum possible value of <math>e</math> is <math>\frac{16}{5}</math>. | ||
+ | == Solution 3== | ||
+ | A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have | ||
+ | <cmath>(a+b+c+d)^2=(8-e)^2,</cmath> and | ||
+ | <cmath>a^2+b^2+c^2+d^2=16-e^2.</cmath> | ||
+ | The second equation times 4, then minus the first equation, | ||
+ | <cmath>(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=4(16-e^2)-(8-e)^2.</cmath> | ||
+ | The rest follows. | ||
+ | |||
+ | J.Z. | ||
== See Also == | == See Also == | ||
{{USAMO box|year=1978|before=First Question|num-a=2}} | {{USAMO box|year=1978|before=First Question|num-a=2}} |
Latest revision as of 15:39, 16 May 2018
Problem
Given that are real numbers such that
,
.
Determine the maximum value of .
Solution 1
By Cauchy Schwarz, we can see that thus Finally, which means so the maximum value of is .
from: Image from Gon Mathcenter.net
Solution 2
Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem. We get the following equations:
If , then according to and according to , so . Setting the right sides of and equal yields . Similar steps yield that . Thus, becomes and becomes . Solving the system yields , so the maximum possible value of is .
Solution 3
A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have and The second equation times 4, then minus the first equation, The rest follows.
J.Z.
See Also
1978 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.