Difference between revisions of "1978 USAMO Problems/Problem 1"
Mathlobster (talk | contribs) (→Solution 2) |
Mathlobster (talk | contribs) (→Solution 2) |
||
| Line 20: | Line 20: | ||
<math>\newline(1)\hspace a+b+c+d+e=8\newline | <math>\newline(1)\hspace a+b+c+d+e=8\newline | ||
| − | (2)\hspace a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline | + | (2)\hspace{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline |
| − | (3)\hspace 0=\lambda+2a\mu\newline | + | (3)\hspace{0.5cm} 0=\lambda+2a\mu\newline |
| − | (4)\hspace 0=\lambda+2b\mu\newline | + | (4)\hspace{0.5cm} 0=\lambda+2b\mu\newline |
| − | (5)\hspace 0=\lambda+2c\mu\newline | + | (5)\hspace{0.5cm} 0=\lambda+2c\mu\newline |
| − | (6)\hspace 0=\lambda+2d\mu\newline | + | (6)\hspace{0.5cm} 0=\lambda+2d\mu\newline |
| − | (7)\hspace 1=\lambda+2e\mu</math> | + | (7)\hspace{0.5cm} 1=\lambda+2e\mu</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 that 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 that maximum possible value of <math>e</math> is <math>\frac{16}{5}</math> | ||
Revision as of 12:58, 30 April 2016
Contents
Problem
Given that 
are real numbers such that
,
.
Determine the maximum value of 
.
Solution 1
Accordting to Cauchy-Schwarz Inequalities, we can see 
thus, 
Finally, 
that mean, 
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:
$\newline(1)\hspace a+b+c+d+e=8\newline (2)\hspace{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline (3)\hspace{0.5cm} 0=\lambda+2a\mu\newline (4)\hspace{0.5cm} 0=\lambda+2b\mu\newline (5)\hspace{0.5cm} 0=\lambda+2c\mu\newline (6)\hspace{0.5cm} 0=\lambda+2d\mu\newline (7)\hspace{0.5cm} 1=\lambda+2e\mu$ (Error compiling LaTeX. Unknown error_msg)
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 that maximum possible value of is
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. 
