Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1978 USAMO Problems/Problem 1"

(→‎Solution 2)
(→‎Solution 1)
Line 9: Line 9:
  
 
== Solution 1==
 
== Solution 1==
−
Accordting to '''Cauchy-Schwarz Inequalities''', we can see <math>(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2</math>
+
By Cauchy Schwarz, we can see that <math>(1+1+1+1)(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2</math>
−
thus, <math>4(16-e^2)\geqslant (8-e)^2</math>  
+
thus <math>4(16-e^2)\geq (8-e)^2</math>  
−
Finally, <math>e(5e-16) \geqslant 0</math> that mean, <math>\frac{16}{5} \geqslant e \geqslant 0</math>
+
Finally, <math>e(5e-16) \geq 0</math> which means <math>\frac{16}{5} \geq e \geq 0</math>
−
'''so''' the maximum value of <math>e</math> is <math>\frac{16}{5}</math>
+
so the maximum value of <math>e</math> is <math>\frac{16}{5}</math>.
  
 
'''from:''' [http://image.ohozaa.com/view2/vUGiXdRQdAPyw036 Image from Gon Mathcenter.net]
 
'''from:''' [http://image.ohozaa.com/view2/vUGiXdRQdAPyw036 Image from Gon Mathcenter.net]
 +
 
== Solution 2==
 
== Solution 2==
 
Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem.
 
Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem.

Revision as of 18:40, 23 February 2018

Problem

Given that $a,b,c,d,e$ are real numbers such that

$a+b+c+d+e=8$,

$a^2+b^2+c^2+d^2+e^2=16$.

Determine the maximum value of $e$.

Solution 1

By Cauchy Schwarz, we can see that $(1+1+1+1)(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2$ thus $4(16-e^2)\geq (8-e)^2$ Finally, $e(5e-16) \geq 0$ which means $\frac{16}{5} \geq e \geq 0$ so the maximum value of $e$ is $\frac{16}{5}$.

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:

$(1)\hspace*{0.5cm} a+b+c+d+e=8\\ (2)\hspace*{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\\ (3)\hspace*{0.5cm} 0=\lambda+2a\mu\\ (4)\hspace*{0.5cm} 0=\lambda+2b\mu\\ (5)\hspace*{0.5cm} 0=\lambda+2c\mu\\ (6)\hspace*{0.5cm} 0=\lambda+2d\mu\\ (7)\hspace*{0.5cm} 1=\lambda+2e\mu$

If $\mu=0$, then $\lambda=0$ according to $(6)$ and $\lambda=1$ according to $(7)$, so $\mu \neq 0$. Setting the right sides of $(3)$ and $(4)$ equal yields $\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b$. Similar steps yield that $a=b=c=d$. Thus, $(1)$ becomes $4d+e=8$ and $(2)$ becomes $4d^{2}+e^{2}=16$. Solving the system yields $e=0,\frac{16}{5}$, so the maximum possible value of $e$ is $\frac{16}{5}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_USAMO_Problems/Problem_1&oldid=92283"