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"
m (→‎Solution 4)
 
(2 intermediate revisions by the same user not shown)
Line 2: Line 2:
 
Given that <math>a,b,c,d,e</math> are real numbers such that
 
Given that <math>a,b,c,d,e</math> are real numbers such that
  
−
<math>a+b+c+d+e=8</math>,
+
<cmath>a+b+c+d+e=8</cmath>,
  
−
<math>a^2+b^2+c^2+d^2+e^2=16</math>.
+
<cmath>a^2+b^2+c^2+d^2+e^2=16</cmath>.
  
 
Determine the maximum value of <math>e</math>.
 
Determine the maximum value of <math>e</math>.
Line 39: Line 39:
  
 
J.Z.
 
J.Z.
 +
 +
==Solution 4==
 +
By the [[Principle of Insufficient Reasons]], since <math>a,b,c,d</math> are indistinguishable variables, the maximum of <math>e</math> is acheived when <math>a=b=c=d</math>, so we have <cmath>4a+\max e=8</cmath> <cmath>4a^2+(\max e)^2=16</cmath> <cmath>\implies e=\boxed{\frac{16}{5}}</cmath>. <math>\square</math> ~[[Ddk001]]
 +
 +
*Note: For some reason I think this solution is missing something.
 +
 
== 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 20:34, 6 July 2024

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 $\frac{16}{5}$.

Solution 3

A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have \[(a+b+c+d)^2=(8-e)^2,\] and \[a^2+b^2+c^2+d^2=16-e^2.\] The second equation times 4, then minus the first equation, \[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=4(16-e^2)-(8-e)^2.\] The rest follows.

J.Z.

Solution 4

By the Principle of Insufficient Reasons, since $a,b,c,d$ are indistinguishable variables, the maximum of $e$ is acheived when $a=b=c=d$, so we have \[4a+\max e=8\] \[4a^2+(\max e)^2=16\] \[\implies e=\boxed{\frac{16}{5}}\]. $\square$ ~Ddk001

  • Note: For some reason I think this solution is missing something.

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=221798"