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

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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>.
  
 
== 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.
 
We get the following equations:
 
We get the following equations:
  
−
<math>\newline(1)\hspace a+b+c+d+e=8\newline
+
<math>(1)\hspace*{0.5cm} a+b+c+d+e=8\\
−
(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\\
−
(3)\hspace 0=\lambda+2a\mu\newline
+
(3)\hspace*{0.5cm} 0=\lambda+2a\mu\\
−
(4)\hspace 0=\lambda+2b\mu\newline
+
(4)\hspace*{0.5cm} 0=\lambda+2b\mu\\
−
(5)\hspace 0=\lambda+2c\mu\newline
+
(5)\hspace*{0.5cm} 0=\lambda+2c\mu\\
−
(6)\hspace 0=\lambda+2d\mu\newline
+
(6)\hspace*{0.5cm} 0=\lambda+2d\mu\\
−
(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 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.
 +
 
 +
==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]]
  
−
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>
+
*Note: For some reason I think this solution is missing something.
  
 
== See Also ==
 
== See Also ==

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

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