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

(Solution 2)
 
(4 intermediate revisions by 3 users not shown)
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.
 
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{0.5cm} 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{0.5cm} 0=\lambda+2a\mu\newline
+
(3)\hspace*{0.5cm} 0=\lambda+2a\mu\\
(4)\hspace{0.5cm} 0=\lambda+2b\mu\newline
+
(4)\hspace*{0.5cm} 0=\lambda+2b\mu\\
(5)\hspace{0.5cm} 0=\lambda+2c\mu\newline
+
(5)\hspace*{0.5cm} 0=\lambda+2c\mu\\
(6)\hspace{0.5cm} 0=\lambda+2d\mu\newline
+
(6)\hspace*{0.5cm} 0=\lambda+2d\mu\\
(7)\hspace{0.5cm} 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>.
  
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>
+
== 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 16:39, 16 May 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 $\frac{16}{5}$.

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.

See Also

1978 USAMO (ProblemsResources)
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