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

(→‎Solution 2)
(→‎Solution 2)
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We get the following equations:
 
We get the following equations:
  
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<math>\newline(1)\hspace a+b+c+d+e=8\newline
+
<math>\newline(1)\hspace*{0.5cm} a+b+c+d+e=8\newline
−
(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\newline
−
(3)\hspace{0.5cm} 0=\lambda+2a\mu\newline
+
(3)\hspace*{0.5cm} 0=\lambda+2a\mu\newline
−
(4)\hspace{0.5cm} 0=\lambda+2b\mu\newline
+
(4)\hspace*{0.5cm} 0=\lambda+2b\mu\newline
−
(5)\hspace{0.5cm} 0=\lambda+2c\mu\newline
+
(5)\hspace*{0.5cm} 0=\lambda+2c\mu\newline
−
(6)\hspace{0.5cm} 0=\lambda+2d\mu\newline
+
(6)\hspace*{0.5cm} 0=\lambda+2d\mu\newline
−
(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 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 12:00, 30 April 2016

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

Accordting to Cauchy-Schwarz Inequalities, we can see $(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2$ thus, $4(16-e^2)\geqslant (8-e)^2$ Finally, $e(5e-16) \geqslant 0$ that mean, $\frac{16}{5} \geqslant e \geqslant 0$ 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:

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

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 that 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

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