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

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.

@@ Line 8: / Line 8: @@
 Determine the maximum value of <math>e</math>.
-== Solution ==
+== 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]
+== Solution 2==
+Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem.
+We get the following equations:
+<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\\
+(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</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.
 == See Also ==
 {{USAMO box|year=1978|before=First Question|num-a=2}}

1978 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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All USAMO Problems and Solutions

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