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 | ||
− | < | + | <cmath>a+b+c+d+e=8</cmath>, |
− | < | + | <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 == | + | == Solution 1== |
− | {{ | + | 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)\geq (8-e)^2</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>. | ||
+ | |||
+ | '''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. | ||
+ | |||
+ | ==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}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 21:34, 6 July 2024
Problem
Given that are real numbers such that
,
.
Determine the maximum value of .
Solution 1
By Cauchy Schwarz, we can see that
thus
Finally,
which means
so the maximum value of
is
.
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:
If , then
according to
and
according to
, so
. Setting the right sides of
and
equal yields
. Similar steps yield that
. Thus,
becomes
and
becomes
. Solving the system yields
, so the maximum possible value of
is
.
Solution 3
A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have
and
The second equation times 4, then minus the first equation,
The rest follows.
J.Z.
Solution 4
By the Principle of Insufficient Reasons, since are indistinguishable variables, the maximum of
is acheived when
, so we have
.
~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.