Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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== Solution 4 (Semi-rigorous) == | == Solution 4 (Semi-rigorous) == | ||
Given that <math>A</math>, <math>M</math>, and <math>C</math> are nonnegative integers, it should be intuitive that maximizing <math>AMC</math> maximizes <math>AM + MC + CA</math>. We thus only need to maximize <math>AMC</math>. By the [[AM-GM Inequality]], <cmath>\frac{A + M + C}{3} \geq \sqrt[3]{AMC},</cmath> with equality if and only if <math>A = M = C</math>. Note that the maximum of <math>AMC</math> occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; <math>A + M + C = 12</math> implies that <math>A = M = C = 4</math>, so <math>AMC + AM + MC + CA = 4^3 + 3*4^2 = 112.</math> The answer is thus <math>\boxed{\text{E}}</math>, as required. | Given that <math>A</math>, <math>M</math>, and <math>C</math> are nonnegative integers, it should be intuitive that maximizing <math>AMC</math> maximizes <math>AM + MC + CA</math>. We thus only need to maximize <math>AMC</math>. By the [[AM-GM Inequality]], <cmath>\frac{A + M + C}{3} \geq \sqrt[3]{AMC},</cmath> with equality if and only if <math>A = M = C</math>. Note that the maximum of <math>AMC</math> occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; <math>A + M + C = 12</math> implies that <math>A = M = C = 4</math>, so <math>AMC + AM + MC + CA = 4^3 + 3*4^2 = 112.</math> The answer is thus <math>\boxed{\text{E}}</math>, as required. | ||
+ | |||
+ | == Solution 5 (Double AM-GM) == | ||
+ | We start off the same way as Solution 4, using AM-GM to observe that <math>AMC \leq 64</math>. We then observe that | ||
+ | |||
+ | <math>(A + M + C)^2 = A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144</math>, since <math>A + M + C = 12</math>. | ||
+ | |||
+ | We can use the AM-GM inequality again, this time observing that <math>\frac{A^2 + M^2 + C^2}{3} \geq \sqrt[3]{{(AMC)}^2}</math> | ||
+ | |||
+ | Since <math>AMC \leq 64</math>, <math>3 \sqrt[3]{{(AMC)}^2} \leq 48</math>. We then plug this in to yield | ||
+ | |||
+ | <math>A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144 \geq 48 + 2(AM + MC + AC)</math> | ||
+ | |||
+ | Thus, <math>AM + MC + AC \leq 48</math>. We now revisit the original equation that we wish to maximize. Since we know <math>AMC \leq 64</math>, we now have upper bounds on both of our unruly terms. Plugging both in results in <math>48 + 64 = 112</math> | ||
== Video Solution == | == Video Solution == |
Revision as of 16:07, 15 May 2022
Contents
Problem
Let and
be nonnegative integers such that
. What is the maximum value of
?
Solution 1
It is not hard to see that
Since
, we can rewrite this as
So we wish to maximize
Which is largest when all the factors are equal (consequence of AM-GM). Since
, we set
Which gives us
so the answer is
.
I wish you understand this problem and can use it in other problems.
Solution 2 (Nonrigorous)
If you know that to maximize your result you have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make
and
as close as possible. In this case, they would all be equal to
, so
, giving you the answer of
.
Solution 3
Assume ,
, and
are equal to
. Since the resulting value of
will be
and this is the largest answer choice, our answer is
.
Solution 4 (Semi-rigorous)
Given that ,
, and
are nonnegative integers, it should be intuitive that maximizing
maximizes
. We thus only need to maximize
. By the AM-GM Inequality,
with equality if and only if
. Note that the maximum of
occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing;
implies that
, so
The answer is thus
, as required.
Solution 5 (Double AM-GM)
We start off the same way as Solution 4, using AM-GM to observe that . We then observe that
, since
.
We can use the AM-GM inequality again, this time observing that
Since ,
. We then plug this in to yield
Thus, . We now revisit the original equation that we wish to maximize. Since we know
, we now have upper bounds on both of our unruly terms. Plugging both in results in
Video Solution
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.