Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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== Problem == | == Problem == | ||
− | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be [[nonnegative integer]]s such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
− | + | [katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } [/katex] | |
== Solution 1 == | == Solution 1 == | ||
Line 16: | Line 16: | ||
Which gives us | Which gives us | ||
<cmath>(4+1)(4+1)(4+1)-13=112</cmath> | <cmath>(4+1)(4+1)(4+1)-13=112</cmath> | ||
− | so the answer is <math>\boxed{\ | + | so the answer is <math>\boxed{\textbf{(E) }112}.</math> |
− | |||
== Solution 2 (Nonrigorous) == | == Solution 2 (Nonrigorous) == | ||
− | If you know that to maximize your result you <math>\textit{usually}</math> 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 <math>A,M</math> and <math>C</math> as close as possible. In this case, they would all be equal to <math>4</math>, so <math>AMC+AM+AC+MC=64+16+16+16=112</math>, | + | If you know that to maximize your result you <math>\textit{usually}</math> 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 <math>A,M</math> and <math>C</math> as close as possible. In this case, they would all be equal to <math>4</math>, so <math>AMC+AM+AC+MC=64+16+16+16=112</math>, <math>\boxed{\text{E}}</math>. |
− | == Solution 3 == | + | == Solution 3 (Answer Choices) == |
Assume <math>A</math>, <math>M</math>, and <math>C</math> are equal to <math>4</math>. Since the resulting value of <math>AMC+AM+AC+MC</math> will be <math>112</math> and this is the largest answer choice, our answer is <math>\boxed{\textbf{(E) }112}</math>. | Assume <math>A</math>, <math>M</math>, and <math>C</math> are equal to <math>4</math>. Since the resulting value of <math>AMC+AM+AC+MC</math> will be <math>112</math> and this is the largest answer choice, our answer is <math>\boxed{\textbf{(E) }112}</math>. | ||
== 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 = \boxed{\textbf{(E) }112}</math> | ||
+ | |||
+ | == Solution 6 (Optimization) == | ||
+ | |||
+ | The largest number for our value would be <math>A = M = C.</math> So <math>3A = 12 </math> and <math>A = M = C = 4.</math> <math>4\times4\times4 + 4\times4 + 4\times4 = 112 </math> or <math>\boxed{\textbf{(E) }112}</math> | ||
+ | |||
+ | ~BowenNa | ||
== Video Solution == | == Video Solution == |
Latest revision as of 15:15, 27 July 2024
Contents
[hide]Problem
Let
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
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 , .
Solution 3 (Answer Choices)
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
Solution 6 (Optimization)
The largest number for our value would be So and or
~BowenNa
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.