Difference between revisions of "2000 AMC 12 Problems/Problem 12"

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Problem

Let $A, M,$ and $C$ be nonnegative integers such that $A + M + C=12$ . What is the maximum value of $A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C$ ?

$\mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }$

Solution 1

It is not hard to see that $(A+1)(M+1)(C+1)=$ $AMC+AM+AC+MC+A+M+C+1$ Since $A+M+C=12$ , we can rewrite this as $(A+1)(M+1)(C+1)=$ $AMC+AM+AC+MC+13$ So we wish to maximize $(A+1)(M+1)(C+1)-13$ Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$ , we set $A=M=C=4$ Which gives us $(4+1)(4+1)(4+1)-13=112$ so the answer is $\boxed{\textbf{(E) }112}.$

Solution 2 (Nonrigorous)

If you know that to maximize your result you $\textit{usually}$ 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 $A,M$ and $C$ as close as possible. In this case, they would all be equal to $4$ , so $AMC+AM+AC+MC=64+16+16+16=112$ , $\boxed{\text{E}}$ .

Solution 3

Assume $A$ , $M$ , and $C$ are equal to $4$ . Since the resulting value of $AMC+AM+AC+MC$ will be $112$ and this is the largest answer choice, our answer is $\boxed{\textbf{(E) }112}$ .

Solution 4 (Semi-rigorous)

Given that $A$ , $M$ , and $C$ are nonnegative integers, it should be intuitive that maximizing $AMC$ maximizes $AM + MC + CA$ . We thus only need to maximize $AMC$ . By the AM-GM Inequality, $\frac{A + M + C}{3} \geq \sqrt[3]{AMC},$ with equality if and only if $A = M = C$ . Note that the maximum of $AMC$ occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; $A + M + C = 12$ implies that $A = M = C = 4$ , so $AMC + AM + MC + CA = 4^3 + 3*4^2 = 112.$ The answer is thus $\boxed{\text{E}}$ , as required.

Solution 5 (Double AM-GM)

We start off the same way as Solution 4, using AM-GM to observe that $AMC \leq 64$ . We then observe that

$(A + M + C)^2 = A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144$ , since $A + M + C = 12$ .

We can use the AM-GM inequality again, this time observing that $\frac{A^2 + M^2 + C^2}{3} \geq \sqrt[3]{{(AMC)}^2}$

Since $AMC \leq 64$ , $3 \sqrt[3]{{(AMC)}^2} \leq 48$ . We then plug this in to yield

$A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144 \geq 48 + 2(AM + MC + AC)$

Thus, $AM + MC + AC \leq 48$ . We now revisit the original equation that we wish to maximize. Since we know $AMC \leq 64$ , we now have upper bounds on both of our unruly terms. Plugging both in results in $48 + 64 = \boxed{\textbf{(E) }112}$

Solution 6 (Optimization)

The largest number for our value would be A = M = C. So 3A = 12 and A = M = C = 4.

~BowenNa

Video Solution

https://youtu.be/lxqxQhGterg

@@ Line 39: / Line 39: @@
 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 A = M = C. So 3A = 12 and A = M = C = 4. <math>4\times4\times4 + 4\times4 + 4\times4 = 112 or \boxed{(E) 112}</math>
+~BowenNa
 == Video Solution ==

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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