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Difference between revisions of "2013 AMC 12A Problems/Problem 16"

m (Clarification)
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<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>
 
<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>
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==solution 2==
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Suppose there are <math>A,B,C</math> rocks in the three piles, and that the mean of pile C is <math>x</math>, and that the mean of the combination of <math>B</math> and <math>C</math> is <math>y</math>. We are going to maximize <math>y</math>, subject to the following conditions:
 +
 +
<cmath>40A+50B=43(A+B)</cmath>
 +
<cmath>40A+xC=44(A+C)</cmath>
 +
<cmath>50B+xC=y(B+C)</cmath>
 +
 +
which can be rearranged as:
 +
 +
<cmath>7B=3A</cmath>
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<cmath>(x-44)C=4A</cmath>
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<cmath>(x-y)C=(y-50)B.</cmath>
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 +
Let us test <math>y=59</math> is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes
 +
 +
<cmath>(x-59)C=9B.</cmath>
 +
 +
So <math>15C = (x-44)C - (x-59)C = 4A - 9B</math>, <math>45C=4(3A)-27B=28B-27B</math>, <math>105C=28A-9(7B)=A</math>, therefore,
 +
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<math>A=105C, B=45C, x=4(105)+44=464</math>, which gives us a consistent solution. Therefore <math>y=59</math> is the answer.
 +
 +
(Note: To further illustrate the idea, let us look at <math>y=60</math> and see what happens. We then get <math>7\cdot 16C = 4A-30A<0</math>, which is a contradiction!)

Revision as of 17:32, 12 February 2013

Let pile $A$ have $A$ rocks, and so on.

The mean weight of $A$ and $C$ together is $44$, so the total weight of $A$ and $C$ is $44(A + C)$

To get the total weight of $B$ and $C$, we need to add the total weight of $B$ and subtract the total weight of $A$

$44A + 44C + 50B - 40A = 4A + 44C + 50B$

And then dividing by the number of rocks $B$ and $C$ together, to get the mean of $B$ and $C$,

$\frac{4A + 44C + 50B}{B + C}$

Simplifying,

$\frac{4A + 44C + 44B + 6B}{B + C}$


$44 + \frac{4A + 6B}{B + C}$

Now, to get rid of the $A$ in the numerator, we use two definitions of the total weight of $A$ and $B$

$40A + 50B = 43A + 43B$

$3A = 7B$

$A = \frac{7}{3}B$

Substituting back in,

$44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}$


$44 + \frac{46}{3}*\frac{B}{B + C}$

Note that $\frac{B}{B + C} < 1$, and the maximal value of this factor occurs when $C = 1$

Also note that $\frac{46}{3}$ must cancel to give an integer value, and the only fraction that satisfies both these conditions is $\frac{45}{46}$

Plugging in, we get

$44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59$

solution 2

Suppose there are $A,B,C$ rocks in the three piles, and that the mean of pile C is $x$, and that the mean of the combination of $B$ and $C$ is $y$. We are going to maximize $y$, subject to the following conditions:

\[40A+50B=43(A+B)\] \[40A+xC=44(A+C)\] \[50B+xC=y(B+C)\]

which can be rearranged as:

\[7B=3A\] \[(x-44)C=4A\] \[(x-y)C=(y-50)B.\]

Let us test $y=59$ is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes

\[(x-59)C=9B.\]

So $15C = (x-44)C - (x-59)C = 4A - 9B$, $45C=4(3A)-27B=28B-27B$, $105C=28A-9(7B)=A$, therefore,

$A=105C, B=45C, x=4(105)+44=464$, which gives us a consistent solution. Therefore $y=59$ is the answer.

(Note: To further illustrate the idea, let us look at $y=60$ and see what happens. We then get $7\cdot 16C = 4A-30A<0$, which is a contradiction!)

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