Difference between revisions of "2017 AIME II Problems/Problem 5"
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There are two cases we can consider. Let the elements of our set be denoted <math>a,b,c,d</math>, and say that the largest sums <math>x</math> and <math>y</math> will be consisted of <math>b+d</math> and <math>c+d</math>. Thus, we want to maximize <math>b+c+2d</math>, which means <math>d</math> has to be as large as possible, and <math>a</math> has to be as small as possible to maximize <math>b</math> and <math>c</math>. So, the two cases we look at are: | There are two cases we can consider. Let the elements of our set be denoted <math>a,b,c,d</math>, and say that the largest sums <math>x</math> and <math>y</math> will be consisted of <math>b+d</math> and <math>c+d</math>. Thus, we want to maximize <math>b+c+2d</math>, which means <math>d</math> has to be as large as possible, and <math>a</math> has to be as small as possible to maximize <math>b</math> and <math>c</math>. So, the two cases we look at are: | ||
− | + | <b>Case 1:</b> | |
<cmath>a+d = 287</cmath> | <cmath>a+d = 287</cmath> | ||
Line 25: | Line 25: | ||
<cmath>a+c = 189</cmath> | <cmath>a+c = 189</cmath> | ||
− | + | <b>Case 2:</b> | |
<cmath>a+d = 320</cmath> | <cmath>a+d = 320</cmath> | ||
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Note we have determined these cases by maximizing the value of <math>a+d</math> determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be: | Note we have determined these cases by maximizing the value of <math>a+d</math> determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be: | ||
− | + | <b>Case 1:</b> | |
<cmath>(a,b,c,d) = (\frac{103}{2},\frac{365}{2},\frac{275}{2},\frac{471}{2})</cmath> | <cmath>(a,b,c,d) = (\frac{103}{2},\frac{365}{2},\frac{275}{2},\frac{471}{2})</cmath> | ||
− | + | <b>Case 2:</b> | |
− | <cmath>(a,b,c,d) = (166,68,121, | + | <cmath>(a,b,c,d) = (166,68,121,154)</cmath> |
See the first case has our largest <math>d</math>, so our answer will be <math>471+\frac{640}{2} = \boxed{791}</math> | See the first case has our largest <math>d</math>, so our answer will be <math>471+\frac{640}{2} = \boxed{791}</math> |
Latest revision as of 23:31, 24 December 2022
Contents
[hide]Problem
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are , , , , , and . Find the greatest possible value of .
Solution 1
Let these four numbers be , , , and , where . needs to be maximized, so let and because these are the two largest pairwise sums. Now needs to be maximized. Notice that . No matter how the numbers , , , and are assigned to the values , , , and , the sum will always be . Therefore we need to maximize . The maximum value of is achieved when we let and be and because these are the two largest pairwise sums besides and . Therefore, the maximum possible value of .
Solution 2
Let the four numbers be , , , and , in no particular order. Adding the pairwise sums, we have , so . Since we want to maximize , we must maximize .
Of the four sums whose values we know, there must be two sums that add to . To maximize this value, we choose the highest pairwise sums, and . Therefore, .
We can substitute this value into the earlier equation to find that .
Solution 3
Note that if are the elements of the set, then . Thus we can assign . Then .
Solution 4 ( Short Casework )
There are two cases we can consider. Let the elements of our set be denoted , and say that the largest sums and will be consisted of and . Thus, we want to maximize , which means has to be as large as possible, and has to be as small as possible to maximize and . So, the two cases we look at are:
Case 1:
Case 2:
Note we have determined these cases by maximizing the value of determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be:
Case 1:
Case 2:
See the first case has our largest , so our answer will be
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.