Difference between revisions of "2017 AIME II Problems/Problem 5"
(→Solution 2) |
m (→Solution 3) |
||
Line 13: | Line 13: | ||
==Solution 3== | ==Solution 3== | ||
− | Note that if <math>a > b > c > d</math> are the elements of the set, then <math>a+b > a+c > b+c, a+d | + | Note that if <math>a>b>c>d</math> are the elements of the set, then <math>a+b>a+c>b+c,a+d>b+d>c+d</math>. Thus we can assign <math>a+b=x,a+c=y,b+c=320,a+d=287,b+d=234,c+d=189</math>. Then <math>x+y=(a+b)+(a+c)=2((a+d)+(b+c))-((c+d)+(b+d))=791</math>. |
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=4|num-a=6}} | {{AIME box|year=2017|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:07, 21 January 2018
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 . 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 .
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.