Difference between revisions of "2017 AIME II Problems/Problem 5"
Geometry285 (talk | contribs) |
Geometry285 (talk | contribs) m (→=Solution 4 ( Short Casework )) |
||
Line 15: | Line 15: | ||
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)=\left[(a+d)-(b+d)+(b+c)\right]+\left[(a+d)-(c+d)+(b+c)\right]=791</math>. | 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)=\left[(a+d)-(b+d)+(b+c)\right]+\left[(a+d)-(c+d)+(b+c)\right]=791</math>. | ||
− | ==Solution 4 ( Short Casework )= | + | ==Solution 4 ( Short Casework )== |
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: | ||
Revision as of 13:01, 30 April 2021
Contents
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:
[b]Case 1:[/b]
[b]Case 2:[/b]
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:
[b]Case 1:[/b]
[b]Case 1:[/b]
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.