Difference between revisions of "2017 AIME II Problems/Problem 5"
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==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>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)= | + | 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 )== | ||
+ | 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>b+c = 320</cmath> | ||
+ | <cmath>a+b = 234</cmath> | ||
+ | <cmath>a+c = 189</cmath> | ||
+ | |||
+ | <b>Case 2:</b> | ||
+ | |||
+ | <cmath>a+d = 320</cmath> | ||
+ | <cmath>b+c = 189</cmath> | ||
+ | <cmath>a+b = 234</cmath> | ||
+ | <cmath>a+c = 287</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <b>Case 2:</b> | ||
+ | |||
+ | <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 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}} |
Latest revision as of 00:31, 25 December 2022
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:
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.