Difference between revisions of "2017 AIME II Problems/Problem 5"
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A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are <math>189</math>, <math>320</math>, <math>287</math>, <math>234</math>, <math>x</math>, and <math>y</math>. Find the greatest possible value of <math>x+y</math>. | A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are <math>189</math>, <math>320</math>, <math>287</math>, <math>234</math>, <math>x</math>, and <math>y</math>. Find the greatest possible value of <math>x+y</math>. | ||
| − | ==Solution== | + | ==Solution 1== |
Let these four numbers be <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, where <math>a>b>c>d</math>. <math>x+y</math> needs to be maximized, so let <math>x=a+b</math> and <math>y=a+c</math> because these are the two largest pairwise sums. Now <math>x+y=2a+b+c</math> needs to be maximized. Notice <math>2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))</math>. No matter how the numbers <math>189</math>, <math>320</math>, <math>287</math>, and <math>234</math> are assigned to the values <math>a+d</math>, <math>b+c</math>, <math>b+d</math>, and <math>c+d</math>, the sum <math>(a+d)+(b+c)+(b+d)+(c+d)</math> will always be <math>189+320+287+234</math>. Therefore we need to maximize <math>3((a+c)+(b+d))-(189+320+287+234)</math>. The maximum value of <math>(a+c)+(b+d)</math> is achieved when we let <math>a+c</math> and <math>b+d</math> be <math>320</math> and <math>287</math> because these are the two largest pairwise sums besides <math>x</math> and <math>y</math>. Therefore, the maximum possible value of <math>x+y=3(320+287)-(189+320+287+234)=\boxed{791}</math>. | Let these four numbers be <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, where <math>a>b>c>d</math>. <math>x+y</math> needs to be maximized, so let <math>x=a+b</math> and <math>y=a+c</math> because these are the two largest pairwise sums. Now <math>x+y=2a+b+c</math> needs to be maximized. Notice <math>2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))</math>. No matter how the numbers <math>189</math>, <math>320</math>, <math>287</math>, and <math>234</math> are assigned to the values <math>a+d</math>, <math>b+c</math>, <math>b+d</math>, and <math>c+d</math>, the sum <math>(a+d)+(b+c)+(b+d)+(c+d)</math> will always be <math>189+320+287+234</math>. Therefore we need to maximize <math>3((a+c)+(b+d))-(189+320+287+234)</math>. The maximum value of <math>(a+c)+(b+d)</math> is achieved when we let <math>a+c</math> and <math>b+d</math> be <math>320</math> and <math>287</math> because these are the two largest pairwise sums besides <math>x</math> and <math>y</math>. Therefore, the maximum possible value of <math>x+y=3(320+287)-(189+320+287+234)=\boxed{791}</math>. | ||
Revision as of 16:53, 27 March 2017
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 
.
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. 
