Difference between revisions of "2017 AIME II Problems/Problem 5"
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==Solution 1== | ==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>. | ||
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+ | ==Solution 2== | ||
+ | Let the four numbers be <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, in no particular order. Adding the pairwise sums, we have <math>3a+3b+3c+3d=1030+x+y</math>, so <math>x+y=3(a+b+c+d)-1030</math>. Since we want to maximize <math>x+y</math>, we must maximize <math>a+b+c+d</math>. | ||
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+ | Of the four sums whose values we know, there must be two sums that add to <math>a+b+c+d</math>. To maximize this value, we choose the highest pairwise sums, <math>320</math> and <math>287</math>. Therefore, <math>a+b+c+d=320+287=607</math>. | ||
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+ | We can substitute this value into the earlier equation to find that <math>x+y=3(607)-1030=1821-1030=\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}} |
Revision as of 17:04, 27 March 2017
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 . 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 .
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 |
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