Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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Therefore <math>(a-4)^2 - k^2 = 16</math> and | Therefore <math>(a-4)^2 - k^2 = 16</math> and | ||
<cmath>((a-4) - k)((a-4) + k) = 16.</cmath> | <cmath>((a-4) - k)((a-4) + k) = 16.</cmath> | ||
| − | Let <math>(a-4) - k = u</math> and <math>(a-4) + k = v</math>; then, <math>a-4 = \dfrac{u+v}{2}</math> and so <math>a = \dfrac{u+v}{2} + 4</math>. Listing all possible <math>(u, v)</math> pairs (not counting transpositions because this does not affect (<math>u + v</math>), <math>(2, 8), (4, 4), (-2, -8), (-4, -4)</math>, yields <math>a = 9, 8, -1, 0</math>. These <math>a</math> sum to <math>16</math>, so our answer is <math>\boxed{\textbf{(C) }16}</math>. | + | Let <math>(a-4) - k = u</math> and <math>(a-4) + k = v</math>; then, <math>a-4 = \dfrac{u+v}{2}</math> and so <math>a = \dfrac{u+v}{2} + 4</math>. Listing all possible <math>(u, v)</math> pairs (not counting transpositions because this does not affect (<math>u + v</math>)), <math>(2, 8), (4, 4), (-2, -8), (-4, -4)</math>, yields <math>a = 9, 8, -1, 0</math>. These <math>a</math> sum to <math>16</math>, so our answer is <math>\boxed{\textbf{(C) }16}</math>. |
==Solution 2== | ==Solution 2== | ||
Latest revision as of 23:44, 23 December 2024
Contents
Problem
The zeroes of the function 
are integers. What is the sum of the possible values of 
Solution 1
By Vieta's Formula, 
is the sum of the integral zeros of the function, and so is integral.
Because the zeros are integral, the discriminant of the function, 
, is a perfect square, say 
. Then adding 16 to both sides and completing the square yields
![\[(a - 4)^2 = k^2 + 16.\]](http://latex.artofproblemsolving.com/0/f/0/0f0fc54cac5d6efbe535f2c0592f14c11a50bde5.png)
Therefore 
and
![\[((a-4) - k)((a-4) + k) = 16.\]](http://latex.artofproblemsolving.com/8/0/3/8035bbf221fd1d0b33f061a4dea96f3e9d32f20d.png)
Let 
and 
; then, 
and so 
. Listing all possible 
pairs (not counting transpositions because this does not affect (
)), 
, yields 
. These sum to 
, so our answer is 
.
Solution 2
Let 
and 
be the integer zeroes of the quadratic.
By Vieta's Formulas, ![\[r_1 + r_2 = a\text{ and }r_1r_2 = 2a.\]](http://latex.artofproblemsolving.com/2/4/1/241dc901e5926dcdf5ebe9088377efdd9d039a41.png)
Plugging the first equation in the second,
![\[r_1r_2 = 2 (r_1 + r_2).\]](http://latex.artofproblemsolving.com/2/c/9/2c9944895d19b5830154a28bc2d0135db44595a1.png)
Rearranging gives
![\[r_1r_2 - 2r_1 - 2r_2 = 0 \implies (r_1 - 2)(r_2 - 2) = 4.\]](http://latex.artofproblemsolving.com/1/d/d/1dd9b233057fedffe4679c96548d1456eea50935.png)
These factors 
(ignoring order, because we want the sum of factors), can be 
or 
.
The sum of distinct 
, and these factors give
![\[\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}\]](http://latex.artofproblemsolving.com/2/b/9/2b9905af688f723f4eda9ece62a68848593db943.png)
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/397
Video Solution
~savannahsolver
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
