Difference between revisions of "2020 AIME I Problems/Problem 11"
(Created page with "Note: Please do not post problems here until after the AIME. == Problem == == Solution ==") |
(→Solution 3 (Official MAA)) |
||
(18 intermediate revisions by 11 users not shown) | |||
Line 1: | Line 1: | ||
− | |||
== Problem == | == Problem == | ||
+ | For integers <math>a,b,c</math> and <math>d,</math> let <math>f(x)=x^2+ax+b</math> and <math>g(x)=x^2+cx+d.</math> Find the number of ordered triples <math>(a,b,c)</math> of integers with absolute values not exceeding <math>10</math> for which there is an integer <math>d</math> such that <math>g(f(2))=g(f(4))=0.</math> | ||
− | == Solution == | + | == Solution 1 (Strategic Casework)== |
+ | Either <math>f(2)=f(4)</math> or not. If it is, note that Vieta's forces <math>a = -6</math>. Then, <math>b</math> can be anything. However, <math>c</math> can also be anything, as we can set the root of <math>g</math> (not equal to <math>f(2) = f(4)</math>) to any integer, producing a possible integer value of <math>d</math>. Therefore there are <math>21^2 = 441</math> in this case*. If it isn't, then <math>f(2),f(4)</math> are the roots of <math>g</math>. This means by Vieta's, that: | ||
+ | |||
+ | <cmath>f(2)+f(4) = -c \in [-10,10]</cmath> | ||
+ | <cmath>20 + 6a + 2b \in [-10,10]</cmath> | ||
+ | <cmath>3a + b \in [-15,-5].</cmath> | ||
+ | |||
+ | Solving these inequalities while considering that <math>a \neq -6</math> to prevent <math>f(2) = f(4)</math>, we obtain <math>69</math> possible tuples and adding gives <math>441+69=\boxed{510}</math>. | ||
+ | ~awang11 | ||
+ | |||
+ | == Solution 2 (Bash) == | ||
+ | Define <math>h(x)=x^2+cx</math>. Since <math>g(f(2))=g(f(4))=0</math>, we know <math>h(f(2))=h(f(4))=-d</math>. Plugging in <math>f(x)</math> into <math>h(x)</math>, we get <math>h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)</math>. Setting <math>h(f(2))=h(f(4))</math>, <cmath>16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc</cmath>. Simplifying and cancelling terms, <cmath>240+112a+24b+12a^2+4ab+12c+2ac=0</cmath> <cmath>120+56a+12b+6a^2+2ab+6c+ac=0</cmath> <cmath>6a^2+2ab+ac+56a+12b+6c+120=0</cmath> <cmath>6a^2+2ab+ac+20a+36a+12b+6c+120=0</cmath> <cmath>a(6a+2b+c+20)+6(6a+2b+c+20)=0</cmath> <cmath>(a+6)(6a+2b+c+20)=0</cmath> | ||
+ | |||
+ | Therefore, either <math>a+6=0</math> or <math>6a+2b+c=-20</math>. The first case is easy: <math>a=-6</math> and there are <math>441</math> tuples in that case. In the second case, we simply perform casework on even values of <math>c</math>, to get <math>77</math> tuples, subtracting the <math>8</math> tuples in both cases we get <math>441+77-8=\boxed{510}</math>. | ||
+ | |||
+ | -EZmath2006 | ||
+ | |||
+ | ==Solution 3 (Official MAA)== | ||
+ | For a given ordered triple <math>(a, b, c)</math>, the value of <math>g(f(4)) - g(f(2))</math> is uniquely determined, and a value of <math>d</math> can be found to give <math>g(f(2))</math> any prescribed integer value. Hence the required condition can be satisfied provided that <math>a</math>, <math>b</math>, and <math>c</math> are chosen so that | ||
+ | <cmath>\begin{align*} | ||
+ | 0 &= g(f(4)) - g(f(2)) = \big(f(4)\big)^{\!2} - \big(f(2)\big)^{\!2} + c\big(f(4) - f(2)\big)\\ | ||
+ | &= \big(f(4) - f(2)\big)\big(f(4) + f(2) + c\big)\\ | ||
+ | &= (12 + 2a)(20 + 6a + 2b + c). | ||
+ | \end{align*}</cmath> | ||
+ | First suppose that <math>12 + 2a = 0</math>, so <math>a = -6</math>. In this case there are <math>21</math> choices for each of <math>b</math> and <math>c</math> with <math>-10 \le b \le 10</math> and <math>-10 \le c \le 10</math>, so this case accounts for <math>21^2 = 441</math> ordered triples. | ||
+ | |||
+ | Next suppose that <math>a \ne -6</math> and <math>20 + 6a + 2b + c = 0</math>, so <math>c = -6a - 2b - 20</math>. Because | ||
+ | <math>-10 \le c \le 10</math>, it follows that <math>-30 \le 6a + 2b \le -10</math>, and because <math>-10\le b \le10</math>, it follows that <math>-8 \le a \le 1</math>. Then | ||
+ | <math>-15 - 3a \le b \le -5 - 3a</math>. The number of ordered triples for various values of <math>a</math> are presented in the following table. | ||
+ | |||
+ | <cmath>\begin{tabular}{|c|c|c|r|} | ||
+ | \hline | ||
+ | a & b & c & triples \\ \hline | ||
+ | -8 & \{9,10\} & -6a - 2b - 20 & 2 \\ | ||
+ | -7 & \{6, 7, 8, 9, 10\} & -6a - 2b - 20 & 5 \\ | ||
+ | -6 & \{-10, \ldots, 10\} & \{-10, \ldots, 10\} & 441 \\ | ||
+ | -5 & \{0, \ldots, 10\} & -6a - 2b - 20 & 11 \\ | ||
+ | -4 & \{-3, \ldots, 7\} & -6a - 2b - 20 & 11 \\ | ||
+ | -3 & \{-6, \ldots, 4\} & -6a - 2b - 20 & 11 \\ | ||
+ | -2 & \{-9, \ldots, 1\} & -6a - 2b - 20 & 11 \\ | ||
+ | -1 & \{-10, \ldots, -2\} & -6a - 2b - 20 & 9 \\ | ||
+ | 0 & \{-10, \ldots, -5\} & -6a - 2b - 20 & 6 \\ | ||
+ | 1 & \{-10, -9, -8\} & -6a - 2b - 20 & 3 \\ | ||
+ | \hline | ||
+ | Total & & & 510\\ | ||
+ | \hline | ||
+ | \end{tabular}</cmath> | ||
+ | The total number of ordered triples that satisfy the required condition is <math>510</math>. | ||
+ | |||
+ | == Notes For * == | ||
+ | In case anyone is confused by this (as I initially was). In the case where <math>f(2)=f(4)</math>, this does not mean that g has a double root of <math>f(2)=f(4)=k</math>, ONLY that <math>k</math> is one of the roots of g. So basically since <math>a=-6</math> in this case, <math>f(2)=f(4)=b-8</math>, and we have <math>21</math> choices for b and we <i>still can</i> ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to <math>b-8</math> ensures this, and of course an integer multiplied by an integer is an integer so <math>d</math> will still be an integer. In other words, you have can have <math>b</math> and <math>c</math> be any integer with absolute value less than or equal to 10 with <math>d</math> still being an integer. Now refer back to the 1st solution. | ||
+ | ~First | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=ftqYFzzWKv8&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=8 ~ MathEx | ||
+ | ==See Also== | ||
+ | |||
+ | {{AIME box|year=2020|n=I|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:05, 25 February 2021
Contents
Problem
For integers and let and Find the number of ordered triples of integers with absolute values not exceeding for which there is an integer such that
Solution 1 (Strategic Casework)
Either or not. If it is, note that Vieta's forces . Then, can be anything. However, can also be anything, as we can set the root of (not equal to ) to any integer, producing a possible integer value of . Therefore there are in this case*. If it isn't, then are the roots of . This means by Vieta's, that:
Solving these inequalities while considering that to prevent , we obtain possible tuples and adding gives . ~awang11
Solution 2 (Bash)
Define . Since , we know . Plugging in into , we get . Setting , . Simplifying and cancelling terms,
Therefore, either or . The first case is easy: and there are tuples in that case. In the second case, we simply perform casework on even values of , to get tuples, subtracting the tuples in both cases we get .
-EZmath2006
Solution 3 (Official MAA)
For a given ordered triple , the value of is uniquely determined, and a value of can be found to give any prescribed integer value. Hence the required condition can be satisfied provided that , , and are chosen so that First suppose that , so . In this case there are choices for each of and with and , so this case accounts for ordered triples.
Next suppose that and , so . Because , it follows that , and because , it follows that . Then . The number of ordered triples for various values of are presented in the following table.
The total number of ordered triples that satisfy the required condition is .
Notes For *
In case anyone is confused by this (as I initially was). In the case where , this does not mean that g has a double root of , ONLY that is one of the roots of g. So basically since in this case, , and we have choices for b and we still can ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to ensures this, and of course an integer multiplied by an integer is an integer so will still be an integer. In other words, you have can have and be any integer with absolute value less than or equal to 10 with still being an integer. Now refer back to the 1st solution. ~First
Video Solution
https://www.youtube.com/watch?v=ftqYFzzWKv8&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=8 ~ MathEx
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.