Difference between revisions of "2015 AIME II Problems/Problem 6"
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Simplifying the right side: | Simplifying the right side: | ||
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− | < | + | <cmath> \begin{align*} |
− | + | a^2-2 \cdot \frac{a^2-81}{2} &= a^2-a^2+81\\ | |
− | + | &= 81.\\ | |
+ | \end{align*} </cmath> | ||
So, we know <math>x_1^2+x_2^2+x_3^2 = 81</math>. | So, we know <math>x_1^2+x_2^2+x_3^2 = 81</math>. | ||
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First things first. Vietas gives us the following: | First things first. Vietas gives us the following: | ||
− | < | + | <cmath>\begin{align} |
− | + | x_1+x_2+x_3 = a\\ | |
− | + | x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}\\ | |
− | + | x_1 \cdot x_2 \cdot x_3 = \frac{c}{2} | |
− | + | \end{align}</cmath> | |
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− | <math> | + | From <math>(2)</math>, <math>a</math> must have odd parity, meaning <math>a^2-81</math> must be a multiple of <math>4</math>, which implies that both sides of <math>(2)</math> are even. Then, from <math>(1)</math>, we see that an odd number of <math>x_1</math>, <math>x_2</math>, and <math>x_3</math> must be odd, because we have already deduced that <math>a</math> is odd. In order for both sides of <math>(2)</math> to be even, there must only be one odd number and two even numbers. |
− | So now we have that <math>9<a</math> from (2), <math>a<16</math> from (5), and <math>a</math> is odd from (2). This means that <math>a</math> could equal <math>11</math>, <math>13</math>, or <math>15</math>. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of <math>(1, 4, 8)</math>, <math>(3, 6, 6)</math>, and <math>(4, 4, 7)</math>, of which the last two return equal <math>a</math> values. Then, <math>2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}</math> AWD. | + | Now, the theoretical maximum value of the left side of <math>(2)</math> is <math>3 \cdot \frac{a}{3}^2=\frac{a^2}{3}</math>. That means that the maximum bound of <math>a</math> is where |
+ | <cmath>\frac{a^2}{3}> \frac{a^2-81}{2} \indent (4)</cmath> | ||
+ | which simplifies to <cmath>\sqrt{243}>a</cmath> meaning | ||
+ | <cmath>16>a(5)</cmath> | ||
+ | So now we have that <math>9<a</math> from <math>(2)</math>, <math>a<16</math> from <math>(5)</math>, and <math>a</math> is odd from <math>(2)</math>. This means that <math>a</math> could equal <math>11</math>, <math>13</math>, or <math>15</math>. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of <math>(1, 4, 8)</math>, <math>(3, 6, 6)</math>, and <math>(4, 4, 7)</math>, of which the last two return equal <math>a</math> values. Then, <math>2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}</math> AWD. | ||
==See also== | ==See also== |
Latest revision as of 17:18, 18 October 2020
Problem
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form for some positive integers and . Can you tell me the values of and ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of ." He writes down a positive integer and asks, "Can you tell me the value of ?"
Jon says, "There are still two possible values of ."
Find the sum of the two possible values of .
Solution 1 (Algebra)
We call the three roots (some may be equal to one another) , , and . Using Vieta's formulas, we get , , and .
Squaring our first equation we get .
We can then subtract twice our second equation to get .
Simplifying the right side:
So, we know .
We can then list out all the triples of positive integers whose squares sum to :
We get , , and .
These triples give values of , , and , respectively, and values of , , and , respectively.
We know that Jon still found two possible values of when Steve told him the value, so the value must be . Thus, the two values are and , which sum to .
~BealsConjecture~
Solution 2 (Algebra+ Brute Force)
First things first. Vietas gives us the following:
From , must have odd parity, meaning must be a multiple of , which implies that both sides of are even. Then, from , we see that an odd number of , , and must be odd, because we have already deduced that is odd. In order for both sides of to be even, there must only be one odd number and two even numbers.
Now, the theoretical maximum value of the left side of is . That means that the maximum bound of is where which simplifies to meaning So now we have that from , from , and is odd from . This means that could equal , , or . At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of , , and , of which the last two return equal values. Then, AWD.
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.