Difference between revisions of "2021 AIME I Problems/Problem 5"
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− | Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>b^2=3a+6+\frac{12}{a-2}</math>. | + | Let the terms be <math>a-b</math>, <math>a</math>, and <math>a+b</math>. Then we want <math>(a-b)^2+a^2+(a+b)^2=ab^2</math>, or <math>3a^2+2b^2=ab^2</math>. Rearranging, we get <math>\displaystyle b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>\displaystyle b^2=3a+6+\frac{12}{a-2}</math>. Looking at this second equation, since the right side must be an integer, <math>a-2</math> must equal <math>\pm1, 2, 3, 4, 6, 12</math>. Looking at the first equation, we see <math>a>2</math> since <math>b^2</math> is positive. This means we must test <math>a=3, 4, 5, 6, 8, 14</math>. |
===Solution 1=== | ===Solution 1=== |
Revision as of 16:37, 11 March 2021
Contents
[hide]Problem
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.
Solution
Let the terms be , , and . Then we want , or . Rearranging, we get . Simplifying further, . Looking at this second equation, since the right side must be an integer, must equal . Looking at the first equation, we see since is positive. This means we must test .
Solution 1
Let the common difference be and let the middle term be . Then, we have that the sequence is This means that the sum of the sequence is We know that this must be equal to so we can write that and it follows that
Now, we can treat as a constant and use the quadratic formula to get We can factor pull out of the square root to get Here, it is easy to test values of . We find that and are the only positive integer values of that make a positive integer. gives and , but we can ignore the latter. gives , as well as a fraction which we can ignore.
Since and are the only two solutions and we want the sum of the third terms, our answer is . -BorealBear
See also
2021 AIME I (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.