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2021 AIME I Problems/Problem 5

Revision as of 08:25, 12 March 2021 by Yuanyuanc (talk | contribs) (→‎Solution 2)

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 $a-b$, $a$, and $a+b$. Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$, or $3a^2+2b^2=ab^2$. Rearranging, we get $b^2=\frac{3a^2}{a-2}$. Simplifying further, $b^2=3a+6+\frac{12}{a-2}$. Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\pm1, 2, 3, 4, 6, 12$. Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$. After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=\boxed{31}$. ~JHawk0224

Solution 1

Let the common difference be $d$ and let the middle term be $x$. Then, we have that the sequence is \[x-d,~x,~x+d.\] This means that the sum of the sequence is \[(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.\] We know that this must be equal to $xd^2,$ so we can write that \[3x^2+2d^2=xd^2,\] and it follows that \[3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.\]

Now, we can treat $d$ as a constant and use the quadratic formula to get \[x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.\] We can factor pull $d^2$ out of the square root to get \[x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.\] Here, it is easy to test values of $d$. We find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\sqrt{d^2-24}$ a positive integer. $d=5$ gives $x=5$ and $x=\frac{10}{3}$, but we can ignore the latter. $d=7$ gives $x=14$, as well as a fraction which we can ignore.

Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\boxed{031}$. -BorealBear

Solution 2

Proceed as in solution 1, until we reach \[3a^2+2d^2=ad^2,\] Write

$d^2=\frac{3a^2}{a-2}$, it follows that $a-2=3k^2$ for some (positive) integer k and $k \mid a$.

Taking both sides modulo $k$, $-2 \equiv 0 \pmod{k}$, so $k \mid 2 \rightarrow k=1,2$.

When $k=1$, $x=5$ and $d=5$. When $k=2$, $x=14$ and $d=7$.

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

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