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Difference between revisions of "2021 AIME I Problems/Problem 5"

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==Solution==
 
==Solution==
 
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>b^2=\frac{3a^2}{a-2}</math>. Simplifying further, <math>b^2=3a+6+\frac{12}{a-2}</math>.
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===Solution 1===
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Let the common difference be <math> d </math> and let the middle term be <math> x </math>. Then, we have that the sequence is
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<cmath>x-d,~x,~x+d.</cmath>
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This means that the sum of the sequence is
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<cmath> (x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2. </cmath>
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We know that this must be equal to <math>xd^2,</math> so we can write that
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<cmath>3x^2+2d^2=xd^2,</cmath>
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and it follows that
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<cmath>3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.</cmath>
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Now, we can treat <math> d </math> as a constant and use the quadratic formula to get
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<cmath>x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.</cmath>
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We can factor pull <math> d^2 </math> out of the square root to get
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<cmath>x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.</cmath>
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Here, it is easy to test values of <math> d </math>. We find that <math> d=5 </math> and <math> d=7 </math> are the only positive integer values of <math> d </math> that make <math> \sqrt{d^2-24} </math> a positive integer. <math> d=5 </math> gives <math> x=5 </math> and <math> x=\frac{10}{3} </math>, but we can ignore the latter. <math> d=7 </math> gives <math> x=14 </math>, as well as a fraction which we can ignore.
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 +
Since <math> d=5,~x=5 </math> and <math> d=7, x=14 </math> are the only two solutions and we want the sum of the third terms, our answer is <math> (5+5)+(7+14)=10+21=\boxed{031} </math>.
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2021|n=I|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:35, 11 March 2021

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, $b^2=3a+6+\frac{12}{a-2}$.

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 $(5+5)+(7+14)=10+21=\boxed{031}$.

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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