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>. 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>. After testing these, we see that only <math>a=5</math> and <math>a=14</math> work which give <math>b=5</math> and <math>b=7</math> respectively. Thus the answer is <math>10+21=\boxed{031}</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>. 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>. After testing these, we see that only <math>a=5</math> and <math>a=14</math> work which give <math>b=5</math> and <math>b=7</math> respectively. Thus the answer is <math>10+21=\boxed{031}</math>. | ||
~JHawk0224 | ~JHawk0224 | ||
+ | |||
+ | Note: If you don't understand the simplification of <math>b^2=\frac{3a^2}{a-2}</math>, you can actually just use synthetic division and arrive at the same place ~ Anonymous | ||
==Solution 2== | ==Solution 2== | ||
Let the common difference be <math> d </math> and let the middle term be <math> x </math>. Then, we have that the sequence is | Let the common difference be <math> d </math> and let the middle term be <math> x </math>. Then, we have that the sequence is | ||
<cmath>x-d,~x,~x+d.</cmath> | <cmath>x-d,~x,~x+d.</cmath> | ||
− | This means that the sum of the sequence is | + | This means that the sum of the squares of the 3 terms of the sequence is |
<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> | <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> | ||
We know that this must be equal to <math>xd^2,</math> so we can write that | We know that this must be equal to <math>xd^2,</math> so we can write that | ||
Line 20: | Line 22: | ||
We can factor pull <math> d^2 </math> out of the square root to get | We can factor pull <math> d^2 </math> out of the square root to get | ||
<cmath>x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.</cmath> | <cmath>x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.</cmath> | ||
− | Here, it is easy to | + | Here, it is easy to figure out the values of <math> d </math>. Let <math>\sqrt{d^2-24} = k</math>, then <math>d^2-k^2=24</math> which is <math>(d+k)(d-k)=24,</math> note that <math>d</math>, <math>k</math> are integers. Examining the parity, we find that <math>d+k</math> and <math>d-k</math> are of the same parity. Now, we solve by factoring. We can 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>^{*}</math> <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. |
+ | |||
+ | 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>. -BorealBear, minor edit by Kinglogic | ||
+ | |||
− | + | <math>^*</math>To prove this, let <math>\sqrt{d^2-24} = k</math>, then <math>d^2-k^2=24</math> which is <math>(d+k)(d-k)=24,</math> then remembering that <math>d</math> and <math>k</math> are integers see if you can figure it out. -[[User:PureSwag|PureSwag]] | |
==Solution 3== | ==Solution 3== | ||
Line 36: | Line 41: | ||
-Ross Gao | -Ross Gao | ||
− | ==See | + | ==Solution 4 (Combining Solution 1 and Solution 3)== |
+ | As in Solution 1, write the three integers in the sequence as <math>a-d</math>, <math>a</math>, and <math>a+d</math>. | ||
+ | |||
+ | Then the sum of the squares of the three integers is <math>(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2</math>. | ||
+ | |||
+ | Setting this equal to the middle term times the common difference squared, which is <math>ad^2</math>, | ||
+ | |||
+ | and solving for <math>d^2</math> we get: | ||
+ | |||
+ | <math>3a^2+2d^2 = ad^2 \implies ad^2-2d^2 = 3a^2 \implies d^2(a-2) = 3a^2 \implies d^2 = \frac{3a^2}{a-2}</math> | ||
+ | |||
+ | The numerator has to be positive, so the denominator has to be positive too for the sequence | ||
+ | |||
+ | to be strictly increasing; that is, <math>a>2</math>. | ||
+ | |||
+ | For <math>\frac{3a^2}{a-2}</math> to be a perfect square, <math>\frac{3}{a-2}</math> must be a perfect square as well. | ||
+ | |||
+ | This means that <math>a-2</math> is divisible by 3, and whatever left over is a perfect square. | ||
+ | |||
+ | We can express this as an equation: let the perfect square left over be <math>n^2</math>. Then: | ||
+ | |||
+ | <math>3n^2 = a-2</math>. Now when you divide the numerator and denominator by 3, you are left with | ||
+ | |||
+ | <math>d^2 = \frac{a^2}{n^2} \implies d = \frac{a}{n}</math>. Because the sequence is of integers, d must also be an | ||
+ | |||
+ | integer, which means that <math>n</math> must divide <math>a</math>. | ||
+ | |||
+ | Taking the above equation we can solve for <math>a</math>: <math>3n^2 = a-2 \implies a = 3n^2+2</math>. | ||
+ | |||
+ | This means that <math>3n^2+2</math> is divisible by <math>n</math>. <math>3n^2</math> is automatically divisible by <math>n</math>, so | ||
+ | |||
+ | <math>2</math> must be divisible by <math>n</math>. Then <math>n</math> must be either of <math>\{1,2\}</math>. Plugging back into the | ||
+ | equation, | ||
+ | |||
+ | <math>n = 1 \implies a = 5 \implies d = 5</math>, so <math>a+d = 5+5 = 10</math>. | ||
+ | |||
+ | <math>n = 2 \implies a = 14 \implies d = 7</math>, so <math>a+d = 14+7 = 21</math>. | ||
+ | |||
+ | Finally, <math>10+21 = \boxed{031}</math> | ||
+ | |||
+ | -KingRavi | ||
+ | |||
+ | ==Solution 5== | ||
+ | Following from previous solutions, we derive <math>3x^2+2a^2=xa^2.</math> We divide both sides to get <math>3\left(\frac{x}{a}\right)^2+2=x.</math> Since <math>x</math> is an integer, <math>\frac{x}{a}</math> must also be an integer, so we have <math>x=pa</math>, for some factor <math>p</math>. We then get <math>3p^2+2=pa.</math> We then take this to modulo <math>p</math>, getting <math>2\equiv 0 \pmod p.</math> The only possibilities for <math>p</math> are therefore 1 and 2. We plug these into <math>3p^2+2=pa</math>, for <math>a=5</math> and <math>x=5</math>, giving us the sequence <math>0,5,10</math>, or <math>2a=14</math> and <math>x=14</math>, for the sequence <math>7,14,21.</math> <math>10+21 = \boxed{031}.</math> | ||
+ | |||
+ | -RYang2 | ||
+ | ==Video Solution == | ||
+ | https://youtu.be/92dvTKV1nPc | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://www.youtube.com/watch?v=I43RH5DUa1I | ||
+ | |||
+ | ==Video Solution 2 == | ||
+ | https://youtu.be/M3DsERqhiDk?t=1465 | ||
+ | |||
+ | ==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}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:15, 1 February 2024
Contents
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 1
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
. After testing these, we see that only
and
work which give
and
respectively. Thus the answer is
.
~JHawk0224
Note: If you don't understand the simplification of , you can actually just use synthetic division and arrive at the same place ~ Anonymous
Solution 2
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 squares of the 3 terms 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 figure out the values of
. Let
, then
which is
note that
,
are integers. Examining the parity, we find that
and
are of the same parity. Now, we solve by factoring. We can 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, minor edit by Kinglogic
To prove this, let
, then
which is
then remembering that
and
are integers see if you can figure it out. -PureSwag
Solution 3
Proceed as in solution 2, until we reach Write
, it follows that
for some (positive) integer k and
.
Taking both sides modulo ,
, so
.
When , we have
and
. When
, we have
and
.
Summing the two cases, we have
.
-Ross Gao
Solution 4 (Combining Solution 1 and Solution 3)
As in Solution 1, write the three integers in the sequence as ,
, and
.
Then the sum of the squares of the three integers is .
Setting this equal to the middle term times the common difference squared, which is ,
and solving for we get:
The numerator has to be positive, so the denominator has to be positive too for the sequence
to be strictly increasing; that is, .
For to be a perfect square,
must be a perfect square as well.
This means that is divisible by 3, and whatever left over is a perfect square.
We can express this as an equation: let the perfect square left over be . Then:
. Now when you divide the numerator and denominator by 3, you are left with
. Because the sequence is of integers, d must also be an
integer, which means that must divide
.
Taking the above equation we can solve for :
.
This means that is divisible by
.
is automatically divisible by
, so
must be divisible by
. Then
must be either of
. Plugging back into the
equation,
, so
.
, so
.
Finally,
-KingRavi
Solution 5
Following from previous solutions, we derive We divide both sides to get
Since
is an integer,
must also be an integer, so we have
, for some factor
. We then get
We then take this to modulo
, getting
The only possibilities for
are therefore 1 and 2. We plug these into
, for
and
, giving us the sequence
, or
and
, for the sequence
-RYang2
Video Solution
~MathProblemSolvingSkills.com
Video Solution 1
https://www.youtube.com/watch?v=I43RH5DUa1I
Video Solution 2
https://youtu.be/M3DsERqhiDk?t=1465
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.