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==Solution 1== | ==Solution 1== | ||
Using the logarithmic property <math>\log_{a^n}b^n = \log_{a}b</math>, we note that <cmath>(2x+y)^2 = x^2+xy+7y^2</cmath> | Using the logarithmic property <math>\log_{a^n}b^n = \log_{a}b</math>, we note that <cmath>(2x+y)^2 = x^2+xy+7y^2</cmath> | ||
− | That gives <cmath>x^2+xy-2y^2=0</cmath> upon simplification and division by <math>3</math>. Factoring <math>x^2+xy-2y^2=0</math> | + | That gives <cmath>x^2+xy-2y^2=0</cmath> upon simplification and division by <math>3</math>. Factoring <math>x^2+xy-2y^2=0</math> gives <cmath>(x+2y)(x-y)=0</cmath> Then, <cmath>x=y \text{ or }x=-2y</cmath> |
From the second equation, <cmath>9x^2+6xy+y^2=3x^2+4xy+Ky^2</cmath> If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>. | From the second equation, <cmath>9x^2+6xy+y^2=3x^2+4xy+Ky^2</cmath> If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>. | ||
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Do as done in Solution 1 to get <cmath>x^2+xy-2y^2=0</cmath> <cmath>\implies (\frac{x}{y})^2+\frac{x}{y}-2=0</cmath> <cmath>\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2</cmath> Do as done in Solution 1 to get <cmath>9x^2+6xy+y^2=3x^2+4xy+Ky^2</cmath> <cmath>\implies 6x^2+2xy+(1-K)y^2=0</cmath> <cmath>\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0</cmath> <cmath>\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}</cmath> <cmath>\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}</cmath>If <math>\frac{x}{y}=1</math> then <cmath>1=\frac{-1\pm \sqrt{6K-5}}{6}</cmath> <cmath>\implies 6=-1\pm \sqrt{6K-5}</cmath> <cmath>\implies 7=\pm \sqrt{6K-5}</cmath> <cmath>\implies 49=6K-5</cmath> <cmath>\implies K=9</cmath>If <math>\frac{x}{y}=-2</math> then <cmath>-2=\frac{-1\pm \sqrt{6K-5}}{6}</cmath> <cmath>\implies -12=-1\pm \sqrt{6K-5}</cmath> <cmath>\implies -11=\sqrt{6K-5}</cmath> <cmath>\implies 121=6K-5</cmath> <cmath>\implies 126=6K</cmath> <cmath>\implies K=21</cmath>Hence our final answer is <math>21\cdot 9=\boxed{189}</math> | Do as done in Solution 1 to get <cmath>x^2+xy-2y^2=0</cmath> <cmath>\implies (\frac{x}{y})^2+\frac{x}{y}-2=0</cmath> <cmath>\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2</cmath> Do as done in Solution 1 to get <cmath>9x^2+6xy+y^2=3x^2+4xy+Ky^2</cmath> <cmath>\implies 6x^2+2xy+(1-K)y^2=0</cmath> <cmath>\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0</cmath> <cmath>\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}</cmath> <cmath>\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}</cmath>If <math>\frac{x}{y}=1</math> then <cmath>1=\frac{-1\pm \sqrt{6K-5}}{6}</cmath> <cmath>\implies 6=-1\pm \sqrt{6K-5}</cmath> <cmath>\implies 7=\pm \sqrt{6K-5}</cmath> <cmath>\implies 49=6K-5</cmath> <cmath>\implies K=9</cmath>If <math>\frac{x}{y}=-2</math> then <cmath>-2=\frac{-1\pm \sqrt{6K-5}}{6}</cmath> <cmath>\implies -12=-1\pm \sqrt{6K-5}</cmath> <cmath>\implies -11=\sqrt{6K-5}</cmath> <cmath>\implies 121=6K-5</cmath> <cmath>\implies 126=6K</cmath> <cmath>\implies K=21</cmath>Hence our final answer is <math>21\cdot 9=\boxed{189}</math> | ||
-vsamc<math>\newline</math> | -vsamc<math>\newline</math> | ||
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==Solution 3 (Official MAA)== | ==Solution 3 (Official MAA)== | ||
Because <math>x^2+xy+7y^2=\left(x+\tfrac{y}{2}\right)^2+\tfrac{27}{4}y^2>0,</math> the right side of the first equation is real. It follows that the left side of the equation is also real, so <math>2x+y>0</math> and <cmath>\log_2(2x+y)=\log_{2^2}(2x+y)^2=\log_4(4x^2+4xy+y^2).</cmath> Thus <math>4x^2+4xy+y^2=x^2+xy+7y^2,</math> which implies that <math>0=x^2+xy-2y^2=(x+2y)(x-y).</math> Therefore either <math>x=-2y</math> or <math>x=y,</math> and because <math>2x+y>0,</math> <math>x</math> must be positive and <math>3x+y=x+(2x+y)>0.</math> Similarly, <cmath>\log_3(3x+y)=\log_{3^2}(3x+y)^2=\log_9(9x^2+6xy+y^2).</cmath> If <math>x=-2y\ne 0,</math> then <math>9x^2+6xy+y^2=36y^2-12y^2+y^2=25y^2=3x^2+4xy+Ky^2</math> when <math>K=21.</math> If <math>x=y\ne 0,</math> then <math>9x^2+6xy+y^2=16y^2=3x^2+4xy+Ky^2</math> when <math>K=9.</math> The requested product is <math>21\cdot9=189.</math> | Because <math>x^2+xy+7y^2=\left(x+\tfrac{y}{2}\right)^2+\tfrac{27}{4}y^2>0,</math> the right side of the first equation is real. It follows that the left side of the equation is also real, so <math>2x+y>0</math> and <cmath>\log_2(2x+y)=\log_{2^2}(2x+y)^2=\log_4(4x^2+4xy+y^2).</cmath> Thus <math>4x^2+4xy+y^2=x^2+xy+7y^2,</math> which implies that <math>0=x^2+xy-2y^2=(x+2y)(x-y).</math> Therefore either <math>x=-2y</math> or <math>x=y,</math> and because <math>2x+y>0,</math> <math>x</math> must be positive and <math>3x+y=x+(2x+y)>0.</math> Similarly, <cmath>\log_3(3x+y)=\log_{3^2}(3x+y)^2=\log_9(9x^2+6xy+y^2).</cmath> If <math>x=-2y\ne 0,</math> then <math>9x^2+6xy+y^2=36y^2-12y^2+y^2=25y^2=3x^2+4xy+Ky^2</math> when <math>K=21.</math> If <math>x=y\ne 0,</math> then <math>9x^2+6xy+y^2=16y^2=3x^2+4xy+Ky^2</math> when <math>K=9.</math> The requested product is <math>21\cdot9=189.</math> | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=4|num-a=6}} | {{AIME box|year=2018|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:58, 14 August 2024
Problem 5
For each ordered pair of real numbers satisfying
there is a real number
such that
Find the product of all possible values of
.
Solution 1
Using the logarithmic property , we note that
That gives
upon simplification and division by
. Factoring
gives
Then,
From the second equation,
If we take
, we see that
. If we take
, we see that
. The product is
.
-expiLnCalc
Solution 2
Do as done in Solution 1 to get
Do as done in Solution 1 to get
If
then
If
then
Hence our final answer is
-vsamc
Solution 3 (Official MAA)
Because the right side of the first equation is real. It follows that the left side of the equation is also real, so
and
Thus
which implies that
Therefore either
or
and because
must be positive and
Similarly,
If
then
when
If
then
when
The requested product is
See Also
2018 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.