Difference between revisions of "2018 AIME I Problems/Problem 5"
(Blanked the page) |
Northstar47 (talk | contribs) m (removed the SFFT comment from solution 1. see https://artofproblemsolving.com/community/c3h3122705_sfft_in_factoring_2_variable_polynomials) |
||
(32 intermediate revisions by 19 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem 5== | ||
+ | For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>. | ||
+ | |||
+ | ==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> | ||
+ | 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>. | ||
+ | |||
+ | -expiLnCalc | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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> | ||
+ | -minor edit:einsteinstudent | ||
+ | |||
+ | -style edit: yeaboi | ||
+ | |||
+ | ==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> | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iE8paW_ICxw | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2018|n=I|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Revision as of 12:52, 30 July 2023
Contents
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
-minor edit:einsteinstudent
-style edit: yeaboi
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
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
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.