Difference between revisions of "2003 AIME I Problems/Problem 4"
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By the first equation, we get that <math>\sin(x)*\cos(x)=10^{-1}</math>. We can let <math>\sin(x)=a</math>, <math>\cos(x)=b</math>. Thus <math>ab=\frac{1}{10}</math>. By the identity <math>\sin^2x+\cos^2x=1</math>, we get that <math>a^2+b^2=1</math>. Solving this, we get <math>a+b=\sqrt{\frac{12}{10}}</math>. So we have | By the first equation, we get that <math>\sin(x)*\cos(x)=10^{-1}</math>. We can let <math>\sin(x)=a</math>, <math>\cos(x)=b</math>. Thus <math>ab=\frac{1}{10}</math>. By the identity <math>\sin^2x+\cos^2x=1</math>, we get that <math>a^2+b^2=1</math>. Solving this, we get <math>a+b=\sqrt{\frac{12}{10}}</math>. So we have | ||
− | <cmath>\log\ | + | <cmath>\log\left(\sqrt{\frac{12}{10}}\right)=\frac12(\log(n)-1)</cmath> |
<cmath>2\log\left(\sqrt{\frac{12}{10}}\right)=\log(n)-1</cmath> | <cmath>2\log\left(\sqrt{\frac{12}{10}}\right)=\log(n)-1</cmath> | ||
<cmath>\log\left(\frac{12}{10}\right)+1=\log(n)</cmath> | <cmath>\log\left(\frac{12}{10}\right)+1=\log(n)</cmath> | ||
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From here it is obvious that <math>\boxed{n=012}</math>. | From here it is obvious that <math>\boxed{n=012}</math>. | ||
+ | |||
+ | |||
+ | ~yofro | ||
== See also == | == See also == |
Latest revision as of 12:32, 10 June 2024
Problem
Given that and that
find
Solution 1
Using the properties of logarithms, we can simplify the first equation to . Therefore,
Now, manipulate the second equation.
By the Pythagorean identities, , and we can substitute the value for
from
.
.
Solution 2
Examining the first equation, we simplify as the following:
With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties):
From here, we may divide both sides by and then proceed with the change-of-base logarithm property:
Thus, exponentiating both sides results in . Squaring both sides gives us
Via the Pythagorean Identity, and
is simply
, via substitution. Thus, substituting these results into the current equation:
Using simple cross-multiplication techniques, we have , and thus
.
~ nikenissan
Solution 3
By the first equation, we get that . We can let
,
. Thus
. By the identity
, we get that
. Solving this, we get
. So we have
From here it is obvious that .
~yofro
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.