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Difference between revisions of "2002 AMC 12P Problems/Problem 25"

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== Problem ==
 
== Problem ==
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How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
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Let <math>a</math> and <math>b</math> be real numbers such that <math>\sin{a} + \sin{b} = \frac{\sqrt{2}}{2}</math> and <math>\cos {a} + \cos {b} = \frac{\sqrt{6}}{2}.</math> Find <math>\sin{(a+b)}.</math>
  
−
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
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<math>
 +
\text{(A) }\frac{1}{2}
 +
\qquad
 +
\text{(B) }\frac{\sqrt{2}}{2}
 +
\qquad
 +
\text{(C) }\frac{\sqrt{3}}{2}
 +
\qquad
 +
\text{(D) }\frac{\sqrt{6}}{2}
 +
\qquad
 +
\text{(E) }1
 +
</math>
  
 
== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
Sum to product gives us
 +
 
 +
<cmath>2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{2}}{2}</cmath>
 +
 
 +
<cmath>2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{6}}{2}</cmath>
 +
 
 +
Dividing these equations tells us that <math>\tan(\frac{a+b}{2}) = \frac{1}{\sqrt{3}}</math>, so <math>\frac{a+b}{2} = \frac{\pi}{6} + \pi n</math> for an integer <math>n</math>, so <math>\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}</math>. The answer is <math>\boxed{(C)}</math>.
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 +
~alexanderruan
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=24|after=Last question}}
 
{{AMC12 box|year=2002|ab=P|num-b=24|after=Last question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:02, 3 October 2024

Problem

Let $a$ and $b$ be real numbers such that $\sin{a} + \sin{b} = \frac{\sqrt{2}}{2}$ and $\cos {a} + \cos {b} = \frac{\sqrt{6}}{2}.$ Find $\sin{(a+b)}.$

$\text{(A) }\frac{1}{2} \qquad \text{(B) }\frac{\sqrt{2}}{2} \qquad \text{(C) }\frac{\sqrt{3}}{2} \qquad \text{(D) }\frac{\sqrt{6}}{2} \qquad \text{(E) }1$

Solution

Sum to product gives us

\[2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{2}}{2}\]

\[2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{6}}{2}\]

Dividing these equations tells us that $\tan(\frac{a+b}{2}) = \frac{1}{\sqrt{3}}$, so $\frac{a+b}{2} = \frac{\pi}{6} + \pi n$ for an integer $n$, so $\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}$. The answer is $\boxed{(C)}$.

~alexanderruan

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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