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

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== Problem ==
 
== Problem ==
−
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>
+
<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 ==
−
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>.
+
Given <math>\begin{cases}\sin{a} + \sin{b} = \frac{\sqrt{2}}{2} \dots \textcircled{1}\\
 +
\cos {a} + \cos {b} = \frac{\sqrt{6}}{2} \dots \textcircled{2}\end{cases} </math>
 +
We multiply both sides of the syetem, <math>\textcircled{1} \times \textcircled{2}</math>, then we get <math> (\sin{a}\cos{a}  + \sin{b} \cos{b}  )+( \sin{a}\cos{b}  + \sin{b} \cos{a} )= \frac{\sqrt{3}}{2}</math>. i.e.
 +
<math>(\sin{a}\cos{a}  + \sin{b} \cos{b}  )+\sin{(a+b)}= \frac{\sqrt{3}}{2}</math>.
 +
 
 +
We must get the sum of the first part of the equation, then we calculate <math>\textcircled{1}^2+\textcircled{2}^2 </math>, we will
 +
get <math>\sin{a}\cos{a}  + \sin{b} \cos{b} = 0 </math> as <math> \sin^{2}{a}+\cos^{2}{a} = 1</math> and <math> \sin^{2}{b}+\cos^{2}{b} = 1</math>.  
 +
 
 +
So <math>\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\mathrm{C}}</math>
 +
 
 +
Comment: This problem is pretty much identical to https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_17 except with different numbers.
  
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC12 box|year=2002|ab=P|num-b=24|after=Last question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:07, 10 March 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

Given $\begin{cases}\sin{a} + \sin{b} = \frac{\sqrt{2}}{2} \dots \textcircled{1}\\ \cos {a} + \cos {b} = \frac{\sqrt{6}}{2} \dots \textcircled{2}\end{cases}$ We multiply both sides of the syetem, $\textcircled{1} \times \textcircled{2}$, then we get $(\sin{a}\cos{a} + \sin{b} \cos{b} )+( \sin{a}\cos{b} + \sin{b} \cos{a} )= \frac{\sqrt{3}}{2}$. i.e. $(\sin{a}\cos{a} + \sin{b} \cos{b} )+\sin{(a+b)}= \frac{\sqrt{3}}{2}$.

We must get the sum of the first part of the equation, then we calculate $\textcircled{1}^2+\textcircled{2}^2$, we will get $\sin{a}\cos{a} + \sin{b} \cos{b} = 0$ as $\sin^{2}{a}+\cos^{2}{a} = 1$ and $\sin^{2}{b}+\cos^{2}{b} = 1$.

So $\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\mathrm{C}}$

Comment: This problem is pretty much identical to https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_17 except with different numbers.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
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All AMC 12 Problems and Solutions

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