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Difference between revisions of "2002 AMC 12P Problems/Problem 8"

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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>AB</math> be a segment of length <math>26</math>, and let points <math>C</math> and <math>D</math> be located on <math>AB</math> such that <math>AC=1</math> and <math>AD=8</math>. Let <math>E</math> and <math>F</math> be points on one of the semicircles with diameter <math>AB</math> for which <math>EC</math> and <math>FD</math> are perpendicular to <math>AB</math>. Find <math>EF.</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) }5
 +
\qquad
 +
\text{(B) }5 \sqrt{2}  
 +
\qquad
 +
\text{(C) }7
 +
\qquad
 +
\text{(D) }7 \sqrt{2}
 +
\qquad
 +
\text{(E) }12
 +
</math>
  
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== 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>.
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== Solution==
 +
We can solve this with some simple coordinate geometry. Let <math>A</math> be the origin at let <math>AB</math> be located on the positive <math>x-</math>axis. The equation of semi-circle <math>AB</math> is <math>(x-13)^2+y^2=13^2, y \geq 0.</math> Since <math>E</math> and <math>F</math> are both perpendicular to <math>C</math> and <math>D</math> respectively, they must have the same <math>x -</math> coordinate. Plugging in <math>1</math> and <math>8</math> into our semi-circle equation gives us <math>y=5</math> and <math>y=12</math> respectively. The distance formula on <math>(1, 5)</math> and <math>(8, 12)</math> gives us our answer of <math>\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.</math>
  
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC12 box|year=2002|ab=P|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:24, 19 May 2024

Problem

Let $AB$ be a segment of length $26$, and let points $C$ and $D$ be located on $AB$ such that $AC=1$ and $AD=8$. Let $E$ and $F$ be points on one of the semicircles with diameter $AB$ for which $EC$ and $FD$ are perpendicular to $AB$. Find $EF.$

$\text{(A) }5 \qquad \text{(B) }5 \sqrt{2} \qquad \text{(C) }7 \qquad \text{(D) }7 \sqrt{2} \qquad \text{(E) }12$


Solution

We can solve this with some simple coordinate geometry. Let $A$ be the origin at let $AB$ be located on the positive $x-$axis. The equation of semi-circle $AB$ is $(x-13)^2+y^2=13^2, y \geq 0.$ Since $E$ and $F$ are both perpendicular to $C$ and $D$ respectively, they must have the same $x -$ coordinate. Plugging in $1$ and $8$ into our semi-circle equation gives us $y=5$ and $y=12$ respectively. The distance formula on $(1, 5)$ and $(8, 12)$ gives us our answer of $\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.$

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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