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Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 2"
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==Problem==
 
==Problem==
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Two points <math>A(x_1, y_1)</math> and <math>B(x_2, y_2)</math> are chosen on the graph of <math>f(x) = \ln x</math>, with <math>0 < x_1 < x_2</math>. The points <math>C</math> and <math>D</math> trisect <math>\overline{AB}</math>, with <math>AC < CB</math>. Through <math>C</math> a horizontal line is drawn to cut the curve at <math>E(x_3, y_3)</math>. Find <math>x_3</math> if <math>x_1 = 1</math> and <math>x_2 = 1000</math>.
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Two [[point]]s <math>A(x_1, y_1)</math> and <math>B(x_2, y_2)</math> are chosen on the [[graph]] of <math>f(x) = \ln x</math>, with <math>0 < x_1 < x_2</math>. The points <math>C</math> and <math>D</math> trisect <math>\overline{AB}</math>, with <math>AC < CB</math>. Through <math>C</math> a horizontal [[line]] is drawn to cut the curve at <math>E(x_3, y_3)</math>. Find <math>x_3</math> if <math>x_1 = 1</math> and <math>x_2 = 1000</math>.
  
 
==Solution==
 
==Solution==
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Since <math>C</math> is the trisector of [[line segment]] <math>\overline{AB}</math> closer to <math>A</math>, the <math>y</math>-coordinate of <math>C</math> is equal to two thirds the <math>y</math>-coordinate of <math>A</math> plus one third the <math>y</math>-coordinate of <math>B</math>.  Thus, point <math>C</math> has coordinates <math>(x_0, \frac{2}{3} \ln 1 + \frac{1}{3}\ln 1000) = (x_0, \ln 10)</math> for some <math>x_0</math>.  Then the horizontal [[line]] through <math>C</math> has equation <math>y = \ln 10</math>, and this intersects the curve <math>y = \ln x</math> at the point <math>(10, \ln 10)</math>, so <math>x_3 = 10</math>.
  
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{{solution}}
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==See also==
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{{Mock AIME box|year=2006-2007|n=4|num-b=1|num-a=3|source=125025}}
  
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*[[Mock AIME 4 2006-2007 Problems]]
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*[[Logarithm]]
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*[[Coordinate geometry]]
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 14:25, 8 October 2007

Problem

Two points $A(x_1, y_1)$ and $B(x_2, y_2)$ are chosen on the graph of $f(x) = \ln x$, with $0 < x_1 < x_2$. The points $C$ and $D$ trisect $\overline{AB}$, with $AC < CB$. Through $C$ a horizontal line is drawn to cut the curve at $E(x_3, y_3)$. Find $x_3$ if $x_1 = 1$ and $x_2 = 1000$.

Solution

Since $C$ is the trisector of line segment $\overline{AB}$ closer to $A$, the $y$-coordinate of $C$ is equal to two thirds the $y$-coordinate of $A$ plus one third the $y$-coordinate of $B$. Thus, point $C$ has coordinates $(x_0, \frac{2}{3} \ln 1 + \frac{1}{3}\ln 1000) = (x_0, \ln 10)$ for some $x_0$. Then the horizontal line through $C$ has equation $y = \ln 10$, and this intersects the curve $y = \ln x$ at the point $(10, \ln 10)$, so $x_3 = 10$.

See also

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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