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Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 2"
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==Solution==
 
==Solution==
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Since <math>C</math> is the trisector of [[line segment]] <math>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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*[[Logarithm]]
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Revision as of 17:01, 16 January 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 $AB$ 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 $x_3 = 10$.


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