Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 2"
m |
|||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
− | 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>. | + | 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>\displaystyle 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>. |
Revision as of 17:03, 16 January 2007
Problem
Two points and are chosen on the graph of , with . The points and trisect , with . Through a horizontal line is drawn to cut the curve at . Find if and .
Solution
Since is the trisector of line segment closer to , the -coordinate of is equal to two thirds the -coordinate of plus one third the -coordinate of . Thus, point has coordinates for some . Then the horizontal line through has equation , and this intersects the curve at the point , so .