# Difference between revisions of "2005 AIME I Problems/Problem 10"

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== Problem == | == Problem == | ||

− | Triangle <math> ABC </math> lies in the Cartesian Plane and has an area of 70. The coordinates of <math> B </math> and <math> C </math> are <math> (12,19) </math> and <math> (23,20), </math> respectively, and the coordinates of <math> A </math> are <math> (p,q). </math> The line containing the median to side <math> BC </math> has slope <math> -5. </math> Find the largest possible value of <math> p+q. </math> | + | [[Triangle]] <math> ABC </math> lies in the [[Cartesian Plane]] and has an [[area]] of 70. The coordinates of <math> B </math> and <math> C </math> are <math> (12,19) </math> and <math> (23,20), </math> respectively, and the coordinates of <math> A </math> are <math> (p,q). </math> The [[line]] containing the [[median]] to side <math> BC </math> has [[slope]] <math> -5. </math> Find the largest possible value of <math> p+q. </math> |

== Solution == | == Solution == | ||

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+ | The [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is <math>\left(\frac{35}{2}, \frac{39}{2}\right)</math>. Let <math>A'</math> be the point <math>(17, 22)</math>, which lies along the line through <math>M</math> of slope <math>-5</math>. The area of triangle <math>A'BC</math> can be computed in a number of ways (one possibility: extend <math>A'B</math> until it hits the line <math>y = 19</math>, and subtract one triangle from another), and each such calculation gives an area of 14. This is <math>\frac{1}{5}</math> of our needed area, so we simply need the point <math>A</math> to be 5 times as far from <math>M</math> as <math>A'</math> is. Thus <math>A = \left(\frac{35}{2}, \frac{39}{2}\right) \pm 5\left(-\frac{1}{2}, \frac{5}{2}\right)</math>, and the sum of coordinates will be larger if we take the positive value, so <math>A = \left(\frac{35}{2} - \frac{5}2, \frac{39}{2} + \frac{25}{2}\right)</math> and the answer is <math>\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = 047</math>. | ||

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== See also == | == See also == | ||

+ | * [[2005 AIME I Problems/Problem 9 | Previous problem]] | ||

+ | * [[2005 AIME I Problems/Problem 11 | Next problem]] | ||

* [[2005 AIME I Problems]] | * [[2005 AIME I Problems]] | ||

+ | |||

+ | [[Category:Intermediate Geometry Problems]] |

## Revision as of 14:35, 17 January 2007

## Problem

Triangle lies in the Cartesian Plane and has an area of 70. The coordinates of and are and respectively, and the coordinates of are The line containing the median to side has slope Find the largest possible value of

## Solution

The midpoint of line segment is . Let be the point , which lies along the line through of slope . The area of triangle can be computed in a number of ways (one possibility: extend until it hits the line , and subtract one triangle from another), and each such calculation gives an area of 14. This is of our needed area, so we simply need the point to be 5 times as far from as is. Thus , and the sum of coordinates will be larger if we take the positive value, so and the answer is .