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>
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[[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 of a triangle | median]] to side <math> BC </math> has [[slope]] <math> -5. </math> Find the largest possible value of <math> p+q. </math>
  
 
== Solution ==
 
== Solution ==

Revision as of 13:22, 14 February 2007

Problem

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

Solution

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


See also

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