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 == | ||
+ | |||
+ | 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
.