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

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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>
 
[[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>
  
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== Solution ==
 
== Solution ==
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=== Solution 1 ===
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Use [[determinant]]s to find that the [[area]] of <math>\triangle ABC</math> is <math>\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\  q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70</math> (note that there is a missing [[absolute value]]; we will assume that the other solution for the triangle will give a smaller value of <math>p+q</math>, which is provable by following these steps over again). We can calculate this determinant to become <math>140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix} \Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q = -197 - p + 11q</math>. Thus, <math>q = \frac{1}{11}p - \frac{337}{11}</math>.
  
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The equation of the median can be found by <math>-5 = \frac{q - \frac{12 + 23}{2}}{p - \frac{19 + 20}{2}}</math>, which yields that <math>q = -5p + 107</math>. Setting the two equations equal to each other, we get that <math>\frac{1}{11}p - \frac{337}{11} = -5p + 107</math>, so <math>\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}</math>. Solving produces that <math>p = 15</math>. [[Substitution|Substituting]] backwards yields that <math>q = 32</math>; the solution is <math>p + q = 047</math>.
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=== Solution 2 ===
 
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>.
 
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>.
  
  
 
== See also ==
 
== See also ==
* [[2005 AIME I Problems/Problem 9 | Previous problem]]
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{{AIME box|year=2005|n=I|num-b=9|num-a=11}}
* [[2005 AIME I Problems/Problem 11 | Next problem]]
 
* [[2005 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 16:19, 4 March 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

Solution 1

Use determinants to find that the area of $\triangle ABC$ is $\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\  q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70$ (note that there is a missing absolute value; we will assume that the other solution for the triangle will give a smaller value of $p+q$, which is provable by following these steps over again). We can calculate this determinant to become $140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix} \Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q = -197 - p + 11q$. Thus, $q = \frac{1}{11}p - \frac{337}{11}$.

The equation of the median can be found by $-5 = \frac{q - \frac{12 + 23}{2}}{p - \frac{19 + 20}{2}}$, which yields that $q = -5p + 107$. Setting the two equations equal to each other, we get that $\frac{1}{11}p - \frac{337}{11} = -5p + 107$, so $\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}$. Solving produces that $p = 15$. Substituting backwards yields that $q = 32$; the solution is $p + q = 047$.

Solution 2

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

2005 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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