2005 AIME I Problems/Problem 10

Revision as of 16:19, 4 March 2007 by Azjps (talk | contribs) (solution w/o guesswork, box)

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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