2005 AIME I Problems/Problem 10

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 $p+q.$

$[asy]defaultpen(fontsize(8)); size(170); pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22); draw(A--B--C--A);draw(A--M);draw(B--P--C); label("A (p,q)",A,(1,1));label("B (12,19)",B,(-1,-1));label("C (23,20)",C,(1,-1));label("M",M,(0.2,-1)); label("(17,22)",P,(1,1)); dot(A^^B^^C^^M^^P);[/asy]$

Solution 1

The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$ . The equation of the median can be found by $-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}$ , so $q = -5p + 107$ .

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, $q = \frac{1}{11}p - \frac{337}{11}$ .

Setting this equation equal to the equation of the median, 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 $p + q = \boxed{047}$ .

Solution 2

Using the equation of the median from above, we can write the coordinates of $A$ as $(p,\ -5p + 107)$ . The equation of $\overline{BC}$ is $\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}$ , so $x - 12 = 11y - 209$ . In general form, the line is $x - 11y + 197 = 0$ . Use the equation for the distance between a line and point to find the distance between $A$ and $BC$ (which is the height of $\triangle ABC$ ): $\frac{|1(p) - 11(-5p + 107) + 197|}{1^2 + 11^2} = \frac{|56p - 980|}{\sqrt{122}}$ . Now we need the length of $BC$ , which is $\sqrt{(23 - 12)^2 + (20 - 19)^2} = \sqrt{122}$ . The area of $\triangle ABC$ is $70 = \frac{1}{2}bh = \frac{1}{2}\left(\frac{|56p - 980|}{\sqrt{122}}\right) \cdot \sqrt{122}$ . Thus, $|28p - 490| = 70$ , and $p = 15,\ 20$ . We are looking for $p + q = -4p + 107 = 47,\ 27$ . The maximum possible value of $p + q = \fbox{047}$ .

Solution 3

Again, the midpoint $M$ of line segment $\overline{BC}$ is at $\left(\frac{35}{2}, \frac{39}{2}\right)$ . 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 $\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = \fbox{047}$ .

2005 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2005 AIME I Problems/Problem 10

Problem

Contents

Solution 1

Solution 2

Solution 3

See also