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

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=== Solution 3 ===
 
=== Solution 3 ===
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Again, the [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is at <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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Again, the [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is at <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} = \fbox{047}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:52, 28 August 2016

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 $p + q = 47$.

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 $\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = \fbox{047}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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