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Difference between revisions of "1990 AIME Problems/Problem 7"

(asymptote)
(clean #3)
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D(P--Q--R--cycle);D(U);D(P--U);
 
D(P--Q--R--cycle);D(U);D(P--U);
 
D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm));
 
D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm));
−
D(Q--(U.x,Q.y)--U,dashed);D(R--(R.x,U.y)--U,dashed);
+
D(Q--(U.x,Q.y)--U,dashed);D(rightanglemark(Q,(U.x,Q.y),U,20),dashed);
 
</asy></center>
 
</asy></center>
−
If we draw a triangle using the points <math>Q</math>, the point by which the angle bisector touches <math>QR</math>, and the point directly to the right of <math>Q</math> and the bottom of the aforementioned point, we get another <math>3-4-5 \triangle</math> (this can be shown by proving its [[similar triangle|similarity]] to the triangle drawn using the side of length <math>20</math> as the [[hypotenuse]]). Using this, the lengths of the triangle are <math>\frac{15}{2}, 10, \frac{25}{2}</math>.
+
By the angle bisector theorem as in solution 1, we find that <math>QP' = 25/2</math>. If we draw the right triangle formed by <math>Q, P',</math> and the point directly to the right of <math>Q</math> and below <math>P'</math>, we get another <math>3-4-5 \triangle</math> (since the slope of <math>QR</math> is <math>3/4</math>). Using this, we find that the horizontal projection of <math>QP'</math> is <math>10</math> and the vertical projection of <math>QP'</math> is <math>15/2</math>.  
  
−
Thus, the angle bisector touches <math>QR</math> at the point <math>\left(-15 + 10, -19 + \frac{15}{2}\right) \Rightarrow \left(-5,-\frac{23}{2}\right) = \frac{y + 8}{x - 5}</math> <math>\Longrightarrow -11x + 55 = 2y + 16</math> <math>\Longrightarrow 11x + 2y + 78 = 0</math>. Thus, the solution is <math>11 + 78 = \boxed{089}</math>.
+
Thus, the angle bisector touches <math>QR</math> at the point <math>\left(-15 + 10, -19 + \frac{15}{2}\right) = \left(-5,-\frac{23}{2}\right)</math>, from where we continue with the first solution.  
  
 
== See also ==
 
== See also ==

Revision as of 20:04, 11 April 2008

Problem

A triangle has vertices $P_{}^{}=(-8,5)$, $Q_{}^{}=(-15,-19)$, and $R_{}^{}=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$. Find $a+c_{}^{}$.

[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f); D(P--Q--R--cycle);D(P--T,EndArrow(2mm)); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]

Solution

Use the distance formula to determine the lengths of each of the sides of the triangle. We find that it has lengths of side $15,\ 20,\ 25$, indicating that it is a $3-4-5$ right triangle. At this point, we just need to find another point that lies on the bisector of $\angle P$.

Solution 1

[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]

Use the angle bisector theorem to find that the angle bisector of $\angle P$ divides $QR$ into segments of length $\frac{25}{x} = \frac{15}{20 -x} \Longrightarrow x = \frac{25}{2},\ \frac{15}{2}$. It follows that $\frac{QP'}{RP'} = \frac{5}{3}$, and so $P' = \left(\frac{5x_R + 3x_Q}{8},\frac{5y_R + 3y_Q}{8}\right) = (-5,-23/2)$.

The desired answer is the equation of the line $PP'$. $PP'$ has slope $\frac{-11}{2}$, from which we find the equation to be $11x + 2y + 78 = 0$. Therefore, $a+c = \boxed{089}$.

Solution 2

[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,NE,f);MP("S",S,E,f); D(P--Q--R--cycle);D(R--S--Q,dashed);D(T);D(P--T); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]

Extend $PR$ to a point $S$ such that $PS = 25$. This forms an isosceles triangle $PQS$. The coordinates of $S$, using the slope of $PR$ (which is $-4/3$), can be determined to be $(7,-15)$. Since the angle bisector of $\angle P$ must touch the midpoint of $QS \Rightarrow (-4,-17)$, we have found our two points. We reach the same answer of $11x + 2y + 78 = 0$.

Solution 3

[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); D(Q--(U.x,Q.y)--U,dashed);D(rightanglemark(Q,(U.x,Q.y),U,20),dashed); [/asy]

By the angle bisector theorem as in solution 1, we find that $QP' = 25/2$. If we draw the right triangle formed by $Q, P',$ and the point directly to the right of $Q$ and below $P'$, we get another $3-4-5 \triangle$ (since the slope of $QR$ is $3/4$). Using this, we find that the horizontal projection of $QP'$ is $10$ and the vertical projection of $QP'$ is $15/2$.

Thus, the angle bisector touches $QR$ at the point $\left(-15 + 10, -19 + \frac{15}{2}\right) = \left(-5,-\frac{23}{2}\right)$, from where we continue with the first solution.

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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