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

Revision as of 23:28, 20 July 2022

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.

Solution 4

This solution uses terminology from the other solutions. The incenter is a much easier point to find on the line $PP'$ . Note that the inradius of $\triangle PQR$ is $5$ . If you do not understand this, substitute values into the $[\triangle ABC] = rs$ equation. If lines are drawn from the incenter perpendicular to $PR$ and $QR$ , then a square with side length $5$ will be created. Call the point opposite $R$ in this square $R'$ . Since $R$ has coordinates $(1, -7)$ , and the sides of the squares are on a $3-4-5$ ratio, the coordinates of $R'$ are $(-6, -6)$ . This is because the x-coordinate is moving to the left $4+3=7$ units and the y-coordinate is moving up $-3+4=1$ units. The line through $(-8,5)$ and $(-6,-6)$ is $11x+2y+78=0$ .

@@ Line 52: / Line 52: @@
 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.
+=== Solution 4 ===
+This solution uses terminology from the other solutions. The incenter is a much easier point to find on the line <math>PP'</math>. Note that the inradius of <math>\triangle PQR</math> is <math>5</math>. If you do not understand this, substitute values into the <math>[\triangle ABC] = rs</math> equation. If lines are drawn from the incenter perpendicular to <math>PR</math> and <math>QR</math>, then a square with side length <math>5</math> will be created. Call the point opposite <math>R</math> in this square <math>R'</math>. Since <math>R</math> has coordinates <math>(1, -7)</math>, and the sides of the squares are on a <math>3-4-5</math> ratio, the coordinates of <math>R'</math> are <math>(-6, -6)</math>. This is because the x-coordinate is moving to the left <math>4+3=7</math> units and the y-coordinate is moving up <math>-3+4=1</math> units. The line through <math>(-8,5)</math> and <math>(-6,-6)</math> is <math>11x+2y+78=0</math>.

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
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