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

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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.
 
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.
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=== Solution 4 ===
 
=== 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>.
  
 +
===Solution 5 (Trigonometry)===
 
<center><asy>
 
<center><asy>
 
import graph;
 
import graph;
 
pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10);
 
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);
+
pair P=(0,0),Q=(-7,-24),R=(9,-12),S=(15,-20),T=(4,-22);
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MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f);
+
MP("Q",Q,W,f);MP("R",R,E,f);
−
D(P--Q--R--cycle);D(U);D(P--U);
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D(P--Q--R--cycle);D(P--T,EndArrow(2mm));
−
D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm));
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D((-8,0)--(13,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm));
−
label("$X_1$",(-9.5,1),W);
 
−
label("$X_2$",(-3,1),W);
 
 
</asy></center>
 
</asy></center>
 +
 +
Transform triangle <math>PQR</math> so that <math>P</math> is at the origin. Note that the slopes do not change when we transform the triangle.
 +
 +
 +
We know that the slope of <math>PQ</math> is <math>\frac{24}{7}</math> and the slope of <math>PR</math> is <math>-\frac{4}{3}</math>. Thus, in the complex plane, they are equivalent to <math>\tan(\alpha)=\frac{24}{7}</math> and <math>\tan(\beta)=-\frac{4}{3}</math>, respectively. Here <math>\alpha</math> is the angle formed by the <math>x</math>-axis and <math>PQ</math>, and <math>\beta</math> is the angle formed by the <math>x</math>-axis and <math>PR</math>. The equation of the angle bisector is <math>\tan\left(\frac{\alpha+\beta}{2}\right)</math>.
 +
 +
 +
As the tangents are in very neat [[pythagorean triple|Pythagorean triples]], we can easily calculate <math>\cos(\alpha)</math> and <math>\cos(\beta)</math>.
 +
 +
Angle <math>\alpha</math> is in the third quadrant, so <math>\cos(\alpha)</math> is negative. Thus <math>\cos(\alpha)=-\frac{7}{25}</math>.
 +
 +
Angle <math>\beta</math> is in the fourth quadrant, so <math>\cos(\beta)</math> is positive. Thus <math>\cos(\beta)=\frac{3}{5}</math>.
 +
 +
 +
By the [[Trigonometric identities#Half-angle identities|Half-Angle Identities]], <math>\tan\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}=\pm\sqrt{\frac{\frac{32}{25}}{\frac{18}{25}}}=\pm\sqrt{\frac{16}{9}}=\pm\frac{4}{3}</math> and <math>\tan\left(\frac{\beta}{2}\right)=\pm\sqrt{\frac{1-\cos(\beta)}{1+\cos(\beta)}}=\pm\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\pm\sqrt{\frac{1}{4}}=\pm\frac{1}{2}</math>.
 +
 +
Since <math>\frac{\alpha}{2}</math> and <math>\frac{\beta}{2}</math> must be in the second quadrant, their tangent values are both negative. Thus <math>\tan\left(\frac{\alpha}{2}\right)=-\frac{4}{3}</math> and <math>\tan\left(\frac{\beta}{2}\right)=-\frac{1}{2}</math>.
 +
 +
 +
By the [[Trigonometric identities#Angle addition identities|sum of tangents formula]], <math>\tan\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}=\frac{-\frac{11}{6}}{\frac{1}{3}}=-\frac{11}{2}</math>, which is the slope of the angle bisector.
 +
 +
 +
Finally, the equation of the angle bisector is <math>y-5=-\frac{11}{2}(x+8)</math> or <math>y=-\frac{11}{2}x-39</math>. Rearranging, we get <math>11x+2y+78=0</math>, so our sum is <math>a+c=11+78=\boxed{089}</math>. ~eevee9406
  
 
== See also ==
 
== See also ==

Latest revision as of 10:34, 22 November 2023

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 $11x+2y+78=0$.

Solution 5 (Trigonometry)

[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(0,0),Q=(-7,-24),R=(9,-12),S=(15,-20),T=(4,-22); MP("Q",Q,W,f);MP("R",R,E,f); D(P--Q--R--cycle);D(P--T,EndArrow(2mm)); D((-8,0)--(13,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]

Transform triangle $PQR$ so that $P$ is at the origin. Note that the slopes do not change when we transform the triangle.


We know that the slope of $PQ$ is $\frac{24}{7}$ and the slope of $PR$ is $-\frac{4}{3}$. Thus, in the complex plane, they are equivalent to $\tan(\alpha)=\frac{24}{7}$ and $\tan(\beta)=-\frac{4}{3}$, respectively. Here $\alpha$ is the angle formed by the $x$-axis and $PQ$, and $\beta$ is the angle formed by the $x$-axis and $PR$. The equation of the angle bisector is $\tan\left(\frac{\alpha+\beta}{2}\right)$.


As the tangents are in very neat Pythagorean triples, we can easily calculate $\cos(\alpha)$ and $\cos(\beta)$.

Angle $\alpha$ is in the third quadrant, so $\cos(\alpha)$ is negative. Thus $\cos(\alpha)=-\frac{7}{25}$.

Angle $\beta$ is in the fourth quadrant, so $\cos(\beta)$ is positive. Thus $\cos(\beta)=\frac{3}{5}$.


By the Half-Angle Identities, $\tan\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}=\pm\sqrt{\frac{\frac{32}{25}}{\frac{18}{25}}}=\pm\sqrt{\frac{16}{9}}=\pm\frac{4}{3}$ and $\tan\left(\frac{\beta}{2}\right)=\pm\sqrt{\frac{1-\cos(\beta)}{1+\cos(\beta)}}=\pm\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\pm\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$.

Since $\frac{\alpha}{2}$ and $\frac{\beta}{2}$ must be in the second quadrant, their tangent values are both negative. Thus $\tan\left(\frac{\alpha}{2}\right)=-\frac{4}{3}$ and $\tan\left(\frac{\beta}{2}\right)=-\frac{1}{2}$.


By the sum of tangents formula, $\tan\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}=\frac{-\frac{11}{6}}{\frac{1}{3}}=-\frac{11}{2}$, which is the slope of the angle bisector.


Finally, the equation of the angle bisector is $y-5=-\frac{11}{2}(x+8)$ or $y=-\frac{11}{2}x-39$. Rearranging, we get $11x+2y+78=0$, so our sum is $a+c=11+78=\boxed{089}$. ~eevee9406

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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