AIME 1990 · 第 7 题

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

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

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

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)

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