AIME 1999 · 第 14 题

Solution

Solution 1

Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.

Let $BE = x, CF = y,$ and $AD = z$. We have that

$$\begin{aligned}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{aligned}$$ We can then use the tool of calculating area in two ways

$$\begin{aligned}[ABC]&=[PAB]+[PBC]+[PCA]\\ &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\ &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{aligned}$$ On the other hand,

$$\begin{aligned}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{aligned}$$ We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot:

$$\begin{aligned}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\ z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\ y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{aligned}$$ Adding $(1) + (2) + (3)$ gives

$$\begin{aligned}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{aligned}$$ Recall that we found that $[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84$. Plugging in $13z+14x+15y=295$, we get $\tan\theta=\frac{168}{295}$, giving us $\boxed{463}$ for an answer.

Solution 2

Let $AB=c$, $BC=a$, $AC=b$, $PA=x$, $PB=y$, and $PC=z$.

So by the Law of Cosines, we have:

$$\begin{aligned}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\ y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\ z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{aligned}$$ Adding these equations and rearranging, we have:

$$a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)$$ Now $[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84$, by Heron's formula.

Now the area of a triangle, $[A] = \frac {mn\sin{\beta}}{2}$, where $m$ and $n$ are sides on either side of an angle, $\beta$. So,

$$\begin{aligned}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{aligned}$$ Adding these equations yields:

$$\begin{aligned}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{aligned}$$ Dividing $(2)$ by $(1)$, we have:

$$\begin{aligned}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\ \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{aligned}$$ Thus, $m + n = 168 + 295 = \boxed{463}$.

Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by

$$cot(\theta)=\frac{a^2+b^2+c^2}{4\Delta}$$

Solution 3

Let $\angle{PAB} = \angle{PBC} = \angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have

$$\begin{aligned} b^2 &= a^2 + 169 - 26a \cos x \\ c^2 &= b^2 + 196 - 28b \cos x \\ a^2 &= c^2 + 225 - 30c \cos x. \end{aligned}$$ Add the three equations up and rearrange to obtain

$$(13a + 14b + 15c) \cos x = 295.$$ Also, using $[ABC] = \frac{1}{2}ab \sin \angle C$ we have

$$[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.$$ Divide the two equations to obtain $\tan x = \frac{168}{295} \iff \boxed{463}.~\square$

Solution 4 (Law of sines)

Firstly, denote angles $ABC$, $BCA$, and $CAB$ as $B$, $A$, and $C$ respectively. Let $\angle{PAB}=x$. Notice that by angle chasing that $\angle{BPC}=180-C$ and $\angle{BPA}=180-B$. Using the nice properties of the 13-14-15 triangle, we have $\sin B = \frac{12}{13}$ and $\sin C = \frac{4}{5}$. $\cos C$ is easily computed, so we have $\cos C=\frac{3}{5}$.

Using Law of Sines,

$$\begin{aligned} \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} \end{aligned}$$ hence,

$$\begin{aligned} \frac{BP}{\sin x} &= \frac{13}{\sin B} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} \end{aligned}$$ Now, computation carries the rest.

$$\begin{aligned} \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C} \\ \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\ 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\ 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\ 169 \sin x &= 168 \cos x - 126 \sin x \\ 295 \sin x &= 168 \cos x \\ \tan x &= \frac{168}{295} \end{aligned}$$ Extracting yields $168 + 295 = \boxed{463}$.

Solution 5 (Coordinate Bash)

Plot $\triangle ABC$ on the coordinate plane with $B=(0,0), C=(14,0), A=(5,12).$ Let $\angle PAB=\angle PBC=\angle PCA=a$, and let $P=(x,y)$. For convenience, define $\tan a = k$.

Since $\angle PBC=a$, the slope of $PB$ is $k$, so $PB:\ y=kx.$

Let $\angle ADB$ be the angle that line $AB$ makes with the $x$-axis. Since $\tan(\angle ADB)=\frac{12}{5}$, the slope of $PA$ is

$$\tan(\angle ADB+a)=\frac{\frac{12}{5}+k}{1-\frac{12}{5}k}=\frac{12+5k}{5-12k}.$$ Thus the equation of $PA$ is

$$y-12=\frac{12+5k}{5-12k}(x-5),$$ which simplifies to

$$PA:\ y=\frac{12x+5kx-169k}{5-12k}.$$ Similarly, since $\tan(\angle ECA)=-\frac{4}{3}$, the slope of $PC$ is

$$\tan(\angle ECA+a)=\frac{-\frac{4}{3}+k}{1+\frac{4}{3}k}=\frac{3k-4}{4k+3}.$$ Thus the equation of $PC$ is

$$PC:\ y=\frac{3k-4}{4k+3}(x-14),$$ or equivalently,

$$PC:\ y=\frac{3kx-4x-42k+56}{4k+3}.$$ Setting the equations of $PB$ and $PA$ equal gives

$$\frac{12x+5kx-169k}{5-12k}=kx.$$ Clearing denominators yields

$$12x+5kx-169k=5kx-12k^2x,$$ which simplifies to

$$12x-169k=-12k^2x.$$ Next, setting the equations of $PB$ and $PC$ equal gives

$$\frac{3kx-4x-42k+56}{4k+3}=kx.$$ Clearing denominators yields

$$3kx-4x-42k+56=4k^2x+3kx,$$ or

$$-4x-42k+56=4k^2x.$$ Multiplying by $-3$ gives

$$12x+126k-168=-12k^2x.$$ Equating the right-hand sides of the two derived equations gives

$$12x+126k-168=12x-169k,$$ so $295k=168,$ and hence $k=\frac{168}{295}.$

Because the problem asks for $m+n$, where $k=\frac{m}{n}$, the final answer is $168+295=\boxed{463}.$

~Voidling