Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1999 AIME Problems/Problem 14

Revision as of 13:58, 23 October 2018 by Mathdummy (talk | contribs) (→‎Solution 2)

Problem

Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$

Solution

[asy] real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,NW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); [/asy]

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.

[asy] import olympiad; real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,SSW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); /* constructing D,E,F as foot of perps from P */ pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); D(MP("D",D,NE)--P--MP("E",E,SSW),dashed);D(P--MP("F",F),dashed); D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); [/asy]

Let $BE = x, CF = y,$ and $AD = z$. We have that \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*} We can then use the tool of calculating area in two ways \begin{align*}[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{align*} On the other hand, \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{align*} 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{align}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{align} Adding $(1) + (2) + (3)$ gives \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{align*} 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{align*}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{align*} 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{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*} Adding these equations yields: \begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*} Dividing $(2)$ by $(1)$, we have: \begin{align*}\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{align*} 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 \[cot(\theta)=\frac{a^2+b^2+c^2}{4\Delta}\]

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems/Problem_14&oldid=98246"