2019 AIME I Problems/Problem 13

Revision as of 15:29, 15 March 2019 by Franchester (talk | contribs) (Solution)

Problem 13

Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

Solution 1

Define $\omega_1$ to be the circumcircle of $\triangle ACD$ and $\omega_2$ to be the circumcircle of $\triangle EBC$.

Because of exterior angles,

$\angle ACB = \angle CBE - \angle CAD$

But $\angle CBE = \angle CFE$ because $CBFE$ is cyclic. In addition, $\angle CAD = \angle CFD$ because $CAFD$ is cyclic. Therefore, $\angle ACB = \angle CFE - \angle CFD$. But $\angle CFE - \angle CFD = \angle DFE$, so $\angle ACB = \angle DFE$. Using Law of Cosines on $\triangle ABC$, we can figure out that $\cos(\angle ACB) = \frac{3}{4}$. Since $\angle ACB = \angle DFE$, $\cos(\angle DFE) = \frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\triangle DEF$ to find that $DE = 4\sqrt{2}$.

Let $G$ be the intersection of segment $\overline{AE}$ and $\overline{CF}$. Using Power of a Point with respect to $G$ within $\omega_1$, we find that $AG \cdot GD = CG \cdot GF$. We can also apply Power of a Point with respect to $G$ within $\omega_2$ to find that $CG \cdot GF = BG \cdot GE$. Therefore, $AG \cdot GD = BG \cdot GE$.

$AG \cdot GD = BG \cdot GE$

$(AB + BG) \cdot GD = BG \cdot (GD + DE)$

$AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$

$AB \cdot GD = BG \cdot DE$

$4 \cdot GD = BG \cdot 4\sqrt{2}$

$GD = BG \cdot \sqrt{2}$

Note that $\triangle GAC$ is similar to $\triangle GFD$. $GF = \frac{BG + 4}{3}$. Also note that $\triangle GBC$ is similar to $\triangle GFE$, which gives us $GF = \frac{7 \cdot BG}{5}$. Solving this system of linear equations, we get $BG = \frac{5}{4}$. Now, we can solve for $BE$, which is equal to $BG(\sqrt{2} + 1) + 4\sqrt{2}$. This simplifies to $\frac{5 + 21\sqrt{2}}{4}$, which means our answer is $\boxed{032}$.

Solution 2

Construct $FC$ and let $FC\cap AE=K$. Let $FK=x$. Using $\triangle FKE\sim \triangle BKC$, \[BK=\frac{5}{7}x\] Using $\triangle FDK\sim ACK$, it can be found taht \[3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}\] This also means that $BK=\frac{21}{4}-4=\frac{5}{4}$. It suffices to find $KE$. It is easy to see the following: \[180-\angle ABC=\angle KBC=\angle KFE\] Using reverse Law of Cosines on $\triangle ABC$, $\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}$. Using Law of Cosines on $\triangle EFK$ gives $KE=\frac{21\sqrt 2}{4}$, so $BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}$.

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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

Invalid username
Login to AoPS