### 2019 AIME #13 (problem request)

This problem was requested to be solved by a user. If you would like to request a problem too, please use the form on the left menu of the blog.

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 $a+b+c+d$.

Let $P=AE\cap CF$, $CP=5x$, and $BP=5y$. By $\triangle{CBP}\sim\triangle{EFP}$, we know that $EP=7x$ and $FP=7y$. By $\triangle{CAP}\sim\triangle{DFP}$ we know that $$\frac{6}{4+5y}=\frac{2}{7y} \implies y=\frac{1}{4}.$$ Hence $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$. Note that the similar triangles also give $DP=\frac{5}{3}x$ so $DE=\frac{16}{3}x$. Then, by Stewart's Theorem on $\triangle{FEP}$ and cevian $FD$, we obtain \[\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x \implies x=\frac{3\sqrt{2}}{4}.\] Finally, we get $BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}$ so the answer is $\boxed{032}$ as desired.

hi

ReplyDeletei requested this problem

solution looks short

but its actually rlly rlly hard to come up with it

but thanks for the sol

this is a multi-purpose math olympiad site

better than aops i'd say im ngl

being honest here

their name is just "art of problem solving"

but what you are doing is the REAL art

what they do is just a basic level of what ur doing

ur doing way way wayyyyy deeper

even tho you dont clearly say it

your sols immediately make me understand

how to come up with it

thanks man