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.


  1. schukkayapally1/21/2022

    i 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


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