2019 AIME I Problems/Problem 13

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB. 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

$[asy] unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumcenter(C, B, E); filldraw(A--B--C--cycle, lightcyan, deepcyan); filldraw(D--E--F--cycle, lightmagenta, deepmagenta); draw(B--D, gray(0.6)); draw(C--F, gray(0.6)); draw(circumcircle(C, A, D), dashed); draw(circumcircle(C, B, E), dashed); dot("A", A, dir(A-O1)); dot("B", B, dir(240)); dot("C", C, dir(120)); dot("D", D, dir(40)); dot("E", E, dir(E-O2)); dot("F", F, dir(270)); dot("X", X, dir(140)); label("6", (C+A)/2, dir(C-A)*I, deepcyan); label("5", (C+B)/2, dir(B-C)*I, deepcyan); label("4", (A+B)/2, dir(A-B)*I, deepcyan); label("7", (F+E)/2, dir(F-E)*I, deepmagenta); label("2", (F+D)/2, dir(D-F)*I, deepmagenta); label("4\sqrt{2}", (D+E)/2, dir(E-D)*I, deepmagenta); label("a", (B+X)/2, dir(B-X)*I, gray(0.3)); label("a\sqrt{2}", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$

Notice that $$\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.$$Then, $$DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.$$Let $X=\overline{AB}\cap\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.$$However, since $\triangle XFD\sim\triangle XAC$, $XF=\tfrac{4+a}3$, but since $\triangle XFE\sim\triangle XBC$, $$\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,$$and the requested sum is $5+21+2+4=\boxed{032}$.

(Solution by TheUltimate123)

Solution 2

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 3

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 that $$3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}$$ 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 $BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}$. -franchester

Solution 4 (No <C = <DFE, no LoC)

Let $P=AE\cap CF$. Let $CP=5x$ and $BP=5y$; from $\triangle{CBP}\sim\triangle{EFP}$ we have $EP=7x$ and $FP=7y$. From $\triangle{CAP}\sim\triangle{DFP}$ we have $\frac{6}{4+5y}=\frac{2}{7y}$ giving $y=\frac{1}{4}$. So $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$. These similar triangles also gives us $DP=\frac{5}{3}x$ so $DE=\frac{16}{3}x$. Now, Stewart's Theorem on $\triangle{FEP}$ and cevian $FD$ tells us that $$\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,$$so $x=\frac{3\sqrt{2}}{4}$. Then $BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}$ so the answer is $\boxed{032}$ as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)