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Difference between revisions of "2019 AIME I Problems/Problem 13"

(Note: I put my solution as Solution 1 as I feel like it has a diagram and is formatted well.)
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==Solution 3==
 
==Solution 3==
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Construct <math>FC</math> and let <math>FC\cap AE=K</math>. Let <math>FK=x</math>. Using <math>\triangle FKE\sim \triangle BKC</math>, <cmath>BK=\frac{5}{7}x</cmath> Using <math>\triangle FDK\sim ACK</math>, it can be found taht <cmath>3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}</cmath> This also means that <math>BK=\frac{21}{4}-4=\frac{5}{4}</math>. It suffices to find <math>KE</math>. It is easy to see the following: <cmath>180-\angle ABC=\angle KBC=\angle KFE</cmath> Using reverse Law of Cosines on <math>\triangle ABC</math>, <math>\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}</math>. Using Law of Cosines on <math>\triangle EFK</math> gives <math>KE=\frac{21\sqrt 2}{4}</math>, so <math>BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}</math>.
+
Construct <math>FC</math> and let <math>FC\cap AE=K</math>. Let <math>FK=x</math>. Using <math>\triangle FKE\sim \triangle BKC</math>, <cmath>BK=\frac{5}{7}x</cmath> Using <math>\triangle FDK\sim ACK</math>, it can be found that <cmath>3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}</cmath> This also means that <math>BK=\frac{21}{4}-4=\frac{5}{4}</math>. It suffices to find <math>KE</math>. It is easy to see the following: <cmath>180-\angle ABC=\angle KBC=\angle KFE</cmath> Using reverse Law of Cosines on <math>\triangle ABC</math>, <math>\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}</math>. Using Law of Cosines on <math>\triangle EFK</math> gives <math>KE=\frac{21\sqrt 2}{4}</math>, so <math>BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}</math>.
 
-franchester
 
-franchester
  

Revision as of 22:58, 18 March 2019

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

[asy] size(10cm); pair A, B, C, D, EE, F, X; B=dir(270-aCos(9/16)); C=dir(270+aCos(9/16)); A=intersectionpoint(circle((0, 0), 1), (B+0.01*(1, 3sqrt(7))) -- (B+100*(1, 3sqrt(7)))); D=B-5/16*(sqrt(2)+1)*(A-B); EE=B-(5+21*sqrt(2))/16*(A-B); F=intersectionpoints(circumcircle(A, C, D), circumcircle(B, C, EE))[0]; X=extension(A, B, C, F); draw(B -- C -- A -- EE -- F -- C); draw(D -- F); draw(circumcircle(A, C, D)); draw(circumcircle(C, EE, F)); dot("$A$", A, N); dot("$B$", B, NW); dot("$C$", C, E); dot("$D$", D, SW); dot("$E$", EE, SW); dot("$F$", F, W); [/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

See Also

2019 AIME I (Problems • Answer Key • Resources)
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

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