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

Latest revision as of 16:51, 28 February 2025

Problem

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

$[asy] unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumcenter(C, B, E); filldraw(A--B--C--cycle, lightcyan, deepcyan); filldraw(D--E--F--cycle, lightmagenta, deepmagenta); draw(B--D, gray(0.6)); draw(C--F, gray(0.6)); draw(circumcircle(C, A, D), dashed); draw(circumcircle(C, B, E), dashed); dot("$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, $AG \cdot GD = BG \cdot GE$ .

$AG \cdot GD = BG \cdot GE$ $(AB + BG) \cdot GD = BG \cdot (GD + DE)$ $AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$ $AB \cdot GD = BG \cdot DE$ $4 \cdot GD = BG \cdot 4\sqrt{2}$ $$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)FC $and let$ FC\cap AE=K $. Let$ FK=x $. Using$ \triangle FKE\sim \triangle BKC $, <cmath>BK=\frac{5}{7}x</cmath> Using$ \triangle FDK\sim ACK $, it can be found that <cmath>3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}</cmath> This also means that$ BK=\frac{21}{4}-4=\frac{5}{4} $. It suffices to find$ KE $. It is easy to see the following: <cmath>180-\angle ABC=\angle KBC=\angle KFE</cmath> Using reverse Law of Cosines on$ \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 \boxed{\textbf{032}}$. -franchester

==Solution 4 (No <C = <DFE, no LoC)== Let$ (Error compiling LaTeX. Unknown error_msg)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 $tells us that <cmath>\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,</cmath>so$ 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)

==Solution 5== Connect$ (Error compiling LaTeX. Unknown error_msg)CF $meeting$ AE $at$ J $. We can observe that$ \triangle{ACJ}\sim \triangle{FJD} $Getting that$ \frac{AJ}{FJ}=\frac{AC}{FD}=3 $. We can also observe that$ \triangle{CBJ}\sim \triangle{EFJ} $, getting that$ \frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7} $Assume that$ BJ=5x;FJ=7x $, since$ \frac{AJ}{FJ}=3 $, we can get that$ \frac{AJ}{FJ}=\frac{AB+BJ}{FJ}=\frac{4+5x}{7x}=3 $, getting that$ x=\frac{1}{4}; BJ=\frac{5}{4}; FJ=\frac{7}{4} $Using Power of Point, we can get that$ BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ $Assume that$ DJ=5k,CJ=15k $, getting that$ JE=21k, DE=16k $Now applying Law of Cosine on two triangles,$ \triangle{ACJ};\triangle{FJE} $separately, we can get two equations$ (1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36$$ (Error compiling LaTeX. Unknown error_msg)(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49 $Since$ \angle{CJA}=\angle{FJE} $, we can use$ 15(2)-7(1) $to eliminate the$ cos$term

Then we can get that$ (Error compiling LaTeX. Unknown error_msg)5040k^2=630 $, getting$ k=\frac{\sqrt{2}}{4}$$ (Error compiling LaTeX. Unknown error_msg)BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4} $, so the desired answer is$ \frac{21\sqrt{2}+5}{4} $, which leads to the answer$ \boxed{032}$~bluesoul

==Solution 6==

First, let$ (Error compiling LaTeX. Unknown error_msg)AE $and$ CF $intersect at$ X $. Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that <cmath>\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC</cmath> By the so-called "Reverse Law of Cosines" on$ \triangle ABC $we have <cmath>\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}</cmath> Applying on$ \triangle DFE $gives <cmath>DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)</cmath> <cmath>= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}</cmath> <cmath>=32</cmath> So$ DE = 4 \sqrt{2} $, now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to$ BX $and$ XD $, which are crucial lengths in the problem. Suppose$ BX = r, XD = s $for simplicity. We have:$ \bulletCharking 15:50, 28 February 2025 (EST)\triangle AXC \sim \triangle FXD$$ (Error compiling LaTeX. Unknown error_msg)\bulletCharking 15:50, 28 February 2025 (EST)\triangle BXC \sim \triangle FXE $So <cmath>\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3</cmath> <cmath>\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}</cmath> <cmath>\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}</cmath> <cmath>\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}</cmath> So$ BE = r + s + 4 \sqrt{2} = \frac{5 + 21 \sqrt{2}}{4} $. The requested sum is$ 5 + 21 + 2 + 4 = \boxed{032}$.

~CoolJupiter

Video Solution by MOP 2024

https://youtube.com/watch?v=B7rFw05AYQ0

~r00tsOfUnity

@@ Line 1: / Line 1: @@
-==Problem 13==
+== Problem ==
 Triangle <math>ABC</math> has side lengths <math>AB=4</math>, <math>BC=5</math>, and <math>CA=6</math>. Points <math>D</math> and <math>E</math> are on ray <math>AB</math> with <math>AB<AD<AE</math>. The point <math>F \neq C</math> is a point of intersection of the circumcircles of <math>\triangle ACD</math> and <math>\triangle EBC</math> satisfying <math>DF=2</math> and <math>EF=7</math>. Then <math>BE</math> can be expressed as <math>\tfrac{a+b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers such that <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.
+==Solution 1==
 <asy>
 unitsize(20);
@@ Line 33: / Line 36: @@
 label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta);
 label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta);
-label("$k$", (B+X)/2, dir(B-X)*I, gray(0.3));
+label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3));
-label("$k\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3));
+label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3));
 </asy>
-==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 <cmath>\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.</cmath>By the Law of Cosines, <cmath>\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.</cmath>Then, <cmath>DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.</cmath>Let <math>X=\overline{AB}\cap\overline{CF}</math>, <math>a=XB</math>, and <math>b=XD</math>. Then, <cmath>XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.</cmath>However, since <math>\triangle XFD\sim\triangle XAC</math>, <math>XF=\tfrac{4+a}3</math>, but since <math>\triangle XFE\sim\triangle XBC</math>, <cmath>\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,</cmath>and the requested sum is <math>5+21+2+4=\boxed{032}</math>.
@@ Line 75: / Line 56: @@
 Let <math>G</math> be the intersection of segment <math>\overline{AE}</math> and <math>\overline{CF}</math>. Using Power of a Point with respect to <math>G</math> within <math>\omega_1</math>, we find that <math>AG \cdot GD = CG \cdot GF</math>. We can also apply Power of a Point with respect to <math>G</math> within <math>\omega_2</math> to find that <math>CG \cdot GF = BG \cdot GE</math>. Therefore, <math>AG \cdot GD = BG \cdot GE</math>.
-<math>AG \cdot GD = BG \cdot GE</math>
+<cmath>AG \cdot GD = BG \cdot GE</cmath>
+<cmath>(AB + BG) \cdot GD = BG \cdot (GD + DE)</cmath>
+<cmath>AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE</cmath>
+<cmath>AB \cdot GD = BG \cdot DE</cmath>
+<cmath>4 \cdot GD = BG \cdot 4\sqrt{2}</cmath>
+<math></math>GD = BG \cdot \sqrt{2}<math>
+Note that </math>\triangle GAC<math> is similar to </math>\triangle GFD<math>. </math>GF = \frac{BG + 4}{3}<math>. Also note that </math>\triangle GBC<math> is similar to </math>\triangle GFE<math>, which gives us </math>GF = \frac{7 \cdot BG}{5}<math>. Solving this system of linear equations, we get </math>BG = \frac{5}{4}<math>. Now, we can solve for </math>BE<math>, which is equal to </math>BG(\sqrt{2} + 1) + 4\sqrt{2}<math>. This simplifies to </math>\frac{5 + 21\sqrt{2}}{4}<math>, which means our answer is </math>\boxed{032}<math>.
+==Solution 3==
+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 \boxed{\textbf{032}}<math>.
+-franchester
+==Solution 4 (No <C = <DFE, no LoC)==
+Let </math>P=AE\cap CF<math>. Let </math>CP=5x<math> and </math>BP=5y<math>; from </math>\triangle{CBP}\sim\triangle{EFP}<math> we have </math>EP=7x<math> and </math>FP=7y<math>. From </math>\triangle{CAP}\sim\triangle{DFP}<math> we have </math>\frac{6}{4+5y}=\frac{2}{7y}<math> giving </math>y=\frac{1}{4}<math>. So </math>BP=\frac{5}{4}<math> and </math>FP=\frac{7}{4}<math>. These similar triangles also gives us </math>DP=\frac{5}{3}x<math> so </math>DE=\frac{16}{3}x<math>. Now, Stewart's Theorem on </math>\triangle{FEP}<math> and cevian </math>FD<math> tells us that <cmath>\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,</cmath>so </math>x=\frac{3\sqrt{2}}{4}<math>. Then </math>BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}<math> so the answer is </math>\boxed{032}<math> as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)
+==Solution 5==
+Connect </math>CF<math> meeting </math>AE<math> at </math>J<math>. We can observe that </math>\triangle{ACJ}\sim \triangle{FJD}<math> Getting that </math>\frac{AJ}{FJ}=\frac{AC}{FD}=3<math>. We can also observe that </math>\triangle{CBJ}\sim \triangle{EFJ}<math>, getting that </math>\frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7}<math>
+Assume that </math>BJ=5x;FJ=7x<math>, since </math>\frac{AJ}{FJ}=3<math>, we can get that </math>\frac{AJ}{FJ}=\frac{AB+BJ}{FJ}=\frac{4+5x}{7x}=3<math>, getting that </math>x=\frac{1}{4}; BJ=\frac{5}{4}; FJ=\frac{7}{4}<math>
+Using Power of Point, we can get that </math>BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ<math> Assume that </math>DJ=5k,CJ=15k<math>, getting that </math>JE=21k, DE=16k<math>
+Now applying Law of Cosine on two triangles, </math>\triangle{ACJ};\triangle{FJE}<math> separately, we can get two equations
+</math>(1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36<math>
+</math>(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49<math>
+Since </math>\angle{CJA}=\angle{FJE}<math>, we can use </math>15(2)-7(1)<math> to eliminate the </math>cos<math> term
+Then we can get that </math>5040k^2=630<math>, getting </math>k=\frac{\sqrt{2}}{4}<math>
+</math>BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4}<math>, so the desired answer is </math>\frac{21\sqrt{2}+5}{4}<math>, which leads to the answer </math>\boxed{032}<math>
+~bluesoul
+==Solution 6==
+First, let </math>AE<math> and </math>CF<math> intersect at </math>X<math>. Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that
+<cmath>\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC</cmath>
+By the so-called "Reverse Law of Cosines" on </math>\triangle ABC<math> we have
+<cmath>\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}</cmath>
+Applying on </math>\triangle DFE<math> gives
+<cmath>DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)</cmath>
+<cmath>= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}</cmath>
+<cmath>=32</cmath>
+So </math>DE = 4 \sqrt{2}<math>, now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to </math>BX<math> and </math>XD<math>, which are crucial lengths in the problem. Suppose </math>BX = r, XD = s<math> for simplicity. We have:
-<math>(AB + BG) \cdot GD = BG \cdot (GD + DE)</math>
+</math>\bulletCharking 15:50, 28 February 2025 (EST)\triangle AXC \sim \triangle FXD<math>
+</math>\bulletCharking 15:50, 28 February 2025 (EST)\triangle BXC \sim \triangle FXE<math>
-<math>AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE</math>
+So
+<cmath>\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3</cmath>
+<cmath>\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}</cmath>
+<cmath>\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}</cmath>
+<cmath>\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}</cmath>
+So </math>BE = r + s + 4 \sqrt{2} = \frac{5 + 21 \sqrt{2}}{4}<math>. The requested sum is </math>5 + 21 + 2 + 4 = \boxed{032}$.
-<math>AB \cdot GD = BG \cdot DE</math>
+~CoolJupiter
-<math>4 \cdot GD = BG \cdot 4\sqrt{2}</math>
+==Video Solution by MOP 2024==
+https://youtube.com/watch?v=B7rFw05AYQ0
-<math>GD = BG \cdot \sqrt{2}</math>
+~r00tsOfUnity
-Note that <math>\triangle GAC</math> is similar to <math>\triangle GFD</math>. <math>GF = \frac{BG + 4}{3}</math>. Also note that <math>\triangle GBC</math> is similar to <math>\triangle GFE</math>, which gives us <math>GF = \frac{7 \cdot BG}{5}</math>. Solving this system of linear equations, we get <math>BG = \frac{5}{4}</math>. Now, we can solve for <math>BE</math>, which is equal to <math>BG(\sqrt{2} + 1) + 4\sqrt{2}</math>. This simplifies to <math>\frac{5 + 21\sqrt{2}}{4}</math>, which means our answer is <math>\boxed{032}</math>.
+== See Also ==
-==Solution 3==
-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
-==Solution 4 (No <C = <DFE, no LoC)==
-Let <math>P=AE\cap CF</math>. Let <math>CP=5x</math> and <math>BP=5y</math>; from <math>\triangle{CBP}\sim\triangle{EFP}</math> we have <math>EP=7x</math> and <math>FP=7y</math>. From <math>\triangle{CAP}\sim\triangle{DFP}</math> we have <math>\frac{6}{4+5y}=\frac{2}{7y}</math> giving <math>y=\frac{1}{4}</math>. So <math>BP=\frac{5}{4}</math> and <math>FP=\frac{7}{4}</math>. These similar triangles also gives us <math>DP=\frac{5}{3}x</math> so <math>DE=\frac{16}{3}x</math>. Now, Stewart's Theorem on <math>\triangle{FEP}</math> and cevian <math>FD</math> tells us that <cmath>\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,</cmath>so <math>x=\frac{3\sqrt{2}}{4}</math>. Then <math>BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}</math> so the answer is <math>\boxed{032}</math> as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)
-==See Also==
 {{AIME box|year=2019|n=I|num-b=12|num-a=14}}
+{{MAA Notice}}
 [[Category:Intermediate Geometry Problems]]
-{{MAA Notice}}

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

Page

Toolbox

Search