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

Latest revision as of 19:44, 20 December 2023

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, $AG \cdot GD = BG \cdot GE$ .

$AG \cdot GD = BG \cdot GE$

$(AB + BG) \cdot GD = BG \cdot (GD + DE)$

$AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$

$AB \cdot GD = BG \cdot DE$

$4 \cdot GD = BG \cdot 4\sqrt{2}$

$GD = BG \cdot \sqrt{2}$

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 $BE=\frac{5+21\sqrt 2}{4}\to \boxed{\textbf{032}}$ . -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)

Solution 5

Connect $CF$ 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

$(1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36$

$(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49$

Since $\angle{CJA}=\angle{FJE}$ , we can use $15(2)-7(1)$ to eliminate the $cos$ term

Then we can get that $5040k^2=630$ , getting $k=\frac{\sqrt{2}}{4}$

$BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4}$ , so the desired answer is $\frac{21\sqrt{2}+5}{4}$ , which leads to the answer $\boxed{032}$

~bluesoul

Solution 6

Nice problem!

First, let $AE$ and $CF$ intersect at $X$ . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that $\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC$ By the so-called "Reverse Law of Cosines" on $\triangle ABC$ we have $\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}$ Applying on $\triangle DFE$ gives $DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)$ $= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}$ $=32$ 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:

$\bullet~~~~\triangle AXC \sim \triangle FXD$ $\bullet~~~~\triangle BXC \sim \triangle FXE$

So $\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3$ $\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}$ $\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}$ $\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}$ 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 70: / Line 70: @@
 ==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>.
+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)
@@ Line 92: / Line 93: @@
 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{32}</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 (Stewart's Theorem Bash)==
+==Solution 6==
+Nice problem!
+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>\bullet~~~~\triangle AXC \sim \triangle FXD</math>
+<math>\bullet~~~~\triangle BXC \sim \triangle FXE</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>.
+~CoolJupiter
-Use the diagram in solution 1 for reference, except ignore the labelled lengths (some differ in terms of variable names and quantities in this solution). Anyway, <math>CX = 5x, XE = 7x, BX = 5a, FX = \frac{4+5a}{3}, DX = \frac{5x}{3}, DE = \frac{16x}{3}</math>. Now using Stewart's Theorem twice, on <math>\triangle{CAX}</math> and <math>\triangle{FEX}</math>, you get two equations. I simplified the equation for <math>\triangle{FEX}</math> to get <math>\frac{560x^2}{9} = \frac{161}{3} + \frac{16(25a^2+40a+16)}{9}</math> since <math>x \neq 0</math>. I also simplified the equation for <math>\triangle{CAX}</math> to get <math>4a \cdot (4+5a) + 5 (4+5a) = 36a+20x^2 \implies 4a^2+a+4 = 4x^2</math>.
+==Video Solution by MOP 2024==
+https://youtube.com/watch?v=B7rFw05AYQ0
-~First
+~r00tsOfUnity
 ==See Also==

2019 AIME I (Problems • Answer Key • Resources)
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