Difference between revisions of "2019 AIME I Problems/Problem 13"
Icematrix2 (talk | contribs) |
R00tsofunity (talk | contribs) |
||
(18 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==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>. | 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== | ==Solution 1== | ||
<asy> | <asy> | ||
− | + | unitsize(20); | |
− | pair A, B, C, D, | + | 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); | |
− | D=B | + | F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; |
− | + | X = extension(A, E, C, F); | |
− | F=intersectionpoints( | + | O1 = circumcenter(C, A, D); |
− | X=extension(A, | + | O2 = circumcenter(C, B, E); |
− | + | filldraw(A--B--C--cycle, lightcyan, deepcyan); | |
− | draw(circumcircle(A | + | 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, | + | dot("$A$", A, dir(A-O1)); |
− | dot("$B$", B, | + | dot("$B$", B, dir(240)); |
− | dot("$C$", C, | + | dot("$C$", C, dir(120)); |
− | dot("$D$", D, | + | dot("$D$", D, dir(40)); |
− | dot("$E$", | + | dot("$E$", E, dir(E-O2)); |
− | dot("$F$", F, | + | 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> | </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>. | 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 55: | Line 70: | ||
==Solution 3== | ==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 | -franchester | ||
+ | |||
==Solution 4 (No <C = <DFE, no LoC)== | ==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) | 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== | ==Solution 5== | ||
− | <math>BE^2=\frac{ | + | 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== | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtube.com/watch?v=B7rFw05AYQ0 | ||
+ | |||
+ | ~r00tsOfUnity | ||
==See Also== | ==See Also== |
Latest revision as of 19:44, 20 December 2023
Contents
Problem
Triangle has side lengths , , and . Points and are on ray with . The point is a point of intersection of the circumcircles of and satisfying and . Then can be expressed as , where , , , and are positive integers such that and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Notice that By the Law of Cosines, Then, Let , , and . Then, However, since , , but since , and the requested sum is .
(Solution by TheUltimate123)
Solution 2
Define to be the circumcircle of and to be the circumcircle of .
Because of exterior angles,
But because is cyclic. In addition, because is cyclic. Therefore, . But , so . Using Law of Cosines on , we can figure out that . Since , . We are given that and , so we can use Law of Cosines on to find that .
Let be the intersection of segment and . Using Power of a Point with respect to within , we find that . We can also apply Power of a Point with respect to within to find that . Therefore, .
Note that is similar to . . Also note that is similar to , which gives us . Solving this system of linear equations, we get . Now, we can solve for , which is equal to . This simplifies to , which means our answer is .
Solution 3
Construct and let . Let . Using , Using , it can be found that This also means that . It suffices to find . It is easy to see the following: Using reverse Law of Cosines on , . Using Law of Cosines on gives , so . -franchester
Solution 4 (No <C = <DFE, no LoC)
Let . Let and ; from we have and . From we have giving . So and . These similar triangles also gives us so . Now, Stewart's Theorem on and cevian tells us that so . Then so the answer is as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)
Solution 5
Connect meeting at . We can observe that Getting that . We can also observe that , getting that
Assume that , since , we can get that , getting that
Using Power of Point, we can get that Assume that , getting that
Now applying Law of Cosine on two triangles, separately, we can get two equations
Since , we can use to eliminate the term
Then we can get that , getting
, so the desired answer is , which leads to the answer
~bluesoul
Solution 6
Nice problem!
First, let and intersect at . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that By the so-called "Reverse Law of Cosines" on we have Applying on gives So , 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 and , which are crucial lengths in the problem. Suppose for simplicity. We have:
So So . The requested sum is .
~CoolJupiter
Video Solution by MOP 2024
https://youtube.com/watch?v=B7rFw05AYQ0
~r00tsOfUnity
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.