Difference between revisions of "2019 AIME I Problems/Problem 13"
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− | ==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 | ||
==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); | |
− | + | 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> | </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>. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=12|num-a=14}} | {{AIME box|year=2019|n=I|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:25, 16 February 2021
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)
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.