Difference between revisions of "2019 AIME II Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (Added in diagram.) |
MRENTHUSIASM (talk | contribs) (I decided to undo: The changes in the diagram section SHOULD NOT label more than the problem statement says. We should leave for the readers to assign points.) |
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==Diagram== | ==Diagram== | ||
− | [ | + | <asy> |
− | ~MRENTHUSIASM | + | /* Made by MRENTHUSIASM */ |
+ | size(350); | ||
+ | |||
+ | pair A, B, C, D, E, F, G, H, I, J, K, L; | ||
+ | B = origin; | ||
+ | C = (220,0); | ||
+ | A = intersectionpoints(Circle(B,120),Circle(C,180))[0]; | ||
+ | D = A+1/4*(B-A); | ||
+ | E = A+1/4*(C-A); | ||
+ | F = B+1/4*(A-B); | ||
+ | G = B+1/4*(C-B); | ||
+ | H = C+1/8*(A-C); | ||
+ | I = C+1/8*(B-C); | ||
+ | J = extension(D,E,F,G); | ||
+ | K = extension(F,G,H,I); | ||
+ | L = extension(H,I,D,E); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(J+9/8*(K-J)--K+9/8*(J-K),dashed); | ||
+ | draw(L+9/8*(K-L)--K+9/8*(L-K),dashed); | ||
+ | draw(J+9/8*(L-J)--L+9/8*(J-L),dashed); | ||
+ | draw(D--E^^F--G^^H--I,red); | ||
+ | dot("$B$",B,1.5SW,linewidth(4)); | ||
+ | dot("$C$",C,1.5SE,linewidth(4)); | ||
+ | dot("$A$",A,1.5N,linewidth(4)); | ||
+ | dot(D,linewidth(4)); | ||
+ | dot(E,linewidth(4)); | ||
+ | dot(F,linewidth(4)); | ||
+ | dot(G,linewidth(4)); | ||
+ | dot(H,linewidth(4)); | ||
+ | dot(I,linewidth(4)); | ||
+ | dot(J,linewidth(4)); | ||
+ | dot(K,linewidth(4)); | ||
+ | dot(L,linewidth(4)); | ||
+ | label("$55$",midpoint(D--E),S,red); | ||
+ | label("$45$",midpoint(F--G),dir(55),red); | ||
+ | label("$15$",midpoint(H--I),dir(160),red); | ||
+ | label("$\ell_A$",J+9/8*(L-J),1.5*dir(B--C)); | ||
+ | label("$\ell_B$",K+9/8*(J-K),1.5*dir(C--A)); | ||
+ | label("$\ell_C$",L+9/8*(K-L),1.5*dir(A--B)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
==Solution 1== | ==Solution 1== | ||
Line 34: | Line 74: | ||
~ Nafer | ~ Nafer | ||
+ | |||
+ | ==Solution 3== | ||
+ | [[File:2019 AIME II 7.png|450px|right]] | ||
+ | Notation shown on diagram. By similar triangles we have | ||
+ | <cmath>k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},</cmath> | ||
+ | <cmath>k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},</cmath> | ||
+ | <cmath>k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.</cmath> | ||
+ | So, <cmath>\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,</cmath> | ||
+ | <cmath>\frac{FY}{BC} = \frac{FE'}{AC} = \frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.</cmath> | ||
+ | <cmath>k = \frac{ZY}{BC} = \frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)</cmath> | ||
+ | <cmath>k = 2 - k_1 - k_2 - k_3 = 2 - \frac{1}{4} - \frac{1}{4} - \frac{1}{8} = \frac{11}{8}.</cmath> | ||
+ | <cmath>\frac{ZY+YX +XZ}{BC +AB + AC} = k \implies ZY + YX + XZ =\frac{11}{8} (220 + 120 + 180) = \boxed {715}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==See Also== | ==See Also== |
Revision as of 22:25, 4 November 2022
Problem
Triangle has side lengths , and . Lines , and are drawn parallel to , and , respectively, such that the intersections of , and with the interior of are segments of lengths , and , respectively. Find the perimeter of the triangle whose sides lie on lines , and .
Diagram
~MRENTHUSIASM
Solution 1
Let the points of intersection of with divide the sides into consecutive segments . Furthermore, let the desired triangle be , with closest to side , closest to side , and closest to side . Hence, the desired perimeter is since , , and .
Note that , so using similar triangle ratios, we find that , , , and .
We also notice that and . Using similar triangles, we get that Hence, the desired perimeter is -ktong
Solution 2
Let the diagram be set up like that in Solution 1.
By similar triangles we have Thus
Since and , the altitude of from is half the altitude of from , say . Also since , the distance from to is . Therefore the altitude of from is .
By triangle scaling, the perimeter of is of that of , or
~ Nafer
Solution 3
Notation shown on diagram. By similar triangles we have So, vladimir.shelomovskii@gmail.com, vvsss
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.