Difference between revisions of "2019 AIME II Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (Added in diagram.) |
MRENTHUSIASM (talk | contribs) (→Diagram: Converted diagram to asy.) |
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==Diagram== | ==Diagram== | ||
− | [ | + | <asy> |
+ | /* 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 (by Geometry Expressions) | ~MRENTHUSIASM (by Geometry Expressions) | ||
Revision as of 10:11, 1 October 2021
Contents
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 (by Geometry Expressions)
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
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.