Difference between revisions of "2019 AIME II Problems/Problem 7"
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+ | ==Solution== | ||
+ | [[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> | ||
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==See Also== | ==See Also== |
Revision as of 07:40, 11 June 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
Notation shown on diagram. By similar triangles we have
So
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.