Difference between revisions of "2019 AIME II Problems/Problem 7"
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Revision as of 14:38, 4 January 2024
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
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
Solution 4 (Way too short)
Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.
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.