Difference between revisions of "2019 AIME II Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (→Diagram: Converted diagram to asy.) |
MRENTHUSIASM (talk | contribs) m (→Diagram) |
||
Line 44: | Line 44: | ||
label("$\ell_C$",L+9/8*(K-L),1.5*dir(A--B)); | label("$\ell_C$",L+9/8*(K-L),1.5*dir(A--B)); | ||
</asy> | </asy> | ||
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM |
==Solution 1== | ==Solution 1== |
Revision as of 11:11, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.