Difference between revisions of "2019 AIME II Problems/Problem 7"
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== Solution 2 == | == Solution 2 == | ||
+ | |||
+ | Let the diagram be set up like that in Solution 1. | ||
+ | |||
+ | By similar triangles we have | ||
+ | <cmath>\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30</cmath> | ||
+ | <cmath>\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30</cmath> | ||
+ | Thus <cmath>HI=AB-AH-IB=60</cmath> | ||
+ | |||
+ | Since <math>\bigtriangleup IHZ\sim\bigtriangleup ABC</math> and <math>\frac{HI}{AB}=\frac{1}{2}</math>, the altitude of <math>\bigtriangleup IHZ</math> from <math>Z</math> is half the altitude of <math>\bigtriangleup ABC</math> from <math>C</math>, say <math>\frac{h}{2}</math>. Also since <math>\frac{EF}{AB}=\frac{1}{8}</math>, the distance from <math>ell_C</math> to <math>AB</math> is <math>\frac{7}{8}h</math>. Therefore the altitude of <math>\bigtriangleup XYZ</math> from <math>Z</math> is | ||
+ | <cmath>\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h</cmath>. | ||
+ | |||
+ | By triangle scaling, the perimeter of <math>\bigtriangleup XYZ</math> is <math>\frac{11}{8}</math> of that of <math>\bigtriangleup ABC</math>, or | ||
+ | <cmath>\frac{11}{8}(220+180+120)=\boxed{715}</cmath> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=6|num-a=8}} | {{AIME box|year=2019|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:16, 4 January 2020
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
.
Solution
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
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.