Difference between revisions of "2019 AIME II Problems/Problem 7"
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==Problem== | ==Problem== | ||
Triangle <math>ABC</math> has side lengths <math>AB=120,BC=220</math>, and <math>AC=180</math>. Lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> are drawn parallel to <math>\overline{BC},\overline{AC}</math>, and <math>\overline{AB}</math>, respectively, such that the intersections of <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> with the interior of <math>\triangle ABC</math> are segments of lengths <math>55,45</math>, and <math>15</math>, respectively. Find the perimeter of the triangle whose sides lie on lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math>. | Triangle <math>ABC</math> has side lengths <math>AB=120,BC=220</math>, and <math>AC=180</math>. Lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> are drawn parallel to <math>\overline{BC},\overline{AC}</math>, and <math>\overline{AB}</math>, respectively, such that the intersections of <math>\ell_A,\ell_B</math>, and <math>\ell_C</math> with the interior of <math>\triangle ABC</math> are segments of lengths <math>55,45</math>, and <math>15</math>, respectively. Find the perimeter of the triangle whose sides lie on lines <math>\ell_A,\ell_B</math>, and <math>\ell_C</math>. | ||
+ | |||
+ | ==Diagram== | ||
+ | [[File:2019 AIME II Problem 7.png|center|750px]] | ||
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
==Solution 1== | ==Solution 1== |
Revision as of 13:54, 3 July 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 |
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