Difference between revisions of "2019 AIME I Problems/Problem 3"
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==Problem 3== | ==Problem 3== | ||
In <math>\triangle PQR</math>, <math>PR=15</math>, <math>QR=20</math>, and <math>PQ=25</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, and points <math>E</math> and <math>F</math> lie on <math>\overline{PR}</math>, with <math>PA=QB=QC=RD=RE=PF=5</math>. Find the area of hexagon <math>ABCDEF</math>. | In <math>\triangle PQR</math>, <math>PR=15</math>, <math>QR=20</math>, and <math>PQ=25</math>. Points <math>A</math> and <math>B</math> lie on <math>\overline{PQ}</math>, points <math>C</math> and <math>D</math> lie on <math>\overline{QR}</math>, and points <math>E</math> and <math>F</math> lie on <math>\overline{PR}</math>, with <math>PA=QB=QC=RD=RE=PF=5</math>. Find the area of hexagon <math>ABCDEF</math>. | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | dot((0,0)); | ||
+ | dot((15,0)); | ||
+ | dot((15,20)); | ||
+ | draw((0,0)--(15,0)--(15,20)--cycle); | ||
+ | dot((5,0)); | ||
+ | dot((10,0)); | ||
+ | dot((15,5)); | ||
+ | dot((15,15)); | ||
+ | dot((3,4)); | ||
+ | dot((12,16)); | ||
+ | draw((5,0)--(3,4)); | ||
+ | draw((10,0)--(15,5)); | ||
+ | draw((12,16)--(15,15)); | ||
+ | draw(rightanglemark((0,0)--(15,0)--(15,20))); | ||
+ | </asy> | ||
==Solution 1== | ==Solution 1== |
Revision as of 22:23, 5 March 2020
Contents
[hide]Problem 3
In ,
,
, and
. Points
and
lie on
, points
and
lie on
, and points
and
lie on
, with
. Find the area of hexagon
.
Diagram
Solution 1
We know the area of the hexagon to be
. Since
, we know that
is a right triangle. Thus the area of
is
. Another way to compute the area is
Then the area of
. Preceding in a similar fashion for
, the area of
is
. Since
, the area of
. Thus our desired answer is
Solution 2
Let be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that
, and
. Using the shoelace theorem, the area is
.
Shoelace theorem:Suppose the polygon
has vertices
,
, ... ,
, listed in clockwise order. Then the area of
is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
.
Solution 3 (Easiest, uses only basic geometry too)
Note that has area
and is a
-
-
right triangle. Then, by similar triangles, the altitude from
to
has length
and the altitude from
to
has length
, so
, meaning that
.
-Stormersyle
Solution 4
Knowing that has area
and is a
-
-
triangle, we can find the area of the smaller triangles
,
, and
and subtract them from
to obtain our answer. First off, we know
has area
since it is a right triangle. To the find the areas of
and
, we can use Law of Cosines (
) to find the lengths of
and
, respectively. Computing gives
and
. Now, using Heron's Formula, we find
and
. Adding these and subtracting from
, we get
-Starsher
Video Solution
https://www.youtube.com/watch?v=4jOfXNiQ6WM
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.