Difference between revisions of "2019 AIME I Problems/Problem 3"
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Note that <math>\triangle{PQR}</math> has area <math>150</math> and is a 3-4-5 right triangle. Then, by similar triangles, the altitude from <math>B</math> to <math>QC</math> has length 3 and the altitude from <math>A</math> to <math>FP</math> has length 4, so <math>[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30</math>, meaning that <math>[ABCDEF]=\boxed{120}</math>. | Note that <math>\triangle{PQR}</math> has area <math>150</math> and is a 3-4-5 right triangle. Then, by similar triangles, the altitude from <math>B</math> to <math>QC</math> has length 3 and the altitude from <math>A</math> to <math>FP</math> has length 4, so <math>[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30</math>, meaning that <math>[ABCDEF]=\boxed{120}</math>. | ||
-Stormersyle | -Stormersyle |
Revision as of 15:21, 6 October 2019
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
.
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 3-4-5 right triangle. Then, by similar triangles, the altitude from
to
has length 3 and the altitude from
to
has length 4, so
, meaning that
.
-Stormersyle
Solution 4
Knowing that has area 150 and is a 3-4-5 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.