Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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+ | ==Video Solution 2 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=6I3ZNpI7qwE | ||
== See Also == | == See Also == |
Revision as of 00:06, 25 February 2023
Contents
Problem
Regular hexagon has side length
. Let
be the midpoint of
, and let
be the midpoint of
. What is the perimeter of
?
Diagram
~MRENTHUSIASM
Solution 1
Let the center of the hexagon be .
,
,
,
,
, and
are all equilateral triangles with side length
. Thus,
, and
. By symmetry,
. Thus, by the Pythagorean theorem,
. Because
and
,
. Thus, the solution to our problem is
.
~mathboy100
Solution 2
Consider triangle . Note that
,
, and
because it is an interior angle of a regular hexagon. (See note for details.)
By the Law of Cosines, we have:
By SAS Congruence, triangles
,
,
, and
are congruent, and by CPCTC, quadrilateral
is a rhombus. Therefore, the perimeter of
is
.
Note: The sum of the interior angles of any polygon with sides is given by
. Therefore, the sum of the interior angles of a hexagon is
, and each interior angle of a regular hexagon measures
.
Solution 3
We use a coordinates approach. Letting the origin be the center of the hexagon, we can let Then,
and
We use the distance formula four times to get
Thus, the perimeter of
.
~sirswagger21
Note: the last part of this solution could have been simplified by noting that
Solution 4
Note that triangles and
are all congruent, since they have side lengths of
and
and an included angle of
By the Law of Cosines,
Therefore,
-Benedict T (countmath1)
Video Solution 1
~Education, the Study of Everything
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=6I3ZNpI7qwE
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.