Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | == See Also == | ||
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+ | {{AMC10 box|year=2022|ab=B|num-b=18|num-a=20}} | ||
+ | {{AMC12 box|year=2022|ab=B|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Revision as of 19:07, 18 November 2022
Contents
[hide]Problem
Let be a rhombus with
. Let
be the midpoint of
, and let
be the point
on
such that
is perpendicular to
. What is the degree measure of
?
Solution (Law of Sines and Law of Cosines)
Without loss of generality, we assume the length of each side of is 2.
Because
is the midpoint of
,
.
Because is a rhombus,
.
In , following from the law of sines,
We have .
Hence,
By solving this equation, we get .
Because ,
In , following from the law of sines,
Because , the equation above can be converted as
Therefore,
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Extend segments and
until they meet at point
.
Because , we have
and
, so
by AA.
Because is a rhombus,
, so
, meaning that
is a midpoint of segment
.
Now, , so
is right and median
.
So now, because is a rhombus,
. This means that there exists a circle from
with radius
that passes through
,
, and
.
AG is a diameter of this circle because . This means that
, so
, which means that
~popop614
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.