Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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− | Extend segments <math>AD</math> and <math>BE</math> until they meet at point <math>G</math>. | + | Extend segments <math>\overline{AD}</math> and <math>\overline{BE}</math> until they meet at point <math>G</math>. |
− | Because <math>AB \parallel ED</math>, we have <math>\angle ABG = \angle DEG</math> and <math>\angle GDE = \angle GAB</math>, so <math>\triangle ABG \sim \triangle DEG</math> by AA. | + | Because <math>\overline{AB} \parallel \overline{ED}</math>, we have <math>\angle ABG = \angle DEG</math> and <math>\angle GDE = \angle GAB</math>, so <math>\triangle ABG \sim \triangle DEG</math> by AA. |
Because <math>ABCD</math> is a rhombus, <math>AB = CD = 2DE</math>, so <math>AG = 2GD</math>, meaning that <math>D</math> is a midpoint of segment <math>\overline{AG}</math>. | Because <math>ABCD</math> is a rhombus, <math>AB = CD = 2DE</math>, so <math>AG = 2GD</math>, meaning that <math>D</math> is a midpoint of segment <math>\overline{AG}</math>. | ||
− | Now, <math>AF \perp BE</math>, so <math>\triangle GFA</math> is right and median <math>FD = AD</math>. | + | Now, <math>\overline{AF} \perp \overline{BE}</math>, so <math>\triangle GFA</math> is right and median <math>FD = AD</math>. |
So now, because <math>ABCD</math> is a rhombus, <math>FD = AD = CD</math>. This means that there exists a circle from <math>D</math> with radius <math>AD</math> that passes through <math>F</math>, <math>A</math>, and <math>C</math>. | So now, because <math>ABCD</math> is a rhombus, <math>FD = AD = CD</math>. This means that there exists a circle from <math>D</math> with radius <math>AD</math> that passes through <math>F</math>, <math>A</math>, and <math>C</math>. |
Revision as of 16:23, 17 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)