Difference between revisions of "2022 AMC 10B Problems/Problem 20"
Ghfhgvghj10 (talk | contribs) (→Solution 5 (Similarity & Circle Geometry)) |
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Lets call <math>\angle CFE = \theta</math>. | Lets call <math>\angle CFE = \theta</math>. | ||
− | <math>\angle | + | <math>\angle CDG</math> also intercepts arch <math>CG</math>, but it's vertical angle (<math>\angle ADH</math>), also intercepts an arch. So <math>\angle CDG = 2\angle CFE</math>. |
− | <math>\angle | + | <math>\angle CDG = 2\theta</math>. |
− | Notice how <math> | + | Notice how <math>CDG</math> and <math>ADC</math> are supplementary to each other. |
This concludes that, <math>2\theta=180-\angle ADC</math>. | This concludes that, <math>2\theta=180-\angle ADC</math>. |
Revision as of 02:16, 20 November 2022
Contents
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
Solution 3
Let meet
at
, then
is cyclic and
. Also,
, so
, thus
by SAS, and
, then
, and
~mathfan2020
Solution 4
Observe that all answer choices are close to . A quick solve shows that having
yields
, meaning that
increases with
.
Substituting,
~mathfan2020
Solution 5 (Similarity & Circle Geometry)
Let's make a diagram, but extend and
to point
.
We know that , and
.
By SAS Similarity, with a ratio of
.
This means that, and
.
.
This also can prove that is the midpoint of
.
Now, let's redraw our previous diagram, but construct a circle with radius or
centered at
and by extending
to point
, which is on the circle.
Notice how and
are on the circle and that
intercepts with arch
.
Lets call .
also intercepts arch
, but it's vertical angle (
), also intercepts an arch. So
.
.
Notice how and
are supplementary to each other.
This concludes that, .
.
Realize how
Which means the answer is
.
~ghfhgvghj10 (feel free to make minor edits)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing
~ pi_is_3.14
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.