Difference between revisions of "2022 AMC 10B Problems/Problem 20"
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</cmath> | </cmath> | ||
− | We have <math>\angle | + | We have <math>\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC</math>. |
Hence, | Hence, | ||
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~mathfan2020 | ~mathfan2020 | ||
+ | |||
+ | A little bit faster: <math>AOFB</math> is cyclic <math>\implies \angle OFE = \angle BAO</math>. | ||
+ | |||
+ | <math>AB \parallel CD \implies \angle BAO = \angle OCE</math>. | ||
+ | |||
+ | Therefore <math>\angle OFE=\angle OCE \implies OECF</math> is cyclic. | ||
+ | |||
+ | Hence <math>\angle CFE=\angle COE=\angle CAD = 67^\circ</math>. | ||
+ | |||
+ | ~asops | ||
==Solution 4== | ==Solution 4== | ||
Observe that all answer choices are close to <math>112.5 = 90+\frac{45}{2}</math>. A quick solve shows that having <math>\angle D = 90^\circ</math> yields <math>\angle BFC = 135^\circ = 90 + \frac{90}{2}</math>, meaning that <math>\angle BFC</math> increases with <math>\angle D</math>. | Observe that all answer choices are close to <math>112.5 = 90+\frac{45}{2}</math>. A quick solve shows that having <math>\angle D = 90^\circ</math> yields <math>\angle BFC = 135^\circ = 90 + \frac{90}{2}</math>, meaning that <math>\angle BFC</math> increases with <math>\angle D</math>. | ||
− | Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math> | + | Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math>. |
~mathfan2020 | ~mathfan2020 | ||
==Solution 5 (Similarity and Circle Geometry)== | ==Solution 5 (Similarity and Circle Geometry)== | ||
− | + | This solution refers to the <b>Diagram</b> section. | |
+ | |||
+ | We extend <math>AD</math> and <math>BE</math> to point <math>G</math>, as shown below: | ||
<asy> | <asy> | ||
/* | /* | ||
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By AA Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>. This implies that <math>2AD=AG</math> and <math>AD \cong DG</math>, so <math>AG=2AD=2\cdot2=4</math>. That is, <math>D</math> is the midpoint of <math>AG</math>. | By AA Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>. This implies that <math>2AD=AG</math> and <math>AD \cong DG</math>, so <math>AG=2AD=2\cdot2=4</math>. That is, <math>D</math> is the midpoint of <math>AG</math>. | ||
− | Now, let's redraw our previous diagram, but construct a circle with radius <math>AD</math> or <math>2</math> centered at <math>D</math> and by extending <math>CD</math> to point <math>H</math>, which is on the circle | + | Now, let's redraw our previous diagram, but construct a circle with radius <math>AD</math> or <math>2</math> centered at <math>D</math> and by extending <math>CD</math> to point <math>H</math>, which is on the circle, as shown below: |
<asy> | <asy> | ||
/* | /* | ||
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~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits). | ~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits). | ||
+ | |||
+ | == Solution 6 (Simplification/Reduction) == | ||
+ | |||
+ | If angle <math>ADC</math> was a right angle, it would be much easier. Thus, first pretend that <math>ADC</math> is a right angle. <math>ABCD</math> is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line <math>AE</math>, which is <math>\sqrt{5}/2</math>. We want the measure of angle <math>BFC</math>, so to work closer to it, we should try finding the length of line <math>BF</math>. Angle <math>FAB</math> and angle <math>ABF</math> are complementary. Angle <math>ABF</math> and angle <math>FBC</math> are also complementary. Thus, <math>\sin FAB=\cos ABF=\sin FBC</math>. <math>\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}</math>. Since <math>\sin FAB=1\sqrt{5}</math>,and <math>AB=1</math>, <math>FB=\sin FAB</math>. It follows now that <math>FE=3*\sqrt{5}/10</math>. | ||
+ | |||
+ | Now, zoom in on triangle <math>BEC</math>. To use the Law of Cosines on triangle <math>FBC</math>, we need the length of <math>FC</math>. Use the Law of Cosines on triangle <math>EFC</math>. Cos <math>E=1/\sqrt{5}</math>. Thus, after using the Law of Cosines, <math>FC=\sqrt{2/5}</math>. | ||
+ | |||
+ | Since we now have SSS on <math>BEC</math>, we can get use the Law of Cosines. <math>\cos BFC=1/-\sqrt{2}</math>. <math>\arccos 1/-\sqrt{2}</math> is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. <math>180-45=135</math>. Angle <math>BFC</math> is <math>135^\circ</math>. | ||
+ | |||
+ | Realize that, around point F, there will always be 3 right angles, regardless of what angle <math>ADC</math> is. There are only two angles that change when <math>ADC</math> changes. Break up angle <math>BFC</math> into angle <math>BFB'</math>, which is always 90 degrees, and angle <math>B'FC</math>, which we have discovered to to be half of <math>ADC</math>. Thus, when angle <math>ADC</math> is 46 degrees, then <math>B'FC</math> will be 23. <math>23+90=113</math>. Angle <math>BFC</math> is <math>\boxed{\textbf{(D) }113}</math> degrees. | ||
+ | |||
+ | |||
+ | ==Video Solution (⚡️Just 1 min!⚡️)== | ||
+ | https://youtu.be/CriWEtfD5GE | ||
+ | |||
+ | <i>~Education, the Study of Everything</i> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=HWJe96s_ugs&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=6 | ||
==Video Solution== | ==Video Solution== | ||
Line 214: | Line 246: | ||
== Video Solution, by Challenge 25 == | == Video Solution, by Challenge 25 == | ||
https://youtu.be/W1jbMaO8BIQ (cyclic quads) | https://youtu.be/W1jbMaO8BIQ (cyclic quads) | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/5Plt3mmZBC0 | ||
+ | |||
+ | ~Interstigation | ||
== See Also == | == See Also == |
Revision as of 20:06, 8 September 2023
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Law of Sines and Law of Cosines)
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5 (Similarity and Circle Geometry)
- 8 Solution 6 (Simplification/Reduction)
- 9 Video Solution (⚡️Just 1 min!⚡️)
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing
- 13 Video Solution, best solution (family friendly, no circles drawn)
- 14 Video Solution, by Challenge 25
- 15 Video Solution by Interstigation
- 16 See Also
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
?
Diagram
~MRENTHUSIASM
Solution 1 (Law of Sines and Law of Cosines)
Without loss of generality, we assume the length of each side of is
.
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
A little bit faster: is cyclic
.
.
Therefore is cyclic.
Hence .
~asops
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 and Circle Geometry)
This solution refers to the Diagram section.
We extend and
to point
, as shown below:
We know that
and
.
By AA Similarity, with a ratio of
. This implies that
and
, so
. That is,
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, as shown below:
Notice how
and
are on the circle and that
intercepts with
.
Let's call .
Note that also intercepts
, So
.
Let . Notice how
and
are supplementary to each other. We conclude that
Since
, we have
.
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).
Solution 6 (Simplification/Reduction)
If angle was a right angle, it would be much easier. Thus, first pretend that
is a right angle.
is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line
, which is
. We want the measure of angle
, so to work closer to it, we should try finding the length of line
. Angle
and angle
are complementary. Angle
and angle
are also complementary. Thus,
.
. Since
,and
,
. It follows now that
.
Now, zoom in on triangle . To use the Law of Cosines on triangle
, we need the length of
. Use the Law of Cosines on triangle
. Cos
. Thus, after using the Law of Cosines,
.
Since we now have SSS on , we can get use the Law of Cosines.
.
is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value.
. Angle
is
.
Realize that, around point F, there will always be 3 right angles, regardless of what angle is. There are only two angles that change when
changes. Break up angle
into angle
, which is always 90 degrees, and angle
, which we have discovered to to be half of
. Thus, when angle
is 46 degrees, then
will be 23.
. Angle
is
degrees.
Video Solution (⚡️Just 1 min!⚡️)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=HWJe96s_ugs&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=6
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
Video Solution, best solution (family friendly, no circles drawn)
https://www.youtube.com/watch?v=vwI3I7dxw0Q
Video Solution, by Challenge 25
https://youtu.be/W1jbMaO8BIQ (cyclic quads)
Video Solution by Interstigation
~Interstigation
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.