Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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~Math_Wiz_3.14 | ~Math_Wiz_3.14 | ||
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+ | ==Solution 4 (Reflections)== | ||
+ | <asy> | ||
+ | draw(circle((0,0),7.07)); | ||
+ | dot((-7.07,0)); | ||
+ | label("A",(-7.07,0),W); | ||
+ | dot((7.07,0)); | ||
+ | label("B",(7.07,0),E); | ||
+ | dot((0,0)); | ||
+ | label("O",(0,0),N); | ||
+ | dot((-2.24,-6.71)); | ||
+ | label("C",(-2.24,-6.71),SSW); | ||
+ | dot((6.71,2.24)); | ||
+ | label("D",(6.71,2.24),NE); | ||
+ | draw((-2.24,-6.71)--(6.71,2.24)); | ||
+ | dot((6.71,-2.24)); | ||
+ | label("D'",(6.71,-2.24),SE); | ||
+ | draw((4.51,0)--(6.71,-2.24)); | ||
+ | dot((4.51,0)); | ||
+ | label("E",(4.51,0),NNW); | ||
+ | draw((-7.07,0)--(7.07,0)); | ||
+ | draw((0,0)--(-2.24,-6.71)); | ||
+ | draw((0,0)--(6.71,-2.24)); | ||
+ | </asy> | ||
+ | By reflecting <math>D</math> across the line <math>AB</math> to produce <math>D'</math>, we have that <math>\angle BED'=45</math>. Since <math>\angle AEC=45</math>, <math>\angle CED'=90</math>. By the Pythagorean Theorem, our desired solution is just <math>CD'^2</math>. | ||
+ | Looking next to circle arcs, we know that <math>\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45</math>, so <math>\overarc{AC}+\overarc{BD}=90</math>. Since <math>\overarc{BD'}=\overarc{BD}</math>, and <math>\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180</math>, <math>\overarc{CD'}=90</math>. Thus, <math>\angle COD'=90</math>. | ||
+ | Since <math>OC=OD'=5\sqrt{2}</math>, by the Pythagorean Theorem, the desired <math>CD'^2= \boxed{\textbf{(E) } 100}</math>. | ||
+ | |||
+ | ~sofas103 | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
On The Spot STEM: https://www.youtube.com/watch?v=h-hhRa93lK4 | On The Spot STEM: https://www.youtube.com/watch?v=h-hhRa93lK4 |
Revision as of 04:01, 22 February 2020
Contents
Problem
Let be a diameter in a circle of radius
Let
be a chord in the circle that intersects
at a point
such that
and
What is
Solution 1
Let be the center of the circle, and
be the midpoint of
. Let
and
. This implies that
. Since
, we now want to find
. Since
is a right angle, by Pythagorean theorem
. Thus, our answer is
.
~JHawk0224
Solution 2 (Power of a Point)
Let be the center of the circle, and
be the midpoint of
. Draw triangle
, and median
. Because
,
is isosceles, so
is also an altitude of
.
, and because angle
is
degrees and triangle
is right,
. Because triangle
is right,
. Thus,
. We are looking for
+
which is also
. Because
. By power of a point,
so
. Finally,
.
~CT17
Solution 3 (Law of Cosines)
Let be the center of the circle. Notice how
, where
is the radius of the circle. By applying the law of cosines on triangle
,
. Similarly, by applying the law of cosines on triangle
,
. By subtracting these two equations, we get
. We can rearrange it to get
. Because both
and
are both positive, we can safely divide both sides by
to obtain
. Because
,
. Through power of a point, we can find out that
, so
.
~Math_Wiz_3.14
Solution 4 (Reflections)
By reflecting
across the line
to produce
, we have that
. Since
,
. By the Pythagorean Theorem, our desired solution is just
.
Looking next to circle arcs, we know that
, so
. Since
, and
,
. Thus,
.
Since
, by the Pythagorean Theorem, the desired
.
~sofas103
Video Solution
On The Spot STEM: https://www.youtube.com/watch?v=h-hhRa93lK4
Video Solution 2
~IceMatrix
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.