Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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Through power of a point, we can find out that <math>(CE)(DE)=20\sqrt{10} - 20</math>, so <cmath>CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{\textbf{(E) } 100}.</cmath> | Through power of a point, we can find out that <math>(CE)(DE)=20\sqrt{10} - 20</math>, so <cmath>CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{\textbf{(E) } 100}.</cmath> | ||
− | ~Math_Wiz_3.14 | + | ~Math_Wiz_3.14 (legibility changes by eagleye) |
==Solution 4 (Reflections)== | ==Solution 4 (Reflections)== |
Revision as of 08:11, 3 July 2021
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
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution 1 (Pythagorean Theorem)
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, .
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 (legibility changes by eagleye)
Solution 4 (Reflections)
Let
be the center of the circle. By reflecting
across the line
to produce
, we have that
. Since
,
. 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 Solutions
https://www.youtube.com/watch?v=h-hhRa93lK4
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.