2020 AMC 12B Problems/Problem 12
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
Video Solution
On The Spot STEM: 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.