Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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<math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | <math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Pythagorean Theorem) == |
Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>CX=a</math> and <math>EX=b</math>. This implies that <math>DE = a - b</math>. Since <math>CE = CX + EX = a + b</math>, we now want to find <math>(a+b)^2+(a-b)^2=2(a^2+b^2)</math>. Since <math>\angle CXO</math> is a right angle, by Pythagorean theorem <math>a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2\times50=\boxed{\textbf{(E) } 100}</math>. | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>CX=a</math> and <math>EX=b</math>. This implies that <math>DE = a - b</math>. Since <math>CE = CX + EX = a + b</math>, we now want to find <math>(a+b)^2+(a-b)^2=2(a^2+b^2)</math>. Since <math>\angle CXO</math> is a right angle, by Pythagorean theorem <math>a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2\times50=\boxed{\textbf{(E) } 100}</math>. | ||
Revision as of 12:20, 13 February 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
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
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 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.