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== | + | ==Solution 1== |
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(50)=\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(50)=\boxed{\textbf{(E) } 100}</math>. | ||
~JHawk0224 | ~JHawk0224 | ||
+ | |||
+ | ==Solution 2 (Power of a Point)== | ||
+ | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>CD</math>. Draw triangle <math>OCD</math>, and median <math>OX</math>. Because <math>OC = OD</math>, <math>OCD</math> is isosceles, so <math>OX</math> is also an altitude of <math>OCD</math>. <math>OD = 5\sqrt2 - 2\sqrt5</math>, and because angle <math>OEC</math> is <math>45</math> degrees and triangle <math>OXE</math> is right, <math>OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}</math>. Because triangle <math>OXC</math> is right, <math>CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}</math>. Thus, <math>CD = 2\sqrt{15 + 10\sqrt{10}}</math>. We are looking for <math>CE^2</math> + <math>DE^2</math> which is also <math>(CE + DE)^2 - 2 \cdot CE \cdot DE</math>. Because <math>CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}, (CE + CD)^2 = 4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}</math>. By power of a point, <math>CE \cdot DE = AE \cdot EB = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20</math> so <math>2 \cdot CE \cdot DE = 40\sqrt{10} - 40</math>. Finally, <math>CE^2 + DE^2 = 60 + 40\sqrt{10} - (40\sqrt{10} - 40) = \boxed{(E) 100}</math>. | ||
+ | |||
+ | ~CT17 | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:24, 8 February 2020
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
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.