Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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==Solution== | ==Solution== | ||
Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>XC=a</math> and <math>XE=b</math>. This implies that <math>ED = a - b</math>. Since <math>CE = XC + XE = 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 = XC^2 + XO^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>XC=a</math> and <math>XE=b</math>. This implies that <math>ED = a - b</math>. Since <math>CE = XC + XE = 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 = XC^2 + XO^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2(50)=\boxed{\textbf{(E) } 100}</math>. | ||
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+ | ~JHawk0224 | ||
==See Also== | ==See Also== |
Revision as of 23:08, 7 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
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
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.