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> | ||
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| + | ==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>. | ||
==See Also== | ==See Also== | ||
Revision as of 22:08, 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 
.
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. 
