Difference between revisions of "2021 AMC 12A Problems/Problem 17"
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==Problem== | ==Problem== | ||
| − | + | Trapezoid <math>ABCD</math> has <math>\overline{AB}\parallel\overline{CD},BC=CD=43</math>, and <math>\overline{AD}\perp\overline{BD}</math>. Let <math>O</math> be the intersection of the diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>, and let <math>P</math> be the midpoint of <math>\overline{BD}</math>. Given that <math>OP=11</math>, the length of <math>AD</math> can be written in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the quare of any prime. What is <math>m+n</math>? | |
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| + | <math>\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215</math> | ||
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==Solution== | ==Solution== | ||
| − | + | {{solution}} | |
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==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2021|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:13, 11 February 2021
Problem
Trapezoid 
has 
, and 
. Let 
be the intersection of the diagonals 
and 
, and let 
be the midpoint of . Given that 
, the length of 
can be written in the form 
, where 
and 
are positive integers and is not divisible by the quare of any prime. What is 
?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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 | |
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