Difference between revisions of "2021 AMC 12A Problems/Problem 24"
Icematrix2 (talk | contribs) (Created page with "==Problem== These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. ==Solution== The solutions will be posted once the problems are...") |
Sugar rush (talk | contribs) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | + | Semicircle <math>\Gamma</math> has diameter <math>\overline{AB}</math> of length <math>14</math>. Circle <math>\Omega</math> lies tangent to <math>\overline{AB}</math> and intersects <math>\Gamma</math> at points <math>Q</math> and <math>R</math>. If <math>QR=3\sqrt{3}</math> and <math>\angle QPR = 60^{\circ}</math>, then the area of <math>\triangle PQR</math> equals <math>\tfrac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. What is <math>a+b+c</math>? | |
| − | ==Solution== | + | |
| − | + | <math>\textbf{(A) } 110\qquad\textbf{(B) } 114\qquad\textbf{(C) } 118\qquad\textbf{(D) } 122\qquad\textbf{(E) } 126\qquad</math> | |
| − | + | ||
| − | + | ==Video Solution by Punxsutawney Phil== | |
| + | https://youtube.com/watch?v=cEHF5iWMe9c | ||
| + | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2021|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:11, 11 February 2021
Problem
Semicircle 
has diameter 
of length 
. Circle 
lies tangent to and intersects at points 
and 
. If 
and 
, then the area of 
equals 
, where 
and 
are relatively prime positive integers, and 
is a positive integer not divisible by the square of any prime. What is 
?
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=cEHF5iWMe9c
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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. 
