Difference between revisions of "2024 AMC 12B Problems/Problem 20"
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\textbf{(E) }913\qquad | \textbf{(E) }913\qquad | ||
</math> | </math> | ||
| + | |||
| + | ==Solution #1 == | ||
| + | Let midpoint of BC as M, extends AM to D and MD = x, | ||
| + | |||
| + | triangle ACD has 3 sides (40,42,2x) | ||
| + | as such, 2< 2x < 82 | ||
| + | 1<= x <=41 | ||
| + | so p = 1, q=41 | ||
| + | |||
| + | 2*Area(ABC) = 40 * 42 * sin(A) <= 2*840 | ||
| + | so r = max(Area{ABC)) = 840 | ||
| + | which is achieved when A = 90 degree , then angle ACD = 90 degree, | ||
| + | <math>(2x)^2 = 40^2 + 42^2 </math> | ||
| + | x = 29 | ||
| + | s= 29 | ||
| + | p+q+s+r = 1 + 41 + 29 + 840 = <math>\fbox{\textbf{(C) } 911}</math> | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
| + | |||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=B|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
Revision as of 02:51, 14 November 2024
Problem 20
Suppose 
, 
, and 
are points in the plane with 
and 
, and let 
be the length of the line segment from to the midpoint of 
. Define a function 
by letting 
be the area of 
. Then the domain of is an open interval 
, and the maximum value 
of occurs at 
. What is 
?
Solution #1
Let midpoint of BC as M, extends AM to D and MD = x,
triangle ACD has 3 sides (40,42,2x) as such, 2< 2x < 82 1<= x <=41 so p = 1, q=41
2*Area(ABC) = 40 * 42 * sin(A) <= 2*840
so r = max(Area{ABC)) = 840
which is achieved when A = 90 degree , then angle ACD = 90 degree,
x = 29
s= 29
p+q+s+r = 1 + 41 + 29 + 840 = 
See also
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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. 
