Difference between revisions of "2024 AMC 12B Problems/Problem 20"
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==Solution #1 == | ==Solution #1 == | ||
| − | Let midpoint of BC as M, extends AM to D and MD = x, | + | Let midpoint of <math>BC</math> as <math>M</math>, extends <math>AM</math> to <math>D</math> and <math>MD=x</math>, |
| − | triangle ACD has 3 sides (40,42,2x) | + | triangle <math>ACD</math> has <math>3</math> sides <math>(40,42,2x)</math> |
| − | as such, 2< 2x < 82 | + | as such,<cmath>2< 2x < 82</cmath> |
| − | 1 | + | <cmath>1\le x \le41</cmath> |
| − | so p = 1, q=41 | + | so <cmath>p = 1, q=41</cmath> |
| − | 2 | + | <cmath>2\cdot f(x) = 40 \cdot 42 \cdot \sin(A) \le 2\cdot840</cmath> |
| − | so r | + | so <math>r = 840 </math> |
| − | which is achieved when A = 90 | + | which is achieved when <math>A = 90^\circ</math> , then <math>\angle ACD = 90^\circ</math> |
| − | < | + | <cmath>(2x)^2 = 40^2 + 42^2 </cmath> |
| − | x = 29 | + | <cmath>x = 29</cmath> |
| − | s= 29 | + | <cmath>s= 29 </cmath> |
| − | p+q+s+r = 1 + 41 + 29 + 840 = | + | <cmath>p+q+s+r = 1 + 41 + 29 + 840 = \fbox{\textbf{(C) } 911}</cmath> |
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
| − | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2024|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 03:15, 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 
as 
, extends 
to 
and 
,
triangle 
has 
sides 
as such,![\[2< 2x < 82\]](http://latex.artofproblemsolving.com/1/c/1/1c1d5d8b1ef6530386679b3dd0049d675f43e0d9.png)
![\[1\le x \le41\]](http://latex.artofproblemsolving.com/f/3/c/f3cd653ce33f118cfa4d83c09e3ca118e110dfea.png)
so ![\[p = 1, q=41\]](http://latex.artofproblemsolving.com/1/b/e/1be1f3d176d63328e97c2914cc764a2231b461f6.png)
![\[2\cdot f(x) = 40 \cdot 42 \cdot \sin(A) \le 2\cdot840\]](http://latex.artofproblemsolving.com/d/6/e/d6ea1bda17f717c2bfee80fb5ca61814faa5baef.png)
so 
which is achieved when 
, then 
![\[(2x)^2 = 40^2 + 42^2\]](http://latex.artofproblemsolving.com/0/f/2/0f25b3b06e55de62d28714cdc044617950cc62a2.png)
![\[x = 29\]](http://latex.artofproblemsolving.com/7/b/f/7bf275207f89aa0f77a5c0090ca9e8ee2e938ac0.png)
![\[s= 29\]](http://latex.artofproblemsolving.com/c/a/9/ca90f61b155648e1b1669ca4e72821f2c8bf0356.png)
![\[p+q+s+r = 1 + 41 + 29 + 840 = \fbox{\textbf{(C) } 911}\]](http://latex.artofproblemsolving.com/0/7/2/0726463ce21647893e19011787dc28db279a6f47.png)
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. 
