2024 AMC 10B Problems/Problem 14
- The following problem is from both the 2024 AMC 10B #14 and 2024 AMC 12B #9, so both problems redirect to this page.
Contents
Problem
A dartboard is the region B in the coordinate plane consisting of points 
such that 
. A target T is the region where 
. A dart is thrown at a random point in B. The probability that the dart lands in T can be expressed as 
, where 
and 
are relatively prime positive integers. What is 
?
Diagram
![[asy] // By Elephant200 // Feel free to adjust the code size(10cm); pair A = (8, 0); pair B = (0, 8); pair C = (-8, 0); pair D = (0, -8); draw(A--B--C--D--cycle); label("$(8,0)$", A, NE); label("$(0,8)$", B, NE); label("$(-8,0)$", C, SW); label("$(0,-8)$", D, SW); filldraw(circle((0,0),4*sqrt(2)), gray); filldraw(circle((0,0),3*sqrt(2)), white); draw((-12, 0)--(12,0),EndArrow(5)); draw((12, 0)--(-12,0),EndArrow(5)); draw((0,-12)--(0,12), EndArrow(5)); draw((0,12)--(0,-12),EndArrow(5)); [/asy]](http://latex.artofproblemsolving.com/4/1/2/412b9706714cbc57ae34580e9386a5f1bc2bac96.png)
~Elephant200
Solution 1
Inequalities of the form 
are well-known and correspond to a square in space with centre at origin and vertices at 
, 
, 
, 
.
The diagonal length of this square is clearly 
, so it has an area of
![\[\frac{1}{2} \cdot 16 \cdot 16 = 128\]](http://latex.artofproblemsolving.com/5/a/6/5a66fdc982311cb313225cf592a68b0cc98e45f3.png)
Now,
![\[(x^2 + y^2 - 25)^2 \le 49\]](http://latex.artofproblemsolving.com/0/a/5/0a51bc9364a32d91a5f2fc59c5d48fd738cfd599.png)
Converting to polar form,
![\[r^2 - 25 \le 7 \implies r \le \sqrt{32},\]](http://latex.artofproblemsolving.com/6/9/7/6974d2bba1243fb5f765acd6432a696a68b3aa93.png)
and
![\[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\]](http://latex.artofproblemsolving.com/4/f/8/4f861463709428b4887973a736f196b5ed62c65e.png)
The union of these inequalities is the circular region 
for which every circle in has a radius between 
and 
, inclusive. The area of such a region is thus 
The requested probability is therefore 
yielding 
We have 
-anonymous, countmath1
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 10 Problems and Solutions | ||
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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. 
