Difference between revisions of "2024 AMC 12A Problems/Problem 2"
(Redirected page to 2024 AMC 10A Problems/Problem 2) (Tag: New redirect) |
(Undo vandalism) (Tags: Removed redirect, Undo) |
||
| Line 1: | Line 1: | ||
| − | # | + | {{duplicate|[[2024 AMC 12A Problems/Problem 2|2024 AMC 12A #2]] and [[2024 AMC 10A Problems/Problem 2|2024 AMC 10A #2]]}} |
| + | |||
| + | ==Problem== | ||
| + | Define <math>\blacktriangledown(a) = \sqrt{a - 1}</math> and <math>\blacktriangle(a) = \sqrt{a + 1}</math> for all real numbers <math>a</math>. What is the value of <cmath>\frac{\blacktriangledown(20 + \blacktriangle(2024))}{\blacktriangledown(\blacktriangle(24))}~?</cmath> | ||
| + | |||
| + | <math>\textbf{(A)}~ 1 \qquad \textbf{(B)}~ 2 \qquad \textbf{(C)}~ 4 \qquad \textbf{(D)}~ 8 \qquad \textbf{(E)}~ 16</math> | ||
| + | |||
| + | ==Solution== | ||
| + | The value of the expression is <cmath>\frac{\sqrt{20+\sqrt{2024+1}-1}}{\sqrt{\sqrt{24+1}-1}}=\frac{\sqrt{20+\sqrt{2025}-1}}{\sqrt{\sqrt{25}-1}}=\frac{\sqrt{20+45-1}}{\sqrt{5-1}}=\frac{\sqrt{64}}{\sqrt{4}}=\frac{8}{2}=\boxed{\textbf{(C)}~4}.</cmath> | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://youtu.be/zaswZfIEibA?t=540 | ||
| + | |||
| + | ~IceMatrix | ||
| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=1|num-a=3}} | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:53, 21 September 2025
- The following problem is from both the 2024 AMC 12A #2 and 2024 AMC 10A #2, so both problems redirect to this page.
Problem
Define 
and 
for all real numbers 
. What is the value of ![\[\frac{\blacktriangledown(20 + \blacktriangle(2024))}{\blacktriangledown(\blacktriangle(24))}~?\]](http://latex.artofproblemsolving.com/8/1/a/81af8e2c799a34ed81e7b1443b2715d505b1f940.png)
Solution
The value of the expression is ![\[\frac{\sqrt{20+\sqrt{2024+1}-1}}{\sqrt{\sqrt{24+1}-1}}=\frac{\sqrt{20+\sqrt{2025}-1}}{\sqrt{\sqrt{25}-1}}=\frac{\sqrt{20+45-1}}{\sqrt{5-1}}=\frac{\sqrt{64}}{\sqrt{4}}=\frac{8}{2}=\boxed{\textbf{(C)}~4}.\]](http://latex.artofproblemsolving.com/4/a/2/4a22553e98eeb8b724a776158373ff133a462d97.png)
Video Solution by TheBeautyofMath
https://youtu.be/zaswZfIEibA?t=540
~IceMatrix
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 | |
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
