Difference between revisions of "2024 AMC 12A Problems/Problem 2"

(Undo revision 245630 by Maa is stupid (talk))
(Tags: New redirect, Undo)
m
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
#redirect[[2024 AMC 10A Problems/Problem 2]]
+
{{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}}

Latest revision as of 00:52, 16 April 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 $\blacktriangledown(a) = \sqrt{a - 1}$ and $\blacktriangle(a) = \sqrt{a + 1}$ for all real numbers $a$. What is the value of \[\frac{\blacktriangledown(20 + \blacktriangle(2024))}{\blacktriangledown(\blacktriangle(24))}~?\]

$\textbf{(A)}~ 1 \qquad \textbf{(B)}~ 2 \qquad \textbf{(C)}~ 4 \qquad \textbf{(D)}~ 8 \qquad \textbf{(E)}~ 16$

Solution

The value of the expression is \[\frac{\sqrt{20+\sqrt{2024+1}-1}}{\sqrt{\sqrt{24+1}-1}}=\frac{\sqrt{20+\sqrt{2025}-1}}{\sqrt{\sqrt{25}-1}}=\frac{\sqrt{20+45-1}}{\sqrt{5-1}}=\frac{\sqrt{64}}{\sqrt{4}}=\frac{8}{2}=\boxed{\textbf{(C)}~4}.\]

Video Solution by TheBeautyofMath

https://youtu.be/zaswZfIEibA?t=540

~IceMatrix

See also

2024 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC Logo.png