Difference between revisions of "2024 AMC 12A Problems/Problem 2"
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+ | {{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 and
for all real numbers
. What is the value of
Solution
The value of the expression is
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.