Difference between revisions of "2024 AMC 12A Problems/Problem 23"
(new solution) |
(→Solution 9 (Vietas)) |
||
Line 212: | Line 212: | ||
<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})=\boxed{68}</cmath> | <cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})=\boxed{68}</cmath> | ||
+ | ==Alternate proof of the two tangent squares formula== | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:17, 23 December 2024
Contents
- 1 Problem
- 2 Solution 1 (Trigonometric Identities)
- 3 Solution 2 (Another Identity)
- 4 Solution 3 (Complex Numbers)
- 5 Solution 5 (Transformation)
- 6 Solution 6 (Half angle formula twice)
- 7 Solution 7(single formula)
- 8 Solution 8(just do it ✅)
- 9 Solution 9 (Vietas)
- 10 Alternate proof of the two tangent squares formula
- 11 See also
Problem
What is the value of
Solution 1 (Trigonometric Identities)
First, notice that
Here, we make use of the fact that
Hence,
Note that
Hence,
Therefore, the answer is .
~tsun26
Solution 2 (Another Identity)
First, notice that
Here, we make use of the fact that
Hence,
Therefore, the answer is .
Solution 3 (Complex Numbers)
Let . Then, Expanding by using a binomial expansion, Divide by and notice we can set where . Then, define so that
Notice that we can have because we are only considering the real parts. We only have this when , meaning . This means that we have as unique roots (we get them from ) and by using the fact that , we get Since we have a monic polynomial, by the Fundamental Theorem of Algebra, Looking at the term in the expansion for and using vietas gives us Since and Therefore
Solution 5 (Transformation)
Set x = , 7x = - x , set C7 = , C5 = , C3 = , C= , S2 = , S6 =
First, notice that
Solution 6 (Half angle formula twice)
So from the question we have:
Using
Using
Using
~ERiccc
Solution 7(single formula)
We use for
vladimir.shelomovskii@gmail.com, vvsss
Solution 8(just do it ✅)
Since A is too big and E is too small, There is only 3 options left, you can make a guess now, however, estimating the value of it is also okay. You will get 68 for the final answer.
Solution 9 (Vietas)
As the above solutions noted, we can factor the expression into .
Before we directly solve this problem, let's analyze the roots of , or equivalently using tangent expansion formula, , which implies . Now note that the roots of this equation are precisely , so the second symmetric sum of these four numbers is by Vieta's. Thus, we have Upon further inspection, using the fact that . Hence, we have
Now, we return to the problem statement, where we see a similar squared sum. We use this motivation to square our equation above to obtain
Then, use the fact that to get Hold on; the first term is exactly what we are solving for! It thus suffices to find . Fortunately, this is just (Where is the nth symmetric sum), with relation to roots of . By Vieta's, this is just .
Finally, we plug this value into our equation to obtain
Alternate proof of the two tangent squares formula
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.