Difference between revisions of "2024 AMC 12A Problems/Problem 18"
(→Solution 2) |
Technodoggo (talk | contribs) (🧀) |
||
Line 53: | Line 53: | ||
==Solution 3(In case you have no time and that's what I did) == | ==Solution 3(In case you have no time and that's what I did) == | ||
tan 15=sin15/cos15=1/(2+sqrt3) and it eliminates all options except 6 and 12. After one rotation it has turned 30degrees, so to satisfy the problem, divide 180 by 30 and you get 6 | tan 15=sin15/cos15=1/(2+sqrt3) and it eliminates all options except 6 and 12. After one rotation it has turned 30degrees, so to satisfy the problem, divide 180 by 30 and you get 6 | ||
+ | |||
+ | ==Solution 4 (cheese core)== | ||
+ | |||
+ | This problem is, of course, infinitely cheesable: it is easy to see that the answer will either be <math>6</math> rotations or no valid rotations whatsoever (A or E). In general, the answer is almost never "none of the above" (or the like), so it makes sense to go with <math>\boxed{\textbf{(A) }6}.</math> | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2024|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:14, 9 November 2024
Contents
[hide]Problem
On top of a rectangular card with sides of length and , an identical card is placed so that two of their diagonals line up, as shown (, in this case).
Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled in the figure?
Solution 1
Let the midpoint of be .
We see that no matter how many moves we do, stays where it is.
Now we can find the angle of rotation () per move with the following steps:
Since Vertex is the closest one and
Vertex C will land on Vertex B when cards are placed.
(someone insert diagram maybe)
~lptoggled, minor Latex edits by eevee9406
Solution 2
AC intersects BD at O,
we want to find
since ,
so each time we rotate BD to AC for , and we need to rotate n = times to overlap with B (from one of A,B,C,D) ( should not be n =
note: if you don't remember
Solution 3(In case you have no time and that's what I did)
tan 15=sin15/cos15=1/(2+sqrt3) and it eliminates all options except 6 and 12. After one rotation it has turned 30degrees, so to satisfy the problem, divide 180 by 30 and you get 6
Solution 4 (cheese core)
This problem is, of course, infinitely cheesable: it is easy to see that the answer will either be rotations or no valid rotations whatsoever (A or E). In general, the answer is almost never "none of the above" (or the like), so it makes sense to go with
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.