Difference between revisions of "2024 AMC 12A Problems/Problem 18"
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==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= | + | <math>\tan{15}=\frac{\sin{15}}{\cos{15}}=\frac{1}{2+\sqrt3}</math> and it eliminates all options except <math>6</math> and <math>12</math>. After one rotation it has turned <math>30^{\circ}</math>, so to satisfy the problem, divide <math>\frac{180}{30}</math> and get <math>\boxed{\textbf{A. }6}</math>. |
==Solution 4 (cheese core)== | ==Solution 4 (cheese core)== | ||
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==Solution 6 (the simplest solution ever)== | ==Solution 6 (the simplest solution ever)== | ||
Look at the picture and draw the next one and continue draw down the line and then when you first hit B, count how many rectangles you’ve drawn (excluding the first which hasn’t been rotated), and you should get <math>\boxed{\textbf{(A) or 6}}</math> as the answer. | Look at the picture and draw the next one and continue draw down the line and then when you first hit B, count how many rectangles you’ve drawn (excluding the first which hasn’t been rotated), and you should get <math>\boxed{\textbf{(A) or 6}}</math> as the answer. | ||
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+ | ~EaZ_Shadow | ||
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==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}} |
Latest revision as of 22:21, 10 November 2024
Contents
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)
and it eliminates all options except and . After one rotation it has turned , so to satisfy the problem, divide and get .
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
Solution 5
Memorize that in a 15-75-90 right triangle, the ratios of the side lengths are , , and . So, we have that the diagonal of the rectangle forms a 15 degree angle. Drawing it out we see the answer is , and this makes sense because 15 times 6 is 90, thus rotating a vertex back to B.
Solution 6 (the simplest solution ever)
Look at the picture and draw the next one and continue draw down the line and then when you first hit B, count how many rectangles you’ve drawn (excluding the first which hasn’t been rotated), and you should get as the answer.
~EaZ_Shadow
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.