Difference between revisions of "2024 AMC 12A Problems/Problem 18"
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Vertex C will land on Vertex B when <math>\frac{150}{30}+1=\fbox{(A) 6}</math> cards are placed. | Vertex C will land on Vertex B when <math>\frac{150}{30}+1=\fbox{(A) 6}</math> cards are placed. | ||
− | ~minor Latex edits by eevee9406 | + | ~lptoggled, minor Latex edits by eevee9406 |
==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 18:17, 8 November 2024
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.
~lptoggled, minor Latex edits by eevee9406
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 |
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