Difference between revisions of "2023 AMC 12B Problems/Problem 25"
Isabelchen (talk | contribs) m (→Supplement (Calculating sin54 from Scratch)) |
|||
Line 150: | Line 150: | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=VLYc7Dscs9M | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2023|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2023|ab=B|num-b=24|after=Last Problem}} | ||
{{AMC12 box|year=2023|ab=B|num-b=24|after=Last Problem}} | {{AMC12 box|year=2023|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:37, 16 November 2023
- The following problem is from both the 2023 AMC 10B #25 and 2023 AMC 12B #25, so both problems redirect to this page.
Contents
[hide]Problem
A regular pentagon with area is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
Solution 1
Let the original pentagon be centered at . The dashed lines represent the fold lines. WLOG, let's focus on vertex .
Since is folded onto , where is the intersection of and the creaseline between and . Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry.
Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by
Option 1: Knowledge
Remember that .
Option 2: Angle Identities
Let the inner pentagon be .
-Dissmo
Solution 2
Let and be the circumradius of the big and small pentagon, respectively. Let be the apothem of the smaller pentagon and and be the areas of the smaller and larger pentagon, respectively.
From the diagram: ~Technodoggo
Solution 3
Interestingly, we find that the pentagon we need is the one that is represented by the intersection of the diagonals. Through similar triangles and the golden ratio, we find that the side length ratio of the two pentagons is Thus, the answer is . ~andliu766
Supplement (Calculating sin54/cos36 from Scratch)
Construct with , and with , . WLOG, let , , .
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=VLYc7Dscs9M
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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.