2023 AMC 12B Problems/Problem 25
- The following problem is from both the 2023 AMC 10B #25 and 2023 AMC 12B #25, so both problems redirect to this page.
Contents
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
We can find the area of the red pentagon by taking the area of the total pentagon and subtracting the area outside the red pentagon.
The area outside the red pentagon is the sum of the larger isosceles triangles, but this double counts the overlapping regions of the small isosceles triangles, so we have to subtract those out.
We have
Lets focus on finding the area of each individual triangle:
Notice that we have no information about the side length, so instead we let the side length be . Now we can drop an altitude from to the base of the triangle, and we know this altitude must split the base of the pentagon in half, so we can create a right triangle. Furthermore, draw a line from to . This must bisect angle which is degrees, so we create triangles. Specifically, we know , , and because is isosceles and we know the vertex angle is . We encode this information in the diagram below:
Since is isosceles, the area of is half the area of . Similarly, the area of is half that of . Thus:
We also know that since we dropped an altitude from to , the area of must be half of a fifth of the total area of the pentagon. Therefore we can rewrite the above equation as
Now notice that by AA similarity. Therefore, if we can write the areas of the latter two triangles as a ratio of the first triangle, we can express the whole equation in terms of , and by extension , which we know. To find these ratios, we can find the side length ratios and square them because the triangles are similar.
We already know , so let's try to find it's analogous side for and . These sides are and , respectively.
First, , so . Then notice that because we have to fold to hit , so the folding crease has to be exactly halfway between and . Therefore,
Now the ratio between the area of two similar triangles is the square of the ratio of their analogous side lengths. Thus
Now let's move on and calculate the ratio of the other side length. Calculating is slightly tricker. First, we find : . Now since is isosceles, and .
Now . Now note that because opposite over adjacent cancel each other out in a right triangle. Thus,
Now,
Now we go back to our first equation and plug in our values:
Note .
Also note that . Thus .
Now all that remains is to find . We can use the tan addition formula to find the general form of or remember question 25 from this year's AMC 12A. We have that .
Plug in . Then we have Now let . We have the equation Where we let . Using the quadratic formula, we have Now since , , which is what we were looking for. Notice that and , so is between and , and so is it's square. Thus , not the other root.
Finally:
Therefore,
~KingRavi
Solution 3
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 4
Interestingly, we find that the pentagon we need is the one that is represented by the intersection of perpendicular bisectors of the connection from the center of the pentagon to one vertex. Through similar triangles and the golden ratio, we find that the side length ratio of the two pentagons is Thus, the answer is . ~andliu766
Solution 5 (answer choices (not rigorous))
After drawing a decent diagram, we can see that the area of the inner pentagon is quite a bit smaller than half the area of the larger pentagon.
Then, we can estimate the values of the answers and choose one that seems the closest to the smallest answer.
We know that , so we'll use for our estimations. The area of the original pentagon is , so half of it is roughly .
A: clearly, this is wrong because it is greater than half the area of the pentagon.
B: This answer could be right.
C: This too.
D: This answer is wrong, as it assumes that the area of the inner pentagon is exactly half the area of the larger one.
E: This answer could be right.
But, from our diagram, assume that the area of the pentagon is significantly less than the area half of the larger pentagon, so we choose the smallest answer choice, giving us . ~erics118
Supplement (Calculating sin54/cos36 from Scratch)
Method 1:
Construct golden ratio triangle with , and with , . WLOG, let , , .
Method 2:
As explained here,
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=ROVjN3oYLbQ
Video Solution 2 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.