Look at rules in 1 and 2 and 3
Solve the equation 4
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Feb 5, 2020
0
#3
huh? You just rearrange the equation into slope point form:
5y + 10 = 4x - 12
5y = 4x -22
y = 4/5(x) - 22/5
the slope is the number before the x... in this case 4/5 (or 0.8).
Feb 5, 2020
0
#4
why do I get the feeling we are doing his homework for him. Around.... 8th grade?
No, I'd say 7th
6th
As shown in the figure below, a regular dodecahedron (the polyhedron consisting of 
congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?
![[asy] import graph; unitsize(5cm); pair A = (0.082, 0.378); pair B = (0.091, 0.649); pair C = (0.249, 0.899); pair D = (0.479, 0.939); pair E = (0.758, 0.893); pair F = (0.862, 0.658); pair G = (0.924, 0.403); pair H = (0.747, 0.194); pair I = (0.526, 0.075); pair J = (0.251, 0.170); pair K = (0.568, 0.234); pair L = (0.262, 0.449); pair M = (0.373, 0.813); pair N = (0.731, 0.813); pair O = (0.851, 0.461); path[] f; f[0] = A--B--C--M--L--cycle; f[1] = C--D--E--N--M--cycle; f[2] = E--F--G--O--N--cycle; f[3] = G--H--I--K--O--cycle; f[4] = I--J--A--L--K--cycle; f[5] = K--L--M--N--O--cycle; draw(f[0]); axialshade(f[1], white, M, gray(0.5), (C+2*D)/3); draw(f[1]); filldraw(f[2], gray); filldraw(f[3], gray); axialshade(f[4], white, L, gray(0.7), J); draw(f[4]); draw(f[5]); [/asy]](https://images.chesscomfiles.com/proxy/latex.artofproblemsolving.com/0/8/5/08541e5d1cec59e7dd49ea0bbdb45fe328dbecbb/https/d9506650aa.png)
Hijacking this thread a bit, but this problem was too big brainned for me during the amc10 lol
When was the AMC 10? I only remember the AMC 8 lol.
Oh ok.
JeshwantVana wrote:
6th
lol!
Feb 6, 2020
0
#14
can you clarify the end of the problem please? Since you already said that no face can be visited more than once, of course you can't go back to the top face. Does it mean that once you are on the bottom you can't move back up at all? Because that is not how it is worded. Please clarify.
createsure wrote:
can you clarify the end of the problem please? Since you already said that no face can be visited more than once, of course you can't go back to the top face. Does it mean that once you are on the bottom you can't move back up at all? Because that is not how it is worded. Please clarify.
yeah i think the wording is what confused me during the actual test, but i think it means that once you get to the bottom ring of 5 pentagons, you can no longer go to the top one anymore
tldr can't go up a "layer" of the dodecahedron
Feb 6, 2020
0
#16
alright, let me work on it. I should have an answer for you later today.
Feb 6, 2020
0
#17
createsure wrote:
can you clarify the end of the problem please? Since you already said that no face can be visited more than once, of course you can't go back to the top face. Does it mean that once you are on the bottom you can't move back up at all? Because that is not how it is worded. Please clarify.
It sounds like you need to count the number of paths starting on the top face and ending on the bottom face, with the other restrictions indicated. After you reach the bottom face, the path ends. Since you can only move to the bottom face once, the path must end as soon as you are on the bottom face.
Feb 6, 2020
0
#18
For what it's worth, I also misread the question and didn't see it til it was pointed out. Counting without the top/bottom ring restriction seems quite difficult.
Feb 6, 2020
0
#19
I believe it is 810. I counted out almost all of them, then estimated the last few parts, and came up with about 790. If I took the time to calculate each of the 9-step and 10-step paths, I assume I would get 60 more paths instead of the 40 that I estimated.
Also, what they meant by their horribly written problem, is that you can never go back up at all... so once you are on the lower half of the shape you cannot go back up (even if you never went to that spot on the way down.) In other words, you can only move down or sideways during any path. Otherwise the number of paths would be insanely higher.
Feb 6, 2020
0
#20
50% of mathematics is figuring out what the author intended.
@createsure is correct, the answer is 810.
Start by placing an analog clock on the top face to determine clockwise and counterclockwise directions.
Moving from the top to the upper ring, there are 5 faces to move to.
After selecting a starting face on the upper ring, you can follow 9 different paths on the upper ring. Move 1 to 4 faces in the clockwise direction, move 1 to 4 faces in the counterclockwise direction, or only use 1 face in the upper ring.
After determining the final face in the upper ring, there are 2 adjacent faces to move to in the lower ring.
After selecting a starting face on the lower ring, you can follow 9 different paths on the lower ring.
Multiply the number of possibilities to determine the total number of paths.
(5 upper faces)(9 upper ring paths)(2 lower faces)(9 lower ring paths) = 810 total paths
