## Poll

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34 members have voted

CrystalMath Joined: May 10, 2011
• Posts: 1898
Thanks for this post from: August 27th, 2020 at 6:57:16 PM permalink
Quote: Gialmere An amusement park wants to restore its Merry-Go-Round and hires a group of artists for the job.

One artist is assigned to paint the carousel floor which is formed by two concentric circles (an annulus). She wants to determine the area of the floor (shown in yellow in the figure below), so she will know how much paint to buy.

Because of all the machinery in the middle, she is unable to measure the radii of the two circles. However, she finds the length of a special chord to be 70 feet. This special chord is a chord of the larger circle and a tangent to the smaller circle. (See diagram below). Can she determine the area of the Merry-Go-Round which she needs to paint using just that one measurement?

Surprisingly, yes.

R = sqrt(35^2 + r^2)
A = pi(R^2 - r^2) = pi*(35^2 + r^2 - r^2) = 1225pi
I heart Crystal Math.
charliepatrick Joined: Jun 17, 2011
• Posts: 2444
Thanks for this post from: August 28th, 2020 at 1:00:13 AM permalink You use Pythagoras.
The area of the large circle is Pi r" 2.
The area of the small circle is Pi r' 2.
You want to know the difference.
r" = OB, r' = OA, OB2 = OA2 + AB2.
So AB2=OB2-OA2.
So the area being painted = Pi AB2.

In this case AB = 35 ft so the area (using Pi=22/7 as one would be allowed to in the exams in the old days and is why the number is divisible by 7) (approx) = 35*35*22/7 = 3850 sq ft.
Gialmere Joined: Nov 26, 2018
• Posts: 2112
August 28th, 2020 at 7:48:26 AM permalink
Quote: CrystalMath

Surprisingly, yes.

R = sqrt(35^2 + r^2)
A = pi(R^2 - r^2) = pi*(35^2 + r^2 - r^2) = 1225pi

Quote: charliepatrick You use Pythagoras.
The area of the large circle is Pi r" 2.
The area of the small circle is Pi r' 2.
You want to know the difference.
r" = OB, r' = OA, OB2 = OA2 + AB2.
So AB2=OB2-OA2.
So the area being painted = Pi AB2.

In this case AB = 35 ft so the area (using Pi=22/7 as one would be allowed to in the exams in the old days and is why the number is divisible by 7) (approx) = 35*35*22/7 = 3850 sq ft.

Correct!

You have to admire the simple elegance of this one.
--------------------------------

I caught a cold on a carousel once.

There must have been something going around.
Have you tried 22 tonight? I said 22.
Gialmere Joined: Nov 26, 2018
• Posts: 2112
August 28th, 2020 at 11:08:32 AM permalink Once again you are messing around with an old computer Tic-Tac-Toe game.

This time you program it to randomly place an X on the grid, and then randomly place an O, and then an X and so on.

Played in this manner, what are the odds that...

...X will win the game?
...O will win the game?
...it will end in a draw?
Have you tried 22 tonight? I said 22.
Gialmere Joined: Nov 26, 2018
• Posts: 2112
September 2nd, 2020 at 10:36:02 AM permalink
Quote: Gialmere Once again you are messing around with an old computer Tic-Tac-Toe game.

This time you program it to randomly place an X on the grid, and then randomly place an O, and then an X and so on.

Played in this manner, what are the odds that...

...X will win the game?
...O will win the game?
...it will end in a draw?

X wins: (.5849206...)
O wins: (.2880952...)
Draw: (.1269841...)
Have you tried 22 tonight? I said 22.
Gialmere Joined: Nov 26, 2018
• Posts: 2112
September 2nd, 2020 at 10:58:10 AM permalink There is a coin shortage going on in the US and you happen to have plenty of change on hand. I ask you to give me change for \$1.

How many different ways can you change an American dollar bill into coins?
Have you tried 22 tonight? I said 22.
charliepatrick Joined: Jun 17, 2011
• Posts: 2444
September 2nd, 2020 at 12:37:33 PM permalink
Only one way, you give me the dollar and I give you the coins - but that's not what you're looking for!
Only an idea....
The dollar can either be broken into coins of (i) 50c + 50c (ii) 50c + two quarters (iii) four quarters.

50c (not being 25c) can either be {50} or {10 10 10 10 10} = 2 ways = A B
25c can be {25} {10 10 5} {10 5 5 5} {5 5 5 5 5} {10 10 1s} {10 5 5 1s} (10 5 1s} {10 1s} {5 5 5 5 1s} {5 5 5 1s} {5 5 1s} {5 1s} {1s} = 13 ways = a b c d e f g h i j k l m

(i) 50+50 can be AA,BB,AB = 3 ways
(ii) 50+25+25 can be (A or B) 2 ways, with (pair e.g. aa 13 ways) or (dispair e.g. ab) 13*12/2 = 78 ways; = 2 * (13+78) = 182 ways
(iii) 25+25+25+25 can be any permutations of four of (a...m)
(A) Four same (aaaa) = 13 ways
(B) Three/One (aaab) = 13*12 = 156 ways
(C) Two/Two (aabb) = 13*12/2 = 78 ways
(D) Two/One/One (aabc) = 13* 12*11/2 = 858 ways
(E) One/One/One/One (abcd) = 13*12*11*10/4/3/2/1 = 715 ways
Total = 13+156+78+858+715 = 1820 ways

So grand total = 3 + 182 + 1820 = 2005 ways.
After thought: I think my way over estimates things as it double counts various combinations!
Last edited by: charliepatrick on Sep 2, 2020
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4959
Thanks for this post from: September 2nd, 2020 at 1:16:17 PM permalink

If you use only nickels and pennies, there are 21 ways: 100 pennies, 95 pennies + 1 nickel, 90 pennies + 2 nickels, ..., 5 pennies + 19 nickels, 20 nickels.

Each dime added removes two of the methods; you can have up to 10 dimes, so there are 21 + 19 + 17 + ... + 1 = 121 ways with just pennies, nickels, and dimes.

With one quarter, you can have:
no dimes and 0, 1, 2, ..., 15 nickels
1 dime and 0, 1, ..., 13 nickels
2 dimes and 0, 1, ..., 11 nickels
3 dimes and 0, 1, ..., 9 nickels
4 dimes and 0, 1, ..., 7 nickels
5 dimes and 0, 1, ..., 5 nickels
6 dimes and 0, 1, 2, 3 nickels
7 dimes and 0 or 1 nickels
There are 16 + 14 + 12 + ... + 2 = 72 ways with one quarter

With two quarters, you can have:
no dimes and 0, 1, 2, ..., 10 nickels
1 dime and 0, 1, 2, ..., 8 nickels
2 dimes and 0, 1, 2, ..., 6 nickels
3 dimes and 0, 1, 2, 3, 4 nickels
4 dimes and 0, 1, 2 nickels
5 dimes
There are 11 + 9 + 7 + 5 + 3 + 1 = 36 ways with two quarters
For each way with two quarters, replace the two quarters with a half dollar; there are 36 ways of this

With three quarters, you can have:
no dimes and 0, 1, 2, 3, 4, 5 nickels
1 dime and 0, 1, 2, 3 nickels
2 dimes and 0 or 1 nickels
There are 6 + 4 + 2 = 12 ways with three quarters
For each way, replace two of the quarters with a half dollar; this is another 12 ways

Finally, there are:
Four quarters
One half dollar, two quarters
Two half dollars
One dollar coin

The total is 121 + 72 + 36 + 36 + 12 + 12 + 4 = 293

Ace2 Joined: Oct 2, 2017
• Posts: 1070
September 2nd, 2020 at 2:16:23 PM permalink
This is a partition problem and there is a way to compute the answer using generating functions.

Generating functions are not my forte.
It�s all about making that GTA
charliepatrick Joined: Jun 17, 2011
• Posts: 2444
Thanks for this post from: September 2nd, 2020 at 2:22:21 PM permalink
I agree with ThatDonGuy
You can either use just 25c or higher, else get 10c and lower involved.
(I) \$; 50+50; 50+25+25; 25+25+25+25 = 4 Ways.
(II) else use 25c+ to use up None: 0, One: 25, Two: 50/25 25, Three: 50 25/25 25 25 quarters' worth.
(a) No quarters could be
10x10c (no 5c) - 1 way
9x10c (2x5c) - (could be 2 0, 1 5, 0 10) - 3 ways . . .
0x10c (20x5c) - (could be 20 0, 19 5, ... ,0 100.) - 21 ways
1+3+...+19+21 = 121 Ways.
(b) One quarter could be
7x10c (1 5c) - 2 ways . . .
0x10c (15 5c) - 16 ways
2+4+...+14+16 = 72 Ways.
(c) Two quarters (or 1 50c) could be
5x10c (no 5c) - 1 way . . .
0x10c (10x5c) - 11 ways
1+3+5+7+9+11 = 36 : then there are two ways to be here so 72 Ways.
(d) Three quarters (or 50c 25c) could be
2x10c (1 5c) - 2 ways . . .
0x10c (5 5c) - 6 ways
2+4+6 = 12 : then there are two ways to be here so 24 Ways.

Ways: 4 121 72 72 24 = 293 WAYS.