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**46 members have voted**

From the diagram above we can see that the number of tiles on the perimeter, 16, exceeds the number of tiles on the inside, 8.

How many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?

Quote:GialmereHow many rectangles exist for which the number of tiles on the perimeter are equal to the number of tiles on the inside?

The following are the dimensions of the inner rectangle:

3x10

4x6

Let H and W be the dimensions, with H >= W

There are a total of HW squares, of which (H-2)(W-2) are interior

The number of interior = the number of exterior when HW = 2 (H-2)(W-2) = 2HW - 4H - 4W + 8

This is HW = 4H + 4W - 8

H (W - 4) = 4W - 8

H = (4W - 8) / (W - 4)

H = (4W - 16 + 8) / (W - 4)

H = 4 + 8 / (W - 4)

Since H is an integer, (W - 4) is a factor of 8

The only factors of 8 are 1, 2, 4, and 8, with resulting H values of 12, 8, 6, and 5, but the latter two have W > H

The only H values where H > W are 12 and 8; the corresponding W values are 5 and 6

12 x 5 and 8 x 6 are the only solutions

Quote:Wizard2

The following are the dimensions of the inner rectangle:

3x10

4x6

Quote:ThatDonGuy

Let H and W be the dimensions, with H >= W

There are a total of HW squares, of which (H-2)(W-2) are interior

The number of interior = the number of exterior when HW = 2 (H-2)(W-2) = 2HW - 4H - 4W + 8

This is HW = 4H + 4W - 8

H (W - 4) = 4W - 8

H = (4W - 8) / (W - 4)

H = (4W - 16 + 8) / (W - 4)

H = 4 + 8 / (W - 4)

Since H is an integer, (W - 4) is a factor of 8

The only factors of 8 are 1, 2, 4, and 8, with resulting H values of 12, 8, 6, and 5, but the latter two have W > H

The only H values where H > W are 12 and 8; the corresponding W values are 5 and 6

12 x 5 and 8 x 6 are the only solutions

Correct!

---------------------------------------

The Wi-Fi at my house went out this Thanksgiving.

There are 148 tiles in my bathroom.

You go to visit your pool playing friend and shoot a few games. He's a very superstitious player and had a special 16-ball made to replace his table's 8-ball.

He also insists on racking the balls in a certain way. He arranges them so that each ball is the difference of the two balls above it. He begins to show you how to do it when his phone rings and he leaves the room.

Can you figure out how to finish racking the balls?

Quote:GialmereCan you figure out how to finish racking the balls?

It's not so much hard as exhaustive, as it's a bit of a brute force problem, but the solution is:

5, 14, 16, 3, 15

9, 2, 13, 12

7, 11, 1

4, 10

6

Quote:ThatDonGuy

It's not so much hard as exhaustive, as it's a bit of a brute force problem, but the solution is:

5, 14, 16, 3, 15

9, 2, 13, 12

7, 11, 1

4, 10

6

Correct!

--------------------------------

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Total = 140 combos.

So P=140/2208.

Quote:charliepatrickTotal perms =25*24*23/6 = 2208

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Total = 140 combos.

So P=140/2208.

...having three in the same "line" that are knight's moves from each other?

i.e. if you label the rows 1-5 from top to bottom and the columns A-E from left to right, then if the pegs are in A1, C2, and E3, that's not a triangle?

There appear to be 12 of these:

A1-C2-E3, A2-C3-E4, A3-C4-E5

A3-C2-E1, A4-C3-E2, A5-C4-E3

A1-B3-C5, B1-C3-D5, C1-D3-E5

A5-B3-C1, B5-C3-D1, C5-D3-E1

Quote:ThatDonGuyQuote:charliepatrickTotal perms =25*24*23/6 = 2208

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Total = 140 combos.

So P=140/2208.

...having three in the same "line" that are knight's moves from each other?

i.e. if you label the rows 1-5 from top to bottom and the columns A-E from left to right, then if the pegs are in A1, C2, and E3, that's not a triangle?

There appear to be 12 of these:

A1-C2-E3, A2-C3-E4, A3-C4-E5

A3-C2-E1, A4-C3-E2, A5-C4-E3

A1-B3-C5, B1-C3-D5, C1-D3-E5

A5-B3-C1, B5-C3-D1, C5-D3-E1

Nice, can't you go a step beyond the knight's move & do a 1 diagonal and 3 spaces up movement or 2 diagonal & 2 spaces up movement to get a triangle as well?

Wait looks like Charlie is only counting the permutations of non-triangle outcomes. So he is accounting for all other outcomes to give a triangle as a result.

Total combinations = 25!/(25-3)!3! = 2,300

Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.

Vertical / Horizontal - 10 combos per 10 lines = 100 combos.

Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300

Quote:USpapergamesOk, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3!)3! = 2,300

Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.

Vertical / Horizontal - 10 combos per 10 lines = 100 combos.

Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300

You missed the same extra diagonals that Donguy pointed out above. For example, A1-C2-E3

Quote:GialmereIt's toughie Tuesday...

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?

I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%

Quote:unJonQuote:USpapergamesOk, so I have the actual answer! Looks like you guys need help with combination analysis.

Total combinations = 25!/(25-3!)3! = 2,300

Total permutations = 25*24*23 = 13,800

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical. It's ironic that Charlie knows this but still got the wrong answer.

Vertical / Horizontal - 10 combos per 10 lines = 100 combos.

Diagonal - 10 combos per 2 (5 hole) lines + 4 combos per 4 (4 hole) lines + 1 combo per 4 (3 hole) lines = 40 combos.

So P = 140/2,300

You missed the same extra diagonals that Donguy pointed out above. For example, A1-C2-E3

It's funny because I thought so also but I believe Donguy is wrong since I am only counting outcomes that don't result in a triangle outcome. So all the knight moves are definitely counted in terms of successful triangle outcomes.

Quote:ChesterDogQuote:GialmereIt's toughie Tuesday...

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

What is the probability that the three pegs will form a triangle?

I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%

Please share your work, this can't be right. Your saying I missed 12 outcomes that don't result in a triangle, and that's just hard for me to believe because from what I have seen on this form, I am the Combo King.

Quote:USpapergames...

Please share your work, this can't be right. Your saying I missed 12 outcomes that don't result in a triangle, and that's just hard for me to believe because from what I have seen on this form, I am the Combo King.

I found 12 combinations congruent to A1-B3-C5.

2 Vertical/Horizantal | 1 Diagonal = 6

1 Vertical/Horizantal | 2 Diagonal = 6

Quote:ChesterDog

I found 12 combinations congruent to A1-B3-C5.

Your right, I completely misunderstood the point Donguy was making. The entire time I thought he was attempting to count all the triangle outcomes and was forgetting to include all the triangles from knight moves lol. So exclusion method doesn't work unless you are aware of all possible straight line combinations. Ok so then there are 12 that we missed :(

It's possible I was too narrow-minded after agreeing with charliepatrick but I think the real culprit was being under the impression the question would be easy. Coming up with the answer in less than 2 minutes felt right to me. No need to overthink a simple question lol. It's times like this that I realize my arrogance can be a huge weakness if I were to ever underestimate a problem that could cost me a fortune.

Quote:charliepatrick...this is wrong...

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Knight moves (each one only have 1 perm - e.g. A1 C2 E3) = 12 combos

(i) West to East - going up/North - three

(ii) West to East - going down/South - three

(iii) North to South - going right/East - three

(iv) North to South - going left/West - three

Total = 152 combos.

So P=152/2300.

Quote:USpapergamesNow here is an easy math puzzle ;) How far can you walk into a forest?

Of course if you didn't see it you might walk straight into a tree - gosh that hurt! And if you had come from a long way away, then it might be miles.

btw your "toes" will have come "far" to make it to "a forest" (anag).

The question isn't about how the traveler would know when he has reached the middle but rather how far is it possible for the traveler to be within the forest.

You make me laugh long timeQuote:USpapergamesOk, so I have the actual answer! Looks like you guys need help with combination analysis.

Quote:ChesterDog

I counted 152 combinations that are colinear out of a total of 2300 leaving 2148 combinations to make triangles.

Probability of a triangle = 2148 / 2300 = 537 / 575 or approximately 93.39%

Quote:charliepatrickTotal perms =25*24*23/6 = 2300

For a triangle to have zero area the holes must be in the same line, diagonal or horizontal/vertical.

Diagonal 3-length lines - 4 of them, 1 perm = 4 combos

Diagonal 4-length lines - 4 of them, 4 perms = 16 combos

Diagonal 5-length lines - 2 of them, 10 perms = 20 combos

Vertical/Horizontal 5-length lines - 10 of them, 10 perms = 100 combos

Knight moves (each one only have 1 perm - e.g. A1 C2 E3) = 12 combos

(i) West to East - going up/North - three

(ii) West to East - going down/South - three

(iii) North to South - going right/East - three

(iv) North to South - going left/West - three

Total = 152 combos.

So P=152/2300.

Correct!

(With an assist from ThatDonGuy)

-----------------------------------

Quote:Ace2You make me laugh long time

Glad to know my confusion was amusing ;) And I was assuming all Charlie did wrong was calculating the combinations wrong and title it as permutations. Even assumed Donguy was attempting to solve the problem by calculating all the possible triangle combos before realizing he was just finding more exclusion outcomes (or more straight lines).

Question, what type of math puzzles come to mind when they are easy? To me, the definition of an easy math puzzle is a puzzle that requires no hidden information to solve, where all parts of the puzzle are straightforward & obvious. The puzzle could even take a very long time to solve as long as the average high school student could complete the puzzle with no help.

P.S. It's not like my combinatorial analysis was wrong either ;) Just didn't account for the extra straight-line outcomes which weren't obvious to me within the 2 minutes it took for me to complete my analysis lol. Don't forget I'm the only person to realize Charlie's permutation value was wrong & to come up with the correct combinations & permutation values for the problem.

The digits 0-9 are written down on ten 3x5 cards. The cards are then separated into five groups of two such that each group makes a prime number. For example...

How many different sets of five primes can you form this way?

Quote:Gialmere...How many different sets of five primes can you form this way?...

Quote:Gialmere

How many different sets of five primes can you form this way?

Each card must contain an odd and an even number, with the even preceding the odd.

05 is the only prime you can create with the number 5, so we only need to look at the ways to place the numbers 2,4,6,8 with 1,3,7,9.

We have the following possible primes:

41, 61

23, 43, 83

47, 67

29, 89

Because 41/61 choice will determine what number we use for 47/67, then we can never use the number 43.

This gets us down to:

41, 61

23, 83

47, 67

29, 89

So, it looks to me like there are only 4 choices:

05, 41, 67, 23, 89

05, 41, 67, 83, 29

05, 61, 47, 23, 89

05, 61, 47, 83, 29

Quote:CrystalMathQuote:Gialmere

How many different sets of five primes can you form this way?

Each card must contain an odd and an even number, with the even preceding the odd.

05 is the only prime you can create with the number 5, so we only need to look at the ways to place the numbers 2,4,6,8 with 1,3,7,9.

We have the following possible primes:

41, 61

23, 43, 83

47, 67

29, 89

Because 41/61 choice will determine what number we use for 47/67, then we can never use the number 43.

This gets us down to:

41, 61

23, 83

47, 67

29, 89

So, it looks to me like there are only 4 choices:

05, 41, 67, 23, 89

05, 41, 67, 83, 29

05, 61, 47, 23, 89

05, 61, 47, 83, 29

Your first assumption is incorrect:

Quote:GialmereHow many different sets of five primes can you form this way?

0 cannot be the second digit as no primes end in 0, so there are four two-digit and one one-digit primes (i.e. 2, 3, 5, or 7).

The only primes < 100 that contain a 5 are 5, 53, and 59

4, 6, and 8 have to be the first digit of a prime as otherwise it would be an even number > 2

If 2 is not the one-digit prime, it must also be the first digit of a prime

The two-digit primes that begin with 2 are 23 and 29

The two-digit primes that begin with 4 are 41, 43, and 47

The two-digit primes that begin with 6 are 61 and 67

The two-digit primes that begin with 8 are 83 and 89

If 5 is one of the primes, then the other four begin with 2, 4, 6, 8 and end with 1, 3, 7, 9

The possibilities are:

05, 23, 41, 67, 89

05, 23, 47, 61, 89

05, 29, 41, 67, 83

05, 29, 47, 61, 83

If 53 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 7, 9

The possibilities are:

02, 53, 41, 67, 89

02, 52, 47, 61, 89

If 59 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 3, 7

The possibilities are:

02, 59, 41, 67, 83

02, 59, 47, 61, 83

There are a total of 8 sets of five primes

Quote:unJonQuote:CrystalMathQuote:Gialmere

How many different sets of five primes can you form this way?

Your first assumption is incorrect:You can make a card with 02 which would then let you make a 53 card or a 59 card.

d'oh

Quote:ThatDonGuy

0 cannot be the second digit as no primes end in 0, so there are four two-digit and one one-digit primes (i.e. 2, 3, 5, or 7).

The only primes < 100 that contain a 5 are 5, 53, and 59

4, 6, and 8 have to be the first digit of a prime as otherwise it would be an even number > 2

If 2 is not the one-digit prime, it must also be the first digit of a prime

The two-digit primes that begin with 2 are 23 and 29

The two-digit primes that begin with 4 are 41, 43, and 47

The two-digit primes that begin with 6 are 61 and 67

The two-digit primes that begin with 8 are 83 and 89

If 5 is one of the primes, then the other four begin with 2, 4, 6, 8 and end with 1, 3, 7, 9

The possibilities are:

05, 23, 41, 67, 89

05, 23, 47, 61, 89

05, 29, 41, 67, 83

05, 29, 47, 61, 83

If 53 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 7, 9

The possibilities are:

02, 53, 41, 67, 89

02, 52, 47, 61, 89

If 59 is one of the primes, then 2 is another one, and the other three begin with 4, 6, 8 and end with 1, 3, 7

The possibilities are:

02, 59, 41, 67, 83

02, 59, 47, 61, 83

There are a total of 8 sets of five primes

Correct!

(Although, I spy a composite number typo in there.)

--------------------------------------

A fair die is rolled ten times.

What is the probability that the sequence of rolls is non-decreasing?

That is, each roll is greater than or equal to the previous roll.

Quote:GialmereWhat is the probability that the sequence of rolls is non-decreasing?

This looks like a Markov chain problem - suitable more for computing than math.

Let P(a,b) be the probability of success after a rolls where the most recent roll is b

P(10,n) = 1 for all n

P(a,n) = 1/6 (P(a+1,n) + P(a+1,n+1) + ... + P(a+1,6)) for all n, a <= n <= 6

The solution is P(0,1)

Rolls | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

9 | 1 | 5 / 6 | 2 / 3 | 1 / 2 | 1 / 3 | 1 / 6 |

8 | 7 / 12 | 5 / 12 | 5 / 18 | 1 / 6 | 1 / 12 | 1 / 36 |

7 | 7 / 27 | 35 / 216 | 5 / 54 | 5 / 108 | 1 / 54 | 1 / 216 |

6 | 7 / 72 | 35 / 648 | 35 / 1,296 | 5 / 432 | 5 / 1,296 | 1 / 1,296 |

5 | 7 / 216 | 7 / 432 | 7 / 972 | 7 / 2,592 | 1 / 1,296 | 1 / 7,776 |

4 | 77 / 7,776 | 35 / 7,776 | 7 / 3,888 | 7 / 11,664 | 7 / 46,656 | 1 / 46,656 |

3 | 11 / 3,888 | 55 / 46,656 | 5 / 11,664 | 1 / 7,776 | 1 / 34,992 | 1 / 279,936 |

2 | 143 / 186,624 | 55 / 186,624 | 55 / 559,872 | 5 / 186,624 | 1 / 186,624 | 1 / 1,679,616 |

1 | 1,001 / 5,038,848 | 715 / 10,077,696 | 55 / 2,519,424 | 55 / 10,077,696 | 5 / 5,038,848 | 1 / 10,077,696 |

Result = 1,001 / 20,155,392 = about 1 / 20,135

Quote:ThatDonGuy

This looks like a Markov chain problem - suitable more for computing than math.

Let P(a,b) be the probability of success after a rolls where the most recent roll is b

P(10,n) = 1 for all n

P(a,n) = 1/6 (P(a+1,n) + P(a+1,n+1) + ... + P(a+1,6)) for all n, a <= n <= 6

The solution is P(0,1)

Rolls 1 2 3 4 5 6 9 1 5 / 6 2 / 3 1 / 2 1 / 3 1 / 6 8 7 / 12 5 / 12 5 / 18 1 / 6 1 / 12 1 / 36 7 7 / 27 35 / 216 5 / 54 5 / 108 1 / 54 1 / 216 6 7 / 72 35 / 648 35 / 1,296 5 / 432 5 / 1,296 1 / 1,296 5 7 / 216 7 / 432 7 / 972 7 / 2,592 1 / 1,296 1 / 7,776 4 77 / 7,776 35 / 7,776 7 / 3,888 7 / 11,664 7 / 46,656 1 / 46,656 3 11 / 3,888 55 / 46,656 5 / 11,664 1 / 7,776 1 / 34,992 1 / 279,936 2 143 / 186,624 55 / 186,624 55 / 559,872 5 / 186,624 1 / 186,624 1 / 1,679,616 1 1,001 / 5,038,848 715 / 10,077,696 55 / 2,519,424 55 / 10,077,696 5 / 5,038,848 1 / 10,077,696

Result = 1,001 / 20,155,392 = about 1 / 20,135

Correct!

Very good.

-------------------------------

There's another way to calculate this. If we only count permutations that have at least one of each side 1-6, then the answer would be combin(9,5) = 126 (out of 6^10). This is because there are 126 ways to insert 5 dividers between 10 values so that 6 bins are created. This, for example, would represent 1123334556: xx/x/xxx/x/xx/xQuote:GialmereA fair die is rolled ten times.

What is the probability that the sequence of rolls is non-decreasing?

That is, each roll is greater than or equal to the previous roll.

However, a bin can contain zero values, so we add 1 for each of the 6 bins and "overcount" each side by 1. But not exactly since the value we are overcounting is a zero. x/xxx/x/xxxxxx/xxx/xx represents 2244444556, for instance.

Therefore the answer becomes combin(15,5) = 3,003 (out of 6^10). This is because there are 3,003 ways to insert 5 dividers between 16 values so that 6 bins are created.

Oh no! You're driving your car through town when you see you're going to be stuck behind a big Budget Rental Truck at the next traffic light.

A Budget Rental Truck is 12 feet 8 inches tall (from the road to the top of the Van). The cab of the truck is 8 feet long and the van is 24 feet long (see diagram below).

If the front of the truck is 8 feet from the stoplight suspended 18 feet above the road, how far behind the truck must your car be so that you can still see the stoplight?

Assume your eyes are 4 feet above the road.

The top of the truck is 5 1/3 feet below the stoplight, and 40 feet to the right of the stoplight

Let X be the distance you are behind the truck

Your eyes are 14 feet below the stoplight, and (40 + X) feet to the right of the stoplight

The minimum distance where you can see the stoplight is where the two height/length ratios are equal

40 / (5 1/3) = (40 + X) / 14

15 / 2 = (40 + X) / 14

210 = 80 + 2X

X = 65 feet

ray from light from signal to top right corner of rear of truck's van continues on to your eye.

similar triangles give (in feet)

x/(8+8+24) =((12+2/3) - 4) / (18 - (12+2/3))

x= (40 ft) * (104 in) / (64 in) =

x= 65 ft

Though this assumes your eyes are located at the front of your car. If we assume a car length of maybe 12 ft and driver's seat is say 4 ft from the front bumper....

x = 61 ft

EDIT: i.e. the car's front bumper is at 61 ft from rear of truck....your eyes are still at 65ft

Quote:ThatDonGuy

The top of the truck is 5 1/3 feet below the stoplight, and 40 feet to the right of the stoplight

Let X be the distance you are behind the truck

Your eyes are 14 feet below the stoplight, and (40 + X) feet to the right of the stoplight

The minimum distance where you can see the stoplight is where the two height/length ratios are equal

40 / (5 1/3) = (40 + X) / 14

15 / 2 = (40 + X) / 14

210 = 80 + 2X

X = 65 feet

Quote:chevy

ray from light from signal to top right corner of rear of truck's van continues on to your eye.

similar triangles give (in feet)

x/(8+8+24) =((12+2/3) - 4) / (18 - (12+2/3))

x= (40 ft) * (104 in) / (64 in) =

x= 65 ft

Though this assumes your eyes are located at the front of your car. If we assume a car length of maybe 12 ft and driver's seat is say 4 ft from the front bumper....

x = 61 ft

EDIT: i.e. the car's front bumper is at 61 ft from rear of truck....your eyes are still at 65ft

Correct!

-----------------------------------

A man was waiting at a red light when an alligator and a parrot approached his car window.

The man rolled down his window to see what they wanted.

The alligator said "Excuse me sir, we have reason to believe you've taken hallucinogenics."

The ratio of boys to girls at a high school dance is 9:10.

An extra 17 boys arrive and the ratio changes to 8:7.

How many girls are there at the dance?

Quote:GialmereHow many girls are there at the dance?

B=# of Boys, G=# of Girls)

Originally : B = (9/10) G

Later : B+17 = (8/7) G

Solving

(9/10) G +17 = (8/7) G

G=70 girls (and B= 63 boys)

Quote:Wizard70

Quote:chevy

B=# of Boys, G=# of Girls)

Originally : B = (9/10) G

Later : B+17 = (8/7) G

Solving

(9/10) G +17 = (8/7) G

G=70 girls (and B= 63 boys)

Correct!

----------------------------

Trust me ... You can dance.

Love,

Alcohol

A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.

What is the radius of the circle?

Quote:Gialmere

......

What is the radius of the circle?

Let r = the radius of the circle.

Let A = the central angle subtending a side of length 2.

Let B = the central angle subtending a side of length 7.

Let C = the central angle subtending a side of length 11.

Apply the "law of cosines" for each angle:

cos A = ( r

^{2}+ r

^{2}- 2

^{2}) / (2 r

^{2}) = 1 - 2

^{2}/ ( 2 r

^{2})

cos B = ( r

^{2}+ r

^{2}- 7

^{2}) / (2 r

^{2}) = 1 - 7

^{2}/ ( 2 r

^{2})

cos C = ( r

^{2}+ r

^{2}- 11

^{2}) / (2 r

^{2}) = 1 - 11

^{2}/ ( 2 r

^{2})

Using "trial and error" to find r such that 2A + 2B + 2C = 360 degrees yields r = 7.