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Quote:djtehch34tQuote:ThatDonGuyHere's a new problem - bad news: it's a "proof" problem.

Interplanetary Express serves 65 different locations throughout the solar system. Every pair of locations is served by a route that is one of four colors - red, blue, green, or yellow. It turns out that, given any three locations, the three routes that connect them to each other are not all the same color.

Somebody wants to add a 66th location, with routes to each of the other 65, and maintaining the "no single-color triangles" policy.

Prove that this is impossible.

(Hint: the fact that you can with 65 is irrelevant; in fact, I am not entirely sure that such a mapping exists.)

link to original postAssume we have a graph that satisfies the condition (call this 4-colorable). Take the 66th vertex and look at its edges. The color C1 that has the maximum number of edges must have at least ceiling(65/4) = 17 edges. Look at the subgraph with the vertices we get from those edges. This must not have any edges with C1 (aka be 3-colorable), or else we get a cycle with color C1.

Pick an arbitrary vertex in the subgraph of size 17 and repeat the argument. Color C2 must have ceiling(16/3) = 6 edges. So, this creates a 2-colorable subgraph of size 6. Repeat the argument once more to get a 1-colorable subgraph of size ceiling(5/2) = 3. But, this is a contraction, as that's the same as a cycle of the same color.

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Exactly.

Call the three paths that connect three locations to each other a "red triangle" if they are all red; similarly for any other color.

Choose a location; call it Location A.

There are 65 paths from A to the other locations. If there were only 16 of each color, that would be only 64 paths; therefore, there must be at least one color with 17 paths from A. Let's assume it is red.

If any of the paths between any of the 17 locations are also red, then those two locations and A form a red triangle; therefore, they can only be blue, green, or yellow.

Choose one of these 17 locations; call it Location B.

There are 16 paths from B to the other locations of these 17; if there were only 5 of each color, that would be only 15 paths, so there must be at least one color with 6 paths from B to the other 16 locations. Let's assume it is blue.

If any of the paths between any of the 6 locations are also blue, then those two locations and B form a blue triangle; therefore, they can only be green or yellow.

Choose one of these 6 locations; call it Location C.

There are 5 paths from C to these five locations; if there were only 2 of each color, that would be only 4 paths, so there must be at least one color with 3 paths from C to these 5 locations. Let's assume it is green, and let CD, CE, and CF all be green.

If DE is green, then CDE is a green triangle; if DF is green, then CDF is a green triangle; if CEF is green, then CEF is a green triangle; if none of them are green, then DEF is a yellow triangle.

For those of you interested in such things, the principle that, if you have N items and each is one of K types, then there must be at least N/K rounded up items of at least one type is called the Pigeonhole Principle (because you are trying to put N pigeons into K pigeonholes).

To clarify, yes, both players may look at their cards.

Either you simply don't like the form of my answer or there is something missing from your description.Quote:WizardTo reply to all posts since my puzzle on the 13 cards, I can't give anyone credit for a proper answer yet.

To clarify, yes, both players may look at their cards.

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Since L2 should be able to deduce all of this, L1 knows it is futile to offer to exchange any other card but a deuce to a logical opponent. The logical opponent will always refuse any exchange unless they are holding a deuce.

Am I missing something?

Quote:MentalEither you simply don't like the form of my answer or there is something missing from your description.

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Sorry, I missed your answer. Credit given for being correct.

The red figure is a square. If you extend its sides, they cross the x-axis at points 5, 7, 12, and 17.

What is the area of the square?

the triangle hitting 7 & 12 is 3-4-5

So right side (from square to 12) is 4

The triangle hitting 5&12 is similar,

so side from (leftmost) square corner to 12 is. 7*(4/5) = 5.6

So side of square is 5.6-4 = 1.6

Area = 2.56

Hmmm

Same method for triangle hitting 7&17

Side is 10*(3/5)=6

side of square is 6-3 = 3

area = 9

I am going something wrong

Never mind…. Doesn’t have to be 3-4-5 triangle

Will look at it further

Similar triangles…y is side of square

7&12 to 7&17

X/5= (x+y)/10…..x=y

5&12to 5&17

Z/7 = (z+y)//12….z=7y/5

Small triangle x^2 + z^2 = 5^2

Y^2 + (49/25* y^2) =25

Wrong again

Trying to do on phone is not helping

……..

Need

Z^2 + (y+(sqrt (25-x^2))^2 = 7^2

Sub in z and x from above…. Solve with paper and pen….. later

Areas of square =Y^2= __________

Note that the lines touching x = 5 and x = 7 are parallel, and the lines touching x = 12 and x = 17 are parallel, so the four triangles that are visible are all similar.

Let x be the length of the side of the square, a the distance from the square to (5,0), and b the distance from the square to (17,0)

(12 - 5) / a = (17 - 5) / (a + x) => 7 a + 7 x = 12 a => a = 7/5 x

(17 - 7) / b = (17 - 5) / (b + x) => 10 b + 10 x = 12 b => b = 5 x

Pythagorean Theorem: (a + x)^2 + (b + x)^2 = (17 - 5)^2

(12/5 x)^2 + (6 x)^2 = 144

1044/25 x^2 = 144

x^2 = 144 * 25 / 1044 = 100 / 29

Since x is a side of the square, the area is x^2, or 100 / 29

Quote:ThatDonGuy

Note that the lines touching x = 5 and x = 7 are parallel, and the lines touching x = 12 and x = 17 are parallel, so the four triangles that are visible are all similar.

Let x be the length of the side of the square, a the distance from the square to (5,0), and b the distance from the square to (17,0)

(12 - 5) / a = (17 - 5) / (a + x) => 7 a + 7 x = 12 a => a = 7/5 x

(17 - 7) / b = (17 - 5) / (b + x) => 10 b + 10 x = 12 b => b = 5 x

Pythagorean Theorem: (a + x)^2 + (b + x)^2 = (17 - 5)^2

(12/5 x)^2 + (6 x)^2 = 144

1044/25 x^2 = 144

x^2 = 144 * 25 / 1044 = 100 / 29

Since x is a side of the square, the area is x^2, or 100 / 29

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I agree!

Quote:WizardThere are ten lillipad (is that the right spelling?) in a row. A frog is sitting on the first one from one side. He can hop to the next pad or jump over it to the next one. How many ways can he get to the last lillipad? Note, he must land on the last one, no jumping over it.

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There are an infinite number of ways, if the frog can jump in either direction.

If all jumps must be from one side to the other, this is the Fibonacci sequence.

I think you can prove it's a fibonacci series...

Let's assume the starting position is called #9, and the home called #0.

Firstly consider the number of ways of getting from #1 to #0, we'll call this f(1) - obviously there is only one way. So f(1)=1.

Next consider getting from #2 to #0. You can either go #2>#1, then f(1), or #2>#0 = 1. So f(2)=f(1)+1 = 2.

Next consider getting from #3 to #0. You can either go from #3 to #2, then f(2); or #3 to #1, then f(1). So f(3)=f(2)+f(1)

f(1) | 1 |

f(2) | 2 |

f(3) | 3 |

f(4) | 5 |

f(5) | 8 |

f(6) | 13 |

f(7) | 21 |

f(8) | 34 |

f(9) | 55 |

Quote:charliepatrick55.

I think you can prove it's a fibonacci series...

Let's assume the starting position is called #9, and the home called #0.

Firstly consider the number of ways of getting from #1 to #0, we'll call this f(1) - obviously there is only one way. So f(1)=1.

Next consider getting from #2 to #0. You can either go #2>#1, then f(1), or #2>#0 = 1. So f(2)=f(1)+1 = 2.

Next consider getting from #3 to #0. You can either go from #3 to #2, then f(2); or #3 to #1, then f(1). So f(3)=f(2)+f(1)

f(1) 1 f(2) 2 f(3) 3 f(4) 5 f(5) 8 f(6) 13 f(7) 21 f(8) 34 f(9) 55

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Agreed!

(i) Camel A goes to MM4, delivers the message and returns to MM3 (at the end of day 5).

(ii) After two days have passed, Camel B goes to MM3 (to reach it at the end of of day 5). Since there are now 2 units of water left, this enables A and B to return (from MM3) to MM2 (at the end of day 6).

(iii) Similarly Camel C goes to MM2 to meet A and B; this uses up 2 units, so they have 3 units to allow A, B and C to travel to MM1 (at the end of day 7).

(iv) Similarly Camel D goes to MM1 to meet A, B and C; this uses up 1 unit, so they have 4 units to allow all four to travel back to the oasis.

Quote:charliepatrickLet's call the camels A B C and D and, to help describe things, assume there are "mile markers" at 1-day, 2-days, 3-days, 4 days from the oasis.

(i) Camel A goes to MM4, delivers the message and returns to MM3 (at the end of day 5).

(ii) After two days have passed, Camel B goes to MM3 (to reach it at the end of of day 5). Since there are now 2 units of water left, this enables A and B to return (from MM3) to MM2 (at the end of day 6).

(iii) Similarly Camel C goes to MM2 to meet A and B; this uses up 2 units, so they have 3 units to allow A, B and C to travel to MM1 (at the end of day 7).

(iv) Similarly Camel D goes to MM1 to meet A, B and C; this uses up 1 unit, so they have 4 units to allow all four to travel back to the oasis.

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Well done Charlie! I worked out your solution and agree it work. Mine is different, as below.

Day 1: All four camels advance to point 1. Camel A gives camels B, C, D one day of water each. At this point the locations of the four camels are (1,1,1,1). The water supply of each is (1,5,5,5).

Day 2: Camel A returns home. B, C, D advance to 2. B gives C and D one day of water each. At this point the locations are (0,2,2,2). The water supply of each is (0,2,5,5)

Day 3: Camel B goes back to 1. C and D advance to 3. C gives D one day of water. At this point the locations are (0,1,3,3). The water supply of each is (0,1,3,5)

Day 4: Camel B returns home. Camel C returns to 2. Camel D advances to 4 and delivers the letter. At this point the locations are (0,0,2,4). The water supply of each is (0,0,2,4)

Day 5: Camel C returns to 1. Camel D returns to 3. At this point the locations are (0,0,1,3). The water supply of each is (0,0,1,3).

Day 6: Camel C returns home. Camel D returns to 2. At this point the locations are (0,0,0,2). The water supply of each is (0,0,0,2).

Day 7: Camel D returns to 1. After arrival he has one day of water left.

Day 8: Camel D returns to 0 with no water.

I do not require an exact answer. An expression for the answer or to six significant digits will suffice.

You probably will need to be familiar with Euler's Constant (not to be confused with Euler's Number) to solve this one.

(i) Camel A goes from 0 to 4 and back, so uses 8 units of water.

(ii) Camel B goes from 0 to 3 and back, so uses 6 units of water.

(iii) Camel C goes from 0 to 2 and back, so uses 4 units of water.

(iv) Camel D goes from 0 to 1 and back, so uses 2 units of water.

Both solutions engineer how the water is delivered to other camels to enable them to finish their journey.

I think my solution meets returning camels and yours lets each camel complete the return/second part of the journey unaided.

Thus in your solution when A and B leave MM2, together they have 10 units of water, and at MM3, A lands up with 5 units and B 3 units; so B can get home and A can deliver the letter and then get home.

Similarly at MM1 ABC together had 15 units of water, so could get 12 units to MM2, allowing C to get home and AB to continue.

Thus the question changes to how long does the series 1 + 1/2 + 1/3 ... need to get to 100cm.

Well this is a very large number!! (e.g. End after 10000000000 numbers PrevTot = 23.603066594897502 ThisTot = 23.6030665949975 , so a simple program isn't going to find the answer)

Your hint suggests googling the Euler constant and I saw ( https://en.wikipedia.org/wiki/Harmonic_series_(mathematics) ) a formula.

H(n) = ln (n) + Euler's constant + 1/2n - e(n) (where e(n)<1/8n

^{2}).

Using excel and assuming e(n)=the value, this gives 15 092 688 622 100 000 000 000 000 000 000 000 000 000 000 (or about 1.5E43)

After day 1, the ant has 99 cm to go - at least before the rubber band's length is expanded by 2/1

After day 2, after the ant moves and the rubber band is then expanded by 3/2, the ant has (99 x 2 - 1) x 3/2 cm to go

After day 3, it is ((99 x 2 - 1) x 3/2 - 1) x 4/3

...

After day N, the value is 99 N - (N + 1) (1/2 + 1/3 + ...+ 1/N); the solution is the value of N for which this = 0

The answer is the value N where (N + 1) (1/2 + 1/3 + ...+ 1/N) / N = 99

I am working on a numeric approximation, but it appears to be well over 20 billion.

Quote:charliepatrickOn the first day the ant walks 1 centimeter (of the 1m long band), and before the second day the band is enlarged to 2m. So on the second day the ant walks 1 centimeter (of the 2m long band). This is equivilant to walking 1/2 centimeter on a 1m long band. Similarly on other days the ant can be considered as walking 1/N centimeters on a 1m long band.

Thus the question changes to how long does the series 1 + 1/2 + 1/3 ... need to get to 100cm.

Well this is a very large number!! (e.g. End after 10000000000 numbers PrevTot = 23.603066594897502 ThisTot = 23.6030665949975 , so a simple program isn't going to find the answer)

Your hint suggests googling the Euler constant and I saw ( https://en.wikipedia.org/wiki/Harmonic_series_(mathematics) ) a formula.

H(n) = ln (n) + Euler's constant + 1/2n - e(n) (where e(n)<1/8n^{2}).

Using excel and assuming e(n)=the value, this gives 15 092 688 622 100 000 000 000 000 000 000 000 000 000 000 (or about 1.5E43)

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I agree!

There is a one-meter rubber band, with an ant on one end. Each day, the ant travels one centimeter. The rubber band is continuously stretched at a rate of one meter per day. How long will it take for the ant to reach the other side?

An expression for the answer or to six significant digits will suffice.

There are five gopher holes in a row. A gopher is in one of them. It is your task to find the gopher. Every day you may pick one hole to search. If you pick incorrectly, the gopher will move to a neighboring hole, in either direction at night.

What strategy will allow you to find the gopher in the least number of days for certain?

For extra credit, what is the strategy for n gopher holes?

Quote:avianrandyAlso don't forget alextrebek stamp being issued July 22

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I am still trying to find the last time before Trebek that USPS issued a stamp commemorating a person that did not have the person's image on it.

I googled the triangle and stumbled across an interesting article with some triangles length 3 (e.g. 162,243,351) and others length 4. Apparently they normalise solutions so the lowest tip is North and the next SE (otherwise you have six ways for each solution).

One interesting length 4 solution had the property that the numbers were a triangle, and also if you squared the numbers they were. I'll leave it to the reader to work it out. (Note the 3 length triangles use the numbers 1 to 6, and the 4 length 1 to 9.)

This page I read has the solution https://www.cuemath.com/learn/do-you-want-to-know-the-secrets-of-the-magic-triangle/

Quote:WizardI was hoping for more traction on the previous puzzle, but maybe integral calculus isn't everybody's cup of tea.

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(snip)

I'm assuming that's deliberate, and I feel I must groan.