The worlds first riddle!

[size=8]CIRCLES
As I suspected, the answer is zero. The three points are collinear.
Using a formula I found that gives the equation of the line formed by the two intersection points, I was able to determine that it bisectes AA'.[/size]

Mark:

CIRCLES
As I suspected, the answer is zero. The three points are collinear.
Using a formula I found that gives the equation of the line formed by the two intersection points, I was able to determine that it bisectes AA'. '



The three vertices of the ``triangle'' are always on the radical axis of C and C'!

Similarly, if C and C' are disjoint, there are four line segments tangent to both circles at their endpoints; the resulting four midpoints are collinear, the common line being again the radical axis.

Likewise in higher dimensions. For instance, three spheres close enough to intersect usually meet in two points, call them again P and Q, and have two planes tangent to all three spheres. For each of the planes, assume that the three points of tangency are not collinear, and consider the center of the unique circle through all three. Then P, Q, and the two circumcenters lie on the radical axis of the three spheres, and are thus collinear. If the spheres are disjoint, there can be as many as eight planes tangent to all three; this yields eight circumcenters, all on the radical axis and thus collinear.




What pie quiz?


Pi is transcendental. What does it mean for a number to be transcendental

A. It is equal to the ratio of two integers.
B. It cannot be expressed as the solution of any polynomial with integer coefficients.
C. It is Ralph Waldo Emerson's favorite number.
D. Its square root is equal to -1.
E. Its decimal expansion is infinite in length.



If you pick any two integers at random, what is that probability that they will be relatively prime

("relatively prime" means that the two numbers share no divisors except 1)

A. pi/2
B. pi/3
C. 1/pi
D. 6/(pi^2)
E. pi^2/9

[size=8]PIE QUIZ
part 1: B
part 2: This is funny. A, B, and E aren't valid probabilities. I'll guess D based on the probabilities of picking two even numbers, two multiples of 3, two multiples of 5, etc.[/size]

What do you get when you divide the circumference of a pumpkin by its diameter?

"What do you get when you divide the circumference of a pumpkin by its diameter?"

In geometry, a diameter (Greek words diairo = divide and metro = measure) of a circle is any straight line segment that passes through the center and whose endpoints are on the circular boundary, or, in more modern usage, the length of such a line segment. When using the word in the more modern sense, one speaks of the diameter rather than a diameter, because all diameters of a circle have the same length. This length is twice the radius. The diameter of a circle is also the longest chord that the circle has.

The symbol or variable for diameter is similar in size and design to ø, the lowercase letter o with stroke. Unicode provides character number 8960 (hexadecimal 2300) for the symbol, which can be encoded in HTML webpages as ⌀ or ⌀. Proper display of this character, however, is unlikely in most situations, as most fonts do not have it included.


It is important not to confuse a diameter symbol (ø) with the empty set symbol, similar to the uppercase Ø. Diameter is also sometimes called phi (pronounced the same as "fie"), although this seems to come from the fact that Ø and ø look like Φ and φ, the letter phi in the Greek alphabet.


Therefore, in conclusion; I would say:

[size=7]Pass me a slice of pumpkin pi[/size]

Now to get me some shuteye, back later…

Mark:


PIE QUIZ

part 1: B

part 2: This is funny. A, B, and E aren't valid probabilities. I'll guess D based on the probabilities of picking two even numbers, two multiples of 3, two multiples of 5, etc.




Second helpings of pie:

What is another name for pi in Germany

A. el numero bueno
B. die Ludolphsche Zahl
C. Gesundheit
D. die Eulersche Zahl
E. Drei



For a given value of N: if you add up the first N decimals of pi (excluding the initial 3), the sum is 666 (not to imply that pi is at all satanic).
What is the value of N

A. 144
B. 169
C. 148
D. 123
E. 121

[size=8]PIE LEFTOVERS
B
Ludolph Van Ceulen (1540 - 1610) spent most of his life working out Pi to 35 decimal places. Pi is sometimes known as Ludolph's Constant

A
144[/size]

Try: Your pumpkin pi answer is correct.

Mark:

PIE LEFTOVERS
B
Ludolph Van Ceulen (1540 - 1610) spent most of his life working out Pi to 35 decimal places. Pi is sometimes known as Ludolph's Constant

A
144

Try: Your pumpkin pi answer is correct.

(That is the kind of math I like best…edible!


If you are of a nervous disposition please read no further, the next set of questions are the most devilish I have ever posted…you have been warned



We were sitting around recently discussing birthdays (as we often do), and Gus mentioned that once on his father's birthday in 1937 someone (who didn't know it was his birthday) asked his father when he was born. He told him that he turned x years old in the year x^2, and that if you added his current age to the number of the month he was born in, it equalled the day of the month on which he was born.

When was Gus's father born

May 50th, 1892 haha

GUS' FATHER
Are you sure this is worded correctly?

His current age (in 1937) must be 29 or 30. Otherwise, the month/day combination isn't unique.

That makes his birth year 1907 or 1908. 43^2 = 1849 and 44^2 = 1936. In either case, he couldn't have been 29 or 30 in 1937.

Using 2-digit years ('00, '01, etc.), if he were born in '07 ('08), he would have:
turned 1 in '08 ('09)
turned 2 in '09 ('10)
turned 3 in '10 ('11)
turned 4 in '11 ('12)

At this point, the age squared exceeds the year.

Something seems wrong about this problem.

In view of the comments made by Mark, I have referred the question to our human rights panel for adjudication. I hope I will not prejudice the result by saying; Thoh has the right year and close to the right month, although the day may be a little out.

Full explanation after sunup.

The word handed down by the K9 select symposium gives cold comfort to the appellant. However, as we live in a democracy I feel it is only fair to post the answer in full and let the Supreme Court deal with any dissident.


Gus's father was x year old in the year x^2, and we know he was alive in 1937.

Now x can't be 43 since 43^2 is 1849 (making him 131 in 1937) and 45^2 is 2025 (wouldn't be born until 1980). So he must have turned 44 in 44^2 = 1936, so he was born in 1892.

He turned 45 in 1937, so if d is the day and m is the month he was born on/in, we have that 45 + m = d^2. Since 1 le m le 12 gives a range for d^2 of 46 to 57, m must be 4, since the only perfect square in this range is 49. It follows that d is 7 and so he was born on April 7, 1892.

Comments anyone?



Eight players participated in the recent club chess tournament; each player played all of the others exactly once.

The winner of a game received 1 point and a loser 0; draws are allowed, giving each player 1/2 point. Now, it turned out that everyone received a different number of points. Furthermore, Dolly, who came in second, earned as many points as the four bottom finishers put together.

What was the result of the game between Waldo, who came in third, and Basil, who came in seventh

"Comments anyone?"

You changed the freakin' problem!

From the original problem:
"if you added his current age to the number of the month he was born in, it equalled the day of the month on which he was born"

From the "solution":
"if d is the day and m is the month he was born on/in, we have that 45 + m = d^2"

I think the Supreme Court will find you in contempt.

Mark eloquently wrote, "You changed the freakin' problem!"



Evolutionary change has an image problem. Some people are threatened by it and thus oppose it. Many people -- even many of its defenders -- view question change as irrelevant outside of academia. And in a few cases, even though change is perceived as relevant, some see it as responsible for confusion and misfortune. Public non-appreciation of evolutionary questions may depend as much on its perceived irrelevance as anything else. Yet, evolution, especially in riddles, has been fundamental to some social improvements this century, and it promises to be profoundly important to life in the next generation.

So evolutionary questions are relevant to progress because change essentially represents the same sort of basic questions and problems that evolution is dealing with ~ Quid pro quo

Now that was eloquent even if it was BS.

You may call it Bicarbonate Soda if you will. However I charge multinational corporations fortunes for identical reports.


As no one seems to play chess, how about:


Birthdays (second attempt)

I was sitting around with my friend Waldo and his grandfather Mortimer last week, and the topic of birthday surprises came up. Mortimer mentioned that one of the greatest surprises that he has had involved his grandfather, who happens to have had the same birthday that Mortimer has.
One
year the family was celebrating this double birthday, and during the events Mortimer proudly mentioned to his grandfather that not only he had just turned as old as the last two digits of the year he was born in, but he was also a prime number of years old, and each of the two digits making up his age was also a prime. Mortimer was floored when the older man thought for a second, turned to him, and said that the same thing had just happened to him!

What year did this occur, and how old had Mortimer and his grandfather just turned

[size=8]CHESS
Dolly has at most 6 points (6.5 would mean a tie for first).
The bottom four finishers played C(4,2)=6 games among themselves. Therefore, they have at least 6 points.

Combine these to conclude that Dolly has 6 points and the bottom four total 6 points. That means none of the bottom four received points for games played against the top four.

Therefore, Basil beat Waldo.

BIRTHDAYS
The only candidates for ages of the two men are 23, 37, 53, and 73. Since doubling each age needs to result in the same last two digits, their ages must be separated by a multiple of 50.

Therefore, the year is '46 (you pick the century). Mortimer turned 23, and his grandfather turned 73.[/size]

Mark:

CHESS
Dolly has at most 6 points (6.5 would mean a tie for first).
The bottom four finishers played C(4,2)=6 games among themselves. Therefore, they have at least 6 points.

Combine these to conclude that Dolly has 6 points and the bottom four total 6 points. That means none of the bottom four received points for games played against the top four.

Therefore, Basil beat Waldo.



Well I thought an easy way of solving this is to note that if each participant beat each person who finished below them, this would fit the given criteria. Since Waldo finished above Basil, if the puzzle has a definite answer Waldo must have beaten Basil.

To show this more rigorously, let x_i be the score of the person who finished in ith place. Then x_2 ≤ 6, since clearly x_2 cannot be 7, and if x_2 = 6.5 this would imply x_1 = 7, but then person 1 would have beaten everyone, hence x_2 could not be 6.5.

Now players 5, 6, 7 and 8 played exactly 6 games among themselves, hence x_5 + x_6 + x_7 + x_8 ≥ 6. Since we are given that x_2 = x_5 + x_6 + x_7 + x_8, we conclude that x_2 = 6.

To finish, note that since the sum x_5 + x_6 + x_7 + x_8 is exactly 6, neither player 5, 6, 7 or 8 could have beaten or tied any of players 1, 2, 3 or 4, since then this sum would be greater than 6. It now follows that Waldo beat Basil, does it not?


Please feel free to expose my frailties so the world may mock me. Nevertheless, even though you may be ahead 27 - 0 on appeals, I have a good feeling about this one.

I shall return after breakfast (if I can find the answer to the birthday question) however your answer is looking good! :wink:

Mark:

BIRTHDAYS
The only candidates for ages of the two men are 23, 37, 53, and 73. Since doubling each age needs to result in the same last two digits, their ages must be separated by a multiple of 50.

Therefore, the year is '46 (you pick the century). Mortimer turned 23, and his grandfather turned 73.




It's clear that Mortimer had to have been born in the 1900s, and his grandfather in the 1800s. If Mortimer was born in 1900 + x and his grandfather in 1800 + y, then 1900 + 2x = 1800 + 2y (the year this happened), and so y = x + 50. Now any odd prime number plus an odd number must be even and greater than 2 and so not a prime, hence Mortimer must have been 20-something. But 25 and 27 are not prime, so Mortimer must have been 23, and so his grandfather was 73 (which is indeed also prime). It follows the year was 1900 + 2*23 = 1946.





I was sitting around with my friend Waldo, his nephew Spike, and Spike's friend Molly recently. I happened to have two tickets to a new movie in my pocket that I had just purchased, and I mentioned this and noted that there were two four-digit numbers on the tickets and that the sum of all 8 digits was 25.

Waldo asked if any digit appeared more than twice out of the 8, which I answered, and then Spike asked if the sum of the digits of either ticket was equal to 13, which I answered also. Much to my surprise Molly immediately told me what the two numbers were!

What were they

CHESS
Oops. My logic was correct. I transposed the names in the conclusion. We agree on this one.

[size=8]TICKETS
Although implied, but not explicitly stated, the tickets must have been consecutively numbered. Otherwise, there are multiple solutions for each of the four response combinations.

The responses were both negative. The ticket numbers are 1299 and 1300.[/size]