The worlds first riddle!

the leash is shortening at 3.51 feet per second.

LEASH
DrewDad has it right. The rate of decrease doesn't drop to 3.51 f/s until the dog is a bit more than 2.18 feet from the corner.

DrewDad

Hooter: The leash starts shortening when Hooter turns the corner, and is at its minimum when you are both 4 feet from the corner. At the moment when Hooter reaches the corner and turns, the leash is shortening at 6 feet per second.

It sure is.

Let's call the distance between me and the corner x and the distance, after his turn, between Hooter and the corner y. We can set up a coordinate system so that just as Scooter reaches the corner, my coordinate is (0,8). The length of the leash is L, where L2 = x2 + y2

Differentiating with respect to time gives
2L(dL/dt) = 2x(dx/dt) + 2y (dy/dt).

When Hooter reaches the corner we have x = 8 ft, y = 0 ft, L = 8 ft, dx/dt = 6 ft/sec, dy/dt = 6 ft/sec. (The negative sign on dx/dt comes from the fact that x is decreasing.)

Plugging in these terms gives that, when Hooter reaches the corner, the length of his leash is decreasing at a rate of 6 ft/sec. To find the minimum length, set dL/dt = 0 and solve to get x = y = 4, L = 4  5.65 ft.


Three lives: Water (water, vapor, ice)

Burden: a snail



Usamashaker
the river

1700 feet

Voilà! No equations.
When the boats meet the first time, the total distance they have travelled spans the river once. When they meet for the second time, they have spanned the river together twice more. As the boats are travelling at a constant speed, the last distance covered by each of them is twice the distance covered previously.

So one of the boats has travelled 700 + 2700 = 2100 feet. Since the boats are 400 feet from the bank, the river is 2100 400 = 1700 feet wide.


Dont try this at home

A rectangular block of wood has for its three dimensions a different odd prime number of inches. Its volume and (total) surface area are, respectively, a three digit number and a four digit number. What are the dimensions of the block


I go around in circles,
But always straight ahead
Never complain,
No matter where I am led.




I look flat, but I am deep,
Hidden realms I shelter.
Lives I take, but food I offer.
At times I am beautiful.
I can be calm, angry and turbulent.
I have no heart, but offer pleasure as well as death.
No man can own me, yet I encompass what all men must have.

RIDDLE 2
the oceans

BLOCK OF WOOD
3, 5, 61

Mark
RIDDLE 2
the oceans



BLOCK OF WOOD
3, 5, 61


The block is of dimension 3 by 5 by 61.

Let the dimensions be x, y, z with x < y < z and write V for the volume and S for the surface area. Then V = xyz, and S = 2(xy + xz + yz) = 2xyz(1/x + 1/y + 1/z). Since S > V, you have 1/x + 1/y + 1/z > 1/2. As x, y, z are primes and 1/5 + 1/7 + 1/11 < 1/2 you conclude that x = 3.
Now yz < 1000/3 so 999 < 2(3y + 3z + yz) < 2(3y + 3z + 1000/3) which gives 55 < y + z. On the other hand 1/6 < 1/y + 1/z, so if y is greater than 5, then z < 42 which makes y + z too small. So y = 5.

The equation for volume gives 15x < 1000 so z < 66, while the equation for surface area gives 2(15 + 3z + 5z) > 999 so z > 60. The only option is z = 61. There are 29 more solutions if you allow 2 as one of the dimensions. By the way, the number 1 is not considered a prime by (most) mathematicians.




Given any nine points in a unit square, must there be among the triangles having vertices on the given points at least one whose area is no more than 1/8



Four Scouts have to cross a dangerous bridge to get to their campground. It is night time and they have one candle between them which will burn for a mere 17 minutes after it is lit. If the candle goes out once lit, they have no way to re-light it.

The bridge will hold at most two Scouts at a time, and each crossing must take place with the lighted candle. The candle must be walked across with each group; it cannot be thrown, for example, as it would go out. Each Scout walks at a different rate, and any two Scouts walking together must walk at the rate of the slower camper's rate. The Scouts walk at the following rates:


M.J. 1 minute to cross
Ollie 2 minutes to cross
Libin 5 minutes to cross
Kathy 10 minutes to cross

Can they make it across, and if so, how

(There is no "tricky" wording here. Take the meaning of everything at face value!)

1.2=2
1=1
5.10=10
2=2
1.2=2

total 17

I've beers and you have beers. (cans, before drinking it).

1- If I give you one beer, you'll have double of mine.

2- If you give me one, we'll have the same number of beers.

How many beers have each of us?

beers

5.7

So you boozer, you drink more than I!

NINE POINTS
yes

Divide the unit square into quarters. By the pigeonhole principle, at least one quarter must contain at least three points. The largest area a triangle contained within a square can have is half the area of the square, which is 1/2 * 1/4 = 1/8.

Mark:
NINE POINTS
yes

Divide the unit square into quarters. By the pigeonhole principle, at least one quarter must contain at least three points. The largest area a triangle contained within a square can have is half the area of the square, which is 1/2 * 1/4 = 1/8.

Clever.

All nine points lie in the interior of the square, so pick any one and call it A. If any two of the other eight points are colinear with A, they form a traingle of area zero. So you can suppose that no pair is colinear; you next pick any further point, call it B, and label the other points, C, D, E,... clockwise and in order around A starting from B. Now look at the eight distinct and disjoint triangles with vertices ABC, ACD, ADE,... The union of the eight triangles lies in the interior of the square, so the sum of the areas must be less than 1. One of the triangles must therefore have area less than 1/8.



Four brothers have a horrible fight and decide to get as far away from each other as humanly possible. Assuming they stay on the surface of the earth (of course) where should they go, to realize the greatest mutual separation

Notes:
(1) You need not give exact latitude and longitude; only describe their relative positions on the globe.

(2) There are several reasonable interpretations of the meaning of "greatest mutual separation". Pick one



You heard me before,
Yet you hear me again,
Then I die,
Till you call me again.

BROTHERS
How about at the vertices of a tetrahedron to maximize the minimum distance between any two brothers?

RIDDLE
echo?

Mark:

RIDDLE
echo?


BROTHERS
How about at the vertices of a tetrahedron to maximize the minimum distance between any two brothers?




There are several ways to interpret the problem. The most common solution involves putting the four brothers at the vertices of a regular tetrahedron inscribed in the sphere of the earth. This solution is mimicked in nature by the arrangement of four hydrogen atoms surrounding a carbon atom in a molecule of methane. Nice one Mark.




Select two integers (without replacement) from the set of integers between one and one million (inclusive).

Is the sum of the two integers more likely to be even, be odd, or will the two outcomes be equally likely




You are paddling your canoe upsteam at a constant velocity. After paddling for six miles, the wind blows your hat into the stream and the hat begins flowing downstream.

You continue to paddle upstream for two more hours before noticing that your hat is missing, at which time you turn around and paddle downstream at the same rate you had paddled upstream, overtaking your hat just as you return to your original starting point.

What is the speed of the current

as the chance of the 2nd number to be different from the first (i mean with different to be different in type odd or even)500/999 is greater than to be the same 499/999

so
the sum of the two integers more likely to be odd

TWO INTEGERS
P(even) = 499999/999999
P(odd) = 500000/999999

CANOE
1.5 MPH

Do integers exist (a,b,c,d) that satisfy
a^2 - b^2 = c^2
a^2 + b^2 = d^2


Whim

Usamashaker:
"500/999 is greater than to be the same 499/999 so the sum of the two integers more likely to be odd"

Mark:
TWO INTEGERS
P(even) = 499999/999999
P(odd) = 500000/999999

There is a greater chance that the sum will be odd!

There are 500,000 even integers between 1 and one million (inclusive), and, of course, the same number of odd integers. The probability the first number chosen is odd is .5 with the same being true for choosing an even number. The probability the second number chosen has the same parity as the first is 499,999/999,999, while the probability it is of opposite parity is 500,000/999,999. For the sum of two integers to be even, they must have the same parity. Therefore the probability is greater that the sum is odd.

You guys are GREAT.


Mark:
CANOE
1.5 MPH

There are several possible equations to set up.
Let c be the speed of the current, b the speed of the boat (in still water). The time it took for the hat to float back to its starting point is 6/c. The distance the boat traveled (upstream) in the two hour period after losing the hat is 2(b - c) miles, so the total upstream distance traveled is 6 + 2(b - c). During the return trip the boat was moving at a rate of b+c so the time from when the hat was lost until the return of the boat to the starting point was [6 + 2(b - c)]/(b + c) + 2. Setting this quantity equal to 6/c and solving gives c = 1.5, independent of b.

Here's another solution:
Relative to the water, the hat stays still. So, again relative to the water and hat, the rower moves just as fast upstream as downstream. After the hat dropped, the rower rowed for four hours and the hat moved six miles, so the current is flowing at a rate of 6/4 = 1.5 miles per hour.


Hi Whim. good to see you posting again with your 'tricky' problems.

There is no solution to the Diophantine equation A4+B4=C2.

Note that this is the first stage of proving Fermat's Last theorem for n=4.
Here, A2,B2,C are Primitive Pythagorean Triple (assuming without the loss of generality, B2 is even), so
A2=a2-b2
B2=2ab
C=a2+b2
Rearrange the first equation to get A2+b2=a2, so A,b,a is a Pythagorean Triple with b even. Thus
A=c2-d2
b=2cd
a=c2+d2
Substitute the expressions for b and a back to B2=2ab, so we have
B2=4cd(c2+d2)
As c, d, c2+d2 pairwise coprime, each of them must be a perfect square. Therefore,
c=e2, d=f2 and c2+d2=g2
Combine these expression to give
e4+f4=g2
with g  g2 = a  a2 < C
Therefore, assuming C is minimally chosen on the outset, we found a smaller value for C. Thus, an infinite descent is produced - no solution as required.




A teacher writes a cubic equation on the board:
p(x) = x3 + p x2 + q x + r
with p,q,r integers. He tells his students that that polynomial has three integer roots, and that the largest of the roots is twice the smallest of the roots.

A student guesses a value, A, for the largest root which is too low, and for which p(A) is positive. A second student guesses a value, B, for the largest root which is too high and for which p(B) again positive. A third student, who was dozing in the back of the classroom, notices that each of the four numbers mentioned, A, B, p(A), p(B), is prime and one of them is the number 13.

What are the roots of p(x)



How many rectangles (of any size and in any position) are there on a standard 8 x 8 checkerboard

RECTANGLES
1296

Mark:
RECTANGLES
1296

There are a number of solutions. Here is perhaps the simplest: Place a coordinate system on the checkerboard with (0,0) at one corner and (8,8) at the other. To choose a rectangle, you need to identify two of the vertices which sit diagonally opposite one other. Such a pair can be identified by their two distinct x-coordiates and the two distinct y-coordinates; for example, the x-coordinates 1,4 and the y-coordinates 3,7 corresponds to the rectangle whose opposite corners sit at (1,3) and (4,7). From the nine possible x-coordinates -- that is from 0 to 8 -- the number of ways to choose two distinct ones is 9!/2!7! = 36, with the same for the y-coordinates. The total comes to 362 = 1296.



Two players take turns throwing a die. The first one to roll a five wins.
Of course, the first player has an advantage, but just how big

What are the exact winning probabilities for the two players


A company wants to test employees for drug use. They have 5000 employees. The company settles on a drug test that is 99% accurate in that if the person who was tested actually used drugs, the test would be positive 99% of the time, and if the person who was tested actually did not use drugs, the test would be negative 99% of the time.

Research has shown that, on the average, 5% of the population uses the drugs that the company is testing for. Given that an employee has received a "positive" drug test (i. e., the test indicated that this employee used drugs), what is the probability that this employee really is a drug user

What accuracy does the company need its drug test achieve in order to be certain that, 99% of the time, someone who tests "positive" reading on their drug test really is a drug user

DIE
Player 1: 6/11
Player 2: 5/11

DRUGS
83.898%
99.947%