The Big Riddle of Four

After a long dry period, I'm going to publish just one post here, and it contains a riddle for you, without my solution. Today is a very special day for me (well, it's my birthday), so I'm giving you the opportunity to solve something truely difficult.

Use the number 4 twice to get 64.

That's it. Not complicated, yet not very simple. You can use any mathematical operation as long as the final results contains only two 4's. For example, 4² x 4 = 64, but it contains the digit 2, which is unacceptable.

Let's see how long it takes people to solve this, and how old I'll be when it's solved.

The answer is purely mathematic. No word tricks or anything involved.

Post your answers in the comments. Can't wait to see them.

Nadav

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Cosine Simplification

Yahoo Answers is definitely the place everyone goes to find an answer for a math question. Otherwise, there is no explanation for the kinds of questions you can find there. Today's question is about simplification of a trigonometric expression:

Simplify: (cos 3x)(1 - cos 2x + cos 4x - cos 6x)

To solve that, we will need some major identities: the sum of cosines, the difference of cosines, the sum of sines, and the product of sines and cosines. With these formulas, this question can be easily solved.

Let's start with the formulas:
cos α + cos β = 2cos((α + β)/2)cos((α - β)/2)
cos α - cos β = -2sin((α + β)/2)sin((α - β)/2)
sin α + sin β = 2sin((α + β)/2)cos((α - β)/2)
sinαcosβ = 1/2(sin(α + β)sin(α - β))

Now, to avoid complications, let's aim for the average. We can turn 1 into cos 0, so we have the cosines of 0 and 6x and the cosines of 2x and 4x. In both formulas they are divided by two, so they should probably cancel each other.

cos 0 - cos 6x = -2sin((6x + 0) / 2)sin(6x - 0)/2 = -2sin² 3x
cos 4x - cos 2x = -2sin((4x + 2x)/2)sin((4x - 2x)/2) = -2sin 3x * sinx

In conclusion, the problem has just become:
(cos 3x)(-2sin² 3x - 2sin 3x * sin x)

Distribute:
-2sin 3x * sin 3x * cos 3x - 2sin 3x * cos 3x * sinx

Transform into a sum of a cosine and a sine (I picked α and β to be 3x, because there will be a sine of zero, which is zero):
-2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin 3x - 2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin x
-sin 6x * sin 3x - sin 6x * sin x

We are getting really close now, just hold on:
-(sin 6x)(sin 3x + sin x)
-(sin 6x)(2sin((3x + x)/2)cos((3x - x)/2)
-2sin 6x * sin 2x * cos x

And that's it.

Nadav

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Projectile Mathematics

Today I want to have some physics in the blog, something I've been neglecting for a long time. The question is about projectiles, walls, velocities, and directions.

A particle is projected 15m above the ground, and it just clears a wall 26.25m high and 30m away. What was the velocity and direction of the particle when it was projected?

This situation calls for some variables, given information, and physics formulas. I'm going to use the gravitational constant, g, as 10 m/s².

Since the particle is projected 15m above the ground and reaches a height of 26.25m, it only rises 11.25m. Let's call this number h.

When a particle is projected, its velocity, v, is split into two velocities - a horizontal velocity (v_x) and a vertical velocity (v_y). Since air friction is ignored, v_x remains constant and v_y changes according to gravity.

First, to find v_y, we can use the law of conservation of energy, which shows how kinetic energy (at the moment of projection) turns into potential energy (when it clears the wall). The formula is:
mv_y²/2 = mgh

We can make 15m our starting height, so h = 11.25. Also, we can cancel m on both sides. The only variable left is v_y:
v_y²/2 = 10 * 11.25
v_y²/2 = 112.5
v_y² = 225
v_y = 15

The positive number is taken because the particle is projected upwards.

Now we have the vertical velocity at the time of projection. When the particle clears the wall, its vertical velocity is zero. Let's see how much times it takes the particle to get to this stage by the formula v = v₀ + at:
0 = 15 - 10t
10t = 15
t = 1.5 seconds

It takes the particle 1.5 seconds from projection until it clears the wall. Now we can find the horizontal velociy of the particle, using the formula x = x₀ + vt:
30 = 0 + 1.5v_x
v_x = 20 m/s

Now we're left with the direction. When you break the velocity into its components, v_x and v_y, you see that the tangent of the angle of projection is v_y/v_x. To find that angle:
tan θ = 15/20
tan θ = 3/4
θ ≈ 36.87°

Enjoy,
Nadav

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The Impossible Proof

Today, to have some fun, I'm going to prove something wrong. The challenge today won't be to find the right answer, but to find the flaw in the answer.

Prove: 5 = 7

Well, it looks impossible and wrong at first, but let's try anyway.

Let a = b

Multiply both sides by a and add a² - 2a to both sides:
2a² - 2ab = a² - ab

Take 2 as a common factor:
2(a² - ab) = a² - ab

Divide by a² - ab:
2 = 1

Multiply by 2:
4 = 2

Add 3 to both sides:
7 = 5

By the reflexive property of equality:
5 = 7

Q.E.D.

Obviously, there is something wrong with this proof, and detailed reading will find it. One step was eliminated from this proof. As you can see, after the first assignment of variables (a = b), two steps were done at once. When you break down the steps, you find that a² = ab. This leads to a² - ab = 0, which shows us that the step that led to 2 = 1 was division by zero.

Now you can all understand why the first commandment of math is "Thou shalt not divide by zero". Division by zero can prove anything and destroy mathematic foundations. Only divide by non-zero numbers.

Nadav

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Rectangles, Squares, Areas, and Calculus

Rectangles, squares, areas, and calculus are closely related, but do you know how much? Well, see this question. It's very common with numbers, I'm just bringing the general version.

A rectangle with sides of lengths x and y has perimeter P. Prove that the largest area of this rectangle is given when the rectangle is a square.

Well, the question already gives us the variables. All we need is to use them, combine all four terms in the question (rectangles, squares, areas, and calculus) and prove what we need to prove.

So, the perimeter is P. This means that 2(x + y) = P. To get an equation with one variable, let's find y in terms of the other numbers:
x + y = P/2
y = P/2 - x

Now we know two sides of the rectangle: x and P/2 - x. Let's multiply them to find the area function:
A = x(P/2 - x)
A = xP/2 - x²

To find the maximum value for A, let's differentiate:
A' = P/2 - 2x

Equate A' to zero:
0 = P/2 - 2x
2x = P/2
x = P/4

To make sure it's a maximum, let's find the second derivative of A and see if it's negative:
A'' = -2

That's definitely negative, so x = P/4 is the point where the area is maximal. As you can see, that's a quarter of the perimeter. When one side of a rectangle is a quarter of the perimeter, it's a square. To prove that, plug P/4 for x in the equation for y:
y = P/2 - P/4
y = P/4

It's now proven. The biggest area of a rectangle with a constant perimeter is a square.
Nadav

nadavs

Pipes and Swimming Pools

Everybody knows the famous water pipe question - two pipes fill a swimming pool in a certain time, so find how much time each pipe needs to fill the pool. Well, here is one question of this type.

Two pipes fill a swimming pool in 11 1/9 hours (eleven hours and one ninth of an hour) together. One pipe can fill the pool in 5 hours less than the other pipe. Find out how much times it takes each pipe to fill up the swimming pool separately.

First, as you know, we need variables. Let x be the time it takes the faster pipe to fill up the swimming pool. This means it takes the slower pipe x + 5 hours to fill up the swimming pool.

Since it takes the first pipe x hours to fill up the pool, each hour it fills 1/x of the pool. For the other pipe, the rate is 1/(x + 5) per hour. Since they fill up the pool in 100/9 hours together (I changed it into the more comfortable fraction form), let's multiply each rate by 100/9 and add them up to 1 (100% of the pool):
100/9x + 100/(9x + 45) = 1

Multiply by the lowest common denominator: 9x(x + 5):
100(x + 5) + 100x = 9x(x + 5)
100x + 500 + 100x = 9x² + 45x
9x² - 155x - 500 = 0
(x - 20)(9x + 25) = 0
x = 20, -25

Since x is a time, it cannot be negative, so it must be 20.

The fast pipe fills the pool in 20 hours. The slower one fills the pool in 25 hours.

Hope you liked it,
Nadav

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The Ultimate Bus

Buses are a nation's pride in some countries and a disgrace in others. Math questions can handle random behavior like the bus schedule in some countries, but it's much better to deal with a good bus system, with a normal schedule. Here is one question about such system.

Andy walks at a constant speed along a long street. Every 6 minutes he is passed by a bus going in this direction. Every 2 minutes he is also passed by a bus going in the opposite direction. Both directions have the same schedule and all buses go at the same speed. What is the time difference between two consecutive bus departures?

First we need to define some variables and understand a conecpt of of physics. Let's call the speeds of Andy and the buses a and b, respectively.

A relative speed is the speed at which two distinct objects move towards or away from each other. This speed is given by subtracting the two speeds. In the first case, the bus is approaching Andy at a speed of b - a. Since the other buses go in the opposite direction, their speed becomes "negative", and the bus goes towards Andy at a speed of b + a.

Let x be the distance between the buses (they are scheduled on regular intervals, so the distance between buses is the same for all buses)

Using the formula distance = velocity * time, we can say that:
x = (b - a) * 6
x = (b + a) * 2

Equate the x's:
6b - 6a = 2b + 2a
4b = 8a
b = 2a
a = b/2

Either Andy is very fast of the buses are very slow, because the buses go at a speed just twice as Andy. That is definitely weird.

x = (b - b/2) * 6 = b / 2 * 6 = 3b

Since the buses need to travel a distance of 3b to get from the location of one bus to another bus and they do so at a speed of b, a bus must be sent every 3b / b = 3 minutes. That's a great interval for the passengers of this bus system.

Hope you liked it,
Nadav

nadavs