Math Questions Answered

The Big Riddle of Four

2010-03-18T09:45:00.000-07:00

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

nadavs

Cosine Simplification

2008-07-29T05:27:00.000-07:00

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

nadavs

Projectile Mathematics

2008-07-22T02:40:00.000-07:00

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

nadavs

The Impossible Proof

2008-07-09T05:03:00.000-07:00

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

nadavs

Rectangles, Squares, Areas, and Calculus

2008-07-08T06:25:00.000-07:00

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

2008-07-07T08:24:00.000-07:00

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

nadavs

The Ultimate Bus

2008-07-06T09:01:00.000-07:00

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

Iodine Solutions

2008-07-05T09:11:00.000-07:00

Iodine is a great material for disinfection. It kills bacteria very well, but it's a little burning when put on a wound. Mathematic calculations of iodine solutions are not as burning, but they definitely require some more thought.

A pharmacist has two iodine solutions: one with a concentration of 30% and one with a concentration of 80%. How much of each solution the pharmacist needs to create 8 liters of a 50% solution?

First, we need to understand the situation? Why would a pharmacist need 8 liters of iodine solution? After you answer that in your head, go back to math.

We need two conditions to happen at once: the solution must have 8 liters of solution and 50% iodine, which means 4 liters of iodine. How do we solve that? A system of equations!

Let x by the number of liters of the 80% solution
Let y be the number of liters of the 30% solution

The amount of iodine in each solution is the percent times the volume. In this case, the amount of iodine from the 80% solution is 0.8x and the amount of iodine from the 30% solution is 0.3y. Now let's create those equations:
0.8x + 0.3y = 4
x + y = 8

The 4 comes from the explanation above: 50% of 8 is 4, which means there are 4 liters of iodine in that solution (ouch). Let's solve this system:
0.8x + 0.3y = 4 /*10
x + y = 8

8x + 3y = 40
x + y = 8 /*3

8x + 3y = 40
3x + 3y = 24

Subtract the equations:
5x = 16
x = 16/5 = 3.2 liters

3.2 + y = 8
y = 4.8 liters

Check:
80% * 3.2 = 2.56 liters of iodine
30% * 4.8 = 1.44 liters of iodine
Combined: 4 liters of iodine out of 8 liters of liquid. A 50% solution.

The pharmacist needs 3.2 liters of the 80% solution and 4.8 liters of the 30% solution. That's a lot of iodine.

Have a great weekend,
Nadav

nadavs

Tangent Fractions Identities

2008-07-04T07:08:00.000-07:00

Today's question comes from Yahoo Answers, and it includes trigonometric identities, fractions, and some hard work. Stay with me here, it will take some time.

Prove the identity:
(1 - tan x)/(1 + tan x) = cos 2x / (1 + sin 2x)

As you know, when working out an identity, you need to pick a side and work out until you reach the other side. Since the left side is more complex, I'll work from there to the simple right side. Here we go:

(1 - tan x)/(1 + tan x)

Using the tangent quotient (tan x = sin x / cos x), we can write the fraction as:
(1 - sin x / cos x)(1 + sin x / cos x)
(cos x / cos x - sin x / cos x)/(cos x / cos x + sin x / cos x)
((cos x - sin x)/cos x)/((cos x + sin x)/cos x)

Notice that both denominators are cos x, so they can be cancelled:
(cos x - sin x)/(cos x + sin x)

Multiply both parts of the fraction by (cos x + sin x)
(cos x - sin x)(cos x + sin x)/(cos x + sin x)²

Distribute the parentheses:
(cos² x - sin² x)/(cos²2 + 2sinxcosx + sin² x)

As you should know, cos² x - sin² x = cos 2x and cos² x + sin² x = 1, so:
cos 2x / (1 + 2sinxcosx)

As you should also know, 2sinxcosx = sin 2x, so:
cos 2x / (1 + sin 2x)

Q.E.D.

Nadav

nadavs

Dual GCD and LCM

2008-07-03T10:15:00.000-07:00

Today's question is more about understanding mathematic concepts and formulas rather than solving a question with a definite answer. Even with that, it's very good.

LCM(a, b) = 120 and GCD(a, b) = 4. Find all possible a and b.

Let's do it in two ways: by definition and by a well known formula.

First, let's go by the definition. The LCM of two numbers is the smallest number that both numbers divide. A good way to find it is completely factoring two numbers and taking the highest degree of each factor. That means the highest degree of factors in a and b are:
120 = 2³ * 3 * 5 = 2 * 2 * 2 * 3 * 5

This means the two numbers can be made up of any combination of factors you want, which gives 5!/3! = 20 possibilities for each number, a and b. However, the GCD is also limiting us.

The GCD is the biggest number that divides both numbers (when dividing each number by this number, there is no remainder). To find it, you take the lowest degree of each factor of each number and multiply them. This means that both numbers cannot have 3 twos in their factorized form, and they cannot have both 3 and 5 as factors. However, each of them must have 2 * 2 as a factor. Using the GCD and LCM, we can finally find a and b:

a = 2 * 2 = 4
b = 2 * 2 * 2 * 3 * 5 = 120

a = 2 * 2 * 2 = 8
b = 2 * 2 * 3 * 5 = 60

a = 2 * 2 * 3 = 12
b = 2 * 2 * 2 * 5 = 40

a = 2 * 2 * 5 = 20
b = 2 * 2 * 2 * 3 = 24

Any other pair is just a mirror of the previous pair, so these are the four possibilities for a and b (as said, they can be reversed).

Now, let's use the formula. There is a well known formula saying that GCD(a, b) * LCM(a, b) = a * b. The boundries of a and b in this situation are the GCD (if a number is below it, the GCD must be smaller) and the LCM (if a number is above it, the LCM must be bigger). Now let's find the pairs using some trial and error:

a = 4
b = 120
Works

a = 8
b = 60
Works too

a = 16
b = 30
The product is correct, but the GCD of these numbers isn't 4 (30 is not divisible by 4). This pair is rejected.

a = 12
b = 40
Works

a = 24
b = 20
Works as well

a = 48
b = 10
Fails, 10 is not divisible by 4

a = 20
b = 24
Already proven to work

a = 40
b = 12
Another duplicate

a = 80
b = 6
Fails - 6 is not divisible by 4

a = 120
b = 4
Duplicate

As you can see, the two methods give the same answer. Whatever method you choose, you will get the same answer. Take what you're comfortable with, and you will succeed. That's the motto of Super Math Tips.

Enjoy,
Nadav

nadavs

Proof of Trigonometric Inequalities

2008-07-02T06:55:00.000-07:00

After yesterday's inequality induction, it's time for some trigonometric induction. The proof is a little hard to understand at first, but it's definitely worth seeing.

Prove:
xsin(1/x) < 1 when x > 0

This question can definitely fit in the short question-long answer template. However, it doesn't necessarily fit short question-hard answer.

First, we know that x is positive, so we can divide by it without changing the inequality sign:
sin(1/x) < 1/x

Now the argument of the sine function and the right side of the inequality are the same, so let's call them y. All we have to prove now is that sin y < y for every positive y.

First, when y is greater than 1, this is obvious. A sine can never be greater than 1. When y is 1, there is no problem either, because sin 1 is about 0.84 (the argument is in radians, don't forget). The problem begins when 0 < y < 1.

To solve that, let's derivate the inequality we want to prove. We get cos y < 1. As you can see, the rate of growth of sin y is cos y, and the rate of growth of y is linear. When y goes up by 0.5, sin y goes by less than that. This means that sin y is always less than y when 0 < y < 1.

Since y = 1/x, we can conclude that:
sin(1/x) < 1/x (when 1/x > 0, which means x > 0)
xsin(1/x) < 1

Problem solved.
Nadav

nadavs

Inequality Induction

2008-07-01T07:22:00.000-07:00

Induction is an untouched topic in this blog so far, but I finally found a good question about it. It's not hard, but it's rather challenging.

Prove by induction:
(1 * 3 * 5 * 7 * ... * (2n - 1)) / (1 * 2 * 3 * 4 * ... * n) < 2ⁿ/√(2n + 1)

The first step in proving an induction is checking whether it works at all. Let's plug 1 for n:
1 / 1 < 2¹/√(2 + 1)
1 / 1 < 2 / √3 - correct!

Assume that the inequality is correct for n = k (k is natural)
(1 * 3 * 5 * ... * (2k - 1)) / (1 * 2 * 3 * ... * k) < 2^k/√(2k + 1)

Now let's prove that if the inequality is right for n = k, it is right for n = k + 1:
(1 * 3 * 5 * ... * (2k - 1) * (2k + 1)) / (1 * 2 * 3 * ... * k * (k + 1)) < 2^{k + 1}/√(2k + 3)

First, we know that (1 * 3 * 5 * ... * (2k - 1)) / (1 * 2 * 3 * ... * k) is smaller than 2^k/√(2k + 1), so it's definitely smaller than 2^{k + 1}/√(2k + 3). For this reason, we can place 2^k/√(2k + 1) instead of this big expression:
(2^k * (2k + 1))/(k + 1)√(2k + 1) < 2^{k + 1}/√(2k + 3)

Divide by 2^k:
(2k + 1)/(k + 1)√(2k + 1) < 2 / √(2k + 3)

Square both sides and cross multiply (k is natural, all positive):
4k² + 8k + 3 < 4k² + 8k + 3
3 < 4 - always true

We have shown that if n = k is correct, n = k + 1 is also correct. By checking that n = 1 is correct, we have proven that the inequality is correct for all natural numbers.

Enjoy,
Nadav

nadavs

Parabolic Minimums

2008-06-30T08:42:00.000-07:00

Today's question deals with parabolas, lines, and some calculus. It's not very hard, but, as always, it's very interesting.

Two parabolas, y = x²/x + 7 and y = -x²/4 + 3x, are drawn on the same coordinate plane. A vertical line is going through the parabolas. P and Q are the points of intersections with the first and second parabolas, respectively. Find the vertical line and the shortest distance between P and Q.

The first thing to do when you face such a question is to find an equation that represents the quantity you want, this time the distance PQ. Since the two points are on the same vertical line, they both have the same x-coordinate. Let's call it x. That makes the points:
P(x, x²/2 + 7)
Q(x, -x²/4 + 3x)

The distance between them is just the difference in the y-coordinates (because they lie on a vertical line). The distance is:
d = x² + 7 - (-x²/4 + 3x) = x²/2 + 7 + x²/4 - 3x = 3x²/4 - 3x + 7

We want the shortest d possible, and that's achieved when the derivative of d is zero:
d' = 3x/2 - 3
0 = 3x/2 - 3
3 = 3x/2
6 = 3x
x = 2

Now we know the vertical line, x = 2. Plug this value in d and find the shortest distance between P and Q:
d = 3 * 2²/4 - 3 * 2 + 7 = 3 - 6 + 7 = 4

The shortest distance between P and Q is 4.

Have a great week,
Nadav

nadavs

Logarithms and Inequalities

2008-06-29T07:04:00.000-07:00

Being independent from Yahoo Answers feels great. I can go to my source, find a great question, and post it here. The topic are somewhat repeating, but there are many more questions. Today's question is about logarithms and inequalities.

Solve:
x^{log₂ x + log_1/4 4} < 4

The solution here is very simple, yet it requires some mathematical knowledge, like what is log_1/4 4. Well, the answer is -1. When the base and the argument are reciprocals, the logarithm's value is -1. Now we can really start solving.

x^{log₂ x - 1} < 4

As you can see, there is a log with base 2 in the exponent. To get rid of the exponent, let's take log base 2 from each side. Since the base is bigger than 1, the inequality sign remains as it is.
log₂ x^{log₂ x - 1} < log₂ 4

Using the log property that says log_a xⁿ = nlog_a x, we can say that:
(log₂ x - 1)log₂ x < 2

Let t = log₂ x

(t - 1)t < 2
t² - t < 2
t² - t - 2 < 0
(t - 2)(t + 1) < 0
-1 < t < 2

Plug the real value of t back:
-1 < log₂ x < 2

By the definition of logs:
1/2 < x < 4

Hope you liked it,
Nadav

nadavs

Towns With Angles

2008-06-28T11:35:00.000-07:00

Today's question also involves the law of cosines, but with a different purpose. It is a very neat question from a test which is considered "hard". Try it before looking at the answer.

A biker travels from town A to town B at 10 km/h. Another biker is traveling from town B to town C at 12 km/h. The distance from town A to town B is d. Angle ∠ABC = 120°. The bikers are closest after 2.5 hours of riding. Find d.

First, try to draw the situation. Draw the three towns, draw d, draw the angle, and place two points on segment AB and segment BC, which are the bikers.

Let t be the time the bikers are traveling. This means the first biker's distance from town A is 10t and the second biker's distance from town B is 12t. Since the first biker's distance from town A is 10t, his distance from town B is d - 10t.

We need a variable that represents the distance between the bikers. Let's call it x.

As you can see now, we have a triangle with three sides (x, 12t, d - 10t) and an angle of 120° which is opposite to the side with measure x. This situation calls for the law of cosines to find the relation of all variables:
x² = (12t)² + (d - 10t)² - 2(12t)(d - 10t)cos 120°

cos 120° = -0.5, so the relation becomes (after expanding):
x² = 144t² + d² - 20dt + 100t² + 12dt - 120t²

Add like terms and take the square root:
x = √(124t² + d² - 8dt)

We know that x has a minimum when t = 2.5, so if we derive x and plug 2.5 for t, we can find d.

As you should know, when you derive √f(x), you get f'(x)/2√f(x). Since we equate the derivative to zero (at a maximum or a minimum, the derivative is zero), we can now ignore the denominator (since it's always positive and we can multiply by it). Don't ignore it on a test, I'm doing it to save time.

Since we derive by t (dx/dt), the derivative of d² is 0 (d is a constant, not a variable).
0 = 248t - 8d

We know that t = 2.5, so:
0 = 620 - 8d
8d = 620
d = 77.5 km

Towns A and B are 77.5 km apart.

Have a great weekend,
Nadav

nadavs

Geometric Sub-Sequence

2008-06-27T08:07:00.000-07:00

Today I have a question inolving sub-sequences and recursively defined squences. I really like these questions, since when you solve it everything just fits in perfectly. Watch it here:

A sequence a_n is defined as:
a₁ = 11
a_{n + 1} = -0.5a_n + 4.5

b_n is defined for natural n's as b_n = a_n - 3

a) Prove that b_n is a geometric sequence
b) Find the sum of all even positioned terms of b_n.

Let's start with a, of course.

A sequence is a geometric sequence if the ratio of every two consecutive terms is constant. So, for our needs, we can show that b_{n + 1}/b_n is a constant, which proves that b_n is a geometric sequence.

By definition:
b_n = a_n - 3
b_{n + 1} = a_{n + 1} - 3

Using the definition for a_n and a_{n + 1}:
b_n = a_n - 3
b_{n + 1} = -0.5a_n + 4.5 - 3 = -0.5a_n + 1.5

Divide both terms:
(-0.5a_n + 1.5)/(a_n - 3) = -0.5(a_n - 3)/(a_n - 3) = -0.5

The quotient of two consecutive terms is a constant, so b_n is a geometric sequence.

b) Since the ratio of consecutive terms in b_n is -0.5, the ratio of consecutive even-positioned terms is that ratio squared, meaning (-0.5)² = 0.25. Now that we have that quotient, we can find the sum of those even-positioned terms.

The formula for the sum of an infinite converging geometric series is a₁/(1 - q). We have q, and now we need a₁. In this case, the first term (which is given the symbol a₁) is b₂, since it's the first even-positioned term. Let's find it:

b₂ = a₂ + 4.5 - 3

Using the recursive definition:
a_{1 + 1} = -0.5a₁ + 4.5
a₂ = -0.5 * 11 + 4.5 - 3 = -1

So:
b₂ = -1 - 3 = -4

Plug it in the formula:
-4/(1 - 0.25) = -4/0.75 = -4/(3/4) = -16/3

The sum of all even-positioned terms of b_n is -16/3.

If you need math help in anything, go to Super Math Tips, sign up, and you can send any questions you want. They might be answered on this blog!

Hope you like it,
Nadav

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Elliptic Triangles

2008-06-26T07:41:00.000-07:00

As promised, today I have a question involving ellipses. In my opinion, it's a great question.

C is a point on the ellipse x²/16 + y²/9 = 1. A and B are the foci of the ellipse. Prove that angle ∠ACB cannot be 90°.

To do that, we first need to find the foci of the ellipse. As you remember, in an ellipse, a² = b² + c². From the equation of the ellipse, a² = 16 and b² = 9. This means:
c² = a² - b² = 16 - 9 = 7
c = √7

Now that we know c, we know one side of triangle ABC. The side is AB and it equals 2√7. Now we only have to find the other sides and show that ∠ACB cannot equal 90°.

Let's call one side, AC, x. By the definition of an ellipse, the sum of the distances from a point on the ellipse to the foci is a constant, 2a. In this case, 2a = 2√16 = 8. This means the third side, BC, is 8 - x.

We have three sides of the triangle and we want to find an angle. To do that, we can use the law of cosines, which involves all three sides and an angle. Since we want ∠ACB, we will use the side opposing it, AB.

The law of cosines states that:
a² = b² + c² - 2bc * cos∠A
(a, b, c, and ∠A are not related to the ellipse)

Let's plug our known values in there:
(2√7)² = x² + (8 - x)² - 2x(8 - x)cos∠ACB

We want to know what happens when ∠ACB = 90°, which means cos∠ACB = 0. Let's do all the squaring and eliminate the last part, because it's zero:
28 = x² + 64 - 16x + x²
2x² - 16x + 36 = 0
x² - 8x + 18 = 0

When trying to solve this through the quadratic formula, the expression inside the square root equals:
8² - 4*18 = 64 - 72 = -8

There is no real number for x which makes the law of cosines correct for 90°, so angle ∠ACB cannot equal 90° in this ellipse.

There are even cooler things on Super Math Tips, go check them out.

Hope you liked it,
Nadav

nadavs

Missing Middle Digits

2008-06-25T08:59:00.000-07:00

Today's question isn't long and isn't hard, but it makes you think a little. Tomorrow I'll have a nicer question, with an ellipse. Until then, you'll have to work your mind on digits.

In a two digit number, the units digit is bigger than the tens digit by 4. First, you add 1 between the two digits and call it x. Then, you add 5 to the right of the units digit of the original number and call it y. x + y = 692. What is the two digit number?

Let's take this step by step:
First, the original number is _ _.

x is _ 1 _
y is _ _ 5
x + y is 692

Now we can find the values of the units digit, since we know that the units digit + 5 gives 12 (it can't give 2, obviously). This makes the units digit 7.

x is _ 1 7
y is _ 7 5

As you can see, adding the two numbers will leave the hundreds digits untouched. As a result, the two hundreds digits, which are equal, must add up to 6. This makes the hundreds digit 3.

The two numbers are 317 and 375, and the original number is 37.

Short, nice, easy. Tomorrow will be a little harder, I promise.
Nadav

nadavs

Exponential Absolute Values

2008-06-24T08:18:00.000-07:00

Complex numbers are incredible, but in order to use them we have to remember some definitions. Here is a question involving both complex numbers and exponential inequalities. It's not hard, but it requires some thinking.

Solve for x (x is real):
abs(2 + 3^{x² - x - 1} - 12i) > 13

To do that, here is a reminder for the definition of absolute value:
Let z be a complex number.
z = a + bi
abs(z) = √(a² + b²)

In the question:
a = 2 + 3^{x² - x - 1}
b = -12

The absolute value is:
√((2 + 3^{x² - x - 1})² + 12²)
√(4 + 4*3^{x² - x - 1} + (3^{x² - x - 1})² + 144)

Now return to the inequality:
√(4 + 4*3^{x² - x - 1} + (3^{x² - x - 1})² + 144) > 13

Since both sides are always positive, we can square both sides without changing the inequality sign:
4 + 4*3^{x² - x - 1} + (3^{x² - x - 1})² + 144 > 169

Let t = 3^{x² - x - 1}. This gives:
4 + 4t + t² + 144 > 169
t² + 4t - 21 > 0
(t + 7)(t - 3) > 0
t > 3 or t < -7 Since t is an exponent of 3, it can't be negative, so we're left with only one solution: t > 3.

Switch t back:
3^{x² - x - 1} > 3

The base is bigger than 1, so the inequality sign remains the same:
x² - x - 1 > 1
x² - x - 2 > 0
(x - 2)(x + 1) > 0
x > 2 or x < -1

And that's it.

Hope you liked it,
Nadav

nadavs

Save Gas Money with Math

2008-06-23T05:44:00.000-07:00

Today's question is very simple, but it will show you have to save money on gas very easily. Watch and learn:

Mary needs to refuel her car with 60 liters of gasoline and get back home. She has two choices: drive to a gas station 20 km away from her house which charges $1.16/liter or drive to a gas station adjacent to her house which charges $1.24/liter. Mary's car's mileage is 10 km/liter. Which is cheaper for Mary?

This question does not require complicated calculations, so let's do them:
The price for Mary on the close station: 60 * $1.24 = $74.4
The price for Mary on the far station: 60 * $1.16 = $69.6

That's quite a difference. Now we only have to see how much it costs mary to drive back and forth:
She can do 10 km/liter, so she spends a total of 40 / 10 = 4 liters to drive to the gas station and back.

Now let's find how much this trip costs her (using the cheap gas price, of course):
4 * $1.16 = $4.64
$69.6 + $4.64 = $74.24

In total, Mary saves 16¢ per 60 liters, or 0.267¢ per liter (1.008¢ per gallon). Not worth the trouble. If her mileage was double, 20 km/liter, she would've saved about $2 for these 60 liters, which is about 15¢ per gallon. Now that's a difference.

Yours,
Nadav

nadavs

Parametric Parabolas

2008-06-22T07:09:00.000-07:00

It hasn't even been a week since we dealt with parabolas, and they come again, this time with actual numbers. The question itself isn't hard, but once again, some thinking is required.

A parabola goes through the points A(17, 3) B(34, z) C(50, -30). B is the maximum point of the parabola. Find point B and the parabola function in standard form.

As you know, a parabola can be written in two ways. One is standard form, with the a, b, and c coefficients, and one is the vertex form. To solve this question, we'll use the vertex form. This form says that the equation a parabola that has a vertex (h, k) is:
y = a(x - h)² + k

We know h and k, so let's plug them in:
y = a(x - 34)² + z

We have two other points, so let's use them:
3 = a(17 - 34)² + z
2 = a(50 - 34)² + z

3 = 289a + z
-30 = 256a + z

Subtract the second equation from the first:
33 = 33a
a = 1
3 = 289a + z
3 = 289 + z
z = -286

The function of the parabola is:
y = (x - 34)² - 286

Open parenthese and combine like terms to switch to standard form:
y = x² - 68x + 1156 - 286
y = x² - 68x + 870

Nice, isn't it?
Nadav

nadavs

Complex Numbers Brilliance

2008-06-21T07:20:00.000-07:00

Finally, after waiting a very long time, I found a really good question with complex numbers. Usually those questions are really lame, and they are an exact copy of each other. However, today I found a really good one (like ones in super math tips), and the solution someone gave there is pure brilliance.

z is a complex number with modulus 1. Solve:
5z⁴ - 11z³ + 16z² - 11z + 5 = 0

Because of its special coefficient pattern, some people factored it. But doing so takes the fun, and not everyone can spot this pattern. However, someone used the modulus property to create a truely amazing solution.

A complex number can be represented in two ways:
z = a + bi
z = r(cos θ + isin θ)

r is the modulus, and it is also given by √(a² + b²).

Let w be the conjugate of z (which means w = a - bi). Since a² + b² = z * w, z * w = 1. Square that, and you get:
z²w² = 1

Multiply the entire equation by w² (w is non-zero because z is non-zero):
5z⁴w² - 11z³w² + 16z²w² - 11zw² + 5w² = 0

Using what we showed before, this equation turns into:
5z² - 11z + 16 - 11w + 5w² = 0

Now it's just playing with numbers (remember that i² = -1 by definition):
5(z² + w²) = 11(z + w) - 16

Since z = a + bi, w = a - bi:
5((a + bi)² + (a - bi)²) = 11(a + bi + a - bi) - 16
5(a² + 2abi - b² + a² - 2abi - b²) = 11(2a) - 16
5(2a² - 2b²) = 22a - 16
10(a² - b²) = 22a - 16

Since √(a² + ²) = 1, we can conclude that:
a² + b² = 1
b² = 1 - a²

Plugging back into the equation:
10(a² - 1 + a²) = 22a - 16
10(2a² - 1) = 22a - 16
20a² - 10 = 22a - 16
20a² - 22a + 6 = 0
10a² - 11a + 3 = 0
(2a - 1)(5a - 3) = 0
a = 1/2, 3/5

Now it's really easy to find b (notice that each a has two b's):
When a = 1/2:
(1/2)² + b² = 1
b² = 3/4
b = ±√3/2

When a = 3/5:
(3/5)² + b² = 1
b² = 16/25
b = ±4/5

The four solutions of the equation are:
z₁ = 1/2 + i√3/2
z₂ = 1/2 - i√3/2
z₃ = 3/5 + i4/5
z₄ = 3/5 - i4/5

Hope you liked it. I know I did.

Have a great, non-complex weekend,
Nadav

nadavs

Slips and Sums

2008-06-20T09:26:00.000-07:00

Today's question is a pure thinking question. Its answer is very short, but it's a great question nonetheless. Read it carefully, then try to answer. Don't look at the answer just yet, it's easy when you think.

A jar is filled with a thousand paper slips which have the numbers 1 - 1000. Each time two slips are taken out of the jar, and the difference between the numbers on them is written on a new slip of paper. The difference is put back in the jar and the two original numbers are taken out. Will the number on the last slip of paper be odd or even?

Like many other questions, this one looks impossible at first. There are so many papers, and it's hard to keep track on them. Trying out all possibilities will take way too long, but luckily, using a very powerful tool, your brain, will solve this question easily.

Try looking at the sum of all numbers in the jar. At first, it's 500,500 (using the sum formula of natural numbers from 1 to n: n(n + 1)/2). Then you take out two numbers and put back their difference. However, the parity does not change.

If the sum of two numbers is odd, their difference is also odd. The same goes for even sums. When you take out two numbers and put back their difference, the sum goes down, but the parity does not change. You either take out an even sum and put back an even number, or take out an odd sum and put back an odd number.

Since the parity of the sum of the numbers in the jar is always kept even (because the initial sum is even), the last two slips will have an even sum. Because their sum is even, their difference is also even, and the number on the last slip will be even.

Hope you liked it,
Nadav

nadavs

Euler And Series

2008-06-19T05:41:00.000-07:00

Yahoo Answers is a goldmine. There are a lot of great questions in there. Today I have one relating to series and Euler. Yes, this guy again. He always comes back for more.

A sequence a_n is defined as 1/n²(n + 1)². Find the sum of the series from 1 to infinity.

Remember Euler from before? He proved that the sum of 1/n² from 1 to infinity is π²/6. Yes, Euler was a smart man. All we need now is to transofrm a_n to something with 1/n² which includes sums. But how?

Let's start with the obvious and proceed from there. To create that denominator, we can try to add 1/n² and 1/(n + 1)², see what we get, and then decide what else to do. So:
1/n² + 1/(n + 1)² = (n + 1)²/n²(n + 1)² + n²/n²(n + 1)²

Which is:
(n² + (n + 1)²)/n²(n + 1)²

Open parentheses and add:
(n² + n² + 2n + 1)/n²(n + 1)²
(2n² + 2n + 1)/n²(n + 1)²

To leave only the 1 in the numerator (which will give a_n, we need to subtract (2n² + 2n)/n²(n + 1)²:
a_n = (2n² + 2n + 1)/n²(n + 1)² - (2n² + 2n)/n²(n + 1)²

The second term can be also written as 2n(n + 1), which can be cancelled. At the end, we get:
a_n = (2n² + 2n + 1)/n²(n + 1)² - 2/n(n + 1)

2/n(n + 1) can also be written as (2n + 2 - 2n)/n(n + 1), which can be transformed into:
(2n + 2)/n(n + 1) - 2n/n(n + 1) = 2/n - 2/(n + 1)

Since it's the opposite on a_n, we can finally conclude that:
a_n = 1/n² + 1/(n + 1)² + 2/(n + 1) - 2/n

Now it's really easy. Notice how 2/(n + 1) and -2/n cancel each other, except for the first term. Their sum is -2/1 + 2/2 - 2/2 + 2/3 - 2/3 + 2/4 - 2/4 + ... . Since 2/∞ is zero, the only remainder of this part of the series is -2.

As Euler said, the sum of the 1/n² part of the series is π²/6. The 1/(n + 1)² has the exact same sum, but minus one. Since the sums starts from 1/2² and not 1/1², the 1/1² is left outside.

The total sum of a_n from 1 to infinity is:
π²/6 + π²/6 - 1 - 2
S = π²/3 - 3

Since the terms of a_n get very small very quickly, it can be easily approximated for validity.

Hope you like it,
Nadav

nadavs

Magic T

2008-06-18T06:46:00.000-07:00

Remember "Magic C"? The one where you had to put the digits 1 - 5 to create sums? Well, today I have Magic T. The question is nearly identical, but with a little twist (besides the other letter).

The digits 1, 2, 3, 4, 5, 6, and 7 are placed in squares that form a T shape. The top row has 3 squares and the middle column has 5 squares. The intersection square, the one that appears on both the row and the column, is shaded. The digits are placed in such way that the sum of the digits in the row and the sum of the digits of the column are equal.

Show that the shaded square must have an even number to create equal sums, and find an example for each possible digit on the shaded square.

One way to do that is find every possible combination for these squares and see that only even numbers can be in the shaded square. However, there is a mental condition that describes doing such thing: insanity. There are many possible combinations (5040 to be exact), and finding the right ones will take forever. It seems like we need a better method: creative thinking.

First, try making a correct combination. It shouldn't take you too long. A good method you can use is number replacement. Check the difference between the two sums and switch digits accordingly. Notice what happens when you change digits: the difference changes by an even number. Always. That happens because one sum goes up by a number, and the other down goes down by the same number, so combined the difference was changed by twice that number, which is always even.

Now we know that the key here is the difference between the sums, but how does the shaded square relate to it? Since it appears on both sums, it dictates how the differences will relate to each other.

If you place an odd number in the shaded square, you will have three odd numbers left. If you place two of them or none of them at the top row, you will have one or three of them left for the middle column, respectively. This means that the top row will have an odd sum and the column will have an even sum. Not good.

If you place only one of the remaining odd numbers on the row, you will have two left for the column. Once again, this situation is not good. It creates an even row and an odd column, which do not have an even difference.

But if you place an even number in the shaded square, it's perfect: you will either have both sums even or both sums odd, so you can play with the differences and create a zero difference Magic T.

Here are three examples, each with a "shaded square" of 2, 4, and 6 (they are lying on the side so they could be written here):

7
2 - 1 - 3 - 4 - 5
6

7
4 - 1 - 2 - 3 - 6
5

7
6 - 1 - 2 - 3 - 5
4

Hope you liked it.
Nadav

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