Parabolic Minimums

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

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Logarithms and Inequalities

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

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Towns With Angles

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

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

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

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Missing Middle Digits

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

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

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

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

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

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

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

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

nadavs

Areas and Tangent Circles

After many questions about trigonometry, calculus, and parabolas, we finally get a fresh new question about geometry. This question deals with one of the most hated figures in the world of geometry: circles.

Two circles with radii of 5 cm each are externally tangent to each other. Each of these two circles is internally tangent to a circle with a radius of 15 cm. What is the perimeter and area of the shape enclosed between the three circles?

To get the idea of what it looks like, try to draw it. You will eventually see what this whole question is about, and that will make the question much easier for you.

First, we need to get the perimeter. When you connect all the centers of the circles, you get an equilateral triangle. This happens because the side lengths are 5 + 5 and 15 - 5 (externally tangent circles and internally tangent circles, respectively). This triangle is the key to the whole question.

There is a theorem saying that the centers of two tangent circles and the point of tangency all lie on the same line. This means you can extend two sides of the equilateral triangle to create a sector of the big circle. This sector's area and perimeter are 60°/360° = 1/6 of the area and the perimeter of the circle.

Since these extensions go through the centers of the small circles as well, they split them up to 60° and 120° angles (the 60° angle is formed by the equilateral triangle, and the 120° angle is its supplement). Now we can calculate both the perimeter and the area of the shape between the three circles.

The perimeter is given by adding the three arcs which form the shape. Each small circle gives an arc of 120°, and the big circle gives an arc of 60°. Let's find these arc lengths.

Small circles:
2π * 5 * 120°/360° = 10π/3

Large circle:
2π * 15 * 60°/360° = 30π/6 = 5π

There are two small circles, so the total perimeter is:
5π + 2 * 10π/3 = 15π/3 + 20π/3 = 35π/3 cm
This equals about 36.65 cm.

As for the area, we can calculate the area of the sector of the large circle, subract the area of the equilateral triangle, and subtract the area of the two sectors of the small circles. This will give us the area of the shape we want.

Area of the big sector:
π * 15² / 6 = 225π / 6 = 37.5π

Area of the equilateral triangle:
10²√3 / 4 = 25√3

Area of a small sector:
π * 5² / 3 = 25π/3

Remember there are two small sectors, so we can now calculate:
37.5π - 25√3 - 2 * 25π/3 = 225π/6 - 25√3 - 100π/6 = 125π/6 - 25√3 cm²
This equals about 22.15 cm²

That's it for today. Hope you liked it. Visit super math tips for more cool information like that.

Yours,
Nadav

nadavs

Parabola Reflections

Today's question is of a type we didn't have for a long time - pure variables, no numbers. It's very interesting, and although it looks impossible at first, one little step makes it very, very easy.

The parabola y = ax² + bx + c has its vertex on (h, k). The parabola is reflected about the line y = k. The new parabola is: y = dx² + ex + f. Find a + b + c + d + e + f.

Looks impossible, doesn't it? You can try doing some complex calculations showing that -b/2a = h (you should know that), then plug it back, and equate it to y. That's hard. However, parabolas have two forms: standard form and vertex form. Guess what - we have the vertex!

y = a(x - h)² + k

Now we can open the parentheses and find exactly a, b, and c:
y = ax² - 2axh + ah² + k

That means:
a = a
b = -2ah
c = ah² + k

When the line is reflected about y = k, it changes the coordinates. Since the vertex is on y = k, the parabola just switches sides. This means a = -a and b becomes the opposite too (2ah) to keep the vertex.

Notice that the vertex remains the same because the line was reflected about y = k. When you reflect point (x, n) about the line y = z, you end up with (x, 2z - n). If you reflect (h, k) about y = k, you get (h, 2k - k), which is (h, k).

Since each point turns into (x, 2k - y), the new y-intercept, f, becomes (0, 2k - (ah² + k)), which is (0, 2k - ah² - k), or (0, k - ah²).

a and d are opposites, so a + d = 0. b and e behave the same. The only addends left are c and f:
c + f = ah² + k + k - ah² = 2k

This big sum, a + b + c + d + e + f, equals to 2k.

Hope you liked it,
Nadav

nadavs

Coordinated Circles

Today I have a very common question - how to find an equation of a circle from three points, which are a requirement for a unique circle. See how this can be done the hard way and the easy way.

Find the center, radius, and area of a circle that passes through (6, 2), (0, -6), and (7, 1).

The usual way of doing this type of problems is finding the equations of two lines between those points, finding their perpendicular bisectors, find their point of intersection (the center), and go from there. There is one very good adjective to describe this: boring. Since this blog is all about interesting math, you'll see there is a way that cuts all this line of operations up to "finding their perpendicular bisectors". Yes, removing two very annoying steps.

Since the center of the circle is equidistant from all three points, we need to find a point which is equidistant from the first two points, then find a point which is equidistant from the second and third point, and then see if they match. If they do, the point we found is equidistant from the first and third point (by the transitive property).

So, let's equate the distance between (6, 2) and (0, -6) (the generic point is (x, y)):
√((x - 6)² + (y - 2)²) = √((x - 0)² + (y + 6)²)

Square both sides:
(x - 6)² + (y - 2)² = x² + (y + 6)²

Open parentheses and add like terms:
x² - 12x + 36 + y² - 4y + 4 = x² + y² + 12y + 36
-12x - 4y + 4 = 12y
-12x - 16y + 4 = 0
-16y = 12x - 4
y = -3x/4 + 1/4

We didn't get a point, we got a line. This line is the perpendicular bisector of the segment between the two points, because the geometric location of all the points equidistant from two points is a line. This line is the perpendicular bisector of the segment connecting the two points. Now you know how to find a perpendicular bisector of two points in one very easy step.

Let's do the same thing to the second pair of points:
√((x - 0)² + (y + 6)²) = √((x - 7)² + (y - 1)²)
x² + (y + 6)² = (x - 7)² + (y - 1)²
x² + y² + 12y + 36 = x² - 14x + 49 + y² - 2y + 1
12y + 36 = -14x + 49 - 2y + 1
14y = -14x + 14
y = -x + 1

Now we have two sets of points that are equidistant from a given set of points. All we need now is to equate those lines and find the point that is equidistant from all three points:
y = -3x/4 + 1/4
y = -x + 1

Equate the y's:
-3x/4 + 1/4 = -x + 1
x/4 = 3/4
x = 3

So:
y = -3 + 1
y = -2

The point which is equidistant from all three points and thus the center of the circle is (3, -2).

Now everything becomes really easy. To find the radius of the circle, we need to find the distance from the center to one of the points. Let's say to (7, 1):
r = √((7 - 3)² + (1 - (-2))²)
r = √(4² + 3²)
r = √25
r = 5

Using the formula for the area of a circle, we finally answer the question completely with the answer:
S = 5² * π = 25π

Problem solved.
Nadav

nadavs

Lawn Mowing Siblings

After yesterday's easy question I found something better. It doesn't deal with differential equations or have a fancy name, but it just deals with some thinking and setting up equations correctly.

A brother and a sister are mowing the lawn on their backyard. It takes the brother 15 minutes longer to mow the lawn than it takes his sister. Together they mow the lawn in 56 minutes. How long does it take each of them to mow the lawn?

To solve this, we need something, and we need it fast: variables. x sounds like a great variable. Let x be the time it takes the sister to mow the lawn.

Now, since it takes her x minutes to mow the lawn, she does 1/x of the lawn every minute. Her brother takes x + 15 minutes to mow the lawn, so he mows 1/(x + 15) every minute.

Since together they mow the lawn in 56 minutes, 56 times the mowing rate of the brother plus 56 times the mowing rate of the sister should give 1, which is 100% of the lawn. We finally have an equation:
56/x + 56/(x + 15) = 1

This equation is easily solved:
56(x + 15) + 56x = x(x + 15)
56x + 840 + 56x = x² + 15x
x² - 97x - 840 = 0

Now we can solve it by factoring or by the quadratic formula. Since the quadratic formula is boring, let's try to factor. The numbers here are big, so you can start from something like 10 and see how the difference between the factors grow. Finally, you will reach the correct factoring:
(x - 105)(x + 8) = 0

Since the sister can't mow the lawn in -8 minutes, it must be 105 minutes, or 1 hour and 45 minutes. This means the brother needs 120 minutes, or two whole hours to mow this lawn. It must be a big lawn.

To test that, check: 56/105 + 56/120 = 8/15 + 7/15 = 1

I really doubt this answer is correct. If it takes the sister alone almost two hours to mow the lawn, with her brother it will take them about 3-4 hours, and that's before they even start mowing.

There is a new proof added at super math tips, go check it out.

Have a great weekend,
Nadav

nadavs

Differential Equations

Today's question deals with differential equations. It's not very difficult, but it's very nice. Also, it provides a reminder for those who forgot how to solve those equations.

Solve: dy/dx = (2tan y) / x

In order to solve that and get a function, we first need to get each variable to where it belongs, and the appropriate "d notation" on its side. To do that, divide by tan y (to move it to the dy side) and multiply by dx (to move it to the x side):

cot y dy = 2/x dx

Now all the variables are sorted out, and we can easily integrate each side to get a function:
ln sin(y) = 2ln x + c (c is the arbitrary constant of integration)

This also equals:
ln sin(y) = ln x² + c

Since we want a function in the form of y = something x, let's make everything an exponent of e (to eliminate the ln's):
sin y = x² * e^c

Since c is an arbitrary constant, we can call it k and make it simpler:
sin y = kx²

Now, take the arcsin of both sides to eliminate the sine:
y = arcsin(kx²)

Have a great weekend,
Nadav

nadavs

Sinusoid Temperature

Temperatures are very hard to predict. Even weathermen get them wrong most of the times. However, there are some paradises which have a constant temperature range, such as Brazzaville (which is in Congo). Here is the question with the amazing details:

The temperature in Brazzaville is given by the following function:
C = 17 - 6cos(πt/12 + 5)
0 <= t < 24

1. What are the highest and lowest temperatures on Brazzaville?

2. What is the biggest rate of change (biggest drop/rise) in temperature?

To show how people complicate questions, I'm going to answer it using calculus, and then using a much simpler method.

The highest and lowest temperatures are achieved when the derivative of the function is 0, so let's do it:
C' = 6π/12 * sin(πt/12 + 5)
0 = πsin(πt/12 + 5) / 2

Divide by π / 2:
0 = sin(πt/12 + 5)
πt/12 + 5 = π*k (where k is an integer)
πt + 60 = 12πk
πt = 12πk - 60
t = 12k - 60/π

Now all we need to do is find the k values which give t's that match our requirement. Since 60/π is about 19, we need k's that are bigger than 1, so:
t₁ = 24 - 60/π = 4.901
t₂ = 36 - 60/π = 16.901

To verify which number is a maximum and which is a minimum, we can use the second derivative:
C'' = 6π²/144 * cos(πt/12 + 5)

Plug the t's:
t₁ makes it positive, so it's a minimum.
t₂ makes it negative, so it's a maximum.

Plug the two t's in the original equation:

C = 17 - 6cos(πt₁/12 + 5) = 17 - 6 = 11°C
C = 17 - 6cos(πt₂/12 + 5) = 17 - 6(-1) = 23°C

The temperature in Brazzaville ranges between 11°C - 23°C for the entire year. We can only be jealous of them...

To get the biggest rates of change, we need to find the maximum of the derivative, so we'll set the second derivative to zero and find the two points. One will give out a negative change and the other one will give a positive change. They are opposites of each other, since the cosine function is symmetrical on a vertical line going through a maximum or minimum.
C'' = 6π²/144 * cos(πt/12 + 5)
0 = 6π²/144 * cos(πt/12 + 5)
cos(πt/12 + 5) = 0
πt/12 + 5 = ±π/2 + 2πk (again, k is an integer)
πt + 60 = ±6π + 24πk
πt = ±6π + 24πk - 60
t = ±6 + 24k - 60/π

The t values in bounds are:
t₃ = 10.901
t₄ = 22.901

Plug them in C':
C' = π/2 * sin(πt₃/12 + 5) = π/2 = 1.57 °C/hour
C' = π/2 * sin(πt₄/12 + 5) = -π/2 = -1.57 °C/hour

So the sharpest temperature changes are 1.57 °C per hour.

This question can be solved using a much easier way: common sense.

The t inside the function is multiplied by π and divided by 12, so plugging in t values from 0 to 24 will give values from 0 to 2π inside the cosine function. This means it goes through an entire cycle from t = 0 to t = 24. With that, we can figure out that the maximum value for the cosine function is 1 and the minimum value is -1. Plug these two numbers and you get:
C = 17 - 1*6 = 11°C
C = 17 - (-1)*6 = 23°C


Also, the cosine function is the "steepest" when it goes through the x-axis, or more accurately, when the argument makes the cosine zero. All we need to do is find when the argument is zero and plug it in the derivative:
πt/12 + 5 = ±π/2 + 2πk

I already solved that, so I'm not going to do it again.

As you can see, thinking outside the box can really help when it comes to such long and tiring functions.

Yours,
Nadav

nadavs

Maximizing School Profit

Today's question is dealing with schools. As you know, schools need money, so they send dance tickets. However, they need to know how much to charge. Here is a question dealing exactly with that.

A school is holding a dance with a ticket price of $6. 250 plan to attend. The dance team knows that for every $1 increase in the ticket price, 25 less people will come. How much should a ticket cost for maximum revnue? Solve using two different methods.

First, we need an equation to solve, otherwise we won't be able to solve it even with one method. As we know, for each $1 price increase, 25 less people will come. So let x be the number of $1 price increases.

When you increase the price x times, the price goes up by $x ($1 * x). When the price is increased x times, 25x less people will come. The profit is the number of people times the price per ticket. Since the starting price is 6, one factor will be (6 + x). Since 250 plan to attend initially, the second factor is (250 - 25x). This makes the entire profit function:
P = (x + 6)(250 - 25x)
P = -25x² + 100x + 1500

To solve that, we can use two different methods. The first one is using vertex form, which will give us the x coordinate of the vertex, and reveal the maximum point that way (it's an upside-down parabola).
P = -25x² + 100x + 1500
P = -25(x² - 4x - 60)

Complete the square inside:
P = -25(x² - 4x + 4 - 64)
P = -25(x - 2)² + 1600

This means the maximum profit is generated when x = 2 and the ticket price is $8. In this case, 200 people will come, and the profit will be $8 * (250 - 25*2) = $8 * 200 = $1600.

To solve the other way, differentiate P:
P' = -50x + 100

Compare to zero:
0 = -50x + 100
50x = 100
x = 2

Once again, the same result.

Hope you liked it,
Nadav

nadavs

Sphere Differentiation

.diToday's question deals with differentiation and spheres. The question uses some very nice properties of differentiation and the dy/dx notation. It has two parts, but so do most questions from text books (apparently, people at Yahoo Answers aren't that original in making up questions).

A sphere has radius of length r cm, surface area of S cm², and a volume of V cm³ on a given instant, t.

1) Prove: (dV/dt) * (dr/dt) = (dS/dt)² / 16π

2) Find the surface area when dV/dt = π cm³/sec and dS/dt = 2π cm²/sec

Not that bad, is it?

First, there are two important formulas we are going to use in order to solve this problem:
V = 4πr³ / 3
S = 4πr²

Now differentiate each function according to t:
dV/dt = 4πr² dr/dt
dS/dt = 8πr dr/dt

To get dV/dt * dr/dt, multiply the first equation by dr/dt:
dV/dt * dr/dt = 4πr² (dr/dt)²

Now square the second equation:
(dS/dt)² = 64π²r² (dr/dt)²

Divide the second equation by 16π:
64π²r² / 16π = 4πr² (dr/dt)²

Now this is exactly the same as dV/dt * dr/dt.
Q.E.D

The second question:
dS/dt = dS/dV * dV/dt
dS/dV = dV/dt * dS/dt
dS/dV = (8πr dr/dt) / (4πr² dr/dt)
dS/dV = 2/r
dS/dt = 2/r * dV/dt
dS/dt = 2/r * π
dS/dt = 2π
2π = 2π/r
1 = 1/r
r = 1

Plug r = 1 into the surface area function:
S = 4π * 1² = 4π cm²

Problem solved.

Hope you liked it.

Nadav

nadavs

Limiting Euler

Today's question is about limits and Euler's number, e. This number has many interesting properties, and one of them is the derivative of e^x, which is e^x. Today's question does not deal with the derivative of this function, but with some limits.

Find the limit:
lim_{x -> infinity} (e^x + x)^1/x

This question is definitely not easy, mainly because of that annoying x in there and the 1/x exponent. However, we must never give up. We need to fight this question, so let's fight it with simplicity. Define another variable!

y = (e^x + x)^1/x

Simple, isn't it? All we have to do now is... well... find the limit of y. That didn't help. You can see there is a fraction in the exponent, so if we could just bring it down to a normal fraction, we could use L'Hopital's rule and solve this.

To get the exponent down, we can use a very well known technique: logs. There is a great rule in logarithms which states:
log_a bⁿ = nlog_a b

Using that fantastic rule we can bring the 1/x down and use L'Hopital's rule! To get a nice answer, let's take the logarithm of base e of this function, also written as ln (natural logarithm). So:
ln y = ln (e^x + x)^1/x

Using the logarithm rule mentioned earlier, we can say this equals:
ln (e^x + x) * 1/x
ln (e^x + x) / x

Mr. L'Hopital starts to smile. Now we can find the limit of this function and remember this is the logarithm of the function, so we'll need to consider that.

lim_{x -> infinity} ln y = lim_{x -> infinity} ln (e^x + x) / x

Using L'Hopital's rule (lim_{x -> a} f(x)/g(x) = lim_{x -> a} f'(x)/g'(x)), we can conclude that:
lim_{x -> infinity} ln (e^x + x) / x
is equal to:
lim_{x -> infinity} (e^x + 1) * (1 / (e^x + x)) / 1
lim_{x -> infinity} (e^x + 1) / (e^x + x)

Use L'Hopital's rule twice more:
lim_{x -> infinity} e^x / (e^x + 1)
lim_{x -> infinity} e^x / e^x

Now that looks familiar, doesn't it? This limit equals 1.

Remember: lim_{x -> infinity} ln y = 1, so lim_{x -> infinity} y = e

The limit of the entire function is e. Problem solved.

Hope you enjoyed,
Nadav

nadavs

Boating Calculus

Today's question is dedicated to all boating lovers and calculus fans. It is not very hard, it has no use in real life, and yet it is a very beautiful question (like the ones on super math tips).

A boat leaves the dock at 12:00pm and heads west at 40km/h. Another boat travels north at 20km/h and enters the dock at 1:00pm. When is the distance between the boats the shortest and what is that distance?

Nice, isn't it? We have an optimization question here, so we need to create a function to find a variable to optimize. We're dealing here with given speeds, variable times and varying distances, so let's call the time variable t (in hours, to match the speeds) and the distance variable d (in km).

Now, the distance from the first boat to the dock is 40t (it starts at the dock). The distance from the second boat to the dock is 20 when it starts, and 20 - 20t total (since it travels towards the dock).

The boats are heading in perpendicular directions, so they form a right triangle. Using the Pythagorean theorem, we can find the distance between the boats, d:
d² = (40t)² + (20 - 20t)²
d² = 1600t² + 400 - 800t + 400t²
d² = 2000t² - 800t + 400

Since we want to find the minimum value for d and d must be positive, we can also find the minimum value of d² (also must be positive). Let y = d²:
y = 2000t² - 800t + 400

Now differentiate this function:
y' = 4000t - 800

To find an extremety, we need to set y' to zero:
0 = 4000t - 800
4000t = 800
t = 1/5

To see that it is indeed a minimum, let's find the second derivative of y:
y'' = 4000
Since it is positive, y is minimal when t = 1/5

Now calculate y when t = 1/5:
y = 2000 * (1/5)² - 800 * (1/5) + 400
y = 2000 * (1/25) - 160 + 400
y = 80 + 240 = 320

However, y is the distance squared, so let's take the square root of this number:
d = sqrt(y) = sqrt(320) = 8sqrt(5) = 17.888... km.

This means that after 1/5 of an hour, which is 60 / 5 = 12 minutes, the boats will be the closest to each other at 8sqrt(5) (or 17.888) km apart.

Hope you enjoyed.
Nadav

nadavs

Altitudes of Obtuse Triangles

Today's question is simple geometry. It deals with triangles, altitudes, sines, and other mathematical functions and objects. The question is quite long, but its answer is not that long. I guess it's another case for the "question theorem" - the length of the question and the length of the answer are inversely proportional.

In triangle ABC, A = 65 degrees, B = 13 degrees, C = 102 degrees. A line perpendicular to AC intersects the line defined by AC at point P. The perpendicular line also passes through point B. The length of PB is 17. Find the area of triangle ABC.

Get your ruler and draw this. If you read correctly, you should draw an obtuse triangle with an altitude outside the triangle. This altitude forms two right triangles: one is outside triangle ABC and one is including triangle ABC.

Since angle BCP is the supplement of angle C, it is 180 - 102 = 78 degrees. Since the outside triangle is a right triangle, we can calculate side BC of triangle ABC:
sin 78 = 17/BC
BC = 17/sin 78 = 17.379

Well, we have one side. To find the area of ABC, we need another side, and then we can use the area formula of 1/2 * a * b * sin C. It never fails.

Now that we have BC, we have two options. Use the law of sines or use a more creative way: find the hypotenuse of triangle APB. I always vote for creativity, so let's find that side.

Angle B is 13 degrees, as given. The angle adjacent to B, CBP, is 12 degrees (the complement of 78 degrees). This makes angle ABP a total of 25 degrees. Now that we have an angle and a side in a right triangle, we can find the other sides. Let's find side AB, because it uses the cosine function, which is... the cosine function:

cos 25 = 17 / AB
AB = 17 / cos 25
AB = 18.757

Now we have everything for the area formula: we have side c (AB), side a (BC), and angle B. Let's plug them all in:
S = 1/2 * c * a * sin B
S = 36.667 sq. units.

Problem solved. Wasn't that hard, was it?

Hope you enjoyed. There are more nice things like this at Super Math Tips.
Nadav

nadavs

Inverse Quadratic Function

Today I have another question in the genre of "short question, long answer". Today's question deals with inverse functions, more accurately the inverse of a quadratic function (or more accurately, what becomes a quadratic function). This is not some simple quadratic function, but rather something more difficult to solve.

Find the inverse of y = 2x + sqrt(x)

Short, isn't it? However, the solution isn't that short.

To find the inverse, we need to substitute x and y and then solve for y. Let's do it:
x = 2y + sqrt(y)

To eliminate the square root, move the 2y to the right side and square:
x - 2y = sqrt(y)
x² - 4xy + 4y² = y

Subtract y from both sides and use the distributive property:
x² - 4xy - y + 4y² = 0
x² - y(4x + 1) + 4y² = 0

This calls for the friend of any math student: the quadratic formula. Notice that squaring may add a solution, but there is only one inverse function, so we'll need to eliminate one:
y = ((4x + 1) ± sqrt((4x + 1)² - 16x²)) / 8

Yes, not a pleasant look. Let's work with that a little so see if it gets any better:
y = ((4x + 1) ± sqrt(16x² + 8x + 1 - 16x²)) / 8
y = ((4x + 1) ± sqrt(8x + 1)) / 8

That's the best it gets, but there's still a problem: there are two possible inverse functions here, but only one is right. To find out which one is right, remember that we reached the following conclusion:
x - 2y = sqrt(y)

This means x - 2y must be non-negative. Let's plug the two possible y values to see what we get:
x - ((4x + 1) + sqrt(8x + 1)) / 4 = 4x / 4 - ((4x + 1) + sqrt(8x + 1)) / 4
= (4x - ((4x + 1) + sqrt(8x + 1))) / 4 = (-1 - sqrt(8x + 1)) / 4
In one word: negative. Not good.

Let's plug the other one now to see if it works:
x - ((4x + 1) - sqrt(8x + 1)) / 4 = 4x / 4 - ((4x + 1) - sqrt(8x + 1)) / 4
= (4x - ((4x + 1) - sqrt(8x + 1))) / 4 = (-1 + sqrt(8x + 1)) / 4

As you can clearly see, sqrt(8x + 1) > 0 if x > 0, just like the domain of the original function. That means the inverse of the original function is:
y = ((4x + 1) - sqrt(8x + 1)) / 8

Here is a little confirmation test:
Plug 9 for x in the original function. You get y = 2*9 + sqrt(9) = 21. Now plug 21 in the inverse function and see if it gives back 9:
y = ((4 * 21 + 1) - sqrt(8 * 21 + 1)) / 8
y = (85 - sqrt(168 + 1)) / 8
y = (85 - sqrt(169)) / 8
y = (85 - 13) / 8
y = 72 / 8 = 9

Yes, it is the inverse.

You can find many more interesting math tricks like that on super math tips.

Hope you liked it.
Nadav

nadavs

Wire Optimization

Finally, after waiting for a good question, it finally came: an optimization question. This one is from Yahoo Answers, and believe me, it's very good.

A 6 meter long wire is cut into 12 pieces. From these pieces, eight have the same length and the other four also have an equal length. These pieces form a frame of the box. How long should each piece for the box to have a maximum volume?

First, we should define ourselves some variables, otherwise we'd be lost.

Let x be the length of one of the 8 pieces. That means the total length of the 8 equal pieces is 8x. That means the other four pieces have a total length of 6 - 8x, and 3/2 - 2x meters each.

Since these wires form a box, the sides of the box have lengths of x, x, and 3/2 - 2x. That makes the volume of the box x * x * (3/2 - 2x), or 3x²/2 - 2x³.

To find the maximum value, we need to differentiate the volume function and set it to zero. Then we need to find the values that we get from solving the equation, plug them in the second derivative, see which one is a maximum and say "Problem Solved".

So:
V = 3x²/2 - 2x³
V' = 3x - 6x²
0 = 3x - 6x²
3x(1 - 2x) = 0
x = 0, 1/2

Now, let's find the second derivative and see when it's negative, so we get a maximum:
V'' = 3 - 12x
V''(0) = 3 - 0 = 3 > 0 - minimum
V''(1/2) = 3 - 12 * 1/2 = 3 - 6 = -3 < x =" 1/2" 1 =" 1/2">3 = 1/8 m³

Problem solved.
Nadav

nadavs

Chairs, Tables, and Ratios

Today I have a very easy question, but once again, two solutions. They are different in their approach, and anyone can choose what works best for him.

The ratio of the price of a table to the price of a chair is 5:3. A table costs $500 more than a chair. How much does each cost?

First, here is one answer (not mine):

Let x be the price of chairs.
That means x + 500 is the price of tables. From the ratio, we can solve the following equation and get each price:
5/3 = (x + 500) / x
5x = 3x + 1500
2x = 1500
x = 750

That makes tables cost $1250. And indeed, when you figure out the ratio, 1250:750 = 25:15 = 5:3.

However, there is another way to answer this question (my way):
Since the ratio is 5:3, you can also call it 5x:3x, where 5x is the price of a table and 3x is the price of a chair. Now we can solve a simpler equation:
5x = 3x + 500
2x = 500
x = 250

Notice that x is not a price of a chair or a table, it's just the unit that gives the price of each. That means the price of a table is $250 * 5 = $1250 and chairs cost $250 * 3 = $750.

This is another way of solving simple math questions using creative thinking.

Hope you liked it,
Nadav

nadavs

Two Way Limits

It does not happen often, but when it does, it's simply beautiful: a question with two completely different yet right answers. I saw such question on Yahoo Answers today. Although I didn't answer it, the two different answers were so interesting and different in their approach, I had to bring it here.

The question is quite simple for people who know limits, but the two different answers are just amazing.

Find the limit:
lim_{x -> 4} (3 - (5 + x)^1/2) / (1 - (5 - x)^1/2)

As you can clearly see, plugging x = 4 gives a zero in the denominator, which is why the question asks for a limit. You can plug in numbers and get a rough estimation, but we want a definite answer.

The first method is using L'Hopital's rule. This rule says that in order to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). All we need now is to differentiate the numerator and denominator and see what we get:
lim_{x -> 4} (-1/2 * (5 + x)^-1/2) / (1/2 * (5 - x)^-1/2)

The halves cancel, and by using the law that says aⁿ / bⁿ = (a/b)ⁿ, we can show that:
lim_{x -> 4} -((5 + x) / (5 - x))^-1/2

Now plug in 4 and you will get -((5 + 4)/(5-4))^-1/2
Which is -9^-1/2 = -1/3

This is one very good solution, and most people who know calculus would choose that. However, there is another solution to this problem. To do that, multiply the numerator and denominator by the conjugate of the denominator:
lim_{x -> 4} (3 - (5 + x)^1/2) / (1 - (5 - x)^1/2) * (1 + (5 - x)^1/2) / (1 + (5 - x)^1/2)

Using the law that says (a + b)(a - b) = a² - b² and the distributive property, we can conclude that:
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (1 - (5 - x))
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (x - 4)

Now multiply both parts of the fraction by the conjugate of the original numerator:
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (x - 4) * (3 + (5 + x)^1/2) / (3 + (5 + x)^1/2)
lim_{x -> 4} (9 - (5 + x)) * (1 + (5 - x)^1/2) / ((x - 4) * (3 + (5 + x)^1/2))
lim_{x -> 4} (4 - x) * (1 + (5 - x)^1/2) / ((x - 4) * (3 + (5 + x)^1/2) )

That is really nice. (4 - x) / (x - 4) = -1, so we can cancel out two terms and turn them into a nice little minus sign:
lim_{x -> 4} -(1 + (5 - x)^1/2) / (3 + (5 + x)^1/2)

Now we can safely plug 4 for x and we get:
-(1 + (5 - 4)^1/2) / (3 + (5 + 4)^1/2)
-(1 + 1^1/2) / (3 + 9^1/2)
-(1 + 1) / (3 + 3)
-1/3

Again, we get -1/3 as an answer.

Math can be very easy if you think outside the box. Try doing it as often as you can.
Nadav

nadavs

Multiplication and Remainders

Today's question is a very nice question I found on Yahoo Answers. There was a trigonometry question I was planning for today, but this question is more unique (and besides that, this blog has already many trigonometry questions).

When an integer N is divided by D, it gives a remainder of 7. When N is multiplied by 5 and divided by D, it gives a remainder of 10. What is D?

At first, you may think "I need N to do that". Well, if you had N, the solution would be obvious. That's why we need some creative thinking to find what this number D is.

First, we know that D must be greater than 10. If a/b gives a remainder of m, then b must be greater than m (if you divide something by 7, you can't get a remainder of 8, since that means the result of division must be greater by 1).

Now, since N/D gives a remainder of 7:
N = xD + 7

Also, we know that 5N/D gives a remainder of 10:
5N = yD + 10

Where x and y are integers.

Now multiply the first equation by 5:
5N = 5xD + 35
And equate it to the second one:
5xD + 35 = yD + 10
yD - 5xD = 25
D(y - 5x) = 25
D = 25 / (y - 5x)

Since D must be an integer, 25 must be divided by one of its factors: 1, 5, or 25. However, dividing it by 25 or 5 will give 1 or 5, which is less than 10. For this reason, D must be 25.

You can now take any number for x and create N. For example, let's see what happens when x = 3:
N = 3*25 + 7
N = 75 + 7 = 82
N/25 = 3 with remainder 7 (not surprising, as we went from the definition backwards).

Now multiply this N by 5:
5N = 410
Divide by 25:
410 / 25 = 16 with remainder 10

Nice, isn't it?
Have a great week,
Nadav

nadavs

Percentages and Copy Machines

Today's question is relatively easy, but very unique. It does not require complex mathematical calculations, but rather a simple mathematical thinking.

A copy machine has 7 buttons to increase or decrease the area of the photocopied area. They are 50%, 75%, 80%, 100%, 120%, 125%, and 150%. Which buttons can be eliminated from the machine in a way that saves the machines original area choices. Notice: when two buttons are pushed consecutively, the effect of the second button is on the area that remains after the first button (for example, two pushes on 50% will photocopy 25%).

It seems easy, and it really is. All you have to do is turn these percents into fractions and see how you can multiply them to get the others.

50% = 1/2
75% = 3/4
80% = 4/5
100% = 1
120% = 6/5
125% = 5/4
150% = 3/2

50% cannot be created by any combination, so it must remain in the machine.
75% = 3/4 = 1/2 * 3/2
80% cannot be created by any combination, so it also stays in the machine.
100% = 1 = 5/4 * 4/5
120% = 4/5 * 3/2
125% is another number that cannot be created from others.
150% is a component in 75% and 120%, so it can't go away.

However, this question has another answer, where 150% = 6/5 * 5/4, and then 120% stays in the machine and 150% goes out.

Easy, isn't it?
Nadav

nadavs