Constructing a Square Root

Today's question is not from Yahoo Answers, but rather from my head. Well, not exactly from my head. I read about the possibility to construct a square root of a number, but I didn't find any instructions about how to do it. So, here is what I found out.

There is a line of length x. Construct a line of length sqrt(x)

If you remember anything about averages, there are two common types. There is a simple arithmetic average, simply adding two numbers and halving the result. There is also a geometric average, which is multiplying two numbers and taking the square root of the result.

This average is called a geometric average because it happens in, surprisingly, geometry. In a right triangle, the altitude from the 90 degree angle to the hypotenuse is equal to the square root of the two parts of the hypotenuse created from the altitude. If the altitude creates segments of length x and y, then the altitude's length is sqrt(xy). We can use that to find sqrt(x). All we have to do is make y = 1.

So to construct this segment of length sqrt(x) we construct a segment with length x and a segment with length 1 (the 1 is relative to x. If x is 5, the segment of length 1 must be 5 times shorter). Now we created the hypotenuse. Now construct the altitude to the hypotenuse from the point the two segments are joined.

We are in a little trouble now (a trouble that made me think for almost 30 minutes). We need to find the exact point on the altitude that will create a right angle with the endpoints of the hypotenuse. However, we have no way of knowing where is that point, since we construct a square root. Unless it's a square root of a perfect square, we cannot count units of 1 and create the right triangle from there.

After almost 30 minutes of thinking, it hit me. When a right triangle is inscribed inside a circle, the hypotenuse is a diameter. Now we can create an exact right triangle and an exact square root. Find the midpoint of the hypotenuse and create a circle around it.

Now it's easy. Continue the altitude until it intersects the circle. Then you can complete the triangle and take the altitude. Square root found.

Hope you liked it. Now go test it - find the square roots of numbers like 4 and 9 and see if you get 2 and 3. If you do, you construct well.

Yours,
Nadav

nadavs

Trigonometric Derivation

Today I have a question about differentiation with the definition of differentiation, which means using the lim_{h -> 0} (f(x + h) - f(x)) / h. However, today it is going to be about trigonometric functions (it's always about trigonometry, isn't it?).

Differentiate cos(x) using the definition of differentiation.

First, the most important limit to remember is this: lim_{x -> 0} sin(x) / x = 1. When you have this in mind, you can solve anything (well, many things).

Let's write what we need to find:
lim_{h -> 0} (cos(x + h) - cos(x)) / h

Using the identity of the cosine of sums of angles, we can get:
lim_{h -> 0} (cos(x)cos(h) - sin(x)sin(h) - cos(x)) / h

Now follow closely:
lim_{h -> 0} ((cos(x)cos(h) - cos(x)) / h - sin(x)sin(h) / h)
lim_{h -> 0} (cos(x)cos(h) - cos(x)) / h - lim_{h -> 0} sin(x)sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1) / h - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1)(cos(h) + 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos²(h) - 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(-sin²(h))/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h

Now watch the beauty of mathematics reveal itself:
L'Hopitals rule says that to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). Let's do that on the first limit:
lim_{h -> 0} -sin²(h) / h
= lim_{h -> 0} (-sin²(h))' / h'
= lim_{h -> 0} -2sin(h)cos(h) / 1
As h approaches 0, sin(h) also approaches zero, so the entire limit is zero, and thus the whole first part of the first big limit.

This leaves us with:
- sin(x) * lim_{h -> 0} sin(h) / h
As you can see, we have lim_{h -> 0} sin(h) / h = 1
That means:
-sin(x) * lim_{h -> 0} sin(h) / h = -sin(x) * 1 = -sin(x)

So the entire derivative equals -sin(x).

That's why super math tips exists. Or you can simply use Wikipedia's list of trigonometric derivatives (but where is the fun in that?).

Nadav

nadavs

Cubic Sequences

Today I have something similar to the cubic function discovery I did exactly two weeks ago. It's somewhat more complex, and therefore more fun. Long? Yes. Fun? Oh, yeah!

Find the nth term: 2, 8, 20, 40, 70, 112, 168, ...

Once again, the genre of "short question, super long answer". Those are the most fun, so let the fun begin.

When you get a sequence and you need to find the nth term, go for differences between numbers. Those differences hide a big secret within them. If the differences are the same, it's a simple arithmetic series. If the differences are not the same, do the differences of the differences. If now they're equal, the relation between the numbers goes through a quadratic expression. If the differences are still not the same, take the differences again and again until they are equal. The number of times you take the differences is the degree of the expression that connects the numbers.

If you see that the differences have a certain ratio to them (for example, try to find the differences for 4, 10, 28, 82, 244, 730), don't bother going more than once or twice. You are dealing with an exponential relation here.

So, let's find the differences of the sequence we're given:
2, 8, 20, 40, 70, 112, 168
6 12 20 30 42 56
6 8 10 12 14
2 2 2 2

Jackpot. We reached equal differences after three times, so the numbers are connected via a cubic expression. To find that expression, we need to set up a system of 4 equations with 4 variables (if you don't know or don't remember, open the link at the top of the post).

Let's call the points that this "function" goes through (1, 2) (2, 8) (3, 20) (4, 40). Now let's set up the system:
a + b + c + d = 2
8a + 4b + 2c + d = 8
27a + 9b + 3c + d = 20
64a + 16b + 4c + d = 40

Subtract the equations from each other to eliminate d:
7a + 3b + c = 6
19a + 5b + c = 12
37a + 7b + c = 20

Subtract them again to eliminate c:
12a + 2b = 6
18a + 2b = 8

And once again:
6a = 2
a = 1/3

Let's start plugging:
12/3 + 2b = 6
4 + 2b = 6
2b = 2
b = 1

Again:
7/3 + 3 + c = 6
c = 2/3

Finally:
1/3 + 1 + 2/3 + d = 2
d = 0

Well, that makes it easier. The nth term is given by:
n³/3 + n² + 2n/3

Go ahead, plug your number for n and see that it works. For all terms, even the ones which were not included in the calculation.

Enjoy your new knowledge,
Nadav

nadavs

Integration and Substitution

Today I have a simple question of integration by substitution. Although it's simple, it is a question many people find difficult.

Integrate (5x + 10)/(3x² + 12x - 7) by substitution

Unlike most short questions, this one also has a short answer. First, we need to define a variable to be one part of the function. This part, when differentiated, must be divisible by another part, or we'll be left with two variables, which is not fun.

As you can see, when you differentiate the denominator, you get 6x + 12, which is similar to the numerator, 5x + 10 (take 6 and 5 as a common factor, respectively). Now all we need is to define a variable and we're set to go:
z = 3x² + 12x - 7
z' = 6x + 12

This can also be written as:
dz/dx = 6x + 12

Make dx the subject:
dx = dz/(6x + 12)

We want to find:
integral((5x + 10)/(3x² + 12x - 7) dx)

Substitute z for the denominator and also replace dx with what we found:
integral((5x + 10)/z dz/(6x + 12))

Take out common factors:
integral(5(x + 2)/z * dz/6(x + 2))

Cancel (x + 2):
integral(5/6z dz)
5/6 * ln(z) + c

Now substitute back the value of z:
5/6 * ln(3x² + 12x - 7) + c

And that's the integral.

Simple, yet somewhat long, isn't it?
Yours,
Nadav

nadavs

Circular Coordinate Geometry

Today I have a question involving some coordinate geometry, or analytic geometry. It also involves some circles. It's quite easy, but somewhat long. Here it comes.

What is the sum of the radii of all circles going through the points (1, 9) and (8, 8) and also tangent to the x-axis?

First, you should know that if a circle is tangent to the x-axis, its radius is the y-coordinate of its center. Let's call the center (x, y), so its radius is y.

Since it's a circle, the distance between the center and each of the points must be y. Let's use the distance formula to show that:

y = sqrt((x - 1)² + (y - 9)²)
y = sqrt((x - 8)² + (y - 8)²)

Square and open parentheses:
y² = x² - 2x + 1 + y² - 18y + 81
y² = x² - 16x + 64 + y² - 16y + 64

y^2 cancels on all equations (I also moved the y terms):
18y = x² - 2x + 82 /*8
16y = x² - 16x + 128 /*9

144y = 8x² - 16x + 656
144y = 9x² - 144x + 1152

9x² - 144x + 1152 = 8x² - 16x + 656
x² - 128x + 496 = 0
(x - 4)(x - 124) = 0
x = 4, 124

Now let's see what's y (the radius) for each x:
16y = 4² - 16*4 + 128 = 80
y = 5

When x = 124:
16y = 124² - 16*124 + 128 = 13520
y = 845

So the two circles have radii of 5 and 845, so the sum of the radii is 850.

Hope you liked it.
Nadav

nadavs

Modular Arithmetic and the Greatest Common Divisor

Modular arithmetic is a fascinating part of math (also discussed in super math tips). You do it all the time: if I tell you that the time is now 9 o'clock and ask what will be the time 5 hours from now, you'll tell me 2 o'clock without hesitation. You are using modular arithmetic here.

Here is a nice question regarding modular arithmetic and the greatest common divisor. Apparently, they are both tightly related:

Given that a ≡ b (mod m), prove that gcd(a, m) = gcd(b, m)

To find the greatest common divisor you can use a nice little trick. Factor out the number completely and take the lowest degree of each factor. For example, gcd(24, 32):
24 = 2^3 * 3
32 = 2^5

The lowest degree of 2 is 3 and the lowest degree of 3 is 0 (3^0 = 1, so it does not count as a factor). This means gcd(24, 32) = 2^3 = 8.

Since a ≡ b (mod m) (read as "a is congruent to b modulo m"), the difference between a and b must be a whole multiple of m:
a - b = km (k is an integer)
a = b + km
Since k can be any integer, we can also write this as:
a = b - km

Let's use the GCD trick on a and m:
a = 2^n1 * 3^n2 * 5^n3 * ...
m = 2^x1 * 3^x2 * 5^x3 * ...

When we add km to a, we don't lower any exponent of any factor, we can just raise it, so:
gcd(a, m) = gcd(a + km, m)

But if you look at the identity written earlier, we can see that:
a = b - km
a + km = b

So:
gcd(a, m) = gcd(a + km, m) = gcd(b, m)

Q.E.D.

Nadav

nadavs

Trigonometric Identities in Fractions

Today I have another trigonometric identity. I also found a nice questions in modular arithmetic, but it will have to wait until tomorrow, since this trigonometry question is much better. The question is short, which means it has a long solution. Here it is, try before you peek:

A + B + C = pi (A, B, and C are in radians).
Prove:
(cot A + cot B) / (tan A + tan B) + (cot B + cot C) / (tan B + tan C) + (cot A + cot C) / (tan A + tan C) = 1

Yeah, that's all you need to prove. That given the first condition, this long sum equals 1. And no, you can't multiply everything by the denominators. That's not how you prove an identity (and besides that, I'm not sure you want to multiply three binomial terms).

All we're left to do is hard work. Lots of fun. Let's see what happens when you take each fraction and try to simplify it a little (I'll write cot A as cotA for simplicity, same for B and C):

First, let's start with the numerator:
cotA + cotB = cosA/sinA + cosB/sinB
Make a common denominator and add:
(cosAsinB + cosBsinA) / (sinAsinB)
The numerator here looks like a trigonometric identity which says:
sin(A + B) = sinAcosB + cosAsinB
So the numerator of the first fraction is:
sin(A + B) / (sinAsinB)

Now the denominator:
tanA + tanB = sinA/cosA + sinB/cosB
Again, make common denominator and add:
(sinAcosB + sinBcosA) / (cosAcosB)
Looks familiar, doesn't it?
sin(A + B) / (cosAcosB)

Now you have the numerator and the denominator of the first fraction. Let's see what we can get out of them:
sin(A + B)/(sinAsinB) / sin(A + B)/(cosAcosB)

As you can see, sin(A + B) cancels (thank god) and we're left with (cosAcosB)/(sinAsinB). As you should know, this equals to cotAcotB. Well, now we have something we can work with.

I'm not going to show work on the other two fractions, I just assume you'll understand how I make this transition now:
cotAcotB + cotBcotC + cotAcotC = 1

Now that's more like it. Let's take cotC out as a common factor. Considering that C = pi - A - B, this can make our life easier. Follow closely now:
cotAcotB + cotC(cotA + cotB)
cotAcotB + cot(pi - A - B)(cotA + cotB)
Since tan(pi - x) = -tan(x), cot(pi - x) = -cot(x), so:
cotAcotB + cot(A + B)(cotA + cotB)

tan(A + B) = (tanA + tanB) / (1 - tanAtanB), so cot(A + B) = (1 - tanAtanB)/(tanA + tanB).

cotAcotB - (1 - tanAtanB)/(tanA + tanB) * (1/tanA + 1/tanB)
Make a common denominator and add, again:
cotAcotB - (tanA + tanB)/(1 - tanAtanB) * (tanA + tanB)/(tanAtanB)

Luckily, we can cancel (tanA + tanB):
cotAcotB - (1 - tanAtanB)/(tanAtanB)
cotAcotB - 1/(tanAtanB) + (tanAtanB)/(tanAtanB)

The first fraction equals cotAcotB and the second one equals 1:
cotAcotB - cotAcotB + 1
= 1

Proved.

That was long, but worth it, wasn't it?

Tomorrow, if there's nothing better, we'll have some modular arithmetic.

Yours,
Nadav

nadavs