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

Geometric Construction and the Sinus Function

Today I have a fantastic question from Yahoo Answers about geometric construction. The idea of geometric construction is to use a compass and a ruler (with no measurements) to construct different geometric shapes and figures.

Notice: to answer this question you need to know basic constructions: copying a segment, creating a segment X times larger than another, find a perpendicular bisector, and copy an angle. Without those you'll be lost.

Here is the question:

A triangle has sides a, b, and c. The ratio a/b = 7/4. You are given side c and the radius of the circumcircle, r. Construct the triangle.

This may seem simple, but it's a little more sophisticated than that. To construct that triangle, we need the law of sines.

The law of sines says that in a triangle, a/sin A = b/sin B = c/sin C = 2R (side a and angle A are opposite, R is the radius of the circumcircle). Let's play with this law a little:
a/sin A = b/sin B
That means:
a/b = sin A / sin B

In our triangle, a/b = 7/4. So all we need to do is find two angle whose ratio of sines is 7/4. So how do we do that?

The definition of sine A is the y-coordinate on the unit circle with the angle measure of A (read here about the unit circle). First, construct two segments, a' and b' in a way that a'/b' = 7/4. Now create a circle with a radius of more than a' and two perpendicular axes that intersect at the center of the circle. Position a' in the circle in such a way that it's perpendicular to the x-axis and touches the circle in one point (yet not tangent to it). Construct the line from the origin to the point of intersection of a' and the circle. Call the angle between the x-axis and that line angle A. Do the same thing for b' and angle B.

Since we used the definition of sine, we now have two angles with a sine ratio of 7/4. Now we need to create the triangle we want.

First, construct segment c. Since the circumcenter, the center of the circumcenter, is on the point of intersection of all perpendicular bisectors, construct the perpendicular bisector of segment c. Now, since the radius is given, use the end point of segment c and the perpendicular bisector to find the circumcenter and draw the circumcircle.

Now, when segment c is inside the circle, copy angle A to one of its sides and angle B to the other side. Complete the triangle, and you are done.

Feel free to send in more questions!
Nadav

nadavs

Exponential Equations Fury

Yesterday a friend called me and asked me to help her with a system of exponential equations. At first I thought the system is impossible to solve. But then, when I thought all is lost, I started over and used another property of exponentiation. Then it all became clear.

This system is special, so if you know how to solve exponential equations try to solve it first. It gives a great feeling of satisfaction when solved:

2^{x - 1} * 3^{y + 1} = 72 * 6^{y - x + 2}
3^{x - 1} * 2^{y + 1} = 3 * 6^{x - y - 2}

At first, like most people, you'll be tempted to get everything in terms of powers of 2 and powers of 3 (since 6^x = 2^x * 3^x). However, this will get you nowhere. Instead, notice that both equations have terms with equal exponents, so let's take advantage of that and divide the first equation by the second (I put the factors in an order that will allow me to manipulate them later):

2^{x - 1} / 3^{x - 1} * 3^{y + 1} / 2^{y + 1} = 72 / 3 * 6^{y - x + 2} / 6^{x - y - 2}

Now let's use the exponentation rule that says aⁿ / bⁿ = (a / b)ⁿ. We also need the rule which says aⁿ / a^m = a^{n - m}:
(2/3)^{x - 1} * (3/2)^{y + 1} = 24 * 6^{y - x + 2 - (x - y - 2)}
Using the exponentation rule of 1/aⁿ = a^-n, we can turn the 2/3 into 3/2 and take the opposite of the exponent. Also, let's get the final version of the exponent of 6:
(3/2)^{1 - x} * (3/2)^{y + 1} = 24 * 6^{2y - 2x + 4}

Using the rule that says aⁿ * a^m = a^{n + m}, we can add the exponents of 3/2:
(3/2)^{y - x + 2} = 24 * 6^{2y - 2x + 4}

Look at the exponent of 6, looks familiar, right? Using the rule of (aⁿ)^m = a^nm, we can say:
6^{2y - 2x + 4} = 6^{2(y - x + 2)} = (6²)^{y - x + 2} = 36 ^{y - x + 2}

Now we have a really nice equation with equal exponents on both sides:
(3/2)^{y - x + 2} = 24 * 36^{y - x + 2}

Divide by 36^{y - x + 2}. Using the rules of exponentations mentioned above, this equals:
((3/2) / 36)^{y - x + 2} = 24
(1/24)^{y - x + 2} = 24¹
24^{x - y - 2} = 24¹
x - y - 2 = 1
x - y = 3
x = y + 3

Finally, a connection between x and y. Now let's plug it in one of the original equations to get the real values of x and y:
3^{x - 1} * 2^{y + 1} = 3 * 6^{x - y - 2}
3^{y + 3 - 1} * 2^{y + 1} = 3 * 6^{y + 3 - y - 2}
3^{y + 2} * 2^{y + 1} = 3 * 6¹
3² * 3^y * 2¹ * 2^y = 18
18 * 3^y * 2^y = 18
6^y = 1 = 6⁰
y = 0

Finally, a value. Let's plug that value in the easy equation: x = y + 3. That means x = 3. Problem solved.

Final answer:
x = 3
y = 0

That was long, but it was worth it. Now it's your turn - solve the two towns question using calculus and you will be in the hall of fame. Go ahead, try it.

Nadav

nadavs

More Trigonometric Identities Revealed

For today there is another question from Yahoo Answers, also about trigonometric identities. This time it's not about simple ones like before, but a more complex one, like the ones I sometimes answer on super math tips. Here it is:

Prove:
(cos 4x + cos 2x) / (sin 4x + sin 2x) = cot 3x

At first glance this seems irritating and impossible, but with some creative thinking it's very possible.

Now, what do 2, 3, and 4 have in common? Of course, 3 is the average, and we're going to use that property heavily.

Since we want to have cot 3x at the end, let's change all sines and cosines to 3x + x or 3x - x and hope they cancel:
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x.

Well, they don't cancel so fast, so we'll have to expand all expressions (remember what's sin (A + B)?) and then hope that we can cancel out some terms. Let's start (I put the arguments in parentheses so you can see them better):
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x
Let's start from the numerator:
cos(3x)cos(x) - sin(3x)sin(x) + cos(3x)cos(x) + sin(3x)sin(x)
2cos(3x)cos(x)

Now, the denominator:
sin(3x)cos(x) + cos(3x)sin(x) + sin(3x)cos(x) - cos(3x)sin(x)
2sin(3x)cos(x)

Now divide those two terms and see what happens:
2cos(3x)cos(x) / 2sin(3x)cos(x)
The 2 and cos(x) cancel:
cos(3x) / sin(3x) = 1/(sin(3x) / cos(3x)) = 1/tan(3x) = cot (3x)

Job complete.

Besides solving these questions, here is a challenge for you: try to solve the previous question (with the towns) using calculus. I haven't solved it myself yet (except for the geometry part, which is easy).

The first one to solve (on comments or email) will be written in the hall of fame on the blog's list (which I'll create) with a link of his choice and the right to post here any question.

Go on to solving!
Nadav

nadavs

Geometry and Road Design

Today I have a very interesting question involving geometry. Its solution is so simple that it will make you think "How didn't I think about it?!" when you see the answer. Here it is, try to solve yourself first (as I said, it's easy):

Two towns, town 1 and town 2, are located to the south of road A going in east-west direction. No roads are connecting the towns to each other or to road A. The people of both towns decided to build two roads: one from each town to road A, so they could travel between the towns. What is the shortest route to build these roads?

Well, the shortest route is to build these roads from town to town, but the people of the towns want a connection to road A, so let's give them that.

To solve this question, let's shift our thinking a little bit: let's imagine that town 1 moves to the north of road A, but still keeps the same distance from it (if it was 100 miles to the south, now it's 100 miles to the north).

The shortest distance between two points is on the straight line between them, and so is the road between the towns now. Notice that this road intersects road A.

Now move town 1 back to where it was, and create the road from town 1 to where the roads between the towns intersected road A. It's the same distance as when town 1 was at north, so it must be the shortest distance. Problem solved!

This principle is also known in physics in the field of optics and mirrors. Who said math has no use in real life?

Nadav

nadavs

Reversing a Number

After a pretty boring question yesterday, I found a good one today. Not just good - great. It's a general mathematics question with almost no hidden hints. All it takes to answer that question is pure thinking. Here it is:

There is a number abcde (each letter represents a different digit). When multiplied by 4, this number becomes edcba. What is the number?

Try thinking for a minute. What is the boldest clue in here? Of course, the number of digits. This information alone gives us two digits.

Since there are five digits on both numbers, the first digit, a, must be 1 or 2 (if it's 3 and above, the big number would have six digits). Now, because digit a appears at the units digit of a number divisible by 4, it must be a possible digit of a multiplication by 4. No integer will end in 1 when multiplied by 4, but ending with 2 is possible (for example: 3*4 = 12), so a = 2 and the number is:
2bcde

Now, because this number is multiplied by 4 and starts with 2, it must end with 8 or 9. 4 * 9 = 36, ends with 6, not good. 4 * 8 = 32 - perfect for us, since it ends with 2. So now the number is:
2bcd8

Notice how the bigger number starts with 8 and not 9. This means that b * 4 < 10, so b must be 1 or 2. Since 2 is taken, b = 1:
21cd8

In the bigger number, 8dc12, the tens digit is 1, but 8 * 4 = 32. This means we need a number that when multiplied by 4 gives 8 (to add to the 3). This means d is either 2 or 7. 2 is taken, so d = 7:
21c78

Now two things can be done: one, guessing all possible values of c. Two: calculating c. Since I don't think guessing is a legitimate reason, let's calculate:
c is in the hundreds place, so the number is 21078 + 100c, and the bigger number is 87012 + 100c. That gives the equation:
4(21078 + 100c) = 87012 + 100c
84312 + 400c = 87012 + 100c
300c = 2700
c = 9

Finally, the number is: 21978. Multiply that by 4 and you'll get 87912. Job complete.

Want to see more of these tricks? Join super math tips, there are some really nice things over there.

Hope you liked it,
Nadav

nadavs

Trigonometry on the Table

Today I have a very simple question from Yahoo Answers. Hopefully tomorrow I'll be able to find something a little more interesting:

Prove that tan² x + 1 = sec² x

Every trigonometry student must remember this identity, but here is the reason why the trig students must know this identity:

As you all should know, sin² x + cos² x = 1 (if you don't know that, blame your math teacher). Now all we have to do is use some more trigonometric identities and rules of exponents, and we're all set:

tan x = sin x/cos x
sec x = 1/cos x
(a/b)ⁿ = aⁿ/bⁿ

Now we take the basic trigonometric identity and divide it by cos² x:
sin² x / cos² x + cos² x / cos² x = 1 / cos² x
(sin x / cos x)² + 1 = (1/cos x)²
tan² x + 1 = sec² x

Simple, nice, and easy.

Bring in more questions!
Nadav

nadavs

Playing with Logarithms

Logarithms, or logs, are very useful and powerful tools in mathematics. They can be used for many different problems, and all their properties make them very easy to manipulate.

Today I have another question from Yahoo Answers, but I'll take only the good part:

log_a x = c
log_b x = d
Find log_ab x in terms of c and d

This question looks simple. After all, logs have so many properties, there must be one with multiplication of base.

Well, there isn't.

In second thought, there is, but it's not written in any textbook, and to get to it we need to go through many log properties. Shall we start?

First, we can use the log property of reversal of base and argument:
log_a x = 1/log_x a

Now we can write this for both definitions above:
log_x a = 1/log_a x = 1/c
log_x b = 1/log_b x = 1/d

Using the addition property of logs that says: log_x ab = log_x a + log_x b, we can say:
log_x ab = 1 / c + 1 / d = d/cd + c/cd = (c + d) / cd.

Now all we need is to reverse the log back to the one we want:
log_ab x = cd / (c + d)

As you can see, logs don't require too much complicated mathematics, but rather original thinking and creativity.

Do you have any log problems?

Yours,
Nadav

nadavs

Magic C

For today's question, I took another Yahoo Answers great question to give you the opportunity to solve a nice question. Here it comes:

The digits from 1 to 5 should be placed on a C shaped form like this:

x-x
x
x-x

(Each x represents a different digit)
The two horizontal rows (x-x) and the vertical row with three x's should have the same sum of digits.
In how many ways can this be done (reflections don't count as different), and why there are no other possibilities?

We can see at first that in order to fill in the digits, we need to find two pairs of digits that add up to the same number (the two x-x). There are three numbers that work like that: 5 (1+4, 2+3), 6 (2+4, 1+5), and 7 (3+4, 2+5).

However, 5 is impossible as a sum, since it leaves the digit 5 outside, which means it'll be in the middle x, creating a sum definitely greater than 5. We are left with only 6 and 7 as sums. After some playing with arrangements, we can see that the two solutions are:

For 6:
1-5
3
2-4

For 7:
2-5
1
4-3

There are no other possibilities because all other sums (except 5) have a maximum of one pair that works for them, and we already showed why 5 doesn't work.

Now it's your turn. Send in the riddles!

Have a great weekend,
Nadav

nadavs

Digit In-A-Box

For today I have another question from Yahoo Answers. It's very original and interesting, and even worth trying out before going over to the answer.

Joe and Joana are playing a game. Joana is picking up between 2-9 boxes and puts a digit in each box, so the number that comes out when the digits are read from left to right is divisible by the number of boxes. For example, if Joana picks 4 boxes, she can use the digits 1,2,7,9 to create the number 1972, but not 7129. Joe must switch any two boxes that he wants. If the number is still divisible by the number of boxes, Joe wins. Otherwise, Joana wins. What number of boxes Joana must pick to always win?

This calls out for some divisibility rules. Let's see how Joe can always win for a number of boxes, and we'll start with 9:

9: it doesn't matter which boxes Joe moves, he wins. When a number is divisible by 9, the sum of its digits is also divisible by 9. Since addition is commutative, it doesn't matter which order the digits are: their sum will remain the same and they'll be divisible by 9.

8: the divisibility of a number by 8 depends on its last 3 digits. As long as Joe doesn't touch the last three digits, the number will be divisible by 8.

7: This is quite tricky, because it doesn't use the divisibility rule of 7 (which you can find on super math tips and tricks), but simple math. Since Joana uses 7 boxes here, there are 7 digits, and the number can be written as a sum:
1000000a + 100000b + 10000c + 1000d + 100e + 10f + g
When we replace the first and last digit, we subtract g and add 1000000g and also subtract 1000000a and add one a. This means the total operation is:
+1000000g - g - 1000000a + a = +999999g - 999999a = 999999(g - a).
Luckily for Joe, 999,999 is divisible by 7, so 999999(g - a) is also divisible by 7. That way, Joe adds or subtracts from a number divisible by 7 another number divisible by 7, and doesn't change its divisibility.

6: a number is divisible by 6 if it's divisible by 3 and 2, which means the sum of its digits must by divisible by 3 and it must be even. As long as Joe doesn't touch the last digit, he wins.

5: a number is divisible by 5 if the last digit is either 5 or 0, so as long as Joe doesn't touch the last digit, he wins.

4: a number's divisibility by 4 is determined by its last two digits, so Joe must replace only the first two to win.

3: a number is divisible by 3 if the sum of its digits is divisible by 3, so just like 9, Joe can switch any boxes he wants.

2: a number is divisible by 2 if it ends with 2, 4, 6, 8, or 0. However, since Joe MUST switch two boxes, Joana may pick an even number with one odd digit, like 74. When Joe switches the boxes, he'll get 47, which is odd, so Joana wins.

The final answer: Joana must use a 2-digit number to win.

And a final notice: I added a sign up form for Super Math Tips on the right column of this blog, for easy access.

Have a great weekend,
Nadav

nadavs

Cubic Funcion Discovery

Today I have another question from Yahoo Answers (super math tips people, send some questions too!). The question is simple, but its answer is very, very long.

There is a cubic function that passes through the following points: (-2, -24) (3, -4) (5, 18) (6, 56). Find out what the function is.

(The original question had a point different than (6,56), but it makes it shorter, so the fun is gone).

At first, it seems impossible. But then, you remember that a cubic function has the following form:
y = ax³ + bx² + cx + d

We have x and y, so we can plug them in to find a, b, c, and d! All we have to do now is set up a system of four equations and solve them. All of them (here comes the long part):

I -8a + 4b - 2c + d = -24
II 27a + 9b + 3c + d = -4
III 125a + 25b + 5c + d = 18
IV 216a + 36b + 6c + d = 56

Now we need to get three equations out of those four. We do that be subtracting the equations from each other and getting rid of d. If done smart, it can even simplify the process in the future (if we can get small enough coefficient). There are six possible outcomes, and the smartest thing to do is to choose the best ones:

A. 91a + 11b + c = 38 (IV - III)
B. 98a + 16b + 2c = 22 (III - II)
C. 35a + 5b + 5c = 20 (II - I)

Now we need to eliminate another variable and create two equations. The chosen one is c. First, we multiply equation A by 2:
A2: 182a + 22b + 2c = 76
B2: 98a + 16b + 2c = 22

After we subtract the two equations we get:
84a + 6b = 54
Divide by 6 to simplify:
14a + b = 9

Now multiply equation A by 5:
A5: 455a + 55b + 5c = 190
C5: 35a + 5b + 5c = 20

Subtract them to get:
420a + 50b = 170
Divide by 10:
42a + 5b = 17

Now we have two more equations:
i. 42a + 5b = 17
ii. 14a + b = 9

Multiply equation ii by 5:
70a + 5b = 45

So now we have:
i21: 42a + 5b = 17
ii21: 70a + 5b = 45

Subtract the two equations:
28a = 28
a = 1

Now everything is going to be super fast. We just need to plug in the numbers we get, this time in equation ii (14a + b = 9):
14 * 1 + b = 9
14 + b = 9
b = -5

Another coefficient! They go down like flies! Now let's use equation A, since it has c with a coefficient of 1:
91a + 11b + c = 38
91 - 55 + c = 38
c = 2

So tiny, with so many problems. One coefficient to go, d, the y-intercept. Let's plug it in equation II, where all coefficients are positive but small:
27a + 9b + 3c + d = -4
27 - 45 + 6 + d = -4
d - 12 = -4
d = 8

THE END!

We now have all coefficients for this cubic function, and we can proudly say that it is:
y = x³ - 5x² + 2x + 8

Who said cubics are hard?

Nadav


nadavs

The Riddle of Four

Yesterday I saw an amazing riddle on Yahoo Answers, and I had to share it, but it was after I published the calculus question, and I didn't want to post too much. Even though I answered it with detail, the asker promised to give the best answer to the first one who answers, and my long answer took away my 10 points.

Without further explaining, here is the riddle:

A number has six digits, and the last one (the units digit) is 4. When the 4 is removed from the units place and placed on the beginning of the number (the hundred thousands), the number is multiplied by 4. What is the number?

It sounds difficult and tempting to solve with complex algebra and calculus methods, but the solution is much simpler: use your head.

First, we know that when the 4 is moved to the beginning of the number, it's multiplied by 4. Since it was and remains 6 digits, the first digit must be 1. Now we have two digits:
1 _ _ _ _ 4

When we multiply 4 by 4, we get 16. Since the tens digit becomes the units digit and the bigger number must end in 6, the tens digit is 6. Now we're half way there!
1 _ _ _ 6 4

Now look at the end of this number: 64. When multiplied by 4, we get 4 * 64 = 256. Notice it ends in 6, which is perfect for us, because it'll fit right in the bigger number. That means we have two more digits, and only one is missing!
1 _ 2 5 6 4

To find the last digit, we need some more thinking. The bigger number looks like 4 1 _ 2 5 6. Notice that the 1 is not changed, and so does the 2. That means we need a digit that does not change when multiplied by 4, and the winner is zero.

The number is 102564. When you multiply it by 4, you get 410246. Riddle solved!

Nadav

nadavs

Buggy Calculus

Today I have a question of a higher level from the previous two, but it's much easier from the previous ones for people who know calculus (seriously, much easier).

So, here is the question:

A bug is walking on a wooden log. Its distance from the right edge of the log after t minutes is given by the function x(t) = t³ - 9t² + 500 inches.

A. What is the bug's velocity after 4 minutes?
B. When does the bug change direction?
C. When does the bug's acceleration equal zero?

If you don't know calculus you'd probably say "There's no way to know that", but there is a way, and it's simple.

The velocity function of the bug is the derivative of its distance function. The acceleration of the bug is the derivative of its velocity function. When we get those two functions, the question is very easy.

A. Let's find v(t), the bug's velocity function:
x'(t) = v(t) = 3t² - 18t .
To find the velocity after four minutes, we just plug in 4 instead of t:
4² - 18 * 4 = 16 - 72 = -56 inches/minute. A negative velocity means that the bug is going "backwards", meaning towards the right edge (because we defined a positive distance as going away from the right edge, meaning left).

B. To find when the bug changes direction, we need to find out when its velocity is zero. That means the velocity is changing signs, and by changing signs it means a change of direction. We already have the function, so let's make it equal zero:
3t2 - 18t = 0
t2 - 6t = 0
t(t - 6) = 0
t = 0, 6

The bug cannot change direction at minute zero, since it did not have a direction before. That means the bug changes direction on the 6th minute.

C. To find when the acceleration is zero, we first have to find the acceleration function. We do that by derivating the velocity function:
v'(t) = a(t) = 6t - 18.
Now let's make it equal to zero:
6t - 18 = 0
6t = 18
t = 3

That means the bug's acceleration is zero after three minutes.

Don't understand a word of this post? Don't be shy to ask anything on this blog's comments or the super math tips.

Hope you learned something.
Yours,
Nadav

nadavs

Angles in Tetrahedron and Octahedron

Today we have a question from one of the subscribers to Super Math Tips. Her question goes like this (I changed it a little to be more accurate):

How many angles can the center of a circumsphere around a tetrahedron and an octahedron (separated) form with the vetices of these polyhedra?

First, you should know what octahedron and tetrahedron are. Click on their names to find their pictures on Wikipedia.

And now, for the answer:
Let's start with the tetrahedron (the pyramid), since it's smaller and simpler.

A tetrahedron has four vertices. To form an angle, we need two points to connect to the center of the circumsphere. At first, we have a choice of 4 different points (vertices) to choose from. After we choose one vertex, we have only 3 choices left. So the number of angles is seemingly 4 * 3 = 12. However, when we choose like that, there is an importance to the order of selection, but in angles there is no importance. That is why we divide this number by 2 to get rid of duplicate angles (like vertices 1 and 2 and vertices 2 and 1, which form the same angle). So in a tetrahedron you can form 6 angles with the vertices.

Another way to solve this problem is using combinatorics. One of the functions in combinatorics allows us to choose r objects (in this case, vertices) out of n objects when order does not matter. It goes like this:
nCr = n!/r!(n - r)!
The exclamation mark means factorial.

We want to choose 2 vertices out of 4. We plug in the numbers, and we get 4C2 = 4!/2!2! = 6, the same answer.

The octahedron is solved the same way: it has six vertices, so we can form 6 * 5 / 2 = 15 angles, or we can use combinatorics to calculate 6C2 = 6!/2!4! = 15. However, this number may be wrong.

Three of these angles are straight lines. From top to bottom of the octahedron and the diagonals of the "middle square". If those straight lines count as 180 degree angles, we have 15 angles in an octahedron. If not, we need to subtract 3 angles, so we get 12 possible angles.

Another way to see that is at first we can choose from 6 vertices, but then we can choose only four: we can't choose the same vertex again and we can't choose the vertex opposite the one we chose. This gives us 6 * 4 / 2 = 12 angles.

Keep sending in questions!
Yours,
Nadav

nadavs

Octagonating a Square

Welcome to the math questions blog!

On this blog I will publish interesting questions asked on Yahoo! Answers and I will try my best to answer them as clearly as possible. Also, if you post questions on the comments section, I will try my best to answer those too.

So, to start with, we have a questions about making an octagon from a square:

A square table is 48 inches on each side. You want to re-shape that table into an octagon, and you're going to do it by cutting off the corners. How long will each side of the octagon be?

At first you might be tempted to divide each side of the table by 3 and call it the answer, but you are cutting a diagonal piece of an isosceles right triangle, which means we're dealing with square roots here. So how do we solve that?

First, we call the length of the cut (the diagonal cut), which is also the length of the new octagon side, x. We are cutting an isosceles right triangle off each corner, and their hypotenuse is x (as we defined it). That means the length of the triangle's leg is x / sqrt(2). New term - sqrt means square root (of what's in the parentheses).

Now we know that each side of the table is twice x / sqrt(2) + x. That's because each side of the table has two "triangle legs" and the length of the side of the octagon. The side of the table is also 48 inches long, and that's where we make our equation:
2x / sqrt(2) + x = 48

When we solve that equation, we get that x = 19.882...

So there you have it: if you have a square table with a side length of 48 inches and you wanted to make a regular octagon out of it, every side of the octagon should be approximately 19.882 inches long.

Also, if you want some nice math tricks and tips, visit the Super Math Tips and Tricks page.

Give me questions!
Yours,
Nadav

nadavs