Problem of the Week Day 2: Week of 6/18/12 - 6/22/12

It is now day two of June’s problem of the week. Good luck!

Easy: Take out your answers to h and a, and solve the problem below:

a^5 + {a + √[-(a x h - h^3 + h ÷ a)]} = z

It looks straight forward, but there is a catch. You cannot solve these problems left to right, or in the order that is most convenient to you. There is a specific way of solving these problems, called the order of operations.

You must solve the problem in this order:

1. Solve what is inside any Parentheses or brackets
2. Solve any Exponents or roots
3. From left to right, complete all Multiplication and Division
4. From left to right, complete all Addition and Subtraction

You can remember this order with the phrase “Please Excuse My Dear Aunt Sally,” standing for parentheses, exponents, multiplication, division, addition, subtraction.

For example, look at this problem:

8 + 2^(2 x 2) ÷ 4

In this case, you would have to do the following:

8 + 2^(2 x 2) ÷ 4

8 + 2^4 ÷ 4

8 + 16 ÷ 4

8 + 4

12

It seems a little weird, but it is the correct way of solving it. Good luck. Remember to jot down z as well.

Hard: Now that we finished our trigonometry day, let’s move on to some Algebra. This day will probably be the longest day of the week for the hard problem, so get ready.

(f, 1808.51)

(g, 14)

(h, -6)

If you plot these three points on the Cartesian Plane, what equation goes through all of these points?

To solve this, you use something called systems of linear equations. Say we had to do it with the points (3, 0), (1, -8), and (-2, -5).

First off, you need to figure out how many points there are, and subtract one. This is the degree we are working with. In this case, we have three points, so we have a quadratic equation, or an equation with an x^2 in it.

Next, take the equation y = ax^2 + bx + c (for quadratic). We know some values that will come out for y, right? The three points above! So, we will plug all of those in for x.

0 = a(3)^2 + b(3) + c

0 = 9a + 3b + c

-8 = a(1)^2 + b(1) + c

-8 = a + b + c

-5 = a(-2)^2 + b(-2) + c

-5 = 4a - 2b + c

Now, we will solve for a, b, and c. To do this, we first have to eliminate a variable so it is just two variables. Conveniently, all of the c’s have the same coefficient. So, we will just subtract all of the equations from each other.

(9a + 3b + c) - (a + b + c) = 0 - -8

9a + 3b + c - a - b - c = 8

(9a - a) + (3b - b) + (c - c) = 8

8a + 2b + 0 = 8

8a + 2b = 8

(4a - 2b + c) - (a + b + c) = -5 - -8

4a - 2b + c - a - b - c = 3

(4a - a) + (-2b - b) + (c - c) = 3

3a - 3b + 0 = 3

3a - 3b = 3

With two variables, you only need two equations. Now, we must eliminate another variable. To do this, we must create a common coefficient. Since there is none, we will multiply both equations by something to do so.

3(8a + 2b) = 3(8)

24a + 6b = 24

-2(3a - 3b) = -2(3)

-6a + 6b = -6

Now, we will subtract just like before.

(24a + 6b) - (-6a + 6b) = 24 - -6

24a + 6b + 6a - 6b = 30

(24a + 6a) + (6b - 6b) = 30

30a + 0 = 30

30a = 30

And now, we have an equation we can solve.

30a = 30

a = 1

Since we have found a, we can plug that back into one of the two variable equations to get b.

3a - 3b = 3

3(1) - 3b = 3

3 - 3b = 3

-3b - 0

b = 0

Since we now have b, we can plug that into one of the original equations to get c.

-8 = (1) + (0) + c

-8 = 1 + c

-9 = c

So, we have:

a = 1

b = 0

c = -9

If we plug that into our y = ax^2 + bx + c, we get:

y = 1x^2 + 0x + -9

y = x^2 - 9

And there is your equation. Go through the same exact process as this, but with the other points. Record down a, b, c, and the equation you came up with. You will need it tomorrow.

Problem of the Week Day 1: Week of 6/18/12 - 6/22/12

Today, we will officially begin 2012’s first problem of the week. Before I start, please do not comment with your answers (please feel free to comment with any tips, or cool stuff with the types of problems I’m giving) because I want everyone to have the chance to figure it out themselves. If you have any questions or want to know if your answers are correct, please email me directly at Ethan@EthanMath.com.

You may think that since this is a five day long problem, it must be extremely hard; even the easy one. That is not the case. The easy problem is definitely doable for anyone who has taken fifth-sixth grade math and understands the directions. The hard problem is on the difficult side, but still definitely possible. With some extra effort, you will be able to get through it.

Easy Problem: I like to start the week with some triangle calculations, because the things you can do with them really are fascinating. With triangles, you can measure the height of a building without measuring the building, find the weight of a nearby planet, and many other incredible calculations. For the easy problem, we will keep it down to the basics of triangles.

Look at the triangle above. It has side lengths 2, √5, and √5. You will need to do two things with the triangle: determine the height h, and then find the area a.

For the height, you will use something called the Pythagorean Theorem. If a right triangle’s shortest side is a, it’s middle side is b, and it’s longest side (which is called the hypotenuse) is c, then a^2 + b^2 = c^2.

Say you have a triangle with sides 6, 5, and 5. To figure out the height, it must be a right triangle. However, we can make it a right triangle by spitting it down the middle. This gives us sides:

a = 6/2 = 3

b = height

c = 5

3^2 + b^2 = 5^2

9 + b^2 = 25

b^2 = 16

b = 4

To figure out the area, you do:

(base x height)/2

In this case, it would be:

(6 x 4)/2

24/2

12

So, the area of this is 12 square units. Do the same thing with the other triangle.

Once you’ve finished, make sure you jot down a and h for tomorrow’s problem.

Hard Problem:

Look at the triangle above. In this problem, two of the sides are missing. You will also have to figure out the missing angle. To do this, you can use something called cosines. Cosines say that if you divide the two sides next to the angle (the shorter one is called the adjacent and the longer is called the hypotenuse), that you will get the cosine of that angle.

adjacent/hypotenuse = cosine

If you know the angle you are looking for, you can take out a calculator and type in the angle, followed by the “cos” button. This will give you the cosine.

Then, you can plug in these values to figure out the next part.

For this triangle, you will be able to find the sides and angles with cosines as well as the Pythagorean Theorem which is in the easy problem’s directions. Remember to jot down the answers to f, g, and h because you will need them for tomorrow’s problem.

The Math of Your Locker

Since school just ended, my mind is kind of on clearing out my locker. So, I thought that for today’s post, I would talk about some math that deals with lockers.

First off, picking lockers. We usually have a locker combination to prevent thieves from breaking in. Some thieves can enter a locker combination in a second.

Let’s say you have three dials, each with a number from one to six on it. Your combination is three numbers between one and six. So, there are 6 x 6 x 6 possible combinations.

6 x 6 x 6 = 216

Pretend this thief went to your locker and tried every single combination. If they can do a second per combination, they will be getting in in under four minutes.

216 ÷ 60 = 3.6 = 3:36

Because of probability, it should only take half this time, meaning that a thief could break into your locker in under two minutes.

Don’t get scared. For a four dial locker with numbers from 1-10, it would take over an hour to break in. I just find it cool how small the numbers are for the three dial locker.

Secondly, I would like to share a problem I learned in fifth grade. As usual, I will post the answer in a month. This will be good practice for the problem of the week beginning on Monday.

There is a school with 1000 students and 1000 lockers. Student one runs through the school and opens every locker. Student two runs through the school and closes every second locker. Student three runs through the school and opens/closes every third locker and so on. If all 1000 students go through the school the same way, which lockers are left open. Is there a pattern they follow?

Our problem of the week is back! If you were not following Cool Math Stuff last summer, let me explain. In the summer, each month, I will create two very long problems: an easy problem and a hard problem. Problems so long that it takes five days to figure out the answer. Each day, I will post the next part of the problems for you to solve. After the five days, you will have a month to determine the answers. I will post the answers on the Saturday's post following the next month's problem of the week.

Last year, there was guidance with the problem. This year, I am going to make it a little more challenging. For June, there will be the same level of guidance as last year. July, I will be taking it down a notch. You will have some hints (maybe the important formulas, or some calculator buttons to look for), but no in-depth instruction like before. In August, it's all you. There will be a problem, and an answer the next September.

June's problem of the week will be given from 6/18 - 6/22.

Also, I gave you a KenKen puzzle a month ago (maybe a little more). Here is the answer:

Simple Math Facts Finally Proven

Before I begin, remember that I posted about the mathematical game of KenKen a few weeks ago. If you didn't see it, go to bit.ly/Obc0rI. I gave a sample puzzle, which I will give the answer to next week. Please make sure to try it, so you can see if you are correct. You can also go to KenKen.com for more puzzles.

There are some things in math that are so basic that we commonly just take it for granted. For example, that an odd number + an odd number = an even number.

1 + 1 = 2
3 + 5 = 8
31 + 87 = 118

Or that an even x an even = an even.

2 x 2 = 4
8 x 6 = 48
14 x 18 = 252 (to do this in your head, check out the very first blog post)

But why are these facts true? Well, let's try to prove them with some handy-dandy algebra,

First off, odd + odd = even. An odd number can be written algebraically as:

2n + 1 (assuming that n has no decimal, or in other words, n is an integer)

This is because 2n is even (an integer times two is always even), and adding one to an even number always gives you an odd number.

So, we have:

(2n + 1) + (2m + 1) =

Pretend m is another integer. Since it is all addition, we can ignore the parentheses (this is known as the associative law) and add the two ones together. This gives us:

2n + 2m + 2 =

We can factor out a 2 to get:

2(n + m + 1) =

Since n, m, and 1 are all integers, they must be an integer when added together. Therefore, we can say that:

2(integer) =

As we said before, 2 times an integer is an even number, so we have just seen that odd + odd = even.

What about the second fact, that even x even = even. Let's use the same exact logic.

(2n) x (2m) =
4mn =
2(2mn) =

Since n, m, and 2 are all integers, when multiplied together, it gives you an integer. So we can say:

2(integer)

Since that is an even number, we have proven it.

After learning these facts in third or fourth grade, it is cool to be able to understand the reasoning behind them. They seem almost so simple that there isn't any need for a proof, but I think the proofs are cool nonetheless.

Our problem of the week is back! If you were not following Cool Math Stuff last summer, let me explain. In the summer, each month, I will create two very long problems: an easy problem and a hard problem. Problems so long that it takes five days to figure out the answer. Each day, I will post the next part of the problems for you to solve. After the five days, you will have a month to determine the answers. I will post the answers on the Saturday's post following the next month's problem of the week.

Last year, there was guidance with the problem. This year, I am going to make it a little more challenging. For June, there will be the same level of guidance as last year. July, I will be taking it down a notch. You will have some hints (maybe the important formulas, or some calculator buttons to look for), but no in-depth instruction like before. In August, it's all you. There will be a problem, and an answer the next September.

June's problem of the week will be given from 6/18 - 6/22.

Fibonacci Day: Some Additional Fibonacci Fun

I don't know if you noticed, but today is a Fibonacci day. It is the second of June, and 2 is the third Fibonacci number.

Something mentioned frequently in the mathematical world is that the Fibonacci numbers often appear around nature. It is also in science, architecture, we even found it in literature.

Rather than explaining some applications, I thought I would show you a neat video I found instead.

I find all of the applications fascinating, as well as the fact that we can easily draw our own golden rectangle. You don't even have to be an artist to do it.

Bonus: In the video, they mentioned that if you take the square out of the golden rectangle, the remaining rectangle is a golden rectangle. You can prove this 2 ways.

First off, the sides of a golden rectangle can be two consecutive Fibonacci numbers. Say they are 55 and 89.

If you cut off a 55x55 square, you are left with a 34x55 rectangle. Since these are two consecutive Fibonacci numbers as well, it is a golden rectangle.

The more interesting one, however, is to look at the golden ratio itself. If you do 1/1.618034..., you get 0.618034...

In other words, phi:1 = 1:phi-1. So, if we cut off a 1x1 square from the rectangle with side ratio phi:1, we are left with a side ratio of 1:phi-1, which is the same as before. I found this really cool about Fibonacci numbers.