Circles and Triangles (And the Snowflake which Results)

Today we discuss circles and triangles, particularly inscribed circles and triangles.

For the past few days at work, while I’ve been waiting for the internet to load, I’ve been working on a problem.

A few weeks ago, my boss made me clean the supply cabinet and I found a very nice compass set. When I asked my boss about it, and told her we could probably have a garage sale and someone would love to buy it, she said, “Those things have been marked out of stock, and we cannot sell them.”

“What!” I thought to myself. “That’s so wasteful! It's going to sit in this cabinet forever and no one will ever use it or even know it's there!”

Anyway, this is not about unethical work habits, but MATH (which in this case made me cry because it was sitting unused in the supply cabinet).

So, when I found this compass, along with many other things which are completely useless in our workplace, I took it to my desk and called it the Shipping Manager’s (mine).

I may or may not be obsessed with snowflakes, so as I sit around, waiting for the internet to load, I like to amuse myself by drawing snowflakes. I like to make my snowflakes realistic, so when I draw or cut out snowflakes, I make them six sided. This is a little harder than the typical eight sided when cutting, and quite a bit harder when drawing. This compass however, proves miraculous when making equilateral triangles.

For those of you who don’t know how to construct an equilateral triangle with a compass, it is simple. Create a circle. Create another circle, using any point of the first circle as the center, and using the same radius as the first circle. Like so:

By the way, the software I use for geometry is called WinGeom, and is a free software I downloaded from the internet. It is similar to Geometer’s Sketchpad, but free (though every time I see Geometer’s Sketchpad lying on the shelves at work I think about buying it).

You may then proceed to draw another equilateral triangle of the correct size, and you get something that looks like the Star of David, and you may proceed by chiseling out whatever holes you like in your snowflake. In my love of snowflakes, I already see so many ways to begin just from the work we did to construct such a beautifully perfect set of triangles. 

So, now to the work problem.

I was sitting in my desk, looking at my compass and thinking about snowflakes, and how the two circles use so much space for such a small triangle. I wanted something better. I decided to make an equilateral triangle inscribed in a circle. This would give me more creative ability when I got to the snowflake part as I would have more room on the paper.

And I came up with the drawing below. Begin with a circle. Chose a point on the circle, C. Draw ray AC. Bisect the segment AC. Draw circle AE. Mark point G. Draw tangent GH. Mark points IJ. Make Triangle! and then you have an inscribed circle too!

Needless to say, today was about the most amazing day of my life even though I didn’t figure this out until I got home and did math with dry erase markers on my windows. (I used some trigonometry to discover this method. I began with a poorly drawn circle and a poorly drawn equilateral inscribed triangle. If you draw segments from the center to the vertices you get three isosceles triangles. Each triangle has two 30 degree angles and a 120 degree angle. If you divide these triangles in half, you get six 30-60-90 triangles which have something in common with the unit circle, and then sine and cosine and then TADA!*)

Then, I sat down to my computer and started constructing said triangles and circles.

Now, consider the pink circle (I have a close up below). In this circle, F0 bisects segment XD0, H0 bisects F0X, and G0 bisects D0F0.

In this pink circle is green triangle I0C0B0. Point Z slides along ray YD0.

Now, I knew that a triangle formed with the diameter of the circle and any point on the circle will produce a right angle at the point on the circle no matter where the point on the circle is. I wondered: if I have a chord C0B0 and move it to some random location in the circle, and then add point I0 to make a triangle, if I move I0 along the circle, will the angle at I0 be constant? Yes! This added even more excitement to my day. 

(This is the beauty of geometrical software. It allows you to see if a hypothesis is true before you do all the work to prove it is true only to find it was false.)

Now I just have to prove why…

*My purpose is not to make you cry, so please leave me a message if you would like further explanations of my process. 
Also, I tend to do things the hardest way possible. Be warned. 

Ending a Proof

Some people end a proof with a square or a check mark. I like to end it with Tada!

Tip of the Day

Chapter Zero may seem like easy review, but the "easy" axioms you already know will come in handy for the rest of the book. If you're like me and like to slide by with the least amount of work, don't skip Chapter Zero. You will need this information for the rest of your life. (Well, for as long as you struggle through the Contemporary Abstract Algebra book.)

My Story

I never believed my AP Calculus teacher when she told us she sobbed over her homework , but a few years later I had my own tears to shed. I was coming up fast on senioritis, and struggling far behind my mathematical peers. Yes, I went to class every day and took notes, but somehow I managed to fail every homework assignment. And that's when it became hopeless.

 

I tried turning to the internet to help me, but wasted hours searching for the answers. Now, as a bored post-graduate, I feel it my duty to help fellow sufferers and reduce the tears shed over ridiculously difficult math. 

In these pages you will find solutions to the Math I have studied laid out in an easy to understand fashion. I do my best to explain things as I go, even at the risk of insulting your intelligence. I believe it is important to understand the math and not simply have a solution, so I explain thoroughly. 

Now I like to think of myself as a MathMagician, though you are likely much smarter than I. 

Chapter 0 1-10

1.         For n = 5, 8, 12, 20, 25, find all positive integers less than n and relatively prime to n.

For numbers to be relatively prime, or coprime, the two numbers must have no common factors other than 1. So we have:

n = 5

1, 2, 3, 4

n = 8

1, 3, 5, 7

n = 12

1, 5, 7, 11

n = 20

1, 3, 7, 9, 11, 13, 17, 19

n = 25

1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24

2.      Determine gcd(2^4*3^2*5*7^2, 2*3^3*7*11) and lcm(2^3*3^2*5, 2*3^3*7*11).

If we let c=gcd(a, b) we know c is the biggest number such that c divides a and c divides b.

So, gcd(2^4*3^2*5*7^2, 2*3^3*7*11) will be all numbers common between 2^4*3^2*5*7^2 and 2*3^3*7*11. We see the two numbers share 2^1, 3^2, and 7^1. Hence our answer is 2*3^2*7.

If we let d=lcm(2^3*3^2*5, 2*3^3*7*11) we know d must have one of each factor in the two numbers. So d = 2*3*5*7*11.

note: d is not 2*3*5*2*3*7*11 as we only need to take the distinct factors of each number. In essence, we do not want to double up on any numbers when finding lcm.

3.      Determine 51 mod 13, 342 mod 85, 62 mod 15, 10 mod 15, (82*73) mod 7, (51+68) mod 7, (35*24) mod 11, and (47+68) mod 11.

To find 51 mod 13, we may divide 51 by 13 and find there is a remainder of 12. You may also do it the hard way as I did at first and say 13 is too small +13=26+13=39+13=52 which is too big, so 51 mod 13 is 12.

(82*73) mod 7 is the same as (82 mod 7)*(73 mod 7).

(51+68) mod 7 is the same as (51 mod 7)+(68 mod 7).

4.      Find integers s and t such that 1=7*s+11*t. Show that s and t are not unique.

There exist more than one s and t such that 1=7*s+11*t. I found the answer through the wonderful guess and check method.

5.      In Florida, the fourth and fith digits from the end of a driver’s license number give the year of birth. The last three digits for a male with birth month m and birth date b are represented by 40(m-1)+b. For females the digits are 40(m-1)+b+500. Determine the dates of birth of people who have last five digits 42218 and 53953.

We see the last three digits of the first one are 218, so we see the person is male. If the person were female, the last three digits would be larger than 500. So we use the equation

40(m-1)+b=218

40m-40+b=218

40m+b=258

I trust in your ability to solve it from here. There are twelve months, so in the worst case scenario, you may try them each, starting with January.

We see the last three digits of the second one are greater than 500, so the person is a female. We use the equation 40(m-1)+b+500=953 and follow the same steps as above to arrive at the answer.

6.      For driver’s license numbers issued in New York prior to September 1992, the three digits preceding the last two of the number of a male with birth month m and birth date b are represented by 63m+2b. For females the digits are 63m+2b+1. Determine the dates of birth and sex(es)corresponding to the numbers 248 and 601.

Use the method from question 5 to solve this problem.

7.      Show that if a and b are positive integers, then ab=lcm(a, b)*gcd(a, b).

a and b are positive integers (given). Let a=p^m1*p^m2*…*p^mr and b=q^n1*q^n2*…q^ns where all ps are unique primes and all qs are unique primes. So we have both a and b broken down into primes.

Now, gcd(a, b) is the product of all ps and qs common between a and b. So if p^m1 were 2^4 and q^n3 were 2^7, gcd(a, b) would contain 2^4.

And the easiest way to find lcm(a, b) (that I could think of) would be to take a (you could take b and the outcome would be the same) and multiply a by every q which is not equal to some p in a.

Say for example we have 12 and 9. Lcm(12, 9) is 36. To find this we take the prime factors of 12 (2, 2, 3) and the prime factors of 9 (3, 3). To find lcm(12, 9) we take 9 and multiply it by the two 2s which are factors of 12. This gives us 36. Or we could take 12 and multiply by the one 3 which is “extra” in 9. This also gives us 36.

Now, when we look at the gcd(a, b) we see it is the product of all common prime factors, and lcm(a,b) would be (WLOG-without loss of generality) a times everything in b which is not also in a. With the two together we have everything in a and b, or ab.

8.      Suppose a and b are integers that divide the integer c. If a and b are relatively prime, show that ab divides c. Show by example, that if a and b are not relatively prime, then ab need not divide c.

We know by the problem that a and b are relatively prime. This means they share no common prime factors. We know from the above problem that to find the lcm(a, b) we take either a or b, let’s take a, and multiply by every factor of b which is not in a. This is every factor, so we have a*b. And as we just found the least common multiple, any common multiple will contain a and b.

9.      If a and b are integers and n is a positive integer, prove that a mod n=b mod n if and only if n divides a-b.

So, say we have the set {c, d, e, f, g, h, i, j, k}.

This was my first, not so good proof. It may help you understand the concept, but will not suffice as a real answer:

We shall begin with the case where a mod n = b mod n and prove that this implies a-b divides n. In the set above, let’s say a mod n = b mod n = f. This means that a is the same as b + some multiple of n which could be written as a=b+yn where y is some integer. Using algebra we can change this to a-b=yn. In other words, a-b is a multiple of n, which was what we wanted to prove.

My second proof is better.

We shall begin by writing a=q1n+r1 and b=q2n+r2. This comes from the division algorithm, and from it we also see that amodn=r1 and bmodn=r2 (Think of going through the set n times, and the remainder is what’s left over, giving you amodn). And we write a-b=q1n+r1-(q2n+r2). Algebra gives us a-b=n(q1-q2)+(r1-r2).

Now, for the first part we let amodn=bmodn and will prove this implies n divides a-b.

From above we know amodn=r1 and bmodn=r2. This means we have a-b=n(q1-q2)+(amodn-bmodn). Since amodn=bmodn we have a-b=n(q1-q2). We can let q1-q2=k where k is some integer and a-b=nk, so n divides a-b.

For the other direction we let n divide a-b and prove this implies amodn=bmodn.

If n divides a-b, we have a-b=nk for some integer k, and the remainder is 0. So this must mean the remainder, r1-r2=0, or r1=r2. And since we know amodn=r1 and bmodn=r2, amodn=bmodn.

Tada!

10.  Let a and b be integers and d=gcd(a, b). If a=da’ and b=db’, show that gcd(a’, b’)=1

The important thing for this problem is to make sure you are familiar with the theorems in the book. When I first attempted this problem, I thought I could do it on my own, but only after I went back and read the theorems did I realize how to solve the problem. Read the book—what a novel idea.

Also, if you do not solve this problem on your test, your professor will know you did not read the book. Reading the book is pretty crucial to difficult math classes, and failing to read implies a failing grade.

We will use Theorem 0.2 from page 5. This theorem states: For any nonzero integers a and b, there exist integers s and t such that gcd(a, b)=as+bt. Moreover, gcd(a, b) is the smallest positive integer of the form as+bt.

Now that we have read this, the problem comes down to simple algebra.

We have

d=as+bt for some integers s and t, and we can substitute a=da’ and b=db’ to get

d=da’s+db’t

d=d(a’s+b’t)

1=a’s+b’t

As seen from above, this is the exact form stated in the theorem (Let gcd(a’ b’)=d. Then d’=a’s+b’t.), so we know gcd(a’, b’)=1.

Tada!