Divine Lunacy

Come Friday, a new POW. The title of it is Diminishing Squares

The Figure begins with a square of side length 2 units. Squares are continually added which have a side length which is one-half the previous square. Suppose this process continues to infinity. What is the area of the Figure? Explain your answer.

Well, the answer is most definitely not infinite, certainly not 5 and definitely not 6. Depending how far one wants to take this, it can involve some calculus, or just some simple arithmetic, but either way, it will involve plenty of exponents. Well anyway, if you give up, the answer can be found here ( a rough draft of my report )

Now for something more enfuriating. Battle of the Bands, which was scheduled for 1800-2200, yesterday was cancelled due to a tyrannical Principle. In our school, we have this annoying little thing called AI. Anyone who is on AI cannot participate in extra-curricular school events. Battle of the Bands is not a school event—In fact, the only association it has with the school is that it was scheduled to be in the auditorium. The insane principle went as far as to call an outside agency that was sponsoring it and tell them it was cancelled. Thats right!!! He wasn’t even running it, but he decided to make the called. If this was something like talent show, which is school run, then this would be ok, but they were paying to rent the auditorium and everything!!!! Talk about over using your power….

Are you one of those persons whom tries to argue the irrationality of 1/3? Do you think that .7234478243278432785478 is an irrational number? If either one of those applies to you, then you better read below. I have way too much time, but if one more person tries to tell me that 1/3 is irrational, I will explode.

What is an irrational number? An irrational number is a number which has exacly 0 exact numerical representations. An irrational number cannot be represened as P over Q, it has no exact form. An irrational number is a number with a never-terminating, non-repeating decimal expansion, it goes on forever and it doesn’t have a repeating pattern throughout. 1/3 is rationally represented in EXACT form as 1/3, .3 repeating, .1 in trinary. ( although there are a million other representations ). pi is irrational, it never terminates and there is no repeating throughout, however it is exactly represented with the symbol, which is understood to be the ratio of a circle’s circumference to its diameter. Not satisfied? look it up on mathworld.

Seems like I am working in backwords order almost. Well, I asked Mr. Horne about number 16 today, he wasn’t quite sure about it, which was not very comforting, but After a few minutes of thought in the stimulating environment of a school bus, I saw the light. Before I go any further, I should probably Write the problem.

Given a triangle ABC, how would you use only a compass and a straight edge to find a point P such that triangles ABP, ACP, and BCP have equal permiters? (Assume that triangle ABC is constructed so that a solution does exist.)

Amazingly, I did this with no help from outside sources, and Even without a compass….All the bus had to offer was a Coke Bottle Cap. Here goes: Triangle ABC is a 45-45-90 triangle, Drawn Such that AC is the hypotenuse. Take your compass and place it on point C, such that the radius of the circle to be drawn is the length CB (equal to AB). Draw this circle, then place the compass on point A and to the same. You will notice that your Two circles Intersect at two points: B , and a new point (opposite B), which is Labeled P!. You have just created a perfect square! Of course, this is not nearly as trivial if the triangle is not a right isosceles triangle… in such case, one would need to find the isoperemetric point which exists at the center of the outer soddy circle.

While I do lack 7/3 of a life, I have enough of a life to know that trying to find a formula to solve this one is not worth it (although, In my combinatorial studies next year I will… [<–written next year]), so I just used another brute force program….

Consider a function for which f(n) returns the number of ones required to count from 1 to n. f(13)=6, notice that f(1)=1. Find the Next largest number that satisfies f(n)=n.

The program is comprised of a loop that iterates until it finds something, then it exits.

199981
Query took 7.47828507423 seconds.

The php source can be found here.

I finally got around to completing #1 on the GLAT today. The google folks really need to be more specific about what they want. Number one basically says: Solve this cryptic equation, noting however, that values for m and e are interchangeable. The equation is: WWWDOT-GOOGLE=DOTCOM It does not specify weather or not a number can appear more than once..(i.e..can T=O?). If you were to try solving this by hand, you would have to make such conjectures as: If two numbers are subtracrted to form a third number comprised of the same number of digits as the original two numbers, then 1^st Digit in the second number subtracted from the 1^st digit in the first number will equal the 1^st digit in the third number. This however, is not nesessarily true, but one can solve this in this manner. #1 also states that no leading zeros are allowed. This means that W can not be 0, D can not be 0, G cannot be 0, and logically, O cannot be 0. Solving this by hand using the method I stated above would lead you to things such as: w=2*O, T=O….and equal values for e and m also mean they are interchangable….so T=2*M, w=4*M…then M is 1 or 2, and w is 4 or 8, respectively. Now, If google did specify that no digit could be repeated, then the only sane way to solve this would be via a computer program, which I also did. I made the brute-force program in php (ran it from command line) and it gave me 144 permutations…2 of which had unique digits…the only difference being interchanged valued for m and e. here they are:
777589-188103=589486
777589-188106=589483

That is probably what google is looking for, but they really should specify. The Brute Force Program took only 126 seconds to run, all the possible answers it generated can be found here.

First off, I must address the title. If I hear the words “How about them Sox?” one more time, I will enter into a state of irrevoverable insanity. That said, how about them stars?

Ah, yes, those stars…Let ABCDE Be a regular pentagon. Drawing a line contecting angle x to the next non-adjacent angle going around until the original angle is reached will form a 5 pointed star. What is the degree measure of the point of the star? There are numberous ways to go about solving this, but I will go with what I believe to be the more direct algebraic way. It may be noted that when Drawing line AC, or any such line, an Isoseles Triangle is formed. Drawing two of these “star lines” in a row will for the point of the star. The point of the star is the original angle measure of the regluar polygon, minus 2 times the base measure of the isoseles triangle (they are conguent)…I will try to get some pictures of that formula up…The base angles of the isoseles triangles are the quantity 180 minus an original angle divided by two. in the pentagon, this is 36…then 108 - 2(36) is also 36. yay! i don’t know why I post on weird things like this…