Divine Lunacy

I had another geometry report due on friday :). I wrote it three times…Twice in prose, once in poetry. The poetry version came out quite interesting, but mixing poetry with math was a really really weird thing to do. I handed all three of them in.

With parallelograms, there is a rule
of equal sides, and angles too
with this in mind, i have come to prove

the perimeter of HARP is fifty-six,
and earl half that—added to twelve, of course

See, it is just weird. Well, I have another due this coming friday. I wrote it today, and i think it came out quite wonderful. Probably one of my best yet. 2.5 pages long. The problem was easy again, It was just a pythagorean theorem problem.

Just so the whole world knows, I have a top-secret development project underway, then a less secret one…But more on them later…I don’t even know much about them yet =-o

Pinch me…I have just “heard” cussing such as I have never heard it before. Such awful, Grisly statements tearing at the very substance of my being…here is the conversation:

Tim: i was using mozilla firefox and avant browser…
Tim: i thought i deleted explorer
Tim: all set
<anti -firefox> dude that bad
Tim: no
Tim: that good
<anti -firefox> mozilla is a hacker network
<anti -firefox> and a spam net.
<anti -firefox> you cant do shit wit mozilla
Me: ?!?!?!?!?!?!??!?!!?!?!?!?!?!??!?!?!?!?!?!?!?!?!?!?!?!
Tim: ya
Me: : How Much Pot is he smoking??@!?@?@?@!?!@?%^*@%#??*^$%?&?%$
Me: Where does he live? Where can I buy a shotgun?
Tim: lol
<anti -firefox> oh he greg no see i dont like mozilla cause you cant use flash and macromedia on some websites.

Let Death Come quickly and Smite me! Such a vulgar display of utter ignorance as I have never before seen!!! ‘Tis unreal! ‘Tis untrue! I am not even going to bother to give reasons or counter any of those arguments right now. JUST USE FIREFOX!!! It is standards compliant! It displays Flash fine, and as far as java goes, you can just install Java by Sun Micro Systems (hehe…better than the microshit Java VM anyway…). In the name of linus, the almighty. I use Firefox everyday for like everything. Did I mention all the Flash and Java stuff I surf? Did I mention that I have all this going for me on my linux box, which is all I use? Well, now I have mentioned that. And if it works fine for me in linux, then there is no excuse for not using it in windows, or any other OS at that.

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.

mmm! Time for another fresh compiled kernel just like Mom used to make! On the 18^th, Kernel 2.6.9 was released. I am currently running Kernel 2.6.8. Rather than downloading a patch file, I for some reason, would rather just download the whole thing and be forced to repeat the config. Hey, I’m weird… If you haven’t ever configured a linux Kernel, It is kind of boring, though not so bad when you use one of the Graphical Config Tools. I was running Mandrake 10.0 before I compiled 2.6.8. Traces of Mandrake still remain, but most I have since replaced most of my software packages with distro-neutral things. Now, if all goes well, my next post will be from a fresh 2.6.9 kernel :). I need to try and compile KDE now. Last time I attempted, I had some Freak compile Error with Library qt-mt :(…But I am determined to Rid my system of every last bit of mandrake, not that I have anything against it…

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…

The Problem of the Week for Geometry is due tomorrow…I spent most of my hours writing that. It was really easy. During the course of doing it, I tried the Microshit Equation Editor thing…it really sucks. The Math writer in openoffice.org is way better. For the problem, I had to create a formula to find the measure of a point angle of a star containing n points. While I am on the Subject of math, I believe I have the anwer to number one on the GLAT, but I must check all the work before I can be certain.

I finally got the content area here formatted (as you can see). there is something I just don’t quite like about it though.

I really have nothing else to post about..

It has been, for some time now my suspiscion that Google plans world conquest. With each passing day, my proof becomes more sound. What do you think of when I say google? Most people probably just think of a search engine, a very good one. Just a white page with an image, some buttons and a text box. Oh, but google is oo so much more. Ever notice the links at the “bottom”? Try clicking the link labeled “About Google”…You will be surprised. Don’t get me wrong, if anyone is going to take over the world, I definitely opt for google. They do everything right, and take everything the extra mile, as is the Google Philosiphy.

There is this mystical Force at google that goes by the name of Google Labs. They pump out some pretty crazy stuff, and they do some pretty crazy stuff. Google labs is all about research and Development. Perhaps one of the more interesting items released By google labs is Google Compute, a Distrubited Computing system. In short, google compute allows google to take advantage of your computers idle time to

…worthwhile endeavors, including projects to improve Google and its services…

It is the Google and its services that freaks me out… :-p

Ok, so you want a job at google labs? Google Does alot of problem solving things to find potential Google Labs Employees. The most recent Wonderful peice of insanity is the GLAT (Google Labs Aptitude Test)

None of the problems are that difficult, and most are pretty fun (there are no problems with unix >:o [unless they speak of the women…])…some of them might even give you a laugh…I will post more on “solutions” later.

Around 1800 tonight, I was caught in what seems to have been a ddos (distributed Denial of Service). I suddenly found my internet connection lost, and promtly connected to my router. My outgoing seemed to be near dead, and everything incoming was on eiher port 445 or 3531. Port 445 is commonly used for samba, but many times, it is also used by worms that Exploit LSASS. I am running linux, so that is really no threat to me. All signs pointed to me being pinnged to death. My router could not handle all the information coming in, and hence was forced to deny the LAN its services. I have tracerouted some of the Ip addresses. Most of the ones on port 445 seem to be from verizon, and they are in a relitively tight Ip range. The attacks on port 3531 were more varied in ISP. The attack lasted for 15 minutes, until I was finally able to obtain a new ip address. I am now running ethereal incase another attack does happen, this way I will have more info.

I expect to have this template finished very soon. It is getting very tedious tweaking the finer points…I really hate that.