I started working at Sunset three days ago. I'll be working there 5 days a week this summer instead of four, to make some extra money. And it's the same as ever.

Cut meat for customers, scoop glop into containers for customers, cut cheese for customers, make chips, bag chips, price chips, put chips out on display rack, stock produce with 50lb bags of carrots (finally, something actually useful, a workout), made stickers for little prepackaged containers of glop (good tasting glop, but it's still glop), filled containers with glop, put stickers on them, and put them on shelves, went back to counter, helped customers, some friendly, some not-so-friendly, stare at empty counter, neaten stuff and stir salads so store managers don't get upset at us standing around waiting for customers (deli managers are more understanding than store managers), listen to some of my co-workers fill the air with completely useless and repetitive talk, learn to appreciate the value of silence, serve more customers, get told to do three things at once, attempt to make three people happy, come close to succeeding, mad rush on counter by customers, but hey, I don't care, it's 5:03 and I'm going home. Cya.

## 5.28.2005

### Stuff

Sorry for not having posted in awhile, I had finals, and David visiting, and trip to my girlfriend's house, and then work started. I'll be updating tomorrow with a long description of my trip to Mia's house.

in the meantime entertain yourselves with this: http://www.albinoblacksheep.com/flash/numa.php

in the meantime entertain yourselves with this: http://www.albinoblacksheep.com/flash/numa.php

## 5.14.2005

### We Love Mary!

Mary's going to be here next year, because she did awesome on her language paper and final quiz and didn't fail!

Yay for Mary!

Yay for Mary!

## 5.13.2005

### DONE!!!!

MWAHAHAHAAAAA!!!!! FINALS ARE OVER!!!!

NOW WE GET TO GO CRAZY AFTER STAYING UP ALL NIGHT STUDYING FOR A WEEK!!!!!

YAHHHHHHHH!!!!!!!

*collapses in an exhausted heap*

NOW WE GET TO GO CRAZY AFTER STAYING UP ALL NIGHT STUDYING FOR A WEEK!!!!!

YAHHHHHHHH!!!!!!!

*collapses in an exhausted heap*

## 5.11.2005

### Absurditas!!!

What happens when you have eight freshman at TAC staying up till 2 studying for a Lab final, and you put a talking Yoda doll in the same room?

Will (my roommate): Will this final be easy?

Talking Yoda Doll (which belongs to Will): Clouded, your future is.

W: Um...thanks.

W: Is Aristotle God?

Y: I sense the answer to your question is...No.

W: Is Socrates God?

Y: The answer you seek is...Yes.

W: Is Kant God?

Y:The answer you seek is...Yes.

W:Are Kant and Socrates the same person?

Y:I sense the answer to your question is...No.

W:Are you right?

Y:True (garbled, either "essence", or "I sense"), this is.

W:Are you always right?

Y:Difficult to answer, this question is.

W:Should I stop listening to you?

Y:Questions about the future a disturbance to the Force are.

W (shouting):Shall I break you into a million little pieces becasue you are vexing me?

Y (calmly): You will find the answer you seek when you are calm and at peace.

Other students: (laugh their heads off)

Will (my roommate): Will this final be easy?

Talking Yoda Doll (which belongs to Will): Clouded, your future is.

W: Um...thanks.

W: Is Aristotle God?

Y: I sense the answer to your question is...No.

W: Is Socrates God?

Y: The answer you seek is...Yes.

W: Is Kant God?

Y:The answer you seek is...Yes.

W:Are Kant and Socrates the same person?

Y:I sense the answer to your question is...No.

W:Are you right?

Y:True (garbled, either "essence", or "I sense"), this is.

W:Are you always right?

Y:Difficult to answer, this question is.

W:Should I stop listening to you?

Y:Questions about the future a disturbance to the Force are.

W (shouting):Shall I break you into a million little pieces becasue you are vexing me?

Y (calmly): You will find the answer you seek when you are calm and at peace.

Other students: (laugh their heads off)

### Three down, one to go

I've now taken my lab and Seminar finals.

Lab went pretty well, the balance problem was easy, and a wrote a few pages about the style of Archimedes and Euclid, and the difference of Epigenesis and Evolution present by Dreisch.

Seminar final I talked with Mr Quackenbush aboutt hte line and the image present by Socrates in books VI-VII of Palto's Republic.

And I think it went pretty well, so now I have a day off before my Philosophy final.

Lab went pretty well, the balance problem was easy, and a wrote a few pages about the style of Archimedes and Euclid, and the difference of Epigenesis and Evolution present by Dreisch.

Seminar final I talked with Mr Quackenbush aboutt hte line and the image present by Socrates in books VI-VII of Palto's Republic.

And I think it went pretty well, so now I have a day off before my Philosophy final.

## 5.09.2005

### Theology Final

One down, three to go.

I just finished my theology final, and I wrote 3 pages on Faith and Works, and one page on St. Paul's comparision of Christian marriage to the relationship of Christ and the church.

Now only three finals left to go. Lab tomorrow at 1, an oral Seminar Final Wednesday at 11:20, and a Philosophy final on Friday at 1.

Prayers are appreciated.

I just finished my theology final, and I wrote 3 pages on Faith and Works, and one page on St. Paul's comparision of Christian marriage to the relationship of Christ and the church.

Now only three finals left to go. Lab tomorrow at 1, an oral Seminar Final Wednesday at 11:20, and a Philosophy final on Friday at 1.

Prayers are appreciated.

## 5.06.2005

### It's over.

No more freshman year. We had our last class this afternoon, Latin with Mr. Dragoo.

He brought donuts and milk, and we translated "The Story of Peter Rabbit," from Latin into English.

We read all about the "Lord McGregor" and Peter the Naughty Bunny who ate "ground apples" and hopped in the "trough with wheels."

It was quite fun, and we're all sad now that classes are over, and finals are looming.

And I'm happy because I got a B on my philosophy paper from Mr Richard (a notoriously hard grader). And a good grade on my latin quiz. It was a good last day of classes.

"Greater love hath no philosopher than this; that he lay down his life for Barbara." Mr. Richards

He brought donuts and milk, and we translated "The Story of Peter Rabbit," from Latin into English.

We read all about the "Lord McGregor" and Peter the Naughty Bunny who ate "ground apples" and hopped in the "trough with wheels."

It was quite fun, and we're all sad now that classes are over, and finals are looming.

And I'm happy because I got a B on my philosophy paper from Mr Richard (a notoriously hard grader). And a good grade on my latin quiz. It was a good last day of classes.

"Greater love hath no philosopher than this; that he lay down his life for Barbara." Mr. Richards

### Quod Est Demonstrandum.

I memorized the entire thing starting at 6 am this morning, did it almost perfectly in class, using Greek letters. Cept Tommy pointed out the fact that I accidentally called Tau Theta instead. Oops. i just learned the greek alphabet last night, so I think I should get a break.

Anyways, it was very good to end the year with such a long prop, tieing 5 different things together (To set out the sides of the five figures and compare them with one another.) And I think I did it pretty well too. :)

Euclid's Elements

Book XIII

Proposition 18

To set out the sides of the five figures and compare them with one another.

Set out AB the diameter of the given sphere, and cut it at C so that AC equals CB, and at D so that AD is double DB. Describe the semicircle AEB on AB, draw CE and DF from C and D at right angles to AB, and join AF, FB, and EB. I.11

Then, since AD is double DB, therefore AB is triple BD. In conversion, therefore, BA is one and a half times AD.

java applet or image But BA is to AD as the square on BA is to the square on AF, for the triangle AFB is equiangular with the triangle AFD. Therefore the square on BA is one and a half times the square on AF. V.Def.9

VI.8

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. And AB is the diameter of the sphere, therefore AF equals the side of the pyramid. XIII.13

Again, since AD is double DB, therefore AB is triple BD. But AB is to BD as the square on AB to the square on BF, therefore the square on AB is triple the square on BF. V.Def.9

VI.8

But the square on the diameter of the sphere is also triple the square on the side of the cube. And AB is the diameter of the sphere, therefore BF is the side of the cube. XIII.15

And, since AC equals CB, therefore AB is double BC. But AB is to BC as the square on AB to the square on BE, therefore the square on AB is double the square on BE.

But the square on the diameter of the sphere is also double the square on the side of the octahedron. And AB is the diameter of the given sphere, therefore BE is the side of the octahedron. XIII.14

Next, draw AG from the point A at right angles to the straight line AB, make AG equal to AB, join GC, and draw HK from H perpendicular to AB. I.11

I.3

I.12

Then, since GA is double AC, for GA equals AB and GA is to AC as HK is to KC, therefore HK is also double KC.

Therefore the square on HK is quadruple the square on KC, therefore the sum of the squares on HK and KC, that is, the square on HC, is five times the square on KC.

But HC equals CB, therefore the square on BC is five times the square on CK. And, since AB is double CB, and, in them, AD is double DB, therefore the remainder BD is double the remainder DC.

Therefore BC is triple CD, therefore the square on BC is nine times the square on CD. But the square on BC is five times the square on CK, therefore the square on CK is greater than the square on CD. Therefore CK is greater than CD.

Make CL equal to CK, draw LM from L at right angles to AB, and join MB. I.3

I.11

Now, since the square on BC is five times the square on CK, and AB is double BC, and KL is double CK, therefore the square on AB is five times the square on KL. But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. And AB is the diameter of the sphere, therefore KL is the radius of the circle from which the icosahedron has been described. Therefore KL is a side of the hexagon in the said circle. XIII.16,Cor.

IV.15,Cor.

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK equals LB, therefore each of the straight lines AK and LB is a side of the decagon inscribed in the circle from which the icosahedron has been described. XIII.16,Cor.

And, since LB belongs to a decagon, and ML to a hexagon, for ML equals KL, since it also equals HK being the same distance from the center and each of the straight lines HK and KL is double KC, therefore MB belongs to a pentagon. XIII.10

But the side of the pentagon is the side of the icosahedron, therefore MB belongs to the icosahedron. XIII.16

Now, since FB is a side of the cube, cut it in extreme and mean ratio at N, and let NB be the greater segment. Therefore NB is a side of the dodecahedron. XIII.17,Cor.

And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid, double the square on the side BE of the octahedron and triple the side FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.

Therefore the square on the side of the pyramid is four-thirds of the square on the side of the octahedron, and double the square on the side of the cube, and the square on the side of the octahedron is one and a half times the square on the side of the cube.

The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.

But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides, for they are irrational, the one being minor and the other an apotome. XIII.16

XIII.17

That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus.

Since the triangle FDB is equiangular with the triangle FAB, proportionally DB is to BF as BF is to BA. VI.8

VI.4

And, since the three straight lines are proportional, the first is to the third as the square on the first is to the square on the second, therefore DB is to BA as the square on DB is to the square on BF. Therefore, inversely AB is to BD as the square on FB is to the square on BD. V.Def.9

VI.20,Cor.

But AB is triple BD, therefore the square on FB is triple the square on BD.

But the square on AD is also quadruple the square on DB, for AD is double DB, therefore the square on AD is greater than the square on FB. Therefore AD is greater than FB. Therefore AL is by far greater than FB.

And, when AL is cut in extreme and mean ratio, KL is the greater segment, for LK belongs to a hexagon, and KA to a decagon, and, when FB is cut in extreme and mean ratio, NB is the greater segment, therefore KL is greater than NB. XIII.9

But KL equals LM, therefore LM is greater than NB.

Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron.

Q.E.F.

Remark

I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.

For a solid angle cannot be constructed with two triangles, or indeed planes.

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron, but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six would be equal to four right angles, which is impossible, for any solid angle is contained by angles less than four right angles. XI.21

For the same reason, neither can a solid angle be constructed by more than six plane angles.

By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they would again be four right angles.

By three equilateral and equiangular pentagons the angle of the dodecahedron is contained, but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles would be greater than four right angles, which is impossible.

Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.

Q.E.D.

Lemma

But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.

java applet or image Let ABCDE be an equilateral and equiangular pentagon. Circumscribe the circle ABCDE about it, take its center F, and join FA, FB, FC, FD, and FE. IV.14

Therefore they bisect the angles of the pentagon at A, B, C, D, and E. And, since the angles at F equal four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth. Therefore the remaining angles FAB and ABF consist of one right angle and a fifth.

But the angle FAB equals the angle FBC, therefore the whole angle ABC of the pentagon consists of one right angle and a fifth.

Q.E.D.

Anyways, it was very good to end the year with such a long prop, tieing 5 different things together (To set out the sides of the five figures and compare them with one another.) And I think I did it pretty well too. :)

Euclid's Elements

Book XIII

Proposition 18

To set out the sides of the five figures and compare them with one another.

Set out AB the diameter of the given sphere, and cut it at C so that AC equals CB, and at D so that AD is double DB. Describe the semicircle AEB on AB, draw CE and DF from C and D at right angles to AB, and join AF, FB, and EB. I.11

Then, since AD is double DB, therefore AB is triple BD. In conversion, therefore, BA is one and a half times AD.

java applet or image But BA is to AD as the square on BA is to the square on AF, for the triangle AFB is equiangular with the triangle AFD. Therefore the square on BA is one and a half times the square on AF. V.Def.9

VI.8

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. And AB is the diameter of the sphere, therefore AF equals the side of the pyramid. XIII.13

Again, since AD is double DB, therefore AB is triple BD. But AB is to BD as the square on AB to the square on BF, therefore the square on AB is triple the square on BF. V.Def.9

VI.8

But the square on the diameter of the sphere is also triple the square on the side of the cube. And AB is the diameter of the sphere, therefore BF is the side of the cube. XIII.15

And, since AC equals CB, therefore AB is double BC. But AB is to BC as the square on AB to the square on BE, therefore the square on AB is double the square on BE.

But the square on the diameter of the sphere is also double the square on the side of the octahedron. And AB is the diameter of the given sphere, therefore BE is the side of the octahedron. XIII.14

Next, draw AG from the point A at right angles to the straight line AB, make AG equal to AB, join GC, and draw HK from H perpendicular to AB. I.11

I.3

I.12

Then, since GA is double AC, for GA equals AB and GA is to AC as HK is to KC, therefore HK is also double KC.

Therefore the square on HK is quadruple the square on KC, therefore the sum of the squares on HK and KC, that is, the square on HC, is five times the square on KC.

But HC equals CB, therefore the square on BC is five times the square on CK. And, since AB is double CB, and, in them, AD is double DB, therefore the remainder BD is double the remainder DC.

Therefore BC is triple CD, therefore the square on BC is nine times the square on CD. But the square on BC is five times the square on CK, therefore the square on CK is greater than the square on CD. Therefore CK is greater than CD.

Make CL equal to CK, draw LM from L at right angles to AB, and join MB. I.3

I.11

Now, since the square on BC is five times the square on CK, and AB is double BC, and KL is double CK, therefore the square on AB is five times the square on KL. But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. And AB is the diameter of the sphere, therefore KL is the radius of the circle from which the icosahedron has been described. Therefore KL is a side of the hexagon in the said circle. XIII.16,Cor.

IV.15,Cor.

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK equals LB, therefore each of the straight lines AK and LB is a side of the decagon inscribed in the circle from which the icosahedron has been described. XIII.16,Cor.

And, since LB belongs to a decagon, and ML to a hexagon, for ML equals KL, since it also equals HK being the same distance from the center and each of the straight lines HK and KL is double KC, therefore MB belongs to a pentagon. XIII.10

But the side of the pentagon is the side of the icosahedron, therefore MB belongs to the icosahedron. XIII.16

Now, since FB is a side of the cube, cut it in extreme and mean ratio at N, and let NB be the greater segment. Therefore NB is a side of the dodecahedron. XIII.17,Cor.

And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid, double the square on the side BE of the octahedron and triple the side FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.

Therefore the square on the side of the pyramid is four-thirds of the square on the side of the octahedron, and double the square on the side of the cube, and the square on the side of the octahedron is one and a half times the square on the side of the cube.

The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.

But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides, for they are irrational, the one being minor and the other an apotome. XIII.16

XIII.17

That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus.

Since the triangle FDB is equiangular with the triangle FAB, proportionally DB is to BF as BF is to BA. VI.8

VI.4

And, since the three straight lines are proportional, the first is to the third as the square on the first is to the square on the second, therefore DB is to BA as the square on DB is to the square on BF. Therefore, inversely AB is to BD as the square on FB is to the square on BD. V.Def.9

VI.20,Cor.

But AB is triple BD, therefore the square on FB is triple the square on BD.

But the square on AD is also quadruple the square on DB, for AD is double DB, therefore the square on AD is greater than the square on FB. Therefore AD is greater than FB. Therefore AL is by far greater than FB.

And, when AL is cut in extreme and mean ratio, KL is the greater segment, for LK belongs to a hexagon, and KA to a decagon, and, when FB is cut in extreme and mean ratio, NB is the greater segment, therefore KL is greater than NB. XIII.9

But KL equals LM, therefore LM is greater than NB.

Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron.

Q.E.F.

Remark

I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.

For a solid angle cannot be constructed with two triangles, or indeed planes.

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron, but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six would be equal to four right angles, which is impossible, for any solid angle is contained by angles less than four right angles. XI.21

For the same reason, neither can a solid angle be constructed by more than six plane angles.

By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they would again be four right angles.

By three equilateral and equiangular pentagons the angle of the dodecahedron is contained, but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles would be greater than four right angles, which is impossible.

Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.

Q.E.D.

Lemma

But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.

java applet or image Let ABCDE be an equilateral and equiangular pentagon. Circumscribe the circle ABCDE about it, take its center F, and join FA, FB, FC, FD, and FE. IV.14

Therefore they bisect the angles of the pentagon at A, B, C, D, and E. And, since the angles at F equal four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth. Therefore the remaining angles FAB and ABF consist of one right angle and a fifth.

But the angle FAB equals the angle FBC, therefore the whole angle ABC of the pentagon consists of one right angle and a fifth.

Q.E.D.

## 5.02.2005

### Secret

[Warning: Noone from my section is allowed to read this post until after Friday!]

So, I was studying the Isochehedon and the octohedon in the room where my math class is at 8:#0 and Mr Wodzinski walks in about 15 minutes before class starts. We talked a bit about the days props, and I asked him if, since I had done the first propr of Euclid, if I wouldn't be able to do the last. He didn't say I could, just that he was "dissapointed he couldn't surprise me, and the it "might be a good idea to study hard on Thursday night." :D

And in class he asked for a list of volunteers for props 16 and 17, but said that prop 18 "was pre-ordained."

Yay! I get to do the last Euclid prop!

So, I was studying the Isochehedon and the octohedon in the room where my math class is at 8:#0 and Mr Wodzinski walks in about 15 minutes before class starts. We talked a bit about the days props, and I asked him if, since I had done the first propr of Euclid, if I wouldn't be able to do the last. He didn't say I could, just that he was "dissapointed he couldn't surprise me, and the it "might be a good idea to study hard on Thursday night." :D

And in class he asked for a list of volunteers for props 16 and 17, but said that prop 18 "was pre-ordained."

Yay! I get to do the last Euclid prop!

### Random people sightings

I was just walking over to the library from lunch, and I see Maggie and Don playing chess under a tree, I pass Sage and a few others having a discussion about Aristotle and Aquinas, and finally I see Dom sitting under a tree playing Irish tunes on his tin whistle.

People here are so cool.

I'm going to miss them when I go home for the summer.

People here are so cool.

I'm going to miss them when I go home for the summer.

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