Yesterday's Math contest(s)

[Clayton]

anyone else besides me write the Fermat (or Cayley or Pascal) contest? Any questions you that you had the "OMG wut t3h f**k"?

[BenLi]

cayley here. What school do you attend freakman?

oh and I found out today i got 118, craptacular

[Clayton]

I go to CDDHS, but I didn't know you could find out your scores so quickly, what's the link?

[bugzpodder]

certain large schools tries to figure out answers and mark them in the same day.

[Clayton]

ahh, well, my computer science teacher (a math major, go figure) and I sat down during today's CompSci class and talked over and tried to solve questions 24 and 25 on the Fermat contest. We were getting close with 24 (we think it has something to do with similiar triangles) when the bell rang at the end of the day.

[Null]

I wrote it. On 17 (the one with the unknown x coordinate of a point on the triangle) I got the right answer, then went back, misread the question, erased the right answer and replaced it with a guess.

Yes, I know.

I did ok last year (116), but I'm not happy about this year's. Regardless, it's all a learning experience.

[Clayton]

heh, I had virtually no trouble with any of them up until part C (20 was a pushover question, 6 more points for me ). I was so close to getting an answer to 22 (the billiard table question), I just needed a couple more seconds to test my answer, but I quickly circled in the answer because I think it was right (I got 7.5, anyone wanna verify?)

[BenLi]

Quote:

certain large schools tries to figure out answers and mark them in the same day.

ie. Massey

Oh and don't say" tries", our math department figures out answers and marks... crazyiness

[Clayton]

hmm, the question that drove me the most nuts would have to be 24. basically, a line going from point P crosses through the parabola y = x^2 at points Q and R. PQ = QR. From this, you have to find the y-intercept.

[CodeMonkey2000]

My friends took Cayley (i didn't) and one of the selections was 1337. I think everyone I know picked that for the fun of it.

[bugzpodder]

that parabola is hilarious

drop perpendiculars to the axis, from left to right call it P (same P), A and B

let PA = z AO = x and OB = y, O is origin

it follows that AQ = x^2 and BR = y^2

now by congruent triangles since PQ=QR it follows that z = x+y and y^2 = 2x^2 so y=sqrt(2) x

are you missing any information? given any point on the parabola R i can construct it so that PQ=QR and P is the intercept...

to see this, fix R on the parabola, move P around. near the origin PQ is almost 0 and far away PQ >>>>> QR so somewhere in between we must have PQ = QR

[Clayton]

I was using a touchpad, not a mouse, now enough on the subject.

The only thing that I didn't say was that the answer is an integer. Other than that, that was basically the question. When I get to school tomorrow, I can post it word for word.

[klopyrev]

Question 25 on the Fermat was an interesting question. You have to know Combinatorics to solve it though. You get an equation (b-g)^2 = b+g. You can find 59 solutions to that equation with 4<=g<=b<=2007, which I think were the bounds. If you want the details on how I got that equation, post a reply. I am just too lazy to type it up right now. Also, here are the first couple of solutions to the equation. I hope you can see the pattern:

6 10
10 15
15 21
21 28
28 36
36 45
45 55
55 66
66 78
78 91

[MihaiG]

i got most of them, 21 was real easy...

spent last half our on 22,23,24
i neer understood 25 since we never talked bout it in class