Monday, December 19, 2005


Okay, so I know I kinda failed on planning a holiday party of some sort, but I've been really busy with college admissions stuff, as I know most of us seniors have been. Anyway, Michelle and I have been talking, and we're going to meet up on January 2nd, a Monday. We'll meet at the Starbucks at the Lodge at Bellevue Square (link below) at 11:00 am (tentative time!) and do stuff. Join us if you want, because you know you want to. Anyway, email me back or post something on the blog if you're going to come. We'll still try to plan some sort of trip for later.

Edward Chang

P.S. The Lodge isn't actually in Bellevue Square, but it's pretty much a part of it. If you drive along Bellevue Way, you'll be able to see the Starbucks sandwiched between the PF Changs and the Crate and Barrel.

Tuesday, December 13, 2005

SIMUW 2006 problems

Please don't post solutions to this years problems. I would rather not have some kid google the answers and discover them here and then be accepted to SIMUW with no skill whatsoever. Please just email out answers amongst us...except to those that will be applying to SIMUW again.
Have fun

P.S. PLEASE PLEASE PLEASE email me if you are interested in the SIGMA competition at Skyview High School. I would appreciate it if you could make it.

Another P.S. what happened when the ship carrying blue paint hit the ship carrying red paint?

..................................the sailors were marooned!!!

Monday, December 12, 2005

2005 comc#4b problem and (my)solution

the pciture is the 2005 COMC#4

my solution:
Part A is rather obvious and rather ridiculous, if you realize that the lenngth of external tangents are the same.

Part B:
let: BC=a, AC=b,AB=c.
Lemma(well known): CT= s - c, BT=s-b
let M, N, T be the tangency points of AB, AC, BC with the incircle of the triangle ABC. Then AM = AN, BM = BT, CN = CT. The perimeter is p = 2s = AM + BM + BT + CT + CN + AN = 2AM + 2BM + 2CT = 2AB + 2CT = 2c + 2CT so CT = s - c ,
similarly BT=s-b

the question is asking us to prove:
ABC=r_1 CT+r_2 BT=r_1 (s-c)+r_2 (s-b)=s(r_1+r_2)+(r_1 c+r_2 b)

OQED = OQGF = s{(r_1 + r_2)/2} so the area of the hexagon ODEQGF is twice this or: s(r_1+r_2).
notice that tri ODA is congruent to tri OKA same goes for (OFB, OKB) same goes for :(QEA,QLA) and (QGC,QLC)
from these congruent triangle we get areas of the pentagons ODABF and QEACG are equal to twice the areas of the triangles OAB and QAC respectively so:
ODABF=r_1 c, QEACG=r_2 b.
finally ABC=ODEQGF - (ODABF + QEACG) = (r_1 + r_2) - (r_1 c + r_2 b)

Tuesday, December 06, 2005

The SIMUW Ad Thing

By now, just about everyone should have received in the mail a card that advertises SIMUW 2006, despite the fact that many of us are ineligible. It has Brian and CrystalAnn in!

In view of all the evidence, I have constructed a hypothesis:

We are intended to become local recruiters!

So find someone you want to send, and send them.

Monday, December 05, 2005

For All People that Have a Math Team or Will Organize One

Hey all,
Skyview (my high school) is holding a math competition called Skyview Invitational Games for Mathematical Achievement (SIGMA for short). I would really appreciate it if you could come. Please contact me via email ( if you are intersted. We need names as soon as possible (preferably before December 16th).

P.S. I just got the SIMUW letter thing in the mail that advertises SIMUW and Brian and CrystalAnn are in a picture in it.

Thursday, December 01, 2005


It is snowing in Seattle, and it looks terrific outside!
It has been snowing for over 3 hours, and now it is starting to stick. (For those of you who don't live in Seattle, it does not snow very much here, so snow is all the more exciting.)
I think a snow day may be in order for tomorrow. Yay!