KCA Open prizes

[hdskull]

I know.

I think the Dutch solution to this problem is to just share the price. In the case of candy that means you pass it around. In the case of a won trip to the next European Championship it means........I don't know how they are going to share that, but I heard they are going to do just that.

Haha, there should be an Americas Championship, it'd be fun to see, the best from Canada(Matt Walter, Chris Bouchard, Harris Chan, etc), the best from US(Frank Morris, Macky, Leyan, Hardwick, Dzoans, Patricio, etc), and the best from South America (Pedro Santos, etc, sorry I don't know a lot of them). It's be interesting.

 

[Harris Chan]

Haha, I guess so...btw it's Craig, not Chris lol There's so many Chris's already eh XD

 

[masterofthebass]

Hey, why don't I make the list... Being #4 in NA for 5x5 doesn't get me there. Obviously that list is a good one. I guess we did have somewhat of an Americas Championship. At the US Open there were a bunch of guys from Canada, including Matt Walter.

 

[philkt731]

oooh, for being 5th in NA for 2x2x2 I would get Edison Black + Edison White if we had the same prizes (and if there even was an American Championships (and if we placed in the smae order as now ))

 

[hdskull]

Haha, I guess so...btw it's Craig, not Chris lol There's so many Chris's already eh XD

Lol, sorry typo, i knew that. **too much Chris Dzoan videos.

Hey, why don't I make the list... Being #4 in NA for 5x5 doesn't get me there. Obviously that list is a good one. I guess we did have somewhat of an Americas Championship. At the US Open there were a bunch of guys from Canada, including Matt Walter.

Sorry, I typed etc to include everyone else I didn't mention (like you, Jon Choi, Jason Baum, Andy Tsao, Tyson/Toby Mao, Ravi, Stern, Kang, Krueger the list goes on...) Btw, does Andy Tsao solve petrus ?

It was still US Open, Americas Championship would be different. Who knows, maybe some amazing S. American will come out of the blue.

 

[AvGalen]

Don't change the US Open to a "just all Americans". I would like to beat "the #4 in NA for 5x5" next time.

 

[Pedro]

I know.

I think the Dutch solution to this problem is to just share the price. In the case of candy that means you pass it around. In the case of a won trip to the next European Championship it means........I don't know how they are going to share that, but I heard they are going to do just that.

Haha, there should be an Americas Championship, it'd be fun to see, the best from Canada(Matt Walter, Chris Bouchard, Harris Chan, etc), the best from US(Frank Morris, Macky, Leyan, Hardwick, Dzoans, Patricio, etc), and the best from South America (Pedro Santos, etc, sorry I don't know a lot of them). It's be interesting.

yep, you don't know a lot of them because I'm still the only one that went to a competition

but that will change in saturday...we have 40+ people registered...

they're not very fast, though...(some are like 20-22 avg)

 

[ExoCorsair]

Don't change the US Open to a "just all Americans". I would like to beat "the #4 in NA for 5x5" next time.

If you can beat him, you can surely beat me!

I also disapprove of an 'American championship'. I personally like meeting people from other countries!

 

[masterofthebass]

I never said that we should have an official American Championship. I'm just saying that the US Open had cubers from both the US and Canada. I definitely enjoy having cubers from other countries, and wish that I could go to at least one European competition. Maybe when I have a lot more money...

 

[watermelon]

Sorry, I typed etc to include everyone else I didn't mention (like you, Jon Choi, Jason Baum, Andy Tsao, Tyson/Toby Mao, Ravi, Stern, Kang, Krueger the list goes on...) Btw, does Andy Tsao solve petrus ?

I feel honored to be mentioned in this list . As for Andy Tsao, he solves with Fridrich (judging from his cube simulator solves). Perhaps you're thinking of Anthony Hsu?

 

[hdskull]

I feel honored to be mentioned in this list . As for Andy Tsao, he solves with Fridrich (judging from his cube simulator solves). Perhaps you're thinking of Anthony Hsu?

Amazing cubers deserve to get mentioned .

Ah, that's the one, Anthony Hsu, I just remembered that someone's name that started with an A and is a fast Petrus solver.