Shavlik Randolph: Internet Hero

I got back from boarding and went straight to the junk drawer and everyone was talking about this epic thread. I couldn’t believe it.

Way to be, Shav.

PS—Horny did the wipe the forehead trick (which I used to emmulate back in my Elks Free Throw Hoop Shoot days), but don’t you mean Nate Robinson, Ben? With his video game callout?

that we get a B Rex chat at some point lol

It was really a lot of fun. Also reminded us why questions are usually moderated in chats with players.

Wasn’t Raek LaFrentz one of the “15”? How does that work?

And Ike Diogu was one of the original “15” has he been “replaced” by Ruffin, or did that make the count 16=16? (or maybe 15-1/2 = 16 since their contributions were small?)

I wish we played him more. There are times when we could really use him as a hustle player to sneak a couple offensive rebounds and get to the line. Let’s Blaze: Keep him ready, and keep him hungry. And by that I mean take some of his food and give it to Rudy.

He is ALWAYS ready and ALWAYS hungry to play . . . .I recognize some intangible contributions – - forcing Channing to push harder, D’ing up hard at practice (Shav is quite a shot blocker and forces LMA to elevate to the max on that jumper), encouraging encouraging encouraging (Yeah Brandon – I think you should be MORE aggressive ==> 52 on PHX), probably rubbing off a little on Joel as to his ball handling around the rim…

but I´m not a fan of the tone of this post.
I think it´d be more apropriate to just thank him and Lets Blaze for the great interview. Comments on Shavlik´s importance in our playoff run and shout outs to Michael Ruffin are not the way to go. Maybe I´m seeing things that are not there, if so I apologize for that.
 
By the way, great week at BE, KP´s interview, playoff spot guaranteed and a NBA player interviewed live at the blog.

Common Nate

Draw an isosceles triangle, with points ABC, with line AC
representing the “base” of the triangle. By definition, side AB and
side BC are equal length, and angle BAC and angle BCA are equal to
each other.

 Let the line AB have length 8; by definition, line BC is then also 8.
Also let the distance along line AC be 15. The distance along the path
from A to B to C equals 16.

Next, bisect each of the three sides. Assign points D,E,F to the
midpoints of lines AB, BC, and AC respectively. Draw lines DF and
EF. Now, by definition, line AF and line FC are equal (both equal
to 7.5, half of 15). Also by definition, line AD and line EC are
equal (both half of 8, therefore 4). We already established
that angles BAC and BCA are equal to each other, therefore by the
geometry rule “side-angle-side”, triangles ADF and FEC are congruent
triangles. Therefore, line EF and line DF are also congruent, and
both are equal to 4.

Now, step back and answer the question: What is the total length of
the path along the line ADFEC? Well, it’s four congruent lines,
each with length 4, for a total of 16. This is the same length as
the original path along A to B to C.

Now, bisect lines AD, DF, FE, EC, AF, and FC. Label the midpoints
as follows: midpoint of AD = G. AF = H. DF = I. FE = J. FC =
K. EC = L. Now draw the lines GH, GI, JK, and KL. By the same
geometric rules and problem solving, triangles AGH, HIF, FJK, and
KLC are all congruent, and every line AG, GH, HI, IF, FJ, JK, KL,
and LC is equal to 1/2 of 4 = 2. Now the total path along the
line AGHIFJKLC is still 16.

If you continue this process, making infinitely smaller isosceles
triangles, and still calculating the path along the line of these
triangles from A to C, you still get 16. In the limit, you arrive at two paths along the
route from point A to point C. The original line AC is still 15.
But, the path from point A to point C along the other route is
always equal to 16. Therefore, 15=16.

(proof is from David Bentley at Pomona College)

Damn it Feels Good to be a BLAZER.

but blazersedge – fantasy basketball next year. FOR SURE.