LHIRT POD #2 solution

Solution for POD #2 below (white on white below):

Let the point on the inner circle be T. Because PQ is perpendicular to OT, OTP is a right triangle. Therefore OP²=OT²+PT². Now, the true area of the doughnut is (π/2)(OP²-OT²)= (π/2)(OT²+PT²-OT²)=(π/2)(PQ/2)².

Amazing, huh? Of course the smart-ass answer is that because a formula must exist for this area based on the length of that tangent alone (or else the problem wouldn't have an answer). Since the radius of the inner circle does not matter, make it radius 0, and the area will stay the same or just the area of the remaining circle. In a non-authoritative setting this form of reasoning is invalid, because you can't be sure if the questioner is being honest in implying the existence of that formula.

Another problem will be posted tomorrow.

Answer to POD #1

The answer is in white on white below, highlight to verify your answer. Surprisingly this didn't take me much time, so I'll continue with my questions tomorrow. I will have to qualify by stating that I may pose a problem that I don't know the answer to, and may pass on the problem until I can research it further. This is all but an experimental exercise. :)
Here's how you figure out the expected number of complete rows: there are five possibilities to consider, because you can leave anywhere from 0 to 4 complete rows. You cannot leave 3 or 4 complete rows because of pigeonhole principle: 4 is impossible after removing one, and 3 is impossible after removing 4 (because you can remove an entire row and have one more to remove, destroying another row). So, in all C(12,4) possibilities there are 3 outcomes: leaving 0 rows, leaving 1 row, and leaving 2 rows.

Case 1: leaving 0 rows... you can do this by choosing 1 from each row, or 3*3*3*3 of 81 possibilities.

Case 2: leaving 2 rows... you can do this by first choosing the rows that are complete ( C(4,2) ) then finishing by choosing what to take within the rows that aren't complete ( C(6,4) ) leaving C(4,2)*C(6,4)=6*15=90 possibilities

Case 3: leaving 1 row. First, choose the complete row C(4,1) then choose which 4 of the 9 remaining to remove. But, be careful: of those C(9,4) you must exclude the ones that leave a full row which is [choose first the row remaining then the 4 from the 6 remaining, or C(3,1)*C(6,4)=3*15 or 45] so C(4,1)*(C(9,4)-45)=4*81=324.

To confirm: 81+90+324=495=11*5*9=12*11*10*9/4*3*2*1=C(12,4)

So, now that we have all the possibilities, let's calculate the expected value E=(81*0+90*1+324*2)/495=738/495=1.49090909... so the expected number of rows is about 1 1/2 rows. More likely than not, you'll have 1 complete row remaining.

I had a really fun time tonight, scoring 46 points in the brainmaster's competition at Tigin. I must look up this website; it sounds like a lot of fun. But, it makes for a long night, and your dear author is too tired to continue. Will write again tomorrow.