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Round 1: Problem 5 Solution

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Cross Section

Cross Section

Pictured at the left is the cross-section of the three spheres. To solve the problem, let us note that the spheres of radii 3 and 5 are similar, and they both have the same center. Therefore, any area traced by the radius-1 ball on the outer sphere will project itself onto the inner ball. We must then discern at what ratio the projection will be in. To determine the ratio, consider what type of inking is occurring on the outer sphere. The black ball colors in an area of the outer sphere’s surface. Therefore the amount covered, in this case an area of 1, should be proportionate with the surface areas of the two spheres. For a sphere of radius r the surface area A = 4 π r2. For our purposes then we need only to compare r2 for each of the spheres, arriving at our answer of 9/25.

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March 16, 2009 at 00:04

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Round 1: Problem 4 Solution

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Chess Board

Chess Board

For this problem, we will need first to count all the rectangles on the 8×8 chessboard, and then to subtract out all the squares on the board.

The easiest way to count the total number of rectangles (including squares) is to visualize the chessboard as being bound by 9 vertical lines and 9 horizontal lines. These 9 lines in each direction form the 8×8 grid in their interiors. By visualizing the board in such a way, finding the number of rectangles becomes trivial using a bit of combinatorial thinking. Within the board, a rectangle may be of any size, and start at any point. So we must merely pick at which of these 9 lines to start from, and at which of these 9 lines to end. Doing this for both horizontal and vertical lines then forms a rectangle. For example, if we were to pick the first line and the ninth line both horizontally and vertically, the rectangle that would be formed is the outer frame of the board. In total, then, we can pick horizontal bounds in 9C2 ways, and independently we can pick vertical bounds in 9C2 ways. Therefore the total number of rectangles is simply 9C2 x 9C2 = 1296 rectangles.

To arrive at a final answer, we need to subtract out the number of squares. The number of squares on the board are interestingly enough the square numbers. There are 64 1×1 squares, 49 2×2 squares, 36 3×3 squares, 25 4×4 squares, 16 5×5 squares, 9 6×6 squares, 4 7×7 squares and 1 8×8 square = 204 squares. Try it: start drawing them out and notice the pattern.

Subtracting the two, the answer is then 1092 non-square rectangles.

For a similar problem, check out Project Euler’s Problem 85. On a personal note, I happen to love combinatorics, especially enumerative combinatorics, and most recently I’ve developed an affinity for chess-related problems. For an extensive discussion see the book Across the Board: The Mathematics of Chessboard Problems, which in my humble opinion is a fantastic book of just about everything you ever did or didn’t want to know about math and chess.

As always, please post any comments, corrections or alternate solutions in the comments.

Credit for the (great) Chessboard Photo goes to Otbora.

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March 16, 2009 at 00:03

Round 1: Problem 3 Solution

with 2 comments

For this problem, just try out a few simpler examples and a pattern emerges:

√(11-2) = 3

√(1111-22) = 33

√(1111111111-22222) = 33333

In general, for any two integers x and y, where x is a sequence of 1’s and y is a sequence of 2’s and x has twice as many digits as y, √(x-y) is a sequence of 3’s with the same number of digits as y.

In our case, then, the answer is 333…333 with 1004 3’s.

As always, please post any comments, corrections or alternate solutions in the comments.

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March 16, 2009 at 00:02

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Round 1: Problem 2 Solution

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This problem at first glance appears somewhat cumbersome, but with an easy trick the answer should be found very quickly.

In contrast to the way the problem is written, for simplicity’s sake let us ignore Team B and simply focus on Team A. When Team A wins, we will denote the game with a W. When A loses, an L. Then, we should end up with some combination of W’s and L’s, such as WLWLLWW. Each of these strings of W’s and L’s correspond to a sequence of wins and losses that may occur in the series.

The Principle of Inclusion / Exclusion really should not be needed here. It works, but it’s not necessary. The easiest way to solve the problem is to think about how problematic combinations occur only when a L occurs in the final game. Note that the series can extend to 4, 5, 6, or 7 games. Then to eliminate the problematic combinations, let us force ourselves to pick a W in the final spot. Now all we have to do is consider each series length and pick the other three games that Team A is going to win.

For 4 games, besides the W that was already chosen in the 4th game, there are 3 remaining games, in which we need to place 3 more W’s, which can be done in 3C3 = 1 way.

For 5 games, 4 games remain to place the 3 W’s in 4C3 = 4 ways.

For 6 games, 5 remaining games for the 3 W’s in 5C3 = 10 ways.

Finally, for 7 games, 6 remaining games for the 3 W’s in 6C3 = 20 ways.

In total, that sums to 35 ways. We now multiply by 2 to compensate for the fact that either Team A or Team B can win in any of these ways, giving the final answer of 70 ways for the sequence to occur.

As always, please post any comments, corrections or alternate solutions in the comments. In particular if you’ve worked out the Inclusion / Exclusion, might as well post it.

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March 16, 2009 at 00:01

Round 1: Problem 1 Solution

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There has developed a set of terminology to describe these objects. These tiling objects have been named polyomino(e)s (as generalized forms of dominoes), or in this specific case, pentomino(e)s. This specific arrangement is usually referred to as the F-pentomino, for its resemblance to the letter F (all of the pentominos are named after the letters they look like, though I’d agree that this one is a bit of a stretch).

With that in mind, there are a number of useful sites with information on pentominos and plane tilings. I won’t list all of them because they are relatively easy to find, but dmoz.org links to a whole directory’s worth of sites dedicated to polyominos. It can be found here. Of particular interest is Gerard’s Polyomino Solution Page, with more than 60 polyomino puzzles and solutions.

For our purposes, however, this problem should not pose too much difficulty beyond simply trying out various rotations of the tiles. There’s a simple solution with fewer pieces; an example can be found at the NGfL here (warning: PDF) along with solutions for all of the other pentominos. I’ll link to a nice larger solution as well because, well, I like the colors.

One point of confusion for this problem is that tilings of the plane (or tessellations as they may be called) do not need to be regular (rectangular). In fact there are only three polygons that form regular tessellations. See the page at MathWorld for further info. Rather, the critical point here is that the shape formed can be attached to itself at any end (including top and bottom), meaning that we could continuously attach replications of the shape to itself to tile an infinitely large area.

Please post any comments, corrections, interesting points, or alternate solutions that you may have found in the comments!

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March 16, 2009 at 00:00

Round 1 is nearly over.

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Round 1 is nearly over. Don’t forget to submit your responses on the official site. Unless we’ve made some serious errors I believe we should have the correct answers, so check back at 12:00 A.M. on Monday, March 16 to see if you’ve got what we’ve got.

Also, in case anyone has any problems, it took me about 6 hours to see my submission recognized on the site, so if I may give some advice, submit earlier than the deadline in case something goes wrong and you need to email support to check if your submission was received.

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March 12, 2009 at 16:56

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Round 1: Problem 5

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A spherical ball with radius 1 is covered with ink and is placed in the hollow region between two concentric spheres with radii 3 and 5. The inked ball rolls about in this region, always touching the inside surface of the outer sphere and the outside surface of the inner sphere. Suppose that a region with area 1 on the outer sphere gets painted with ink by the rolling ball. Find the area of the region of the inner sphere that gets inked by the rolling ball.

Solution will be posted on March 16 at 12:00 a.m.

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March 4, 2009 at 20:14

Posted in Round 1