One summer day, Jack, deciding to paint his house, placed his ladder against the side of the house and began to climb. However, having placed the ladder perpendicular to the ground, he began to tip over backwards when just halfway up. It's easy to see that while Jack clung hopelessly to the middle rung, his trip to the ground (Point A on the side of the house to Point B on the ground) followed a path EXACTLY ONE QUADRANT OF A CIRCLE WHOSE CENTER WAS ON THE GROUND NEAR THE FOUNDATION OF HIS HOUSE AND WHOSE RADIUS WAS HALF THE LENGTH OF THE LADDER.
Six months later, Jack had recovered from his broken bones and decided to finish the job he started. Even though it was now middle of winter, Jack hauled out his ladder and placed it just as he had before. Being a slow learner, Jack repeated his original error concerning the angle of the ladder, and again had some bad luck when he reached the halfway point. But now the ground was a frozen sheet of ice and instead of tipping backwards, now the base of his ladder slipped out and the top slid down the side of the house. Jack clinging to the center rung, wound up moving from Point A on the side of his house to Point B on the ground as before. Question: DESCRIBE THE PATH HE TRAVELED THE SECOND TIME!
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Source: 1. Submitted by Bill Coyle. 2. Many, many places.
I [Ken] think this can best be understood by considering two ladders, joined at their centers. Place them against the house, then unfold them, one with its foot against the house, and the other with its top against the house. Obviously the center follows the same path for either ladder for the entire trip to the ground.
2. A Product Square 1. Sum   8  3  4    1  5  9    6  7  2   The middle square has to be 5; then juggle with the other numbers and a solution is quickly found. 2. Product First, if there is a solution for the sumsquare, it gives you a solution for the productsquare: Replace each number x for the sumsquare by g^x. So (with g=2)   256  8  16    2  32  512    64  128  4   is a solution with product 32768 (!) Dividing each number by 2 we get:   128  4  8    1  16  256    32  64  2   with product 4096. So, a solution is already found, but what about the lowest product? Some math: Let P be the product of this square:   A  B  C    D  E  F    G  H  I   Then we have 8 equations: (1) ABC=P (2) DEF=P (3) GHI=P (4) ADG=P (5) BEH=P (6) CFI=P (7) AEI=P (8) CEG=P multiply (1), (2) and (3): (9) ABCDEFGHI = P^3 Multiply (2), (5), (7) and (8): (10) ABCDE^4FGHI = P^4 Divide (10) by (9): (11) E^3 = P This means that P is always a cube, and E = P^(1/3). This means that all the numbers in the square are <=E^2 (=P/E) So, by choosing E (and thus P=E^3), we have to find 9 unique numbers(A through I) all smaller or equal to E^2 and dividing P. E  E^2  P  number satisfying  How many?   1  1  1  1  1  2  4  8  1,2,4  3  3  9  27  1,3,9  3  4  16  64  1,2,4,8,16  5  5  25  125  1,5,25  3  6  36  216  1,2,3,4,6,9,12,18,36  9   But is it possible to fill a square with these numbers? YES !!  18  1 12    4  6  9    3 36  2   So P = 216 is the lowest product that can be made.
When I was looking for a product square, I also first found the simple trick of using the sum square as the exponents of a single number. To look for a smaller product, I assumed there were two numbers, namely 2 and 3, and their repsective exponents would have to sum to a constant. I figured their exponents could run from 00 to 22 to give nine possible combinations, and I placed them into a square as shown. When actual numbers are used with these exponents, Wilfred's solution above is found. 
 12 00 21   20 11 02   01 22 10   