## Ken's POTW

Two Fields
1. A Texas ranchman, who owned more land than he could conveniently farm, leased half of a certain field to a neighbor. This field was 2000 yards long by 1000 yards wide, but because of certain bad streaks which ran through the land it was decided that a fairer division would be obtained by cutting a band around the field than by dividing it in half. Find the width of the strip to cut around the field. [Also find the width for any L x W field.]
2. Two boys start from the same spot on the south side of a river. One boy runs due East for 250 yards, crosses the river via a bridge, then runs due North for 600 yards through a field to a flag. The other boy runs due West to a different bridge, crosses the river, then runs directly for the flag. Both boys run at the same rate and reach the flag at the same time. How far apart are the bridges?

Source: Mathematical Puzzles of Sam Loyd, Volume Two, edited by Martin Gardner, Dover Publications, New York, 1960, #90, #94.

Solutions: Several people sent correct solutions this week, including: Sorin Ionescu, Denis Borris, Jon Abraham, Jeff Soesbe, Richard Brabant.
1. Sam Loyd's answer (attributed to the reapers who would actually perform this) is "One quarter the difference between a short cut cross lots, and round by the road." Mathematically, this is better understood as: from the sum of the two sides, subtract the diagonal and divide the remainder by four. For the dimensions given, this yields 190.983 yards. The smallest rectangle giving integral solutions is 8x6, with a strip of width 1.
2. This is just a simple application of the pythagorean theorem. Sam Loyd, however, avoids the square roots through application of another identity. If Boy1 runs 'b=250' then 'c=600', and Boy2 runs 'a' then 'd=hypotenuse', the distance 'a' is equal to (c/[(c/b)+2]), or (600/[(600/250)+2), or 136 and 4/11 yards. The distance between the bridges is 'a+b' or 380 and 4/11 = 386.36 yards.

Denis Borris extended these problems for use at the Panthera Puzzle Contest:
```To let you know I used your 2 current puzzles as the springboard
for my next 2 puzzles that I'll submit next week to panthera's:

PUZZLE1
Well, I've decided to lower my standards and bore
you with a couple of straight math puzzles.
You gotta huge floor, 1050 by 700.
You paint it this way:
-green strip x wide all around the outer edge
-next to this, pink strip y wide
-next to this, polkadot strip z wide
-you're left with a rectangle at center: babyblue.
Warning: start at center, else you'll be stuck there...
After all's done, you measure to find the areas:
-green: 3/5 of floor
-pink: 1/7 of floor
-polkadot: 1/10 of floor
What are the "floored"(how appropriate!) values of x y z?
Enter like this: x-y-z
And now go to my next one; bring your paintbrush to
help B paint his new bridge...

PUZZLE2
A and B are standing at X on south edge of a river, 100 wide.
There's a bridge on their right (east), 50 from X.
There's also a bridge on their left. Both bridges 100 long.
A and B leave together; A goes right, B goes left (west).
B walks at twice A's speed (main reason: has longer legs).
A walks the 50 to the bridge, then crosses over.
A continues directly north, a distance of 50, to point P.
B reaches the other bridge, then crosses over.
B then walks a straight path, heading toward P.
(if your B is not heading in a general "northly-eastly"
direction, then you better stop and re-read the above!)
When A arrives at P, B still has 50 to go before reaching P.
Ok; in order to make this puzzle possible, B decides to
lengthen the bridge he crossed on, making it cross the river
at an angle, keeping its position on the south side the same.
B does this such that in the future, when the above walkathon
is similarly repeated, he gets to point P at same time as A.
B punches keys on his handheld Texas Instrument EL-733-A to
realise that he has 2 options, that is that he can have
2 bridges (different lenghts) to accomplish his goal.
```

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