Puzzle #11: T.C. Wentworth’s Problem

By Muriel Bristol (Transcriber) | March 9, 2019

T.C. Wentworth’s problem – how to construct a square containing 20 square inches: draw a base line 2 inches long. Draw perpendicular 4 inches long at one end of the baseline. Now draw hypotenuse from the above lines. This hypotenuse will be one side of a square containing exactly 20 square inches. The square of the base plus the square of the perpendicular equals the square of the hypotenuse. – Milton Mills, N.H. (Boston Globe, March 13, 1902).

[Answer to Puzzle #11 to follow in the next Puzzle]

Solution to Puzzle #10: J.O. Porter’s Cork Problem

Followers of the Boston Globe’s Puzzle Problem column of long ago answered:

G.W. Monegan, North Chelmsford, says the cork costs 5 cents; so say Frank E. Witherell, Walter L. Colburn, E.G. Hayden, L.N. Lewis, S.O. Keep, Rowley; D.T. Jardine, Cambridge; James A., Newton (Boston Globe, December 25, 1901).

For those that do not want to simply take their word: $1.10 = $1.00 + 2X; $0.10 = 2X; $0.05 = X.

If the cork is worth 5¢, and the bottle is worth that plus a dollar, then the bottle is worth $1.05. Taken altogether, the total is $1.10.

Author: Muriel Bristol

"Lady drinking tea"

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