# Fun With Fundamentals: Problem 207

July 1, 2000
Problem 207 - A rose by any other name doesn't always smell as sweet, as this month's problem by Leon G. Wilde of Andover, Mass., demonstrates

Problem 207 — A rose by any other name doesn’t always smell as sweet, as this month’s problem by Leon G. Wilde of Andover, Mass., demonstrates.

There was turmoil in the C.I.A. offices at Langley.

“After a week, you haven’t come up with anything yet?” shouted Wilbur Weasel. “We’ve already learned that there are two groups of spies — over a dozen of them — assigned to state capitals in the U.S., but we’ll never intercept them until we know which states they’re headed for!”

“I’m trying,” Fred Ferret replied defensively. “But the microfilm is in some kind of code. Our best cryptographers have been working on it round the clock.”

Group M: IDA STONE Group N:JED CHYMV

“These two people, Ida Stone and Jed What’s-his-name are obviously the leaders. All you have to do is find which states the spies are being assigned to,” said Wilbur.

Just then Wilbur’s secretary, Marilyn came into the office. “What’s all the shouting about?”

“We’re trying to decipher this microfilm,” replied Fred. “It shows the destinations of those foreign spies.”

“What’s so difficult about that?” she asked. “Here, I’ll write them down for you.”

What were the states?

Fun With Fundamentals
POWER TRANSMISSION DESIGN
1100 Superior Ave.
Cleveland, OH 44114-2543

Deadline is August 10. Good luck!

Technical consultant, Jack Couillard, Menasha, Wis.

Solution to last month’s problem 206 — You cross all your Ts and dot all your Is if you answered 30 in.

Here’s how Bluff came close:

To solve this problem, take the general case, where the centerline of the smaller 56.25-in. radius pipe is below the centerline of the 120-in.-ID pipe. The solution requires that the lower I-beam be horizontal (perpendicular to the line through centers “A” and “B.”) for the other two I-beams to be of equal length.

Draw a line through Center “B” parallel to the lower I-beam (horizontal). Where it intersects the 56.25-in. circle draw another 56.25 radius circle. For the problem to have a solution, the following must be true:

BC = CD = 56.25 in.
AC = AD - CD = 63.75 in.
BC
= CD = 56 25
= 120
AC
= AD - CD = 63.75
(56.25)2 + x2 = (63.75)2
x
= 30

Bluff recut the I-beams!

Correction: We published an erroneous solution to the March puzzle.The correct answer is 1.43 hours.

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