Quiz Question

Number 186 of a series

This didn’t really happen, but imagine it. The only clock in my house is one of those grandfather types, with hands and a pendulum, and you have to wind it up. So I got up in the morning, and the clock was stopped. I forgot to wind it. I needed to set the correct time.

My friend Bill lives about five blocks away, and he always has the correct time, but no telephone. I walked over to Bill’s house and talked to him for some time. Then I walked back home and set my clock to very close to the correct time. I walked at the same rate the whole trip, but I have no idea how fast I walked.

How was I able to set my clock so accurately?

Post your answer in the comments section below.

Update and Answer

People provided answers to the Facebook posting, but nobody commented one the blog site. Anyhow, here is my answer.

 

Before I left the house I wound my clock and set the time to noon. Then I walked to Bill’s house and noted the time I arrived and the time I left. When I got home I noted how much time had elapsed on my clock, and computation of the current time was straightforward.

Quiz Question

Number 178 of a series

The Quiz Question from two weeks ago was to compute the volume of the intersection of three cylinders. See the above schematic. That’s a daunting task, so last week I posted a simpler problem. Compute the volume of two intersecting cylinders.

That meant computing the volume of this shape.

Supposing that has been done, to solve the original problem can be accomplished by computing the volume of a cylinder intersecting with the above shape.

What is of interest is the shaded part shown next.

Use R as the radius of the cylinders. Post your answer in the comments section below.

Quiz Question

Number 177 of a series

Last week’s Quiz Question was to determine the volume common to three intersecting cylinders of 3-inch diameter. Mike has suggested two solutions, but I am sure they are not correct. I have not, myself, worked this one.

So this week’s Quiz Question will address possibly ways to attack the problem. Start with the following diagram. It shows two intersecting cylinders.

From above the intersection looks like the above. If you join cylinders thusly, you will see the straight lines crossing in the middle as shown. The common volume is four of the shapes shown in the next diagram, a set of three orthogonal projections of the piece out.

These are what you get when you take a cylinder…

and make two cuts as shown.

 

The diameter of the cylinder is 3 inches. What’s the volume of the cut out piece shown. Post your answer in the comments section below.

Update and solution

Consider the shape depicted in the set of orthographic projections above.

Compute the volume of this shape based on a cut that passes through the center of the cylinder. View this cut from above, and it is apparent the shape of the cut is triangle ABC. The length of triangle base is b, and b is R sin ∝.

Now compute the volume of the shape by summing up all the small pyramids generated as shown. α is the angle of the cut measured from the top edge. R is the radius of the cylinder. Δα is the vanishingly small angular width of such a pyramid.

Revert to integral calculus to compute the total volume. Here are the equations. This is based on the volume of a pyramid of height R and base of area R2 sin ∝ Δα, using the volume of a pyramid being AR/3. The height of the pyramid is the radius R.

 

And I get for the volume of the shape the following:

This is ¼ the volume of the intersection of two cylinders. Compare this to the volume of a sphere of radius R. The volume of the intersection should be more than the volume of the sphere.

Quiz Question

Number 173 of a series

What I have attempted to draw here is a solid sphere with a hole drilled straight through the center. The hole is six inches long. What is the volume left of the sphere after the hole is hollowed out as shown?

Post your answer in the comments section below.

Update and Solution

Steve provided the correct answer:

1/6*pi*L^3 where L is the length of the hole.

In case you don’t notice, that is exactly the volume of a sphere whose diameter is the length of the hole (6 for this problem). Steve, who has deep capabilities in mathematics worked the problem using calculus.

 I got it by doing a volume integral in cylindrical [coordinates].

I obtained my solution by more devious means. Use the statement of the problem to provide the answer. The statement of the problem says nothing about the diameter of the hole, leaving the reader to conclude the diameter does not matter. The remaining volume is the same, no matter what the diameter of the hole.

The let the diameter of the hole approach zero as a limit, and the volume of the remaining volume is simple a sphere of diameter 6. The lazy man’s solution.

Quiz Question

Number 171 of a continuing series

Here is another from Martin Gardner.

THE EFFICIENT switching of railroad cars often poses frustrating problems in the field of operations research. The switching puzzle shown is one that has the merit of combining simplicity with surprising difficulty. The tunnel is wide enough to accommodate the locomotive but not wide enough for either car. The problem is to use the locomotive for switching the positions of cars A and B, then return the locomotive to its original spot. Each end of the locomotive can be used for pushing or pulling, and the two cars may, if desired, be coupled to each other. While working on the problem, assume that both cars are far enough east along the track so that there is ample space between each car and switch to accommodate both the locomotive and the other car. No “flying switch” maneuvers are permitted. For example, you are not permitted to turn the switch quickly just after the engine has pushed an unattached car past it, so that the car goes one way and the engine, without stopping, goes another way. Post your solution as a comment below.

Quiz Question

Number 169 of a continuing series

Martin Gardner died eight years ago, but he left behind a treasure of his writings. Here is from one of his books.

SMITH DROVE at a steady clip along the highway, his wife beside him. “Have you noticed,” he said, “that those annoying signs for Flatz beer seem to be regularly spaced along the road? I wonder how far apart they are.” Mrs. Smith glanced at her wrist watch, then counted the number of Flatz beer signs they passed in one minute. “What an odd coincidence!” exclaimed Smith. “When you multiply that number by ten, it exactly equals the speed of our car in miles per hour.” Assuming that the car’s speed is constant, that the signs are equally spaced and that Mrs. Smith’s minute began and ended with the car midway between two signs, how far is it between one sign and the next?

Gardner, Martin. My Best Mathematical and Logic Puzzles (Dover Recreational Math) (Kindle Locations 347-354). Dover Publications. Kindle Edition.

No fair going to the book for the answer. Post your answer as a comment below.

Quiz Question

Once again I turn to Martin Gardner’s book for a Quiz Question.

AN AIRPLANE FLIES in a straight line from airport A to airport B, then back in a straight line from B to A. It travels with a constant engine speed and there is no wind. Will its travel time for the same round trip be greater, less or the same if, throughout both flights, at the same engine speed, a constant wind blows from A to B?

No fair going to the book to look up the solution. Post your answer in the comments section below.

Update and solution

And Greg has the correct answer. See the comment below. Mike has the correct answer, as well, since the question only asked whether greater, less, or the same.

Quiz Question

Here’s another from Martin Gardner. You can purchase the book and look up the answer, or you can (preferred) work this one yourself.

A SQUARE FORMATION of Army cadets, 50 feet on the side, is marching forward at a constant pace. The company mascot, a small terrier, starts at the center of the rear rank [position A in the illustration], trots forward in a straight line to the center of the front rank [position B], then trots back again in a straight line to the center of the rear. At the instant he returns to position A, the cadets have advanced exactly 50 feet. Assuming that the dog trots at a constant speed and loses no time in turning, how many feet does he travel?

Post your answer in the comments section below.

Quiz Question

This one is from a book by Martin Gardner, so no fair going to the book for the solution.

In the diagram you see four bugs. they are on the corners of a square 10 feet on a side. Each bug looks at a bug on the adjacent corner and crawls directly toward that bug. Of course, the other bug is also moving, so each bug has to constantly change course to keep heading toward the other bug. The bugs crawl at a constant speed, the same for all.

  1. Do the bugs ever meet?
  2. If the bugs meet, how far does each bug travel?

Post your answer in the comments section below.

Update And Solution

The problem was designed to throw people off by means of distraction. Your first thought in attacking this is you’re going to have to compute the length of a spiral. In contrast, the approach is greatly simplified.

Start by taking a bug’s eye view, literally. Each bug sees another bug, at ten feet away at the start, and traveling perpendicular to the bug’s line of sight. That means the bug being looked at is not coming closer nor getting further from the bug traveling at him. And the situation never varies. Each bug is always going straight at another bug, and the closure rate is always due to the bug’s walking speed. So each bug keeps closing the ten-foot distance at a constant rate and travels 10 feet total to reach the other bug. And that’s the solution to the problem.

What might confuse people is that the total number of turns in the spiral is infinite, seeming to pose an intractable problem. It’s one of the interesting aspects of mathematics.

Quiz Question

Another one from the Internet. People, I don’t think these up, myself:

Huge pie. A huge pie is divided among 100 guests. The first guest gets 1% of the pie. The second guest gets 2% of the remaining part. The third guest gets 3% of the rest, etc. The last guest gets 100% of the last part. Who gets the biggest piece?

Submit your answer in the comments section below.

Quiz Question

An easy problem. The area of the large circle is 10. What is the combined area of the two semicircles? Prove it. Post your answer as a comment below.

Update and solution

You can resolve this one without doing any math. The solution is in how the question is posed. It is posed that there is a solution, but the sizes of the two semicircles are not specified. Therefore the answer is the same for all configurations of the two semicircles, including the case where the yellow semicircle takes up half the area of the large circle, and the blue semicircle disappears. Therefore the answer is that the sum of the areas of the two semicircles is ½ the area of the large circle.

Quiz Question

I stole this one off the Internet, and I’m not telling where I got it. No fair using Google to find it.

    There are 12 coins. One of them is false; it weights differently. It is not known, if the false coin is heavier or lighter than the right coins. How to find the false coin by three weighs on a simple scale?

Post your solution in the comments section below.

Quiz Question

The picture says it all. What fraction of each shape is shaded? This one is on the Internet, so no fair hunting it down. Post your answer as a comment below.

Update and Solutions

I’m going to settle this week’s Quiz Question today so I can start looking for one for next week. Here are my solutions, from left to right.

This is the only one that requires some math, despite what top diagram promises. I have drawn an arrow across the width of the blue hexagon to show that it is the same width as the length of a side of the outer hexagon. It’s left to the reader to  determine that the blue hexagon is 1/3 the area of the outer hexagon.

This one is easiest of them all. Slide the blue hexagon down and to the left, where I have drawn in a red hexagon of the same size. This shows that the sides of the blue hexagon are ½ the sides of the outer hexagon, so the area of the blue hexagon is ¼ the area of the outer hexagon.

A little imagination solves the last problem. I have numbered the squares 1, 2, and 3. Now rotate square number 2 45°, and you see that square 2 is ½ the area of square 1, and square 3 is ½ the area of square 2 and therefore ¼ the area of square 1.