Ride Sharing

You and I have to travel from Startville to Endville, but we have only one bicycle between us. So we decide to leapfrog: We’ll leave Startville at the same time, you walking and I riding. I’ll ride for 1 mile, and then I’ll leave the bicycle at the side of the road and continue on foot. When you reach the bike you’ll ride it for 1 mile, passing me at some point, then leave the bike and continue walking. And so on — we’ll continue in this way until we’ve both reached the destination.

Will this save any time? You say yes: Each of us is riding for part of the distance, and riding is faster than walking, so using the bike must increase our average speed.

I say no: One or the other of us is always walking; ultimately every inch of the distance between Startville and Endville is traversed by someone on foot. So the total time is unchanged — leapfrogging with the bike is no better than walking the whole distance on foot.

Who’s right?

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You are. My argument would be sound if each of us simply stood by the bike after dismounting until the pedestrian caught up. But instead we’re investing that extra time in walking, which accounts for the faster progress. Suppose the total trip is 2 miles, and each of us can walk at 4 mph and ride at 12 mph. I ride 1 mile in 5 minutes and leave the bike for you, walking the remaining 1 mile in 15 minutes. You take 15 minutes to reach the bike and then ride to Endville in 5 minutes. We both arrive at the destination in 20 minutes, where it would have taken 30 minutes if we’d walked the whole way.

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May 20, 2015May 31, 2015 | Puzzles

A Passing Wave

A puzzle from J.A.H. Hunter’s Fun With Figures (1956):

A man paddling a canoe upstream sees a glove in the water as he passes under a bridge. Fifteen minutes later, he turns around and paddles downstream. He passes under the bridge and travels another mile before reaching the rock from which he started, which the glove is just passing. If he paddled at the same speed the whole time and lost no time in turning around, what is the speed of the current?

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Say the canoeist paddled at x miles per hour through the water and the current moved at y miles per hour relative to the land. Fifteen minutes of paddling took the canoe x/4 miles from the glove, and the glove took 1/y hours to travel the 1 mile between the bridge and the rock. So after the canoeist turned around, the glove took (1/y – 1/4) or (4 – y)/4y hours to reach the rock. During that time the man moved x(4 – y)/4y miles through the water. We know that he was x/4 miles from the glove at the start of that interval, so x/4 = x(4 – y)/4y and hence 1 = (4 – y)/y and y = 2. The speed of the current is 2 miles per hour. A simpler solution, sent in by reader Peter McLeod: “You need to switch reference frames. Take the reference frame of the river, i.e. that which is moving downstream at the speed of the current. In that frame, the canoeist passes the glove, spends fifteen minutes paddling away from it, then turns and paddles back towards it. How long does he take to reach the glove again? It must be fifteen minutes, since we know his paddling speed is constant. Hence, the canoeist passes the glove half an hour after he left it. We know that in that time, the glove travelled a mile, so the current is going at 2 mph!”

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May 17, 2015May 24, 2015 | Puzzles

Modern Art

modern art puzzle

Which part of this square has the greater area, the black part or the gray part?

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Color the indicated regions dark gray: Now if we add the dark gray regions to the black region, we get triangle BCF, whose area is half that of the whole square (a triangle’s area is one-half base times height). If we add the dark gray areas to the light gray areas, we again produce a region that’s half the area of the square (because it’s the full square minus triangle ABE). So the black part and the light gray part are equal. By V. Proizvolov, from the Russian science magazine Kvant.

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May 13, 2015May 24, 2015 | Puzzles

Black and White

shinkman chess problem

By William Anthony Shinkman. White to mate in two moves.

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1. Rc8! (ha!) and the rook will mate. From the Detroit Free Press, 1905.

May 10, 2015May 17, 2015 | Puzzles

Roll Call

A problem from the 2002 Moscow Mathematical Olympiad:

A group of recruits stand in a line facing their corporal. They are, unfortunately, rather poorly trained: At the command “Left turn!”, some of them turn left, some turn right, and some turn to face away from the corporal. Is it always possible for the corporal to insert himself in the line so that an equal number of recruits are facing him on his left and on his right?

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Yes. When the corporal has taken a place in the line, let l denote the number of recruits to his left and facing him, and let r denote the number of recruits to his right and facing him. If the corporal takes the leftmost place in line, then there’s no one to his left (l = 0). If no one to his right happens to be facing him either, then the problem is solved. Otherwise r > 0. Now suppose the corporal moves gradually rightward, one position at a time. If he passes a recruit who is facing right (that is, whose back had been to him), then l increases by 1 and r doesn’t change. If he passes a recruit who’s facing left, l doesn’t change and r decreases by 1. And if he passes a recruit who’s facing the back of the parade ground, then neither l nor r changes. Now consider what happens to the difference l – r as the corporal moves along the line. At first it’s negative (because r is positive and l is 0). As the corporal moves down the line, l – r changes by at most 1 as he passes each soldier. When he reaches the right end of the line, r will be 0 and so l – r will be nonnegative. So we’ve started with a negative number, set it increasing and decreasing by 1 several times, and arrived at a nonnegative number. At some point, then, the difference must have been 0. At that moment l = r, and the corporal has an equal number of soldiers facing him on both sides.

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May 7, 2015May 17, 2015 | Puzzles

Podcast Episode 56: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits! Solve along with us as we explore some strange scenarios using only yes-or-no questions. Many were submitted by listeners, and most are based on real events.

See full show notes …

May 4, 2015October 25, 2017 | Podcast · Puzzles

Black and White

gold chess problem

By Sámuel Gold. White to mate in two moves.

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1. Ra1 Kxg6 2. Qb1# From Benjamin Glover Laws, The Two-move Chess Problem, 1890.

April 27, 2015May 3, 2015 | Puzzles

Star Power

A puzzle by A. Korshkov, from the Russian science magazine Kvant:

It’s easy to show that the five acute angles in the points of a regular star, like the one at left, total 180°.

Can you show that the sum of these angles in an irregular star, like the one at right, is also 180°?

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Join two of the star’s adjacent vertices, such as A and E. Triangles BMD and AME have an angle in common (the congruent angles at M), so the sum of angles B and D in triangle BMD equals the sum of angles A and E in triangle AME. This means that the sum of the star’s five angles equals the sum of the angles of triangle ACE — which is 180°.

April 25, 2015May 3, 2015 | Puzzles · Science & Math

Hoop Dreams

A memorably phrased puzzle from The Graham Dial: “Consider a vertical girl whose waist is circular, not smooth, and temporarily at rest. Around the waist rotates a hula hoop of twice its diameter. Show that after one revolution of the hoop, the point originally in contact with the girl has traveled a distance equal to the perimeter of a square circumscribing the girl’s waist.”

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Hold the hoop steady and let the girl roll around inside it: Image: Wikimedia Commons Since the ratio of the diameters is 2:1, so is the ratio of the circumferences. This means that a point on the girl oscillates back and forth between two opposite points on the hoop, passing through the hoop’s center on the way and producing a straight line (the “Tusi couple”). “The perimeter of circumscribing square equals four girl diameters or two hula hoop diameters which is the total displacement of initial point of contact between hula hoop and the aforementioned vertical girl.” From L.A. Graham, The Surprise Attack in Mathematical Problems, 1968.

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April 22, 2015May 3, 2015 | Puzzles

Quickie

University of Strathclyde mathematician Adam McBride recalls that in his student days a particular teacher used to present a weekly puzzle. One of these baffled him:

Find positive integers a, b, and c, all different, such that a³ + b³ = c⁴.

“The previous puzzles had been relatively easy but this one had me stumped,” he wrote later. He created three columns headed a³, b³, and c⁴ and spent hours looking for a sum that would work. On the night before the deadline, he found one: 70³ + 105³ = 35⁴.

“This shows how sad a person I was! However, I then realised also how stupid I had been. I had totally missed the necessary insight.” What was it?

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“All you need to do is start with 2³ + 3³ = 35 and multiply by 35³.” (From Adam McBride, “Mathematics: The Greatest Subject in the World,” The Mathematical Gazette, vol. 89, no. 516 [November 2005].)

April 19, 2015April 26, 2015 | Puzzles