Two numbers work: 41312432 and its reverse, 23421314.

Consider the face with the largest number of sides. If that face has m sides, then it’s surrounded by m faces. Any face must have at least 3 sides. So altogether in this group we have m + 1 faces, and each face must have between 3 and m sides. At least two faces must have the same number of sides.

From Arthur Engel’s Problem-Solving Strategies, 2003.

1.Bh1! Kxh1 2.Rxh3#

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.

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!”

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.

1. Rc8! (ha!) and the rook will mate.

From the Detroit Free Press, 1905.

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.

1. Ra1 Kxg6 2. Qb1#

From Benjamin Glover Laws, The Two-move Chess Problem, 1890.