## ________________

## ________________

### World Turns

Coach Newton reminds the team that the Earth is constantly rotating about an axis that connects the North and South Poles. “Umm, yeah,” responds the team. “Ask us something we don’t know.”

*Rounded to the nearest quarter-turn, how many rotations on its axis does the Earth make per year, averaged over many years?*

### Water Slide

Coach Newton sets up ten glasses arranged in a triangle as shown, with the glasses numbered 2, 3, 7, 8, 9, and 10 full of water and the others empty. He challenges the team to rearrange the glasses into the second configuration shown, with full glasses at positions B, C, E, G, H, and I, and in which the triangle points in the opposite direction.

*What is the smallest number of glasses the students can move to perform Coach Newton’s challenge?*

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## Solutions to week 88

**Soccer or Math?** Since Duke is tied exactly at one right angle of the square, and the square won’t fit through the goalposts, Duke is always at the vertex of a right triangle whose two other vertices are the goalposts. In fact, Duke can be at any point which makes a right triangle with the two goalposts, and the goal line will always be the hypotenuse of the right triangle (since Duke is tied up at the right angle). The collection of all of the vertices of right triangles on the same side of a line segment that have that line segment as hypotenuse is a semicircle whose diameter is that line segment (this is covered in many introductory geometry classes; for a quick calculation showing that it must be so, see the box below). So we are just looking for half the perimeter of the circle whose diameter is the 7.32-meter long goal line. That’s π×7.32/2, which turns out to be very close to eleven and a half. So rounding as requested, the desired length is **eleven** meters.

*x*-axis, centered on the origin. In other words, the coordinates of its endpoints will be (-

*a*,0) and (

*a*,0). Let (

*x,y*) be any vertex of a right triangle with the given hypotenuse. By the Pythagorean theorem, the square of the distance from (

*x,y*) to (-

*a*,0) plus the square of the distance from (

*x,y*) to (

*a*,0) must equal the square of the length of the segment (which is 2

*a*). Using the coordinates, that means (

*x*+

*a*)² +

*y*² + (

*x*–

*a*)² +

*y*² = 4

*a*². Simplifying, this is

*x*² + 2

*ax*+

*a*² +

*y*² +

*x*² – 2

*ax*+

*a*² +

*y*² = 4

*a*². The

*ax*terms cancel, and rearranging and dividing by 2, we get

*x*² +

*y*² =

*a*², an equation immediately recognizable as the equation of a circle. Also, Coach Newton would like to thank Prof. Sergey Cherkis of the University of Arizona for this puzzle.

**Belt, Suspenders, and Braces**.In the diagram, the points *A*, *B*, and *C* represent the three trees, and the circles represent the regions in which Duke could roam if he were only chained to that one tree. So with all three chains in place, Duke can only reach points in the intersection of all three circles, namely that shape in the center that looks like a triangle with curvy sides. (It’s actually called a Reuleaux triangle.) If the chain to tree *C* breaks, let’s say, then Duke can reach any point in the intersection of the circles centered at *A* and *B*, which is a sort of pointy oval reaching from *C* down to *D*. So the additional area he can reach when one chain breaks is the region outlined in a heavier stroke. That might seem daunting to calculate the area of, but notice that if you cut off the slice indicated by the dashed line, it just fits into the top depression in the shape as indicated. Therefore, the extra area is simply one sixth of the entire circle centered at *A*, or 36π/6 = **6π** square meters.

*two*generations down, since the only males one generation down are the fathers of the females two generations down (since males have no fathers). Thus, the number of females in a given generation is the number in the next generation plus the number in the generation just after that, which is the characteristic relation defining the Fibonacci sequence.

## Recent Weeks

**Week 88**: Soccer or Math? & Belt, Suspenders, and Braces, solutions to Queen Bee & Coco-Not

**Week 87**: Queen Bee & Coco-Not, solutions to Uneven Odds & Stable Selection

**Week 86**: Uneven Odds & Stable Selection, solutions to Peri-area? & Hypotenuse Partition

**Week 85**: Peri-area? & Hypotenuse Partition, solutions to Potato Puzzle & Cucina Combinations

**Week 84**: Potato Puzzle & Cucina Combinations, solutions to Sour Lemons & One Liner

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

**Come back next week** for answers and more puzzles.

[asciimathsf]