Happy Pythagoras Day! Today is 8/15/17 (or 15/8/17 in other parts of the world), and .

We might as well celebrate today, because the next Pythagoras Day won’t happen for over 3 years. (Bonus points if you can figure out when it will be.)

Happy Pythagoras Day! Today is 8/15/17 (or 15/8/17 in other parts of the world), and .

We might as well celebrate today, because the next Pythagoras Day won’t happen for over 3 years. (Bonus points if you can figure out when it will be.)

*Posted by John Quintanilla on August 15, 2017*

https://meangreenmath.com/2017/08/15/happy-pythagoras-day-4/

I really enjoyed this video about how to trisect angles. Trisection of arbitrary angles is impossible using only a straightedge and compass; however, it is possible by carefully folding a piece of paper.

For further reading:

http://mars.wne.edu/~thull/papers/amer.math.monthly.118.04.307-hull.pdf

*Posted by John Quintanilla on August 13, 2017*

https://meangreenmath.com/2017/08/13/constructing-an-angles-trisectors/

This series of posts concerns solving the following problem from the 2016 University of Maryland High School Mathematics Competition.

A sphere is divided into regions by 9 planes that are passing through its center. What is the largest possible number of regions that are created on its surface?

a.

b.

c. 81

d. 76

e. 74

This series was actually written by my friend Jeff Cagle, department head for mathematics at Chapelgate Christian Academy, as he tried technique after technique to solve this problem. I thought that his resolution to the problem was an excellent example of the process of mathematical problem-solving, and (with his permission) I am posting the process of his solution here. (For the record, I have no doubt that I would not have been able to solve this problem.)

**Reflection**

I didnโt really need the projection into the plane for the solution, but my problem-solving self needed it to be able to count points and regions in slow motion. So, I should present a cleaned-up solution:

**Solution**

Since there are 9 planes, each plane must intersect with every other in a line, creating two points on the surface of the sphere. Thus, there are (9โ8)/2 * 2 = 72 points of intersection, and for n planes, there are ๐(๐ โ 1) points of intersection. With the first plane, there are zero points of intersection and two regions. Suppose we now have n planes and N regions. We add another plane, creating a circle on the sphere. For each segment that the circle intersects, it creates an additional intersection point as it enters, and it divides the region into two parts, adding one additional region. Hence, for each point added, a region is added as well. Since there are two

regions with zero points, there are thus 74 regions with 72 points of intersection.

*Posted by John Quintanilla on August 12, 2017*

https://meangreenmath.com/2017/08/12/solving-a-math-competition-problem-part-9/

This series of posts concerns solving the following problem from the 2016 University of Maryland High School Mathematics Competition.

A sphere is divided into regions by 9 planes that are passing through its center. What is the largest possible number of regions that are created on its surface?

a.

b.

c. 81

d. 76

e. 74

This series was actually written by my friend Jeff Cagle, department head for mathematics at Chapelgate Christian Academy, as he tried technique after technique to solve this problem. I thought that his resolution to the problem was an excellent example of the process of mathematical problem-solving, and (with his permission) I am posting the process of his solution here. (For the record, I have no doubt that I would not have been able to solve this problem.)

OK, so I wanted to prove that each region would be a triangle. So I decided to project the sphere onto a plane.

The projection of four planes:

**Conjecture: **The max number of regions is the number of intersection points plus 2.

**Proof (by induction)**

If we have 1 plane, we have no intersection points and 2 regions. Suppose we have n planes with ๐(๐ โ 1) intersection points and ๐(๐ โ 1) + 2 regions. Now we add the next plane to our figure. The plane creates a circle on the sphere. To maximize the number of regions, we angle the plane so that our circle does not intersect any already-existing intersection points. So the circle goes through a number of segments. Each time it does, it cuts the region bounded by that segment into two. So for each new intersection point, we lose one region and gain two, for a net gain of one region. That is, however many intersection points are added, that will be the number of regions added as well. And since ๐ + 1 planes have (๐ + 1)(๐) intersection points, we will

have (๐ + 1)(๐) + 2 max regions. DONE.

For the original competition problem, we have 9 planes and hence 9*8 + 2 = 74 regions, answer e.

*Posted by John Quintanilla on August 11, 2017*

https://meangreenmath.com/2017/08/11/solving-a-math-competition-problem-part-8/

This series of posts concerns solving the following problem from the 2016 University of Maryland High School Mathematics Competition.

A sphere is divided into regions by 9 planes that are passing through its center. What is the largest possible number of regions that are created on its surface?

a.

b.

c. 81

d. 76

e. 74

This series was actually written by my friend Jeff Cagle, department head for mathematics at Chapelgate Christian Academy, as he tried technique after technique to solve this problem. I thought that his resolution to the problem was an excellent example of the process of mathematical problem-solving, and (with his permission) I am posting the process of his solution here. (For the record, I have no doubt that I would not have been able to solve this problem.)

OK, so I wanted to prove that each region would be a triangle. So I decided to project the sphere onto a plane.

The projection of four planes:

After a while, I had a chart for max possible regions.

- 1 plane: Max regions = 2
- 2 planes: Max regions = 4
- 3 planes: Max regions = 8 (exponential?)
- 4 planes: Max regions = 14 (nope!)
- 5 planes: Max regions = 22 (huh?)

Then, really because I had no other ideas, I tried counting intersection points AND max regions

(remembering that one intersection point is โat infinityโ โ that is, the north pole).

- 1 plane: Intersection Points = 0, Max regions = 2
- 2 planes: Intersection Points = 2, Max regions = 4
- 3 planes: Intersection Points = 6, Max regions = 8
- 4 planes: Intersection Points = 12, Max regions = 14
- 5 planes: Intersection Pointsย 20, Max regions = 22

Oh. My. Goodness. The max regions are simply the number of intersection points plus 2. Could it really REALLY be that simple?

*Posted by John Quintanilla on August 10, 2017*

https://meangreenmath.com/2017/08/10/solving-a-math-competition-problem-part-7/

a.

b.

c. 81

d. 76

e. 74

OK, so I wanted to prove that each region would be a triangle. So I decided to project the sphere onto a plane.

The projection of four planes:

After a while, I had a chart for max possible regions.

- 1 plane: Max regions = 2
- 2 planes: Max regions = 4
- 3 planes: Max regions = 8 (exponential?)
- 4 planes: Max regions = 14 (nope!)
- 5 planes: Max regions = 22 (huh?)

*Posted by John Quintanilla on August 9, 2017*

https://meangreenmath.com/2017/08/09/solving-a-math-competition-problem-part-6/

a.

b.

c. 81

d. 76

e. 74

For a while, I toyed with the situation where we have

- Plane 1 โ equator (this always happens: Just make plane 1 the equator) ๐1(0๐, 0๐ธ).
- Plane 2 โ Prime Meridian ๐2(90๐, 0๐ธ)
- Plane 3 โ Intl Date Line ๐3(90๐, 90๐ธ)
- Plane 4 โ at an angle to all of those ๐4(45๐, 45๐ธ)

Here is our mapping with P1, P2, and P3 on it:

Now, how to represent P4? Aha! The inside of the unit circle is the southern hemisphere, and the outside is the northern. P4 must hit the equator a two points 180 degrees apart, go inside the southern hemisphere, and then outside to the northern. Thus:

The white region is a NONtriangular region created by the intersection of four planes. These are strange-looking regions, and I spent a long time โ several days โ vainly trying to count max regions created when I added P5, P6 etc. But one thing was clear: not all of the regions are triangular, nor can they be. For if a plane (say P4) cuts through a triangular region, it will create a new triangular region and a non-triangular โquadrilateralโ, as in the figure below. So counting triangles from points is NOT the solution here!

*Posted by John Quintanilla on August 8, 2017*

https://meangreenmath.com/2017/08/08/solving-a-math-competition-problem-part-5/

a.

b.

c. 81

d. 76

e. 74

OK, so I wanted to prove that each region would be a triangle. So I decided to project the sphere onto a plane. Thereโs a standard way of doing that, used both by map-makers and mathematicians. Place the sphere with the south pole on the plane at the origin. Then for each point on the sphere, run a line from the north pole through that point to the plane. This gives a 1-1 mapping of sphere to plane. The diagram below shows this mapping, with the points A and B on the sphere mapping to the points Aโ and Bโ on the plane respectively.

Notice that in the mapping above, the south pole is mapped to the origin (โstraight downโ), while the north pole itself cannot be mapped. We call the north point the โpoint at infinity.โ Also notice that the equator gets mapped to a circle. And, any circle around the sphere that goes through the north pole will also go through the south pole, and so becomes a line in the plane.

*Posted by John Quintanilla on August 7, 2017*

https://meangreenmath.com/2017/08/07/solving-a-math-competition-problem-part-4/

a.

b.

c. 81

d. 76

e. 74

AHH! Insight! Each plane must intersect the others because they all pass through the center. And two planes intersect in a line. And the line must intersect the sphere at two points. SO, we can count intersection points: There are 9 planes, and each plane will intersect the other 8, so there are 9 โ 8 = 72 intersection points IF we arrange the planes for maximum regions. More generally, if we have n planes arranged for max intersection points, we will have ๐(๐ โ 1) intersection points.

Wait, letโs do this carefully. There are 9 planes, and they can each intersect 8 different planes; but that counts the intersections of plane A and plane B twice, so there are (9*8)/2 = 36 lines of intersection, but 36 โ 2 = 72 points of intersection with the sphere. So our problem just got narrower: Given 72 intersection points defining various regions on the sphere, how many regions do we get?

And thatโs where the problem stands as of this writing. My preliminary conjecture is that each region will be a โtriangleโ (officially, spherical triangle) on the surface of the sphere, especially if we are maximizing regions. I need to prove that conjecture and then count triangles, which shouldnโt be too hard.

*Posted by John Quintanilla on August 6, 2017*

https://meangreenmath.com/2017/08/06/solving-a-math-competition-problem-part-3/

a.

b.

c. 81

d. 76

e. 74

At this point, various methods suggested themselves. Perhaps we could use recursion: let be the regions created by planes, and then we could examine the number of additional regions formed by planes?

Or, related to this, perhaps we needed to find the number of intersection points of each of the planes, and then relate the number of intersection points to the number of regions. But how to describe the intersection points?

It did occur to me that if we have n planes situated for maximal regions, they will divide the equator up into subintervals, and adding another plane will divide up two of those subintervals into 4. Did that help? Well, it could help count the number of regions touching the equator: two for each subinterval (one north of equator, one south). But what about the regions not touching the equator? Hmph.

One possible way to visualize this problem is to project the plane onto a sphere. I know how to

do that, but counting the regions still seems hard.

For a while, I toyed with the situation where we have

- Plane 1 โ equator (this always happens: Just make plane 1 the equator) ๐1(0๐, 0๐ธ).
- Plane 2 โ Prime Meridian ๐2(90๐, 0๐ธ)
- Plane 3 โ Intl Date Line ๐3(90๐, 90๐ธ)
- Plane 4 โ at an angle to all of those ๐4(45๐, 45๐ธ)

I looked at my daughterโs wall map of the world: P4 goes through Tblisi Georgia and south of French Polynesia.

Where does P4 intersect the others? Could I make a formula to find the intersection points?

*Posted by John Quintanilla on August 5, 2017*

https://meangreenmath.com/2017/08/05/solving-a-math-competition-problem-part-2/