# Halving and Rehalving, as Well as Doubling and Redoubling, as a Calculator-Free Calculating Strategy I don’t like being too dependent on calculators. The future might bring an EMP (electromagnetic pulse) that would fry all such gadgets (and cell phones, cars, computers, etc.), and I want to be ready for a post-calculator world, if that happens.

My overarching strategy for doing math in my head is this: don’t have just one single strategy. Instead, devise one, on the fly, based on the problem you are trying to solve.

I do, however, use a few “go to” strategies for certain things, such as finding 25% of something, or multiplying by eight, or similar problems. This involves looking for, and take advantage of, powers of two, as well as their reciprocals. 25% is 1/4, which is simply halving twice, and multiplying by eight is three doublings, since 8 = 2³. With practice, doubling or halving numbers repeatedly and silently, in one’s head, becomes much faster and easier. If done out loud, it becomes easier still, and on paper, it’s extremely easy.

I intend to do more blog-posts in the future with calculator-free calculation strategies, but not all at once — instead, these techniques will be posted one at a time. However, these postings will stop immediately, in the event of an EMP.

# Zome: Strut-Length Chart and Product Review

This chart shows strut-lengths for all the Zomestruts available here (http://www.zometool.com/bulk-parts/), as well as the now-discontinued (and therefore shaded differently) B3, Y3, and R3 struts, which are still found in older Zome collections, such as my own, which has been at least 14 years in the making. In my opinion, the best buy on the Zome website that’s under \$200 is the “Hyperdo” kit, at http://www.zometool.com/the-hyperdo/, and the main page for the Zome company’s website is http://www.zometool.com/. I know of no other physical modeling system, both in mathematics and several sciences, which exceeds Zome — in either quality or usefulness. I’ve used it in the classroom, with great success, for many years.

# A Graphical Survey of Prime, Perfect, Deficient, and Abundant Numbers From Two to Thirty In this graph, each number on the x-axis (from 2 to 30) is plotted against the sum of all its factors (including one, but excluding the number itself) on the y-axis. Numbers on the blue line y = 1 have no factors other than one and themselves, and are therefore prime numbers. Numbers on the green line y = x are equal to the sum of their factors (including one, but excluding themselves), and are therefore perfect numbers. Perfect numbers are much rarer than prime numbers in the entire set of natural numbers, as well as in this small sample.

If a number’s factor-sum, examined in this manner, is smaller than the number itself, such a number is called a “deficient number.” This applies to all numbers with points below the green line. Numbers which have points on the blue line are deficient numbers, as well as being prime numbers – and this is true for all prime numbers, no matter how large. The numbers represented by points between the green and blue lines are, therefore, both deficient and composite, and can also be called “non-prime deficient numbers.”

A few numbers on this graph, called “abundant numbers,” are represented by points above the green line, because their factor-sum is greater than the number itself. There are only five abundant numbers in this sample: 12, 18, 20, 24, and 30. As an example of how a number is determined to be abundant, consider the factors of 30: 1+2+3+5+6+10+15 = 42, which is, of course, greater than 30.

Of the 29 numbers examined in this sample, here is how they break down by category:

• Abundant numbers: 5 (~17.2% of the total)
• Perfect numbers: 2 (~6.9% of the total)
• Non-prime deficient numbers: 12 (~41.4% of the total)
• Prime numbers: 10 (~34.4% of the total)

These percentages only add up to 99.9%, due simply to rounding. Also, the total number of deficient numbers in this sample (both prime and composite) is 22, which is ~75.9% of the total sample of 29 numbers.

So what happens if this survey is extended far beyond the number 30, to analyze much larger (and therefore more meaningful) samples? Well, for one thing, the information on the graph above would quickly become too small to read, but that is only of trivial importance. More significantly, what would happen to the various percentages, for each category, given above? First, both prime and perfect numbers become more difficult to find, as larger and larger numbers are examined – so the percentages for these categories would shrink dramatically, especially the one for perfect numbers. With smaller percentages of prime and perfect numbers in much larger samples, the sum of the percentages for the other two categories (abundant and non-prime deficient numbers) would, of necessity, grow larger. That has to be true for this sum – but that says nothing about what would happen to its two individual components. My guess is that abundant numbers would become more common in larger samples . . . but since I have not yet examined the data, I’m only calling this a guess, not even a conjecture. As for what would happen to the percentage of non-prime deficient numbers when larger samples are analyzed, I don’t even (yet) have a guess.

## A BASIC Program To Factor Numbers Into Primes

### Image *** *** ***

10 print “For what number do you want the prime factorization”;
20 input n
25 c = 3
30 if n <> int(n) then end
40 if n < 4 then goto 450
50 if n/2 <> int(n/2) then goto 100
60 print ” 2″;
70 n = n/2
75 if n = 1 then goto 400
80 goto 50
100 if n/3 <> int(n/3) then goto 200
110 print ” 3″;
120 n = n/3
125 if n = 1 then goto 400
130 goto 100
200 c = c + 2
205 p = 0
210 for t = 3 to c step 2
220 if c/t = int(c/t) then goto 290
230 if n/t <> int(n/t) then goto 290
240 if p > 0 then goto 290
250 p = t
290 next t
295 if p = 0 then goto 200
300 print ” “;p;
310 n = n/p
320 if n = 1 then goto 400
330 if n/p = int(n/p) then goto 300
340 goto 200
400 print
410 goto 10
450 if n = 2 then print ” 2″;
460 if n = 3 then print ” 3″;
470 goto 400
500 end

## Octagonal Mandala II

### Image If the colors are inverted, but the background remains black, this is what it turns into: ## Starry Dual Polyhedron

### Image This is the dual of the polyhedron seen as the second image in the last post on this blog. If colored differently, so that only parallel faces have the same color, it looks like this (click to enlarge): I used Stella 4d to make these images, and you can find that program at http://www.software3d.com/Stella.php.

## Six Pairs of Parallel Decagons

### Image Each pair is a different color. Because these decagons intersect in space, but do not meet at edges, they do not form a true polyhedron. They are merely a symmetrical configuration of twelve decagons in space, surrounding a central point.

I made this out a “true polyhedron” by hiding all the other faces from view. Before the hiding and recoloring of faces, this looked this way (you can click on it to enlarge it): I used Stella 4d to make these images, and you can find that program at http://www.software3d.com/Stella.php.

## A Box, with Nothing In It

### Image Get it?

## Statement of the Obvious, in Venn Diagram Form

### Image ## I Have Found a (Possibly) “New” Point On the Euler Line — But I Also Need Help Nailing Down Its Properties and Definition.

### Image In this diagram, the original triangle is ABC, and is yellow. The brown line, u, is that triangle’s Euler Line, which contains the triangle’s orthocenter (O), circumcenter (K), centroid (R), and nine-point circle center (circle shown, centered at Q). The point I have found on the Euler Line is at W.

To find W, do the following: reflect triangle ABC over the Euler Line to form triangle A’B’C’ (shown uncolored, and with thick black edges). Both triangles, ABC and A’B’C’, have the same circumcircle (shown in green, with uncolored interior). Because of this, a cyclic hexagon may be formed by joining A, A’, B, B’, C, and C’ with segments, linking each in turn as one encounters them on a mental trip once around this circumcircle (the order in which these six points are encountered can change as A, B, and/or C are moved).

A hexagon has 9 diagonals. Of these, six are the sides of ABC and A’B’C’. Here, they are shown in black, and the other three diagonals are shown in red. These red diagonals are not necessarily concurrent, but any two of them do have to intersect, and those three intersections are points T, V, and W. At least one of those points — W, in this case — must be on the Euler Line. To get the other two points on the Euler Line, make the triangle approach regularity. As this is done, K, R, Q, O, and W converge, making the definition of the Euler Line itself problematical.

Point W needs a better definition. Which two of the three hexagon-diagonals which aren’t sides of the original triangle, nor its reflection, intersect on the Euler Line? I haven’t figured that out yet. Also, a formal proof for most of what I have described here is beyond my present abilities.

Why, then, do I believe the statements to be true? Answer: the evidence provided by experiment. This image is a screenshot from Geometer’s Sketchpad — but I don’t know how to post an animation of what happens when A, B, or C are moved. However, I can move them myself, with the program in operation, and observe how everything changes (this is one of the best features of Sketchpad, in my opinion). As these points are moved around, pairs of the heavy red segments (hexagon sides, and three of its diagonals) sometimes “flip” — a side becomes a diagonal, and that diagonal becomes a side. At that point, T, V, and W must be relabeled. Also, some positions of A, B, and C make the area of triangle TVW approaches zero — it collapses to a point on the Euler Line.

Odd things also happen if you make triangle ABC isosceles, because the Euler Line for an isosceles triangle is the perpendicular bisector of the base, which causes triangle A’B’C’, upon reflection of triangle ABC across the Euler Line, to map onto triangle ABC. When this happens, the hexagon becomes a single triangle, making its diagonals vanish — and point W goes and “hides” at the vertex opposite the base of isosceles triangle ABC, by which I mean W approaches that vertex as scalene triangles get closer to being isosceles.

Also, things change a bit if triangle ABC is obtuse: The Nagel Line (that line which contains the incenter, S, and the centroid, R, where it intersects the Euler Line) has been added, and is shown in purple. As you can see, point W is not on the Nagel Line. With the triangle being obtuse, the earlier all-red convex hexagon is now gone, because two of its sides are black, due to them being sides of triangles ABC and A’B’C’. Point W persists, though, still on the Euler Line, and located in the area between the vertices of these two triangles’ obtuse angles. My hope, in pointing this last fact out, is that it might help define which two hexagon-diagonals’ intersection defines the location of W. It might also be possible to use this to distinguish between the first diagram’s points T, V, and W, for, in the first diagram, W, which is what I am calling the only one of these three points to be on the Euler Line, was the one nearest the largest interior angle of triangle ABC — and the same is also true of triangle A’B’C’, as well. However, the matter of picking W out of the “T, V, and W” set of points may have nothing to do with angle size — it could be, instead, a matter of proximity of A, B, and C, as well as their reflections, to the Euler Line itself. In other words, “Which one is W?” might be answerable simply by examination of which member of the “T, V, and W” set is closest to the member of the set “A, B, and C,” as well as “A’, B’, and C’,” which is, itself, closest to the Euler Line. This matter needs further investigation, with which I would welcome help from anyone.

Also:  there are two easier-to-define points on the Euler Line, unlabeled in the diagrams above, which are the two points where triangles ABC and A’B’C’ intersect. The existence of these points on the Euler Line is simply a consequence of the fact that A’B’C’ was formed by reflecting the original triangle over the Euler Line. These two points could use special names, but nothing is immediately springing to my mind which would be appropriate. Another point on the Euler Line, also a consequence of reflection, appears as the midpoint of segment TV in the first diagram, and the midpoint of BB’ in the second — segments which appear analogous. This also seems to apply to the midpoints of AA’ and CC’. At this stage of the discovery process, though, appearances can be misleading.

I want to work out a better definition for this point, W, on the Euler Line, perhaps as an as-yet-undiscovered member of the large collection of triangle centers. I also need to know if it has already been found. If you have information pertinent either of these things, or to any part of this post, please leave it here, in a comment.