There are some great tools available to anyone who wants to do math, regardless of the level involved. And whether it’s a calculator, spreadsheet or software package, they don’t just cater to time-savers. Many are programmable and therefore thought-provoking. But in the old days of hand-calculations, people did not necessarily surrender to tedium. They often came up with shortcuts, some of which were excellent approximations, while others gave an exact result. Let’s look at an example of each variety.

Whether you’re on a plane, hill or mountain, if you know your altitude, what is the easiest way of calculating the distance to the horizon?

It turns out you don’t have to resort to the tangent to a circle and the Pythagorean theorem. The shortcut consists of multiplying the altitude in feet by 1.5 (its units not shown) and to square root the product. The answer is a close approximation of the distance in miles!

To understand why the shortcut works we need to look at the longer and more accurate way of solving the problem. The distance, d, from an altitude, h, is simply the leg of a right-angled triangle. One angle is 90 degrees because the line of sight is tangent to the radius (R) of the earth, an approximate circle. Finally the hypotenuse is the sum of R and h.

Since d² + R² = (h+R)², after expanding and eliminating the common R² term, we obtain:

d² = h² + 2hR

The radius of the earth is about 3959 miles and there are 5280 feet in a mile, so plugging in R and solving for d we obtain

d = √( h² +  41 807 040 h ) in feet.

But the trick claims that d =√(1.5h) in miles. Converting to  feet ,

d =5280√(1.5h).

Equating the two expressions for d:

√( h² +  41 807 040 h) = 5280√(1.5h).

Squaring both sides:

h² +  41 807 040 h = 5280²(1.5)h.

Dividing through by h and simplifying the right hand side:

h +  41 807 040  = 41 817 600.

For any h value between 1 and 40 000 feet, the left hand sum will differ from the right hand side by a maximum of 0.07%. And so the trick works well. But how was it devised?

My guess is that if someone plotted various values of height in feet versus the distance in miles using the Pythagorean expression, she would have noticed a square root function going through the origin of the form y = a√h, where a is constant. To verify it, she likely calculated successive changes in distance per changes in the square roots of height. I’ve tried that and I obtained values in the neighborhood of 1.224 and 1.225. Then all she had to do was recognize that those values are pretty close to the square root of 1.5.

The second shortcut is a simple algorithm for computing averages mentally.  I came up with it independently when I was a high school student, and needless to say this trick has also occurred to many other people!

Instead of adding up values x1, x2, x3 …. xn and dividing by the number of values, n,  if you have no calculator, we begin with a convenient estimate. Then add up the differences from the estimate, divide the sum by n and add the result to the estimated value. For example, for the average of 78, 82, 97, 60 and 77, we could use an estimate of 80. The differences from the values are -2, +2, +17, -20 and -3. Mentally the numbers are smaller so it’s easy to come up with the sum of -6. Finally the average is -6/5 + 80 = -1.2 + 80 = 78.8, also a very feasible mental calculation.

Here’s why it works. Let E = estimate for the average(Xavg) of an n number of values: x1, x2, x3…xn

The first step involves adding up the differences (call the sum Sd) from E:

x1 – E  + x2 – E + x3 – E… + xn – E = Sd =  x1 + x2 + x3 … + xnnE

The we divide through by the sum by n:

Sd/n = (x1 + x2 + x3 … + xn nE)/n = (x1 + x2 + x3 … + xn)/nnE /n.

But on the right hand side , the first term is the definition of average, so:

Sd/n = Xavg – E

Or  Xavg =  Sd/n + E

And how did I and others devise such a trick? Intuitively, at least that’s how I came up with it, from what I recall. That’s is even more fun than trying to figure out how someone else stumbled upon it.

In the diagram below, in between two circles, each with a radius of 1 meter, three smaller circles are squeezed in, centred at C1 , C2 and C3. If you imagine it was possible to continue drawing and squeezing in more circles until you had a billion of them, how much space would be left between the top of the smallest circle and point A? And what ubiquitous thing’s diameter is very close to that distance?

This is a variation of an old problem, and although the situation is often used as an example of a non-geometric series in calculus books, it could also be solved with some basic algebra and induction. And for that reason, versions of the puzzle have appeared in math competitions for bright kids who have yet to venture into calculus.

Let’s begin by solving for the radius of the first circle (the red one centered at C1). If we come with expressions or numbers for all three sides,  we could apply the Pythagorean theorem to triangle ABC1.

Notice that AC1 ‘s length can simply be obtained by subtracting the radius of the red circle from the 1-meter length of the square’s side, which we obtain from the radius of the large, identical circles.

AC1 = 1 – r1.

Since AB² + AC1 ² = BC1 ²,  where the hypotenuse, BC1 , is simply the sum of the radii, we obtain:

1² + (1 – r1) ² = (1 + r1) ²

Solving for r1 yields  r1 = 1/4 so the red circle’s diameter = d1  = 2 * 1/4 =  1/2 .

Not to bore you, we will walk through the steps only one more time to obtain the diameter, d2, of the blue circle centered at C2.

We add the red circle’s obtained diameter of  1/2 to the radius of  the blue circle, r2 , and then subtract the sum from 1 meter to obtain AC2.

AC2 = 1 – (1/2 + r2 )=   1/2 – r2

Then we apply the Pythagorean theorem to the second triangle, ABC2,

AB² + AC2 ² = BC2 ²,  or:

1² + ( 1/2 – r2) ² = (1 + r2) ²

1 + 1/4 – r2 + r2 ² = 1 + 2r2 + r2 ²

1/4 = 3r2

r2 = 1/12

so the blue circle’s diameter = 2 * 1/12 = d2 = 1/6.

As promised, without going through the similar details, d3 = 1/12 and for a fourth circle that we squeeze in, d4 will be 1/20.

Now a pattern becomes apparent. For the diameter of the nth circle, dn,  dn = 1/n(n+1). But if you did not realize that, you could still solve the problem. After all, we’re interested in the leftover distance between the top of the billionth circle and point A.

After we squeezed in one circle , we had a diameter of  1/2 . After squeezing in a pair, the sum of the two circles’ diameters 1/21/6 = 2/3. Squeeze in three circles and the sum is  1/21/6 + 1/12 = 3/4 .  A fourth insertion yields 1/21/6 + 1/12 + 1/20 = 48/60 = 4/5.

So for squeezing in an n-number of  circles, the sum of the diameters is n/n+1 and since the side of the square is 1 meter in length, the remaining distance will be 1 – n/n+1. Using a common denominator, that expression is equal to n+1 -n/n+1 = 1/n+1.  For a billion circles then, the leftover distance would be

1/1 000 000 000 +1 = 1/1 000 000 001 meters,

which is extremely close to being to a nanometer.

1/1 000 000 000 meters or 1 nanometer, is a little more than three times the size of a single water molecule, and just slightly wider than the ring of one glucose molecule, the fuel for all brains, including those that devised and solved this puzzle. And use up just an extra iota of  glucose to realize that the leftover space would be long enough but not wide enough to accommodate the entire molecule. Finally, imagine a circle for every person alive on Earth in 2050, about 9.4 billion, and the distance remaining between the smallest circle and point A will only as big as the diameter of the smallest and most common atom in the universe.

Whether you’re a researcher, programmer, doctor or professional athlete, most of the satisfaction you derive from work probably comes from uncovering some truth, problem-solving, treating the ill and mastering a sport, respectively. Money is not the main motivator as long as there’s enough to raise a family.  But here’s where things get complicated. If you live in a society where basic needs such as shelter, healthcare and education begin to get prohibitively expensive, people with highly specialized skills will demand much higher salaries and usually obtain them. But those whose skills are more commonplace are not in the same privileged position. They have to settle for less because it’s easier for someone else to come along and accept a compromised offer. The majority of people consequently have an increasingly difficult time coping if the daily essentials can only be attained by precariously accumulating personal debt.

Some governments are fully aware of this dilemma. It’s the reason they tax those with higher incomes. The revenues are used to provide the entire population with affordable education and health care. For the most disadvantaged, housing is also subsidized.  But most governments of the world are not committed to such a path. And even among those who are, the commitment has wavered, resulting in a widening gap between rich and poor and more poor people.  As mentioned by the Conference Board of Canada, some economists argue that institutions have blocked income-disparity mitigation by allowing lower unionization rates, low minimum wages, deregulation and national policies that favor the rich. This pattern is not only of concern from an ethical standpoint, but it’s also socially and economically destabilizing.

Income inequality in Canada has increased from 1990 to 2010. This reverses the trend experienced from 1970 to 1990. Although we are doing better than the United States where inequality has increased consistently since the 1970s, a minority of countries like Norway and Belgium have done better than Canada and bucked the common trend.

But just how is income inequality measured? We begin with an idealized situation of total equality where any percentage of the population equals the percentage of net income earned. In other words if you are looking at half the population, then they would earn half of all income. And in such a society if you considered only a tiny 2% slice, then they would be receiving only 2% of the money-pie. That one-to-one correspondence of income fraction on the y– axis and population fraction on the x -axis generates the simple identity function y = x.

On the other hand, the Lorenz function ( L(x) ) describes the actual distribution of income. For example in the United States in 2008, 80% of the population took home only half of all earnings. Only 3.4% of earnings went to a sizable chunk ( 20% ) of Americans. When several points are plotted, a curve is obtained  representing L(x). A 20th century statistician named Corrado Gini decided to calculate the area between the two functions and divide it by the area under y = x. The result is called the Gini coefficient. The bigger the number, the more income disparity there is.

If you enjoy math, you are probably wondering how the area under the curve is calculated. A simple method lies in googling “power function fit” and an online java program will use the data points including the previously mentioned (0.2, 0.034) and (0.80, 0.50) to generate an equation of the form L(x) =axb . In general terms, the sandwiched area in between the two functions is found by integrating the difference between the identity function and the curve.

The G coefficient according to its definition is:

But since the region under y = x is just a triangle with an altitude of 1 and base of 1, its area = 0.5 units, so

By replacing L(x) with the power function that was fit to the curve, and after evaluating the integral, we obtain a Gini coefficient for the United States in 2014 of 0.390. Among OECD (Organization for Economic Cooperation and Development) countries only Mexico, Chile and Turkey fared worse. Iceland was the fairest country with a Gini coefficient of 0.246. Canada was in the middle of the pack at 0.313. The OECD site also points out that the average Gini coefficient for its 35 countries is the highest on record since they started to measure it over 30 years ago.