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Triangles in a row

My friend Prof Vlasta Dlab (Ottawa-Bzí) sent me his solution (which will appear in a Czech MathPhys Journal) to the following puzzle (well, its generalization from \(4\) to \(n\) triangles):

He was deeply dissatisfied with a solution of this puzzle on YouTube (posted in January 2021) because a part of that derivation is a blind application of a geometric formula (a sine formula for triangular areas) which doesn't really reveal the simplicity of the situation and its independence of angles, precise shapes... I completely agree with Vlasta. While the YouTube solution looks short, it uses an excessively heavy and irrelevant tool, and the person using these methods would probably waste too much extra time if he were supposed to solve a nearly isomorphic problem.

A good mathematician should look what matters and what doesn't matter. And an important point about this calculation of areas is that the angles don't matter at all! Correspondingly, you shouldn't be using the Pythagorean theorem, a sine formula for triangular areas, Heron's formula, and similar tools that deal with angles and, almost equivalently, with ratios of lengths of line intervals that go in different directions. In a good, conceptual solution, the "square roots of three" and similar things shouldn't appear at all (while they appear in the YouTube video).

Fine, here is my solution. First, the verbal formulation of the problem. The area of one equilateral triangle is \(S=6\). You place \(n=4\) of them on a horizontal line, those are triangles \(A_0 A_1 B_1\) up to \(A_{n-1} A_n B_n\). Then you connect \(A_0\) with \(B_n\) with a tilted line, it cuts yellow pieces from \(n-1=3\) triangles, and your task is to calculate the total area of the cut yellow pieces. Obviously, as good enough mathematicians who want to think generally if there is no reason not to, we will be solving it for a general \(n\), not just \(n=4\), and I will also keep the symbol \(S\) for the single triangular area.

Now, my solution starts (Vlasta's solution is similar, dealing with similar triangles etc., and surely independent of angles). First, we need to know what we should get approximately. The yellow area is taking pieces from \(n-1\) or \(3\) equilateral triangles. Which fraction? The fraction of the yellow area depends on the triangle but it is less than \(1/2\). Well, for \(k=1,2,\dots,n\) labeling the component of the yellow area, the fraction goes like \(ck^2\) from \(0\) to \(1\), and because the integral of \(k^2\) from \(0\) to \(1\) is \(1/3\), the total area of the yellow polygons will be about \(nS/3\) for a large enough \(n\).

Oops, I need to post Vlasta's basic diagram to clarify the names of the points that I also use.

We want the precise answer now.

Only here, the real calculation begins. My calculation is straightforward. The yellow triangles are similar to each other; and their linear dimensions are linearly increasing (just look at the distances of the upper vertices \(B_k\) from the last \(B_n\), the center of similarity). Areas are proportional to the squared linear distances. That is why the final result (total yellow area \(S_Y\)) is equal to the product of the largest yellow triangle times\[ \kappa=\frac{0^2 + 1^2 + 2^2 + \dots + (n-1)^2}{(n-1)^2} = \frac{n(n-1/2)}{3(n-1)} \] To find the numerator, Google search for "sum of squares of integers" or prove my formula by mathematical induction. For \(n=4\), you may compute this sum of fractions manually, \((1^2+2^2+3^2)/3^2 = 14/9\).

We need to calculate the area of the largest yellow triangle \(S_{L}\). But that triangle \(L\) is just the difference of the single equilateral triangle \(S\) and the bottom piece triangle \(A_0 A_1 K_1\). But the latter is similar to the triangle \(A_0 A_n B_n\) whose area is \(nS\) because it has the same height but \(n\) times wider base than \(S\). Because \(A_0 A_1 K_1\) has a shorter base, by a factor of \(n\) because it's the same width as the normal triangle, the area of \(A_0 A_1 K_1\) is simply \(nS/n^2 = S/n\). Subtract it from a single \(S\) to get\[ S_{L} = S \cdot \zav{1 - \frac{1}{n}} = S\cdot \frac{n-1}{n}. \] To get the total yellow area, we just multiply\[ S_{\rm yellow,total} = \kappa \cdot S_{L} = S\cdot \frac{n-1/2}{3} \] because \(n\) and \(n-1\) canceled. Oops, I could have canceled them right away and avoid the subtraction from \(S\) as well if my area unit were the smallest yellow triangle and not the largest one LOL. At any rate: Substitute \(n=4\) to get \(7S/6\) or \[ {\Huge {\rm AREA = 7}} \] for \(S=6\).

I would claim that the short derivation in between the two horizontal lines above is everything you need. No trigonometry, sine formulae for areas, Pythagorean theorems, Heron's formulae etc. I could have made the derivation even shorter by reducing the font for the result \(7\) LOL. The simplified \(n=4\), \(S=6\) calculation of the area would be a one-line process (whose justification is explained above)\[ \frac{1^2+2^2+3^2}{3^2} \cdot 6 \cdot \zav{ 1 - \frac 14} = 7 \] The pretty MathJax \(\rm\LaTeX\) mathematics (currently MathJax 3.1.2) is already fastest in the world. The mobile template page is completed within half a second on my PC.

Let me say another thing. It's nice to use the simplest tools and one may find some cleverness, beauty, or pattern recognition in people's methods when they do so. On the other hand, I do believe that it is only a "clever approach to teaching" if we assume that the student will remain stuck at a lower stage of his or her understanding of mathematics. And at that lower stage, or even when he is not stuck, he may brag about a clever elegant idea. But the ability to do things cleverly can't quite be taught. I think that schools should teach powerful tools that the students may get even if these tools will often be more powerful than needed to solve problems such as this one. Powerful tools do belong to mathematics, they may be useful to do really wonderful things, and shooting an ant with a cannon is sometimes fun, too!

Almost as much fun as shooting an innocent, gullible cat with the water from a Kofola bottle.

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