Popular Mechanics - USA (2020-07)

THE

MATHEMATICAL

PROOF

Figure 1

Figure 2

Figure 3

When Slobodan Jaksic

stumbled upon our brain

teaser, the software

engineer living in Serbia

counted the triangles

sideways along the x-axis

and vertically along the

y-axis before assembling

his proof. As such, it

breaks down into hori-

zontal and vertical steps.

“I sent my ‘concep-

tual solution’ using

Mathematical Induction

to Francis Bonahon, Pro-

fessor of Mathematics,

University of Southern

California,” Jaksic says.

“He provided...his solu-

tion to the problem, with

a correct formula, using a

different approach.” Let

that be a reminder that,

even among the pros,

there are always multiple

ways to approach a

problem.

When we published the problem on Pop Mech,
readers started sending in their own solutions.
Soft ware engineer Slobodan Jaksic came up with
a mathematical proof to demystify the triangle
teaser. It goes like this:
Two lines meet at the top vertex denoted by
“A,” and the segment line connecting those two
lines is the “base.” The new polygon is the fun-
damental triangle. We can place an arbitrary
number of points on the base and connect them
to the top vertex A.
Assuming there are “n” vertices on the base,
where n is a natural number greater than or
equal to 2, we will prove that the new shape—let’s
call it “pyramid”—contains a number of trian-
gles equal to S(n) = n(n–1) ½. (See Fig. 1) For n
= 2, S(2) = (2 1) ½ = 1. It is correct.
Assuming the formula S(n) holds for n = k,
with k greater than or equal to 2, S(k) = k(k–1)
½. We will prove that S(n), where n = k + 1, holds
as well. (See Fig. 2)
In the modified pyramid (See Fig. 3), there
are “k” additional new triangles. Therefore,
S(k+1) = S(k) + k = (k(k–1) ½) + k = (k+1)
k ½.
Modify the pyramid by extending both edge
lines, and all lines between them connect to the
top vertex A, beyond the base, in the direction
opposite of the top vertex. Create new bases by
connecting one or more segments parallel to the
starting base.
Assuming there are a total of “m” bases in the
pyramid, we can prove the formula calculating
the number of triangles in the structure is: S(n,
m) = n(n–1) ½ * m.

Let’s prove that the formula holds for m = p +

1. (See Fig. 3) The number of new triangles can
   be given as: (n–1) + ... + 2 + 1 = n(n–1) ½. Con-
   sequently, S (n, p+1) = S(n, p) + (n(n–1) ½) =
   (n(n–1) ½) p + (n(n–1) ½) = (n(n–1) ½) (p

• 1).
  According to the Principles of Mathemati-
  cal Induction, we just proved the total number
  of triangles in the pyramid is given through the
  formula S(n, m) = n (n–1) ½ m, where n and m
  are natural numbers, n is greater than or equal
  to 2, n is an arbitrary number of vertices, and m
  is an arbitrary number of bases.
  Hat tip to Slobodan. Here I was just trying to
  annoy my coworkers.

A

A1 A2 A3 An

A1 A2 A3

A

AK AK+1

A

B1

A1 A2 A3

B2 B3

Y1 Y2 Y3

X1 X2 X3

b1

b2

b3

bp

bpm

Am

Bm

Xm

Ym