Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
1
Puzzle Solving Techniques – Algebra Meets Geometry
Preface:
A telephone company is about to lay the telephone lines for 100 houses and
is determining how much it will have to pay its contractor. The contractor charges
$1,000 for each telephone connection made (from one house to another) and
demands the sum to be paid upfront before work begins. Panicked about how to
determine the number of direct telephone connections to be made, the telephone
company turns to its engineer. How many connections does the engineer find out
need to be made? How much money does the contractor charge the telephone
company?
When facing a problem, I am often overwhelmed and find it hard to choose a
good starting point. Perhaps the reason is that there is no sure-fire method to solve
a problem because, by their very nature, problems are obstacles between data and
solutions. However, having a toolbox filled with a few techniques goes a long way
and if you fill your toolbox well, it is possible to solve almost any problem or puzzle.
Though I am still discovering new techniques and skills, I have a few methods that I
try on problems I am faced with.
One example of a problem that I often come across is a geometric figure with
one or more given data and one or more unknowns. For example, refer to problem
A.3 (Gardner, p.459) in the appendix of Gardner’s text. The problem shows a square
of side length 2, inside which is inscribed a circle, and inside of that is inscribed
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
another square, with sides parallel to the original square (although this last fact is
irrelevant). The problem asks for the area of the smaller square. The figure is as
follows:
2
Figure 1
My go-to method for a problem like this used to be using all the geometric
knowledge I have to deduce a few unknowns, such as the radius of the circle or the
side length of the inside square. To start, I see that because the circle is inscribed in
the square, the side length of the square is equal to the diameter of the circle (2
units). Knowing that the radius of a circle is half the diameter, I find the circle’s
radius to be 1. Up to this point, I could calculate either the area of the outside square
or circle, but those data seem unhelpful. Therefore I try to glean some information
about the inside square from what I already know. I see that the inside square is
inscribed in the circle, so I should be able to use that fact to find something related
to the side length of the inside square (like I did to find the radius of the inside
circle). It turns out that the radius of the circle is one-half the diagonal of the inside
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
square, so the diagonal of the square must be twice the radius, or 2 units. Now I am
pretty close to the answer; I know that the diagonal of the square relates to the side
lengths by Pythagoras’s Theorem (𝑎2 + 𝑏 2 = 𝑐 2 given a right triangle with legs ‘a’
and ‘b’ and hypotenuse ‘c’). The diagonal of the inside square is the hypotenuse and
the two legs are equal, so I rearrange Pythagoras’s Theorem to 2𝑎2 = 𝑐 2 . Because I
want to find the area of the inside square, I want to find ‘a,’ so I solve the equation to
find 𝑎 = √𝑐 2 /2 when ‘c’ = 2 . Now that I have ‘a,’ all I have to do is square it. The
answer to the problem is√
2
4
2
, or just 2 units.
Although I solved the problem correctly, there are a few places where I could
have employed more elegant thinking. I solved for the radius of the circle, but it was
unnecessary in solving the problem. I should have seen that the diagonal of the
inside square is equal to the diameter of the circle. When I solved for ‘a’ in
Pythagoras’s Theorem, the answer to the problem 𝑎2 , (the area of the square), was
staring me in the face. I should have solved for 𝑎2 by dividing both sides of the
equation by 2, plugging in 2 for ‘c,’ and avoiding any square roots. George Polya
discusses the merits of solving algebraically in his well-known text, “How to Solve It”
(Polya, p.14). Polya points out that by solving algebraically, one can skip over
unnecessary steps as well as understand the underlying concepts behind the
problem by manipulating the algebraic symbols. Using our equation 𝑎2 = 𝑐 2 /2, we
can solve for any given c (the side length of the outside square) value. Generalizing a
solution shows understanding of a problem beyond what is required. Furthermore,
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
we discover that it is a coincidence that the side length of the outside square (in
𝑢𝑛𝑖𝑡𝑠1 ) equals the area of the inside square (in 𝑢𝑛𝑖𝑡𝑠 2 ). This is only the case when
2
2
the length equals 2 because 2 ∗ 2 = 22 ( = 2).
2
Gardner’s book features another geometric puzzle that can be easily solved
with simple algebra and can be generalized for any case. Problem 5.4 asks for the
area of the hexagon that has the same perimeter as an equilateral triangle of area 2
(Gardner, p.108). We can use the fact that the two shapes have the same perimeter
to establish that the side length of the triangle is twice that of the hexagon. This is
because the hexagon has twice as many sides as the triangle and needs to evenly
distribute its perimeter among those sides. Let’s call the side length of the triangle 𝑙𝑡
and the side length of the hexagon 𝑙ℎ . We can relate these two variables by the
equation 𝑙𝑡 = 2𝑙ℎ . A regular hexagon is composed of six radially arranged equilateral
triangles and because we know that the sides of each equilateral triangles are half
that of the large equilateral triangle, we can conclude that their heights are also half
that of the large triangle (the triangles are similar). That is, ℎ𝑡 = 2ℎℎ , where ℎ𝑡
represents the height of the large triangle and ℎℎ represents the height of each small
triangle that composes the hexagon.
Now that we have relationships between data established, let’s use the area
1
formula of a triangle (𝑎 = 𝑏ℎ) to find the area of the hexagon. The area of the
2
6
hexagon will be 𝑙ℎ ℎℎ , or 6 times the area of its smaller triangles. Substituting the
2
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
hexagonal measures into the equation for the area of the large triangle leads to 2 ∗
1
2 ∗ 𝑙ℎ ℎℎ , or simply 2𝑙ℎ ℎℎ . Now that we have the areas of both the hexagon and
2
triangle expressed in terms of the same variables, let’s set up a ratio to determine
the area of the hexagon:
3
𝑎𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
𝑎ℎ𝑒𝑥𝑎𝑔𝑜𝑛
=
2𝑙ℎ ℎℎ
3𝑙ℎ ℎℎ
2
= . This can also be expressed
3
2
as 𝑎ℎ𝑒𝑥𝑎𝑔𝑜𝑛 = 𝑎𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 or 𝑎𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝑎ℎ𝑒𝑥𝑎𝑔𝑜𝑛 .
2
3
By dividing away the variables, we discover that the ratio of the area of a
triangle to the area of a hexagon with equal perimeters is 2/3. Therefore, the area of
the hexagon is 3. The reason this result is significant is that, like in the previous
problem, we can relate two seemingly unrelatable data to each other by a simple
number. Let’s say the area of the hexagon is 15; the area of the triangle will be 10.
We can corroborate this fact by a little mathematical reasoning. If the larger
triangle has twice the side length of the small triangles that compose the hexagon,
the large triangle can be divided into 4 smaller triangles. Because the large triangle
is composed of 4 of these small triangles (see Figure 2) while the hexagon is
6
3
4
2
composed of 6, the area of the hexagon will be , or the area of the triangle.
Figure 2
Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
6
Curiously, we just solved the same problem with much less work by drawing
a geometric figure, so perhaps algebra is not the best strategy for this problem. As I
cautioned earlier, not all techinques fit all problems. However, as of yet, algebra is
my favorite technique for solving the following problem.
A triangular number is a number that is defined by WolframAlpha as a
“number that can be represented in the form of a triangular grid of points where the
first row contains a single element and each subsequent row contains one more
element than the previous one.” A rack of 10 bowling pins is the 4th triangular
number; it is equal to 1 pin in the first row plus 2 in the second, 3 in the third, and 4
in the fourth. What is the nth triangular number? That is, how many bowling pins
does a rack of n rows contain? The fifth triangular number would be 10 + 5 and the
sixth, 15 + 6, but this doesn’t help to solve a general case.
To start, let’s think about the series 1 + 2 + 3 + 4 + 5 + 6 + 7. What
properties does it have? If the numbers increase by 1 going up the series, then they
must decrease by 1 coming down the series. 1 + 6 = 7, so 2 + 5 should equal 7 too
because 2 is one more than 1 and 5 is one less than 6. This applies to all numbers in
the series of triangular numbers. The first number plus the nth number equals the
second number plus the n – 1th number and so on. This means that the sum 𝑛 + 1
will be added
𝑛
2
times. But what if ‘n’ is odd? The middle number won’t have a
counterpart to add to create the sum 𝑛 + 1. One way to have every number in the
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
triangular series have a counterpart is to add the list to the reverse of the list, and
add the numbers as follows:
1
2
3
4
5
6
7
+
+
+
+
+
+
+
n-1
n-2
n-3
n-4
n-5
n-6
n
The result of this is n terms of 𝑛 + 1, or 𝑛(𝑛 + 1). To account for the doubling
of the list, we need to divide by 2 and our final expression is
𝑛(𝑛+1)
2
.
Although it may seem simple or uninteresting, our result has some very
interesting implications. How is it that our formula can handle division by 2 if n is
odd and still result in an integer? If n is odd, then n + 1 must be even and if n is even,
n + 1 must be odd. By solving algebraically, we have generalized our result and by
using this formula, anybody could easily calculate the nth triangular number. If you
were at a bowling alley with 6 rows, there would be
7∗6
2
= 21 pins, not at all
surprisingly the 6th triangular number. But if you arrived at the alley and your friend
told you that there were 120 pins, would you be able to surmise how many rows
there were? Thanks to algebra, we can.
Let’s call 𝑝 the total number of pins arranged in triangular formation of 𝑛
rows. We know that 𝑝 =
𝑛(𝑛+1)
2
, so 2𝑝 = 𝑛(𝑛 + 1). Any student who has learned
algebra of quadratic functions would recognize the equation 2𝑝 = 𝑛(𝑛 + 1) as
quadratic, although it is much more obvious in the form 𝑛2 + 𝑛 − 2𝑝 = 0. For those
that haven’t taken algebra of quadratic functions, a quadratic function is simply a
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
function whose highest degree is 2 and is often written in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 =
0. Quadratic functions can be solved by the equation 𝑥 =
−𝑏±√4𝑎𝑐−𝑏2
2𝑎
provided
that √4𝑎𝑐 − 𝑏 2 is rational. (A quick sidenote about the quadratic equation: it can
either produce 2 answers, 1 answer, or 0, in which case the answer is an unreal
number. Unreal numbers, as their name suggests, can get very confusing.) In our
case, 𝑎 = 1, 𝑏 = 2, 𝑐 = 2𝑝, so we can plug in a, b, and c to to obtain a result. After
simplifying the result, 𝑛 =
√8𝑝+1−1
2
; we now have a function that will tell us n as a
function of p. (We ignored the negative result because the triangular series is
positive.) Plugging in 120 for p, we obtain 𝑛 = 15; the pins are arranged triangularly
in 15 rows!
Let’s take a look at the radical sign in our equation 𝑛 =
√8𝑝+1−1
2
. If we know
that n can only be an integer (triangular numbers are arranged in full rows) it must
mean that 8𝑝 + 1 (where p is a triangular number) is a square number because the
square root of 8𝑝 + 1 must result in an integer. Furthermore, √8𝑝 + 1 must be odd,
because 1 is subtracted from it and divided by 2, resulting in an integer. While there
is a geometric way to come up with the formula 𝑝 =
the formula 𝑛 =
√8𝑝+1−1
2
𝑛(𝑛+1)
2
for triangular numbers,
and its many implications are a product of algebra.
Although geometry can very eloquently define spaces and objects, algebra is much
better at defining numbers and how they relate. The reason that triangular numbers
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
are so interesting is that they are the cross section between a geometrically defined
figure and algebraic manipulations of the series that describe the figure.
There are countless subtleties and applications of triangular numbers. Pierre
Fermat, who is famous for Fermat’s Last Theorem (which remained unproven for
358 years after his death), put forth a theorem that goes as follows: “every positive
integer is a sum of at most three triangular numbers, four square numbers, five
pentagonal numbers, and n n-polygonal numbers” (Weisstein). Pentagonal and
hexagonal numbers are just two examples of the wide world of n-polygonal
numbers. It also turns out that any square number can be represented as the sum of
two consecutive triangular numbers, as Figure 3 demonstrates:
15 + 10 = 25
Figure 3
Now that we have conducted a significant exploration of using algebra to
accomplish geometric and non-geometric tasks, let’s return to the preface problem.
Figure 4 demonstrates three simplified versions of the problem:
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
What do you notice about the number of connections made in the 2- and 5household networks? The network with 2 households is connected by 1 wire and
the network with 5 households is connected by 10 wires. That doesn’t seem fairly
significant. If this were a triangular number problem, the 2-household network
would be connected by 3 wires and the 5-household network would be connected
with 15 wires. But 1 and 10 are triangular numbers, of 1 and 4 respectively. Hold on;
1 and 4 are one less than 2 and 5, which are the number of households (or nodes,
mathematically) in the networks. Could it be that the number of wires to be laid is
defined by the triangular number of the number of households minus 1? It certainly
works for 2 and 5, and as it turns out, it works for 12 as well (12 − 1 = 11, 11 ∗
12
2
=
66, the number of connections in the 12-node network).
But how can we explain this? Although this problem involves triangular
numbers, there are no pictures that resemble arranged bowling pins. The key is in
the addition of a new house to a network. With each new house added, the number
of wires laid is equal to the number of existing houses (because the new house
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
needs to be connected to each of the old ones). When a sixth house is added, for
example, 5 new connections need to be made and the new total number of houses is
6. This is, in fact, is proof that the number of connections made between n houses
will be described by the triangular number 𝑛 − 1, or
𝑛(𝑛−1)
2
. From this point on, the
rest of the problem is easy. The telephone engineer has to determine the number of
connections required between 100 houses, so he will use the 99th triangular
number. Thankfully he has a calculator and rapidly computes
99∗100
2
to be 4,950.
These 4,950 connections will cost $1,000 each for a total of $4,950,000. It looks like
cell phones are the way to go.
Throughout this exploration of the crossroads of algebra and geometry, I
have given little indication of which technique I prefer to solve problems. It turns
out that each technique has certain advantages and disadvantages. But by using
algebra and geometry in tandem, it is possible to explore shapes, spaces, and
numbers with incredible ease and elegance. Mankind has always linked algebra and
geometry, from trigonometry, to calculus of several variables, to graphs of functions.
It is because of this fact that the two subjects work so well together and comprise a
formidable puzzle solving technique to add to anybody’s toolbox.
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Chandler Gregg
5/8/13
Professor Winkler
Math 7 – 13S
Works Cited
Figure 1:
http://www.mathteacherctk.com/blog/wpcontent/uploads/2010/06/SquareCirc
leSquare.jpg
Figure 2:
http://www.cs.princeton.edu/courses/archive/fall98/cs341/solutions/f_2_1.jpg
Figure 3:
http://en.wikipedia.org/wiki/File:Square_number_25_as_sum_of_two_triangular_nu
mbers.svg
Figure 4:
http://en.wikipedia.org/wiki/File:Metcalfe-Network-Effect.svg
Gardner, Martin. The Colossal Book of Short Puzzles and Problems. New York: W. W.
Norton & Company, 2006.
Pólya, George. How to Solve It: A New Aspect of Mathematical Method. New York:
Ishi Press, 2009.
Weisstein, Eric W. "Fermat's Polygonal Number Theorem." From MathWorld—A
Wolfram Web Resource.
http://mathworld.wolfram.com/FermatsPolygonalNumberTheorem.html