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Far East Journal of Mathematical Education
© 2013 Pushpa Publishing House, Allahabad, India
Published Online: January 2014
Available online at http://pphmj.com/journals/fjme.htm
Volume 11, Number 2, 2013, Pages 129-161
Received: June 26, 2013; Accepted: August 7, 2013
Keywords and phrases: mathematical proof, problem solving, multiple solutions and proofs,
DGS - dynamic geometry software, teacher education.
*Corresponding author
ONE PROBLEM, MULTIPLE SOLUTIONS:
HOW MULTIPLE PROOFS CAN CONNECT
SEVERAL AREAS OF MATHEMATICS
Moshe Stupel1 and David Ben-Chaim2,*
1Shaanan Religious Teachers’ Training College
Haifa, Israel
2The Technion
Israel Institute of Technology
Haifa, Israel
2Technion R & D Foundation
Canada Building, 1st Floor
Technion City, Haifa 32000, Israel
e-mail: davidb@cc.technion.ac.il
Abstract
Problem solving and proofs have always played a major role in
mathematics. They are, in fact, the heart and soul of the discipline.
Even more so, using a number of different proof techniques - from
within and between several domains of mathematics - for one specific
problem can demonstrate the connections between these domains, as
well as the wealth, beauty, and elegance of mathematics. We present
one specific, interesting geometry problem, and present nine different
proofs for it, using methods from a wide variety of different
mathematical domains, including using dynamic geometry software
Moshe Stupel and David Ben-Chaim
130
(DGS) application. Encouraging high school as well as college
students to derive results in such various ways will enhance their
appreciation of mathematics and give them incentive to derive even
more elegant solutions on their own. After conduction of a case study
that involved a course on this topic as a part of a pre-service
mathematics teacher education program (including student feedback
via questionnaire and interviews), it was concluded that mathematics
educators should be encouraged to introduce many authentic multiple-
proof problems into their teaching program. As a bonus, the problem
we used in this report provided a pleasant surprise in that it revealed a
special triangle with interesting properties.
1. Introduction
This article aims to exhibit the extent to which different high school and
college mathematical topics are connected to each other, specifically, how
Euclidean geometry serves as a base on which are built analytic geometry,
trigonometry, vectors, and complex numbers.
From a didactical-pedagogical view of point, this article presents a
specific mathematical problem that at first glance seems to be one that should
be solved only with Euclidean geometry. We shall subsequently show,
however, that it can also be solved by several other methods, some
geometrical and others based on other mathematical areas. In addition, we
will report on the incorporation of the multiple proofs topic within the
teacher education program of our teacher college.
The existence of such multiple-proof solutions not only demonstrate to
students the beauty and aesthetics of mathematics [12, 41, 42] but also
enhance the students’ mathematical understanding by presenting the material
from different points of view and including using dynamic geometry
software (DGS).
In addition, the problem presented in this paper reveals a surprising
result: a special triangle with interesting properties that, to the best of our
knowledge, has not been presented yet in the professional literature.
Multiple Solutions: Bridging Areas in Mathematics 131
2. Proofs in Mathematics
Proof and reasoning are fundamental in solving mathematical problems:
they help us make sense of the mathematics, aid in communicating
mathematical ideas, and justify the validity of mathematical theorems [31].
One would never doubt the importance of proof in mathematics, in
general, nor in school mathematics [19]. The roles of proof are to prove, to
explain, and to convince [16, 21]. Stylianides [44] defines proof as
“A mathematical argument, a connected sequence of assertions for or
against a mathematical claim, with the following characteristics:
1. It uses statements accepted by the classroom community (a set of
accepted statements) that are true and available without further
justification.
2. It employs forms of reasoning (modes of argumentation) that are
valid and known to, or within the conceptual reach of, the classroom
community.
3. It is communicated with forms of expression (modes of argument
representation) that are appropriate and known to, or within the
conceptual reach of, the classroom community [44, p. 291]”.
Stylianides uses this definition to analyze instruction involving proof and
to illuminate possible actions teachers may take to support proving activities
in their classrooms. Lo and McCrory [29] proposed that in order to
understand proving activities in mathematics courses, a fourth element in the
definition should be added:
4. The proof is relative to objectives within the context (context
dependence) which determine what needs to be proved.
Rav [35] indicates that a proof is valuable not only because it
demonstrates a result but also because it may display fresh methods, tools,
strategies, and concepts that are of wider applicability in mathematics
and open up new mathematical directions. In Rav’s view, proofs are
Moshe Stupel and David Ben-Chaim
132
indispensable to the broadening of mathematical knowledge and are, in fact,
“the heart of mathematics, the royal road to creating analytic tools and
catalyzing growth” [35, p. 6]. Rav stated in his thesis that “proofs, rather than
the statement-form of theorems, are bearers of mathematical knowledge”
[35, p. 20]. For additional discussion of Rav’s paper [35], see Hanna and
Barbeau [18] who examine Rav’s central idea on proof, and then discuss its
significance for mathematics education, in general, and for the teaching of
proofs, in particular.
Hemmi and Löfwall [20], after a study in which they explored
mathematicians’ views on the benefits of studying proofs, concluded that “all
mathematicians in the study considered proofs valuable for students because
they offer students new methods, important concepts, and exercise in logical
reasoning needed in problem solving” [20, p. 201]. It is thus no wonder that
developing students’ ability to prove and reason is one goal of the curricular
standards in many countries. For example, one principle for American school
mathematics is: “reasoning and proof should be a consistent part of students’
mathematical experiences in prekindergarten through grade 12” [32].
Learning mathematics is fundamentally about “acquiring a mathematical
point of view”, “developing mathematical reasoning”, “learning to
communicate mathematically”, “making connections” in mathematics, and
building “connections” with other disciplines and (among mathematical)
experiences [32, p. 56]. Because learning mathematics involves discovery,
proof and reasoning are powerful ways of developing insight, making
connections, and communicating mathematically. NTCM underlines this fact
by claiming: “being able to reason is essential to understanding” [32, p. 56].
This suggests that proficiency in mathematical proof and reasoning is an
integral part of mathematics.
Similar to NCTM, mathematics educational researchers also argue that
they should not undervalue the role of argumentation and proofs in students’
learning and support the idea of including this subject in the school curricula
and in teacher education programs. For example, Ball et al. [3], Dreyfus [11]
and Hanna [17] support this position:
Multiple Solutions: Bridging Areas in Mathematics 133
“Proof is central to mathematics and as such should be a key component
of mathematics education. This emphasis can be justified not only
because proof is at the heart of mathematical practice but also because it
is an essential tool for promoting mathematical understanding” [3, IX
ICME].
“Proof is at the heart of mathematics, and is considered central in many
high school curricula” [11].
“It maintains that proof deserves a prominent place in the curriculum
because it continues to be a central feature of mathematics itself, as the
preferred method of verification, and because it is a valuable tool for
promoting mathematical understanding” [17].
3. One Problem, Multiple Solutions and Proofs
Mathematics educators agree that linking mathematical ideas by using
more than one approach to solving the same problem (e.g., proving the same
statement) is an essential element in the development of mathematical
reasoning [32, 33, 36]. Problem solving in different ways requires and
develops mathematical knowledge [33], and encourages flexibility and
creativity in the individual’s mathematical thinking [24, 28, 39, 45].
In addition to the specific roles of proof in mathematics, we suggest that
attempts to also prove a certain result (or solve a problem) using methods
from several other different areas of mathematics (geometry, trigonometry,
analytic geometry, vectors, complex numbers, etc.) are very important in
developing deeper mathematical understanding, creativity, and appreciation
of the value of argumentation and proof in learning different topics of
mathematics. Our approach, that of presenting multiple proofs to the same
problem, as a device for constructing mathematical connections is supported
by Polya [33, 34], Schoenfeld [37], NCTM [32], Ersoz [13] and Levav-
Waynberg and Leikin [26].
Very similar to our notion of ‘one problem, multiple solutions/proofs’
is the idea of multiple solution tasks (MSTs) presented by Leikin and Lev
[28], Leikin [27] and Levav-Waynberg and Leikin [26]. MSTs contain an
Moshe Stupel and David Ben-Chaim
134
explicit requirement for proving a statement in multiple ways. Leikin [27]
indicates that the differences between the proofs are based on using: (1)
different representations of a mathematical concept; (2) different properties
(definitions or theorems) of mathematical concepts from a particular
mathematical topic; (3) different mathematics tools and theorems from
different branches of mathematics; or (4) different tools and theorems from
different subjects (not necessarily mathematics). In our case, we apply the
third type of differences between the proofs; we shall present various
solutions to a problem using the tools and theorems of Euclidean geometry,
analytic geometry, trigonometry, vectors, and complex numbers.
Adding the concept of multiple solutions/proofs for one problem into the
curriculum of mathematics studies as well as MSTs allows the development
of connected mathematical knowledge not only for students but also for their
teachers as well.
4. The Theorem1
Given: a triangle ABC
(
)
.;; cABbACaBC
=
=
=
From vertex A, the
triangle is divided by constructing three special line segments as follows:
()
hAD is the altitude to BC;
(
)
lAN is the angle bisector of ;A
∠
and
()
mAM is the median to BC. See Figure 1.
According to Figure 1, it is obvious that ΔBAN is an isosceles triangle,
hence:
cANAB === 1 and .xDNBD
=
=
Also, let .tNM
=
Then =MC
.2 tx + Given that ,
α
=
∠
=
∠
=
∠
=
∠MACNAMDANBAD what is the
value of α?
1The original source of the problem presented herein is derived from a query made by a high
school mathematics teacher to the first author of this paper. The authors immediately became
engaged in solving the problem and providing the proofs. Sadly, upon trying to contact the
teacher to inform him of our results, we discovered that he had died in a car accident near the
high school at which he was teaching. We dedicate this article and its surprising result to
Mr. Menachem Orenshtein, the mathematics teacher who initiated this process but did not live
to see its results.
Multiple Solutions: Bridging Areas in Mathematics 135
Figure 1
Theorem 1. If A∠ in ΔABC is divided into four equal angles of measure
α each using the altitude, angle bisector, and median from A, then D
5.22=α
and ΔABC is a right triangle with angle measures of 90°, 67.5° and 22.5°.
At first glance, it seems that the first attempt to cope with this problem
should be the use of the Angle Bisector Theorem. Therefore, the first
presented direct proof is by using this theorem.
A. Direct proof by Euclidean geometry
By the Angle Bisector Theorem:
.:In x
t
h
m
ADM =Δ (1)
.
2
:In ttx
c
b
ANC
+
=Δ (2)
.
2
22
:In xtx
xtx
c
b
ABC
+
=
+
=Δ (3)
By (2) and (3): .2
2xt
xtx
ttx =⇒
+
=
+
(4)
By (1) and (4): .2
2h
x
hx
x
ht
m=== (5)
Using the Pythagorean Theorem in ΔADM:
() ( )
.
22
2
22
2mhtxmhDM =++⇒=+ (6)
Moshe Stupel and David Ben-Chaim
136
By (6) and (5):
[
( )
]
( )
21221 22
2+=⇒=++ xhhhx
.DMh
=
⇒ (7)
By (7): ΔADM is an isosceles and a right-angled triangle
.5.22452 DD =α⇒=α⇒
The final result reveals that ΔABC must be a right-angled triangle, which
is a surprising result.
Assuming that the above result is one direction of a proposition, let us
look at the opposite direction - the converse proposition:
Theorem 2. Given a ΔABC with angles measuring 90°, 67.5° and 22.5°,
suppose that the right angle A
∠
is divided into four equal angles by
segments ,AD ,AN and .AM Then AD is the altitude, AN is the angle
bisector, and AM is the median.
The proof is based on simple angle computations. AN is the angle
bisector as it divides A
∠ into 2α and 2α pieces by assumption. Commenting
on angle sums in ΔABD shows that AD is the altitude. Finally, calculation of
3α shows that ΔAMB and ΔAMC are both isosceles, which gets =
B
M
MCAM = and that AM is the median.
It is interesting to note that in mathematics, especially in geometry, there
are many cases in which a proof of one direction of a theorem is easier to
obtain than that of the opposite one. Following the proofs of the presented
problem in both the directions, a surprising result is achieved: it actually
presents a unique triangle - a right-angled triangle with angles of 22.5°,
67.5° and 90°. Indeed, this is the only triangle in which the three special
segments of a right-angled triangle from the right angle can form 4 equal
angles of 22.5° each.
Note that as both Theorems 1 and 2 are true, one can state the result as a
biconditional “if and only if” theorem:
Multiple Solutions: Bridging Areas in Mathematics 137
Theorem 3. In a ΔABC, the altitude, angle bisector, and median lines at
A quadrisect A
∠ if and only if ΔABC is a 90°-67.5°-22.5° triangle.
In the following sections (B, C and D), we present two additional direct
proofs and one indirect proof of the above problem, using Euclidean
geometry.
B. Direct proof no. 2 by Euclidean geometry
Lemma. If AB is a diameter in a circle, C is some point on the circle,
and K is a point on the diameter such that ,KBKC
=
then K is the center of
the circle.
Proof. D
90=∠ ACB as an inscribed angle on the given diameter AB.
Therefore, the calculation of angles in ΔCKB and =
∠
⇒
Δ
KCAAKC
,90 α−=∠ D
KAC which means that ,KBKCAK
=
=
and hence point K
is the center of the circle.
Use of the lemma in the four-angle problem.
The triangle ΔABC is inscribed in a circle. See Figure 2. GH is a mid-
perpendicular to the chord BC, and therefore, it is a diameter in the circle.
The mid-perpendicular GH and the angle bisector AN meet at the point H on
the circle
()
.since BAHCAHHBHC ∠=∠⇒=
Since
MHADANGHAD
∠
=
∠⇒|| (alternate interior angles).
Therefore, the triangle ΔAMH is an isosceles triangle, in other words,
.MHAM =
The point M is a point on the diameter such that ,MHAM
=
therefore,
according to the above lemma, the point M is the center of the circle and
hence BC is also a diameter and the angle BAC
∠
is equal to 90° as an
inscribed angle on the diameter .5.22904 DD =α⇒=α⇒
Moshe Stupel and David Ben-Chaim
138
Figure 2
C. Direct proof no. 3 by Euclidean geometry
The triangle ΔABC is inscribed in a circle, as shown in Figure 3, with E
the intersection of the circle with the extension of AD and F the intersection
of the circle with the extension of .AM GH is a mid-perpendicular to the
chord BC, and therefore, it is a diameter in the circle.
Figure 3
Multiple Solutions: Bridging Areas in Mathematics 139
The arcs
FCBE = are equal due to the equal inscribed angles resting
against them: .
α
=
∠=∠ FACBAE As arcs BE and FC are equal,
.
CEFBCE ∠=∠
Thus, EC is a transversal that crosses BC and
EF
with equal alternate
interior angles. Hence, it follows that .
BCEF
|
|
From parallelism, it follows that ,90°
=
∠
AEF therefore, the chord AF
is a diameter in the circle (a 90° inscribed angle).
Since GH and AF are two diameters in the circle, the point of their
intersection M is the center of the circle. Hence, it also follows that BC is
a diameter in the circle and therefore, α⇒=α⇒=∠ DD 90490BAC
.5.22 D
=
D. Indirect proof using Euclidean geometry
Three possibilities exist for the position of the circumcenter O when
inscribing the given triangle ΔABC. See Figure 4.
Figure 4
Possibility A: If ,904 D
>α then O is outside the triangle.
Possibility B: If ,904 D
<α then O is inside the triangle.
Moshe Stupel and David Ben-Chaim
140
Possibility C: If ,904 D
=α then the side BC of the triangle must be a
diameter of the circle.
Recall that the circumcenter O is the intersection of the perpendicular
bisectors of ΔABC, and so O lies on the perpendicular bisector of .BC
An indirect proof is used to refute possibilities A and B.
Possibility A
Connect circumcenter O with vertices A and B. See Figure 5. Note that
D
904 >α implies that .3
α
<
∠OAB
()
α−=∠=∠ 39022 D
ACBAOB
(
the central angle equals twice the
inscribed angle resting on the same
)
.arc OBOA
=
(
)
,radii therefore,
ΔOAB is an isosceles triangle, and hence, .3
α
=
∠
OAB
Hence, possibility A: D
90>∠ BAC is refuted.
Figure 5
Multiple Solutions: Bridging Areas in Mathematics 141
Possibility B
As in possibility A, the circumcenter O is connected with vertices A
and B of the triangle. See Figure 6. Note that D
904 <α implies that
.3α>∠ OAB As before, we have seen that ,3
α
=
∠
OAB refuting
possibility B.
Conclusion: The only viable possibility is Possibility C, .904 D
=α
Therefore, .5.22 D
=α It is important to take note that the principle of proof
by contradiction requires identification of a finite number of different
possibilities of which only one can be absolutely true, and that one
possibility can only be proven as true if all of the others can be negated. The
professional literature has pointed out that this type of proof is often a cause
of difficulty to students and is not well understood (see, for example,
Antonini [1], Antonini and Mariotti [2] and Thompson [46]). Hence, it is
very important for the students to acquire experience with this type of proof.
Figure 6
Moshe Stupel and David Ben-Chaim
142
5. Combining Fields in Mathematics in Order to Obtain Additional
Solutions to the Problem
It is the authors’ belief that proving a task using different methods, in
particular, by combining methods from different branches of mathematics,
serves to illustrate the connections between the various areas, reveals the
beauty of mathematics, and illustrates how mathematics is a conjunction of
intertwined fields. Silver et al. [40] claimed that “different solutions can
facilitate connection of a problem at hand to different elements of knowledge
with which a student may be familiar, thereby strengthening networks of
related ideas” [40, p. 228].
Thus, a search was made for solutions to the proposed problem by
methods using other branches of mathematics. In effect, these branches are
related to each other and share concepts with Euclidean geometry, and can
claim to be based upon these interrelations.
E. Proof by trigonometric method
Various proofs applying trigonometry were produced: using the sine
theorem, using the sine theorem along with a calculation of areas, and using
the cosine theorem. In all of these proofs, trigonometric identities were used
that led to a trigonometric equation to be solved. The difference between
these three proofs is the complexity and length of their solutions.
Below we present the simplest of these proofs. See Figure 7.
In this case: .2aBC = The length of median AM is expressed by using
the sine theorem in two triangles and comparing the obtained expressions.
From triangle ΔAMC:
()
.
sin
3cos
sin390sin α
α
=⇒
α
=
α−° a
AM
aAM (1)
From triangle ΔAMB:
()
.
3sin
cos
3sin90sin α
α
=⇒
α
=
α−° a
AM
aAM (2)
Multiple Solutions: Bridging Areas in Mathematics 143
Hence, by (1) and (2):
.cossin23cos3sin2
3sin
cos
sin
3cos αα=αα⇒
α
α
=
α
αaa
Using the double-angle identities leads to: ,2sin6sin
α
=
α
the solution of
which is either kk DDDD 455.2236021806 +=α⇒+α−=α or =α6
.903602 kk DD =α⇒+α The only viable solution is: D
5.22=α and
therefore: .904 D
=α=∠ BAC
Figure 7
Figure 8
F. Proof using vectors
Vector notation: See Figure 8.
,cAB = ,bAC = ,aBC = ,hAD = ,xBN = mAM =
Moshe Stupel and David Ben-Chaim
144
which gives:
.cba
−
=
(1)
Also, cb >
(
since the triangles ΔADB and ΔADC are right-angled
triangles with a common leg, and the other legs satisfy
)
.DBDC >
From the angle bisector theorem in triangle ΔABC, we have
( )
.
b
c
xax ⋅−= Hence:
( )
.
bc c
cb
bc c
ax +
−=
+
⋅= (2)
According to the notation:
,
2
1xch += and by (2)
( )
.
2
1bc c
cbch +
−+=⇒ (3)
Since the vectors a and h are perpendicular, their scalar product is:
.0=⋅ ha
Substituting a and h from (1) and (3) in the last identity, we have:
( ) ( )
.0
2
1=
⎥
⎦
⎤
⎢
⎣
⎡−
+
+− cb
cb c
ccb Simplifying, we have:
()
.
2
222
bbcbcc
cb −+
=⋅ (4)
From triangle ΔADM, we have ,2cos m
h
=α and from the theorem of
the angle bisector in that triangle, we have ,
2
2
x
a
x
m
h
−
= which results in
,
2cos21
2cos
α+
α
⋅= ax and using relation (2):
bc c
+
=
α+
α
2cos21
2cos or .2cos cb c
−
=α (5)
Multiple Solutions: Bridging Areas in Mathematics 145
From the trigonometric identity ,12cos24cos 2−α=α and by (5), we
obtain:
() ()
2
22
2
22
1
2
4cos cb
bcbc
cb
c
−
−+
=−
−
=α
()
.4cos2 2
22 α−=−+⇒ cbbbcc (6)
Based on this last relation (6), we can write the relation (4) as:
()
.4cos
2
2α−⋅=⋅ cb
b
c
cb
But, on the other hand, based on the scalar product of b and :c
.4cos α⋅=⋅ cbcb By equating the two expressions above for ,cb
⋅
we get:
()
.0
2
4cos 2=
⎥
⎦
⎤
⎢
⎣
⎡−⋅−α cb
b
c
bc
The first option: 04cos
=
α gives .904 °
=
α
The second option is:
()
bcb
b=−⋅ 2
2
1 leads to
()
2
22bcb =− and
after taking the square root
(
given
)
,cb > we have: 2bcb =− or
( )
.012 <−=− cb Since ,012 >− the relation is not possible and this
answer is eliminated.
Hence, the triangle ΔABC must be right-angled at A.
Although the proof using vectors is more complicated than the other
proofs, it should be practiced as well, if only to appreciate the relative
simplicity of the other proofs and to apply the material associated with
vectors in solving problems and in connection to plane geometry and
trigonometry.
Moshe Stupel and David Ben-Chaim
146
G. Proof using complex numbers2
A coordinate system is chosen such that the vertex A is the origin and the
altitude AD lies on the x axis (in the positive direction). See Figure 9.
We denote the length of the altitude by .,, cABbACxAD
=
=
=
The
points B, D, N, M and C have the same x coordinate. The x axis is the real
axis and the y axis is the imaginary axis (i). Other notation: ,
1
yDC =
32 ,yDNyDM ==
(
it is clear that DB also equals
)
.
3
y
Figure 9
In the Gauss plane, the following complex numbers correspond to the
points:
.:;:;:;: 33332211 iyxzBiyxzNiyxzMiyxzC −
=
+
=
+
=
+=
()
2
3
z is expressed algebraically and trigonometrically:
() ( ) ( ())
.2 3
2
3
2
2
3
2
3ixyyxiyxz +−=+=
2The essence of this form of the proof was presented to us by one of our pre-service students
(Irena Laryanov), as a result of presenting the problem to them and asking to prove it by
different methods and applying different mathematical topics.
Multiple Solutions: Bridging Areas in Mathematics 147
From De-Moivre’s formula, we obtain:
() ()(())()
.2sin2cos2sin2 cos 2
3
2
2
3
2
3α+α+=α+α= iyxizz
Comparing the real and imaginary values in the two representations, we
obtain:
() ( ())
α+=− 2cos
2
3
2
2
3
2yxyx and
(())
.2sin2 2
3
2
3α+= yxxy
By dividing the two equalities, we obtain the following relation:
()
.
2
2tan 2
3
2
3
yx
xy
−
=α (1)
From the triangle ΔAMD: .2tan 2
x
y
=α
Since the point M is the middle of the interval BC, we have
(
)
.
2
31
2yy
y−+
= (2)
Hence
.
2
2tan 31 xyy −
=α (3)
Equating the two representations of ,2tan
α
(1) and (3), we have:
()
2
3
2
331 2
2yx
xy
xyy
−
=
−
and after algebraic manipulation, we obtain:
.
5
31
31
2
3yy yy
x
y−
−
=
⎟
⎠
⎞
⎜
⎝
⎛ (4)
From the triangle ΔAND, we have: ,tan 3
x
y
=α and by substituting in
relation (4), we obtain
.
5
tan
31
31
2yy yy −
−
=α (5)
Moshe Stupel and David Ben-Chaim
148
From the bisector theorem, we have:
In triangle ΔANC In triangle ΔBAC
32
21 yy yy
MN
CM
AN
AC
c
b−
−
=== and from
the relation (2) ,
2
31
2yy
y−
= we
have .
331
31 yy yy
c
b−
+
=
.
23
31 yyy
NB
CN
c
b−
==
Comparing the two representations of ,
c
b we have: .
32 31
31
3
31 yy yy
yyy −
+
=
−
By algebraic manipulation, we arrive at the quadratic equation:
,016
3
1
2
3
1=+
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛y
y
y
y whose solutions are ,223
3
1±=
y
y and since from
the drawing we know that ,1
3
1>
y
y we obtain that
( )
.223 31 yy += We
substitute the value of 1
y in the expression for α
2
tan (marked as (5)), and
obtain .
12
12
tan2
+
−
=α To find the value of α:
( )
()( )
( ) ( )
2
22
212
12
12
1212
12
12
12
tan −=
−
−
=
−+
−
=
+
−
=α
α−
α
=α⇒−=α⇒ 2
tan1
tan2
2tan12tan
( )
( )
(
)
1
222
122
121
122
2=
−
−
=
−−
−
=
.904452 DD =α⇒=α⇒
Of course, another option is to use a calculator to find the value of the
angle α for ,
12
12
tan2
+
−
=α but obviously, it will not provide the exact
value of α.
Multiple Solutions: Bridging Areas in Mathematics 149
H. Proof using analytic geometry
The vertices of the triangle on a coordinate system are positioned as
shown in Figure 10: Vertex A is on the y axis with the coordinates
()
;,0 1
yA
vertices B and C are on the x axis with the coordinates
(
)
0,
2
xB and
()
.0,
3
xC The origin is at the point O.
Figure 10
From the coordinates of the vertices B and C, the coordinates of points M
and N are obtained as follows:
()
,0,;0,
22
32 xN
xx
M−
⎟
⎠
⎞
⎜
⎝
⎛+ since ΔABN is
an isosceles triangle.
From triangle ΔAOB, the slope of AB is:
()
.tan
tan
1
90tan
0
0
12
2
1
2
1α−=⇒
α
=α−=
−
=
−
−yx
xy
x
y (1)
From triangle ΔAOM, the slope of AM is:
()()
α−−=α+=
+
−
−290tan290tan
2
0
0
32
1xx
y
.
2
2tan
2tan
1
1
32 yxx +
=α⇒
α
−
= (2)
Moshe Stupel and David Ben-Chaim
150
From triangle ΔAOC, the slope of AC is:
()
.3tan
3tan
1
390tan
0
0
13
3
1α=⇒
α
−
=α+=
−
−yx
x
y (3)
By substituting the values of 2
x and 3
x from (1) and (3) into (2), the
result is:
.tan3tan2tan20,
2
3tantan
2tan 1
1
11 α−α=α⇒≠
α+α−
=α y
yyy (4)
By writing α2tan as
(
)
α
−
α
3tan and using the formula of
()
,tan β−α
we obtain: .tan3tan
tan3tan1
tan3tan
2α−α=
αα+
α
−α Dividing both the sides by
,0tan3tan ≠α−α the result is: ,1tan3tan
=
α
α
leads to: −α
α
cos3cos
0sin3sin =αα which is equal to .04cos
=
α
Therefore, .904 °
=
α
I. Alternative solution using analytic geometry
A set of coordinates is selected such that the base of the triangle, BC, is
on the x axis, and the center point of the base, M, is the origin (as shown in
Figure 11). The coordinates are thus as follows:
(
)
,, 11 yxA
()
,0,aB −
()
.0,aC
Figure 11
Multiple Solutions: Bridging Areas in Mathematics 151
The slope of side AC is ,
1
1ax y
− and the slope of median AM is .
1
1
x
y
Therefore, from the formula for the tangent of the angle between two lines:
()
.
1
tantan 2
11
2
1
1
11
2
1
1
1
1
1
yaxx
ay
axx y
x
y
ax y
CAM +−
=
−
+
−
−
=α=∠ (1)
The slope of side AB is:
()
.90tan
1
1ax y
+
=α−
D (2)
By multiplying (1) and (2) accordingly:
() ()
.0
190tantan 22
1
2
11
1
1
2
11
2
1
1=−+⇒
+
⋅
+−
==α−⋅α ayxx
ax y
yaxx
ay (3)
It is obvious that ,0
1≠x since else the median and the altitude would
coincide, leading to an isosceles triangle and four equal angles would not
exist, hence:
()
.1 0
1
1
1
1
22
1
2
1
22
1
2
1−=
+
⋅
−
⇒−−=⇒=−+ ax y
ax y
axyayx
In this case, the product of the slopes of the sides AC and AB is 1,
−
meaning
that ,90D
=∠ BAC and consequently, angle α equals 22.5°.
A different continuation from result (3) will result in
22
1
2
1
222
1
2
10AMyxaayx =+=⇒=−+
(by using the Pythagorean Theorem at ΔADM) .aAM
=
⇒
Hence, the triangle ΔBAC must be right-angled
()
D
90=∠ BAC based on the
theorem that “if the median to a side is equal to half its length, then the
triangle is right-angled”.
Moshe Stupel and David Ben-Chaim
152
This last solution is very surprising, since no use was made of bisector
AN. This leads to the conclusion that if an altitude in a triangle forms an
angle δ (any size) with the side adjacent to it, and the angle formed by the
median from the same vertex with the adjacent side is also δ, then the angle
at that vertex must be a right angle and the triangle, of course, is a right-
angled triangle. Coincidentally, just recently, we came upon a report about
the experience of a mathematics teacher in Israel, who accidentally came
upon this specific finding, which was published in a teachers’ journal in
Hebrew by Sigler [38]. Sigler proved this result by trigonometric methods
and by the application of a formula by Steiner. It is interesting to note that
our finding is not true in the case of an isosceles triangle that is not right-
angled, even though the altitude and the median (which are the same segment
from the vertex head of the triangle) to the base form equal angles with the
equal sides of the triangle.
It should be emphasized that our suggested problem is a special case of
the above finding: if the altitude, the angle bisector, and the median issuing
from the same vertex form 4 equal angles, then the triangle must be a right-
angled one. There is only one type of a triangle that fulfills these conditions,
a right-angled triangle with acute angles of 22.5 and 67.5 degrees. This is a
special triangle!
6. ‘Semi Proof’ and Inductive Reasoning Using a Dynamic Geometry
Environment (DGE)
The introduction of dynamic geometry software (DGS) (such as
Geometers Sketchpad® and recently GeoGebra) into classrooms (high
schools and colleges) creates a challenge to the praxis of theorem acquisition
and deductive proof in the study and teaching of Euclidean geometry.
Students/learners can experiment through different dragging modalities on
geometrical objects that they construct, and consequently, infer properties,
generalities, and conjectures about the geometrical artifacts.
The dragging operation on a geometrical object enables students to
apprehend a whole class of objects in which the conjectured attribute is
Multiple Solutions: Bridging Areas in Mathematics 153
invariant, and hence, the students become convinced that their conjecture
will always be true [9]. Nevertheless, because of the inductive nature of the
DGE, we entitle this process ‘semi proof’. Hence, following the employment
of DGE, the experimental-theoretical gap that exists in the acquisition and
justification of geometrical knowledge becomes an important pedagogical
and epistemological concern [25]. Students must be aware that they still need
to prove rather than rely on the virtual experiment.
Boero, Garuti, Lemut and Mariotti (Garuti et al. [14], Boero et al. [7],
Mariotti et al. [30], Garuti et al. [15] and Boero et al. [5, 6]) conducted a
body of research into students’ behavior in the linkage between the process
of producing conjectures and the process of proving theorems or statements.
In particular, Boero et al. [7] proposed a hypothesis on conjecture
production as follows:
“the conditionality of the statement can (authors’ emphasis) be the
product of a dynamic exploration of the problem situation during which
the identification of a special regularity leads to a temporal section of the
exploration process, that will be subsequently detached from it and then
‘crystallize’ from a logic point of view (‘if…, then …’)” [7, p. 121].
‘Quasi-empirical investigations’ are acquiring more and more
importance, highlighting functions of proof that were traditionally
undermined [8, 10]. Examples of such functions are explanation, insight,
understanding, validation, and discovery. These non-deductive methods of
investigation, which rely on experimental, intuitive and inductive reasoning
[10], are seen to provide more meaningful contexts for teaching-learning
geometry with DGE software than the classical approach of proof as a way of
obtaining certainty.
As a result of the above, we decided to examine how our problem might
be solved within a DGE framework. We used GeoGebra software, with
which we could construct and control the given conditions of the problem (a
triangle with three special segments from one vertex) and drag the
mathematical object at one of the vertices of the triangle until the four angles
Moshe Stupel and David Ben-Chaim
154
formed by the sides of the triangle and the three special segments were equal.
GeoGebra has an advantage in that it indicates the size of the different angles
as they are being changed by the dragging action while, concurrently, the
basic conditions are controlled. At this point, we knew that this should
happen when the angle is 90° and, in actuality, this is the result we acquired.
The experiment is presented in its entirety in the following link3:
http://highmath.haifa.ac.il/stupel/st.doc
The ease with which preliminary conjectures concerning the nature of the
triangle could be arrived at using DGS, led us to conclude that an effectual
course of action is to initially present the problem with GeoGebra software
(or any other suitable DGS), allowing students to construct the triangle with
the three special segments, pose conditions to be controlled, and apply the
dragging of the geometrical object until the result is achieved. Then, once the
conjecture that the angle must be 90° had been obtained, the students should
be asked to prove the result deductively using the accepted theoretical tools.
The DGE software serves as an intermediary tool used to bridge the gap
between the physical model and the formal symbolic proof.
Note: This procedure has been very successfully implemented with our
students in the teacher college. The presentation of this problem, along with
others that could similarly be solved using various mathematical methods,
was a part of the course dealing with the incorporation of technology in
mathematics teaching. Following this step, the students were required to find
as many as possible different solutions - to be discussed in another course
that was dedicated to developing mathematical reasoning and exploring how
to solve mathematical problems in various ways. In many cases, the students
surprised us by coming up with delightful solutions that we had not
considered. Another implementation of the method of multiple solutions has
been done within the seminar course in our teacher education program. The
3We would like to acknowledge the contribution of The National Teachers’ Center for
Mathematics in High School, located at Haifa University, Israel, in preparing the clips using
the GeoGebra software application, with special thanks to academic Head Dr. Varda Talmon
and staff members Geula Sever and Maya Katz.
Multiple Solutions: Bridging Areas in Mathematics 155
students are instructed to choose a problem with a potential of multiple
solutions as a project and to present the solutions with the detailed
description of the pedagogical action for including it within the teaching of
the learners in the educational system. The above courses received a very
favorable rating from the students. The average general rating from the
classes of the students was 5.8 out of 6. The verbal responses of the
participating students also indicated high satisfaction with the courses.
Hence, we believe that the ideas presented in this paper should be of great
interest to teachers of geometry or trigonometry at a college-level or
secondary level.
7. Concluding Remarks
Geometry is a goldmine for multiple solution tasks. Proofs may be
derived by applying different methods within the specific topic of geometry
or within other mathematical areas such as analytic geometry, trigonometry,
and vectors. The authors claim that multiple proofs foster both better
comprehension and increased creativity in mathematics for the student/
learner.
The multiple solutions that were presented herein for one geometry
problem demonstrate the connectivity between different areas of
mathematics, and show how geometry can serve as a base for topics such as
trigonometry, analytic geometry, and vector spaces. For example, solving the
problem by analytic geometry methods (besides the clear employment of
Euclidean geometry) reveals a connection to trigonometry by the slopes of
the sides of the triangle and the resulting need to solve trigonometric
equations. When doing the proof by vectors, the connection to trigonometry
is shown by the requirement to apply the scalar product formula and, again,
the need to solve trigonometric equations in order to arrive at the conclusion.
This occupation with the connection between different mathematical
domains builds among students a vision of mathematics as a linked science
and not as a collection of discrete, isolated topics [22]. In most school
Moshe Stupel and David Ben-Chaim
156
textbooks worldwide, mathematics problems are organized by specific topics
that are presented in the curriculum: students tend to understand that certain
problems are connected to specific topics and, hence, assume that for each
problem there is one, and only one, method for its solution [37]. Also, while
the NCTM standards document emphasizes teachers should find tasks that
exhibit connectivity between different mathematical domains, it also
indicates, nevertheless, that locating such problems is time consuming and
calls for special initiative from the teachers. Certainly, our experience also
shows that while this mission is not an easy one, there is certainly a need to
identify additional problems that can be solved by various methods and
demand the application of proofs from different realms of mathematics. The
authors see it their task to continue searching for such appropriate problems
and encourage their colleagues to do so, too. They believe that mathematics
teachers should present to their students problems that must be solved in
more than one way, requiring as much as possible, the application of
knowledge from different domains in mathematics. In his study, Bingolbali
[4] indicated that applying multiple ways of solving problems has the
potential to develop students’ ‘relational understanding’ (a term attributed
to Skemp [43]) and contribute to the development of their autonomy.
Furthermore, when testing students, teachers should occasionally allow their
students to solve problems by employing proofs from any mathematical area,
and not insist on a proof from the specific subject under study.
In this day and age, it is impossible to ignore the rapid development of
technology and the way it affects almost every facet of life. The education
system is no exception, and one can certainly not disregard the value of
technology in teaching mathematics. The authors firmly believe in the
advantages of incorporating dynamic geometry software (DGS) in
mathematical problem solving and actually demonstrated our problem using
GeoGebra software. In a special issue of PME, Jones et al. [23], writing a
guest editorial entitled “Proof in dynamic geometry environments,” stated
that “This Special Issue provides a range of evidence that working with
Multiple Solutions: Bridging Areas in Mathematics 157
dynamic geometry software affords students possibilities of access to
theoretical mathematics, something that can be particularly elusive with other
pedagogical tools” [23, p. 3]. Inductive exploration with DGS can lead
students to develop their own conjectures about the solution of the problem
and then to deal with the deductive proof. This is in addition to the
contribution of visualizing different graphical representations of concepts
and other related situations to the problem. We further recommend, therefore,
that mathematics teachers initially allow their students to cope with problem
solving by working in such an environment until they reach a solid
conjecture for deductive proof.
In conclusion, our efforts in coping with the presented problem afforded
us a real feeling of working as mathematicians who look for multiple
solutions to a problem, especially those that are short, elegant, and
mathematically aesthetic. By encouraging student/learners to do the same,
they, too, will learn to appreciate the connections between the various
branches of mathematics and discover how one problem can be tackled from
different points of view. In addition, the incorporation of DGE software will
add a complementary technological tool to assist the students in their
investigations, while also providing teachers a means with which to base
pedagogical action and in-class discussion. As a bonus, we revealed a special
and unique triangle, in which the three special line segments from one vertex
form four equal angles with the sides of the triangle and between them.
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