Download Math Heuristics

Assignment #1: Problem 8: squares?
Original statement of the problem:
On the square ABCD lines are drawn:
From A to the midpoint of CD,
From B to the midpoint of DA,
From C to the midpoint of AB, and
From D to the midpoint of BC.
The intersections of these lines give points A*, B*, C*, D* as shown.
What is the ratio of the area of the square ABCD to the area of the quadrilateral
A*B*C*D*?
Addition to original statement: Let the midpoints of AB, BC, CD and AD be named E, F,
G, and H, respectively.
A
E
B
B*
A*
F
H
C*
D*
D
G
C
Method #1:
Since E, F, G, and H are the midpoints of AB, BC, CD, and AD, respectively
AE=EB=BF=FC=GC=DG=HD=AH. Arbitrarily, let each of these segments have length
of 1.
Since ABH is a right triangle, we can find HB using the Pythagorean Theorem:
(HB)^2=(AB)^2+(AH)^2=(2^2)+(1^2)=5
So, HB=sqrt(5).
Similarly, we can use the Pythagorean Theorem to find the length of DF:
(DF)^2=(DC)^2+(FC)^2=(2^2)+(1^2)=5
So, DF=sqrt(5).
The same process and reasoning can be applied to find AG and EC.
AG=sqrt(5) and EC=sqrt(5)
Next, angle FDC can be found by solving the trigonometric equation
tan(angle FDC)=1/2
So, angle FDC=arctan(1/2)=26.57 degrees.
Then, angle ADF= 90-arctan(1/2)=63.43 degrees
Since AD is parallel to BC, angle DFC=63.43 degrees (because alternate interior angles
must be equal).
Since angle BFC=180 degrees and angle DFC=63.43 degrees, angle BFC*=18063.43=116.57 degrees.
Since HBFD has a pair of congruent parallel sides (HD and BF), it is a parallelogram
and thus has opposite angles equal. So, angle HBF=63.43 degrees, angle
BHD=116.57 degrees,
angle ABD=90-63.43=26.57 degrees and angle AHA*=63.43 degrees.
tanDAG=1/2 so angle DAG=26.57 degrees.
Then angle AA*H=180-26.57-63.43=90 degrees and since angle AA*H and angle
B*A*D* are vertical angles, angle B*A*D=angle AA*H=90 degrees.
The same can be done to sow that each angle A*B*C, angle B*C*D, and angle AD*C
are all 90 degrees.
Then, we can say that triangles AHA*, BEB*, CFC* DGD* are congruent by ASA.
From these congruent triangles, AA*=BB*=CC*=DD*.
Similarly, A*H=EB*=C*F=D*G.
Since HB=DF=AG=EC=sqrt(5), A*B*=B*C*=C*D*=A*D* by subtraction.
Since we found A*B*C*D* to have all equal sides and all right angles, it is a square.
Next we can find the lengths of DD*=BB*=CC*=AA* by using trigonometry:
cos(angleD*DG)=cos(26.57 degrees)=(DD*)/1
So, DD*≈0.8944=AA*=BB*=CC*.
D*G=C*F=EB*=A*H can be found as follows:
sin(26.57 degrees)=(D*G)/1=0.44729
So, D*G=0.44729
And since DF=sqrt(5)=DD*+D*C*+C*F=0.8944+D*C*+0.44729, D*C*=0.8944.
The square A*B*C*D* has area (0.8944)^2=0.8.
The square ABCD has area (2)(2)=4.
So, the ratio of ABCD to A*B*C*D is 4:0.8=5:1.
Method #2
We can put the diagram on a Cartesian plane with the following points:
A=(0, 10); B=(10, 10); C=(10, 0); D=(0, 0).
Then by using the midpoint formula, the coordinates of E, F, G, and H can be found to
be the following:
E=((10+10)/2, (10+10)/2)=(5, 10)
F=((10+10)/2, (0+10)/2)=(10, 5)
G=((0+10)/2, (0+0)/2)=(5, 0)
H=((0+0)/2, (0+10)/2)=(0, 5)
We can notice that A*, B*, C*, D* are the intersections of AG and BH, BH and EC, EC
and DF, and DF and AG, respectively.
We can also find the equations of these lines by using slope intercept form:
AG: y=-2x+10
BH: y=(1/2)x+5
EC: y=-2x+20
DF: y=(1/2)x
We can notice that in A*B*C*D*, pairs of opposite sides (AG and EC; BH and DF) have
parallel, equal slope.
Furthermore, pairs of adjacent sides in A*B*C*D* (AG and BH; EC and DF) have
opposite reciprocal slope, so they are perpendicular.
From the two previous statements, we can conclude that A*B*C*D* is some kind of
rectangle.
We can find the area of the interior rectangle with the formula:
Area=(A*B*)(B*C*)
The coordinates for A*, B*, and C* can be found by solving the respective systems of
equations:
y=-2x+10
y=(1/2)x+5
y=-2x+20
y=(1/2)x+5
y=-2x+20
y=(1/2)x
So, A*=(2, 6); B*=(6, 8); C*=(8, 4).
Then, A*B*=sqrt((4)^2+(2)^2)=sqrt(20) and
B*C*=sqrt(((2)^2+(4)^2)= sqrt(20).
We know that A*B*C*D* is a square since these are equal.
Then Area A*B*C*D*=(sqrt(20))^2=20.
Area ABCD=10(10)=100.
Thus, the ration between the two areas is 20:100=1:5.
Method #3
Let AE=EB=BE=EC=CG=GD=DH=AH=1.
Using the Pythagorean Theorem, DE=HB=sqrt((1)^2+(2)^2)=sqrt(5).
The area of square ABCD=(1+1)(1+1)=4.
Furthermore, Area of triangle ABH=Area of triangle CDF=(1/2)(1)(2)=1.
Then Area HBDF=Area ABCD-Area ABH- Area CDE= 4-1-1=2.
B*C* is the height of HBDE (since we already established perpendicularity in previous
methods).
Area HBDE=2=(sqrt(5))(B*C*).
So, B*C*=2/(sqrt(5)).
Then, Area of square A*B*C*D*=(2/(sqrt(5)))^2=4/5.
So the ratio between area of ABCD to area of A*B*C*D* is 4:(4/5)=1:5.
Analysis and Reflection
Our group initially addressed this problem by using the heuristic of looking at a specific
example. More specifically, we arbitrarily set each side of the square ABCD to be 2.
And then using properties that we knew about triangles and quadrilaterals and
trigonometry, we were able to eventually conclude that quadrilateral A*B*C*D* is a
square and furthermore find the length of its sides. We then used the lengths of the
sides to calculate the areas and then compared the two areas to find the ratio. This
worked because although the calculations were based upon arbitrarily giving a measure
to one of the lengths, this length was carried out throughout the problem so that the
relationships between different lengths (and thus between the areas) remained the
same regardless of the fact that the initial side length was chosen arbitrarily. Thus, we
are able to conclude that the ratio of the areas of the squares is the square of the ratio
of the side lengths.
Next, I realized that we would be able to see important relationships among the sides
and angles (such as parallelism and perpendicularity and congruency) if we put the
original square on a Cartesian plane because we would be able to easily calculate slope
and distance using the coordinates of the different points. When we put D at the origin
and chose arbitrary side lengths for the square ABCD, we were able to calculate the
coordinates, slopes, and lengths (and then consequently, the areas) without using
trigonometry.
After coming up with these solutions as a group, Scott came up with a way of looking at
the diagram (Method #3) that allowed us to solve the problem simply by looking at areas
without having to calculate distances or slopes. Most of our group found that the
strategy very ingenious because it involved very little calculations. After discussing with
the entire class, it was clear that there were many, many ways of solving this problem,
many of which I wouldn’t have been able to come up with on my own. It was very
satisfying to solve and when other group members came up with different ways of
solving it, it seemed so clever and interesting that we could look at the problem at a way
that we never thought of. It kind of felt like looking at an optical illusion: when you are
used to seeing the picture one way and when someone finds another way of looking at
the same picture, it just seems like a whole new revelation, like an ah-ha moment.
I came away from this problem solving session realizing the value of different solutions.
I think that in different contexts, these solutions would have different importance and
value. For example, if the contextual unit was focused on trigonometry, the first solution
would be most applicable whereas if the unit was focused on equations of lines and the
relationships between lines (such as slope, parallelism, and perpendicularity), the third
solution would be best useful. From a curricular aspect, I think that certain solutions are
better than others in different curricular contexts; however, I also think that students
must also be given the opportunity to develop their own ideas that may fall out of what is
covered in the curriculum. If teachers allow students to come up with more “creative”
solutions, they are allowing for not only students to think out of the box, but also for
students to develop their own problem solving heuristics, which may be more powerful
and more meaningful to students than simply teaching them directly.
Furthermore, this activity makes it clear to me that groupwork definitely plays an
important role in problem solving. When we work with others, we can “put our heads
together” and possibly come up with ideas that we would have not come up with
otherwise. Another reason why groupwork works well with problem solving is that it
forces us to put our ideas into logical words in order to explain to our groupmates the
reasoning and the thinking process used to arrive at certain conclusions. Since each of
our cognitive processes differ, we must show step-by-step to someone who does not
necessarily think in the same way as we do the routes we took and the reasoning
behind our conclusions. I think this strengthens students’ ability to formulate informal,
yet logical, proofs as well as to learn to articulate their mathematical ideas.
I think that this changed my view of mathematics and problem solving as primarily an
individual process that depends on one’s own mathematical abilities. In contrast, I have
seen the benefit of mathematical discussion, which manifests itself in the classroom
most of the time as groupwork, as not simply a combination of different ideas, but rather
as a synthesis of ideas which has the ability of raising new ideas that each group
member individually would not be able to come up with own his/her own. This reminds
me of the video that we saw in class in which even a Princeton mathematician relied on
the ideas of others to develop his own ideas. This just goes to show the power and the
necessity of communicating mathematical ideas with others: solving problems involves
an exchange of ideas through written and verbal language. Thus interpersonal
communications is not simply a means of conveying math ideas but help to develop
solutions along the way.
Using these ideas, I think that in the future, as an educator, I would try to emphasize the
importance of sound mathematical reasoning by having students work in groups more
often and like the students in Professor Maher’s study, start to justify their thought
processes at younger ages. This way, they would have a better idea of how
mathematical ideas build upon one another which may help them to see the many
relationships between different mathematical concepts. Again, this would introduce
them informally to proof writing and logical thinking, which is beneficial not only to math,
but is an interdisciplinary skill which is used in many other topics.