Geometry POW Solution, October 11 (part 3)
Geometry Problem of the Week
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From: Den...@aol.com
From: Erin Jacobs
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: Problem of the week
Erin Jacobs grade 10
Integrated math III (Miller)
Cheshire, CT
1. In order to divide a regular hexagon into 3 equal parts, a line segmant is
drawn from the center to every other vertex. When this is done, the hexagon
will be divided into 3 rhombi. The geometric properties are that a regular
polygon with an even number of sides can be divided into equal parts by
creating line segments from the center to every other vertex, or to every
vertex. When the line segments are drawn to every other vertex, the polygon
will be divided into the number of equal parts as half of the number of sides
of the polygon.
2. In order to divide a regular hexagon into six equal parts, a line is drawn
from the center to each of the vertices. This divides the hexagon into six
equal triangles. This was done because each line segment creates a new
section of the hexagon. Therefore, a line segment was drawn to each of the
vertices because there are six vertices in a hexagon and six equal parts were
needed. Another way to divide a regular hexagon into six equal parts is to
draw a line segment from the center to every other vertex and then draw a
line segment from each of these vertices to eachother. The parts created in
this instance are also triangles. In each case, six line segments were
needed to divide the hexagon into six equal parts. This shows that as many
line segments as there are sides of the polygon are needed to divide the
polygon into the number of equal parts as the polygon has sides.
3. In order to create six kites from a regular hexagon, a line segment was
drawn from the center to the middle of each of the sides. This was done
because it divided the hexagon into six equal parts but the were
quadrillaterals instead of triangles this time. Again, six line segments
were needed to divide the hexagon into six equal parts.
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From: Den...@aol.com
From: Shane Chang
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: Math Forum-Geometry Problem of the Week 10/7/96
Shane Chang, grade 10
Integrated Math III (Miller)
Cheshire, CT
1.
____
/ \ \
/ \ __ \
\ / /
\/___/
Because the polygon is a regular hexagon, it has six lines of symmetry.
To divide the hexagon, first draw three lines of symmetry with the end
points on the vertices of the hexagon. These three lines of symmetry
intersect at the center of the hexagon and divide it into six triangles.
These triangles are all congruent because each line of symmetry divides a
120 degree angle (the measure of any interior angle in a hexagon) into two 60
degree angles. Thus, when the axes of symmetry create triangles, the base
angles of the triangles are all congruent. Since the hexagon has all equal
sides, the base lines of the triangles are also all equal. Thus, the
triangles are all congruent by law of ASA. Since the triangles are all
congruent, the six triangles can be paired up into three rhombuses that are
also congruent. All of the triangles share sides, the sides of the triangle
are all equal. So when any two at the trianlges are combined to form a
quadrilateral, the four sides are congruent. Thus, a regular hexagon can be
divided into three identical rhombuses.
2. ___
/ \ /\
/__ \/__\
\ / \ /
\/__ \/
This model is much like the solution to problem 1. It divides the hexagon
into six equal triangles. The three lines with end points on vertices of the
hexagon. of symmetry intersect at the center of the hexagon and form six
triangles. Each triangle shares one side with its neighboring triangle.
That side is the same length for both triangles. That bordering line also
splits a 120 degree interior angle of the hexagon into two angles of 60
degrees. Then, as each side of of a regular hexagon is equal, so are the
baselines of the triangles. Thus, the adjacent sides are equal, the
baselines are equal, and the angles between them are both 60 degrees. That
makes the triangles equal by law of SAS.
___
/ | \
/ \ | /\
\ / | \ /
\ _|_/
This model divides the hexagon into six equal kites. The kites are made
by drawing a perpendicular bisector for each side of the hexagon. (Notice
these are also lines of symmetry) The perpendicular bisectors intersect at
the center of the hexagon and divide it into six kites. To prove the kites
are identical, each side facing outward from the center has a length of half
the side of the hexagon. This is because the perpendicular bisectors divide
each side of the hexagon in half.
Since each side is the same, so is each half, so the two outward sides of
each kite are all equal. The bordering sides between the kites are
congruent because each kite shares a side with the neighboring one. Thus, as
all of the sides match up for each kite, they must be identical by SSSS. The
model is not very accurate. Note: for a more detailed explanation, see the
solution of problem 3.
3. ___
/ | \
/ \ | / \
\ / | \ /
\__|_/
This is the same as the second model in problem 2. By drawing a
perpendicular bisector for each side, six kites can be formed. This is
because the perpendicular bisectors all intersect at the center and divide
the sides of the hexagon into halves. Doing this, the perpendicular
bisectors create kites with two 90 degree angles, a 120 degree angle (an
interior angle of the hexagon), and a 60 degree angle from the angle created
by the intersection. Each angle from the intersection of perpendicular
bisectors is a vertical angle, which has an identical angle opposite it so
the angles created by the perpendicular bisectors all share a measure of 60
degrees. Every kite has its corresponding angles congruent. Thus they are
similar in shape. Each kite shares a side with each of its adjacent kites,
so they are equal in both kites. This works for both sides facing
neighboring kites. Also, the perpendicular bisector divides the lines of the
hexagon into halves, so the sides facing outward from the center are also
equal. Thus, with corresponding angles and sides congruent, the kites are
all identical. The diagram does not accurately depict the model.
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From: Den...@aol.com
From: Ben Falit
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: Math forum - Geometry Problem of the Week (10/7 - 10/11)
Ben Falit, Grade 10
Integrated Math III (Miller)
Cheshire, CT
1.) When I split a regular hexagon into three identical parts, I discovered
that each shape is a rhombus (diamond). [Please see diagram A located
below]. Each rhombus is simply a combonation of two triangles (each is equal
to one sixth of the hexagon). This relationship makes perfect sense because
one third plus one third plus one third equals one and six times one sixth
equals one. Also, one sixth plus one sixth equals one third. As one can
clearly see, a regular hexagon can be divided into three equal rhombii, each
with two congruent sides and two congruent angles.
Diagram A: _____
/ / \
/____/ \
\ \ /
\____\/
2.) A regular hexagon can be split into six equal parts many ways. First of
all, it can be done by drawing boundary lines from the center to every
vertex. [Please see diagram B located below]. This forms six equilateral
triangles. The "six equilateral triangles" method is an obvious one for
dividing the figure into six equal parts. This is because each of the
"quadrants" of a regular figure is always going to be congruent to the
others. Each section has three sixty degree angles. Another way to divide
the figure up into six equal parts is to draw lines from the center of the
figure to the midpoints of all of the line segments. This forms six
identical quadrilaterals or kite shapes. The reason this occurs is, the
figure is regular and the same procedure is being done to each side. Each
section has interior angles that add up to three hundred and sixty degrees as
well as exterior angles that add up to three hundred and sixty degrees.
Diagram B: ______
/ \ /\
/ ___\ /___\
\ / \ /
\/____\_/
3.) It is possible to split the hexagon up into six identical kite shapes.
In fact, I divided the figure into six kite shapes when I did my second
division in question number two. [Please see diagram C located below]. I
simply drew lines from the center of the figure to the midpoints of all of
the line segments composing the hexagon. This forms six identical kite
shapes. These congruent figures emerge because the figure is regular and the
same procedure is being done to each side of the hexagon. Each individual
kite shape possesses interior angles, as well as exterior angles that add up
to three hundred and sixty degrees.
Diagram C: _________
/ | \
/ \ | / \
/ \ | / \
\ / | \ /
\ / | \ /
\____|____/
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From: Den...@aol.com
From: Adam Staffaroni
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: problem of the week
Adam Staffaroni Grade 10
Integrated Math 3 (Miller)
Cheshire, CT 06410
______
/\ \ 1.Each of the three divisions of the
/ \ \ comes out to be a rhombus, which has
/ \_____\ four equal sides and four angles, of
\ / / which the opposite angles are equal.
\ / / Opposite sides of the rhombus are
\/_____/ parallel, and the figure has 2 lines of
symmetry.
2.In the first way that I divided the hexagon, the six parts come out to be
equilateral triangles, which have all sides equal and all angles equal. I
found these parts by drawing in all the lines of symmetry which passesd
through the vertices of the hexagon. In my second division of the hexagon
into six parts, the shapes created were all kites, which are four-sided
figures that have two pairs of adjacent, equal sides. I found these shapes by
drawing in all the lines of symmetry of the hexagon which did not pass
through the vertices.
______
/\ /\ This is my first division, into six
/ \ / \ triangles.
/____\/____\
\ /\ /
\ / \ /
\/____\/
______
/ | \ This figure is supposed to look like
/-\ | /-\ six kites, but I was not able to create
/ \_|/ \ lines going directly from the midpoints
\ / |\ / of the sides to the center of the
\_/ | \_/ hexagon, like they should be. Also, the
\ __|__/ vertical line going down the middle of
the hexagon should be centered.
3. I found a way to divide a hexagon into six equal kites, which are
four-sided figures that have two pairs of adjacent, equal sides. I found
these shapes by drawing in all the lines of symmetry of the hexagon which did
not pass through the vertices.
______
/ | \ This figure is supposed to look like
/-\ | /-\ six kites, but I was not able to create
/ \_|/ \ lines going directly from the midpoints
\ / |\ / of the sides to the center of the
\_/ | \_/ hexagon, like they should be. Also, the
\ __|__/ vertical line going down the middle of
the hexagon should be centered.
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From: Den...@aol.com
From: Jake DeGennaro
Grade: 9
School: Cheshire High School, Cheshire, Connecticut
Subject: Problem of the week
Jake DeGennaro-grade 9
Geometry (Miller)
Cheshire, Ct
This weeks problem was not very complicated. All we had to do ws try to
split a regular hexagon into different shapes and amounts of shapes. this
problem was more artistic than last weeks.
In diagram number one, I was told to split the hexagon into three
identicle parts. I made three rhombuses.I chose to do it this way because it
was the first way that came to my mind. That is basically the only reason.If
something else came to my mind first, I probably would have used it
(depending on what it was). As I looked, I notice that it also forms a cube.
This is something that I didn't think of while I wasmaking my sketch.
1. _______
/\ \
/ \ \
/ \ ______\
\ / /
\ / /
\/______/
In diagram 2(a), I divided the hexagon into six segments. They all came
out to be triangles (equillateral) of the same size. The lines that make up
the sides of the triangle all run from thecenter (where they all meet) to a
different vertex.
2. _______
/ \ / \
/ \ / \
/____ \_/____\
\ /\ /
\ / \ /
\_/_____\/
In diagram 2b, I made the lines run from the center to the middle point
on each of the sides of the hexagons. This made six kites. Not only did I
answer question number 2 with this diagram but I also found a solution to
number three. I had another diagram for number three but the lines
overlapped so it is invalid.
2b. _______
/ | \
/ \ | / \
/ \ | / \
/ /|\ \
\ / | \ /
\ / | \ /
\ | /
\___|____/
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From: Den...@aol.com
From: Amy Mousaw
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: Problem of the week
Amy Mousaw grade 10
Geometry
Mrs. Miller
Cheshire Ct.
WE were given visual puzzzles to "play around with" in order to
find answers to three questions. These questions were all about how to split
a rectangle into: 1) three identicle parts, 2) two different sets of six
identicle parts, and 3) six identicle "kites". In order to do this I had to
give reasons, include sketches, and discuss the geometric properties that
each dixision had.
When I saw the first question, I thought I would try to put a dot
in the middle of the hexagon and divide it from there. I saw that there wew
six vertexes, so I diveded that number by three to get two. I then drew a
line at every second vertex. These divisions caused the hexagon to become
divided into three Rhombus's. It ends up looking like this:
________
/ /\
/ 2 / \
/_______/ 1 \
\ \ /
\ 3 \ /
\_______\/
Since a Rhombus is an equalateral quadrilateral, it is completely equal
lengths on all sides. If you take one of these three rhombus's and cut one
out, you can place it perfectly on top of another.This is because they are
equal.
The second problem had two parts, to split the hexagon into six
parts in two different ways. For my first way, I took the information I
obtained from the first problem and used it to answer this question. I used
the same shape as the last one (with the rhombuses) and connected the line
segments at the vertexes. This made six triangles. Two of the sides of the
triangle are equal, and two of the angles are equal. The triangles are all
equal to each other and can be placed on top of each other perfectly. The end
result will look like this.
_____
/ /\
/ / / |\
/ /_ / | \
\\ \ | /
\ \ \ |/
\_ \ \/
The second way I divided the regular Hexagon is again by placing a dot in the
center and drawing a lone to each vertex. Since there are six vertexes it
worked out making six equal triangle. All the sides of these triangles are
equal and so are the angles. It is an equalateral triangle, and the six are
all equal to each other.
The way I dived the regular hexagon to make "six kites" is also
another answer to number two. I figured out this answer by trial and error.
after a few attempts, I placed a line down the center of the Hexagon from one
line segment to the one opposite it. I then drew lines from the center of the
hexagons line segments through the center of the hexagon and to the center of
the line segment opposite it. This made kites that have four sides. Of these
four sides two pairs equal each other.
_______
/ | \
/\ | / \
/ \ / \
\ / \ /
\/ | \/
\___|___/
By dividing up the regular hexagon, i have made four different
geometric shapes including: a rhombus, two different kinds of triangles, and
a kits. all these shapes have been equal to the others tat make up the
hexagon. The illustrations help to prove this point.
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From: Den...@aol.com
From: Lauren Orlowski
Grade: 9
School: Cheshire High School, Cheshire, Connecticut
Subject: Geometry Problem of the Week
Lauren Orlowski, Grade 9
Geometry( Miller)
Cheshire, CT
Given a six sided figure called a hexagon, I needed to find a way to divide
it into three identical segments, find two ways to divide it into six
identical sections and a way to divide the regular hexagon into six kite
shaped figures.
My first challenge was to split the hexagoon into three identical
parts. I figured that since there were six sides and I needed three
sections, I would need to draw a line into the center at every vertex. The
following figure shows that each shape is a rhombus.
----------
/ \ 1 \
/ \ \
/ 3 \------- \
\ / 2 /
\ / /
\ / /
----------
My second problem was comprised of two parts. I had to find two ways
to divide the hexagon into six parts. The first way, I drew a line into the
center from each of the vertexes, making six equalateral triangles. The
diagram looks like this:
---------
/ \ / \
/ 1 \ 2 / 3 \
/ \ / \
---------------
\ / \ /
\ 4 / 5 \ 6 /
\ / \ /
---------
For the second division of a hexagon, I made six lines from the center
of each of the line segments to my point in the center. Each formation was
a quadrilateral in the shape of a kite. This is also the solution for the
thrid problem, dividing the regular hexagon into six kite shaped sections.
I had a lot of trouble trying to draw this diagram.
---------
/ | \
/\ | /\
/ \ | / \
/ \ | / \
/ \ | / \
\ / | \ /
\ / | \ /
\ / | \ /
\/ | \/
\ | /
---------
After analyzing the figures, I realized that three of the divisions
involve four sided figures, while the other involves a three sided figure.
These are the ways to divide the hexagon into three identical sections and
six identical sections.
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From: Den...@aol.com
From: Chris Goodwin
Grade: 9
School: Cheshire High School, Cheshire, Connecticut
Subject: Problem of the Week October 7-11,1996
Chris Goodwin Grade 9
Geometry(Miller)
Cheshire,CT
A regular hexagon has six sides and angles that are all equal. It looks
something like this:
--------
/ \ \
/ \ \
/ \ ------\ Diagram 1
\ / /
\ / /
\ /-------/
If you were to take this hexagonand try to split it into 3 identical
parts, then each part would be a diamond. I did this by finding the center
point of the hexagon. Then I used trial and error to find that it could be
split into 3 identical parts. I went from the far left angle to the center
point and from there to the top right angle. Then I made another line from
the center point to the bottom right angle. You can see what the diagram
looks like in diagram 1. In diagram one you can see three diamonds exactly
the same. If you look at the diagram closely it looks like it is a
3-dimensional cube.
Next if you wanted to take a regular hexagon and split it into six
identical pieces all you would have to do is split the hexagon into 3 pieces
or diamonds then split each diamond in half. This would make six identical
equilateral triangles. I went from the center point down to any angle that I
did not use for the diamonds. You can see what it looks like in diagram 2.
--------
/ \ / \
/ \/ \
/------------\
\ / \ / Diagram 2
\ / \ /
\ / \/
--------- This problem was the
easiest problem out of all three of them because once you figured out thr
first problem all you had to do is cut each diamond in half.
Lastly, I had to do problem three. This problem was the hardest problem
for me because I kept thinking that I had to make lines from the angles.
This, however, was not true. In order to split a regular hexagon into six
kites I had to first find the center point. Then instead of going to any
angles with any lines I had to make the lines go from the center point to the
middle of each line on the hexagon. This would make six equal kites. Kites
have two equal sides and then two other sides that are equal to each other.
You can see what it looks like in diagram 3.
--------
/ | \
/ \ | / \
/ \ | / \ diagram 3(It was very
\ / | / / hard to draw this on
\ / | \ / the computer so it is
\ / very inacurate.)
--------
This was a fun problem of the week. It involved a lot of thinking and
reasoning. This problem of the week was much more fun than any other problem
except the parachute problem.
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From: Den...@aol.com
From: Pat Esposito
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: Problem of the Week
Pat Esposito, 10th Grade
Geometry(Miller)
Cheshire,CT
A regular hexagon can be split into many different
parts and those parts come out to many different shapes. Our task was to
split a regular hexagon into three identical parts, then six parts, then six
kites.
When I split the hexagon into three identical parts the shapes that came
out were quadrilaterals.
------
/ / \
/----/ \
\ \ /
\ \ /
------
This happens because when you put three regular quadrilaterals together they
form a diamond. As you can see the three quadrilaterals form together as a
perfect hexagon.
When I split the hexagon into six identical parts the shapes that came
out were six identical triangles.
--------
/\ /\
/ \ / \
/ \ / \
\------------/
\ /\ /
\ / \ /
--------
This happens because when six identical triangles are put together they form
a regular hexagon. If all the triangles are 60 degrees, when put together
the angles add up to 360 degrees making a regular hexagon.
It is possible to split the regular hexagon into 6 identical kites. If
you draw lines from the middle of each line segment, through the vertex, to
the middle of the line segment directly across from the starting line
segment, the hexagon is split into 6 identical figures, these figures come
out to be kites.
------
/ | \
/\ | /\
/ \ | / \
\ / | \ /
\/ | \/
\ | /
------
Although I did not get to make my diagram the way I wanted to, you can make a
diagram yourself using the steps listed above. This uses the same principle
as the triangles above.
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From: Den...@aol.com
From: Adam Ferrell
Grade: 10
School: Cheshire High School, Cheshire, Connecticut
Subject: Problem of the Week
Adam Ferrell Grade 10
Integrated Math III
Cheshire, CT
1. ----------
/ \ \
/ \ \
/ \ \
/ \---------\
\ / /
\ / /
\ / /
\ / /
-----------
Each part of the three sections of the hexagon is a parallelogram. To get
these you need to draw one line segments from one angle to the center. Then
do the same for every other angle around the hexagram.
2. ---------
/ \ / \
/ \ / \
/ \ / \
/ \ / \
\ / \ /
\ / \ /
\ / \ /
\ / \ /
-----------
One way to divide the hexagon into six equal parts is to draw line segments
from the center of a side to the center of the opposite side. When you are
done you will have three lines passing through the center of the hexagon and
you will have formed six identical parts.
Another way is to draw line segments from one angle to the opposite angle and
you will get the same results only different identical shapes.
3. ----------
/ | \
/ | \
/ | \
/ | \
\ | /
\ | /
\ | /
\ | /
-----------
In order to divide this hexagon into six kites you need to construct three
line segments. Each line segment goes from the center of one of the sides to
the center of the opposite side. The line segments are perpendicular
bisectors to each side.
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From: Kandee...@msn.com
From: Leif Linden
Grade:
School: Germantown Academy, Fort Washington, Pennsylvania
Subject: Math Problem Of The Week
From Leif Linden,
If you split a regular hexagon into three equal pieces, each piece
would be
a
rhombus. In order to do this, you start from the center of the hexagon and
draw three lines to three of the corners, skipping every other corner.
One way to split a regular hexagon into six congruent pieces is to
start
at the center point and draw six lines to each of the corners, resulting in
six little triangles.
Another way to divide the shape into six congruent pieces would be
to start
at the center point again, and this time draw lines to each of the sides of
the hexagon, making quadraleterals that look like kites.
If you draw one line to the midpoint of one side of the hexagon,
and then
repeat this process for the other five lines, the result is six quadraleterals
that resemble kites. Each quadraleteral will have as two of its sides, two
consecutive lines (out of the six) that were drawn from the center to the
midpoints of the sides of the hexagon. The other two sides of the kite will
be made up of the halves of two adjacent sides of the hexagon.