Geometry POW Solution, February 17-21
Geometry Problem of the Week
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Geometry Problem of the Week, February 17-21
This week we're going to play with polygons. Take an equilateral
triangle. Put squares on the three sides (I've drawn it below with only
one square). Connect the outside corners of the squares with three line
segments to form a hexagon. Is the resulting figure equilateral? Is it
equiangular?
Now do the same thing with a regular hexagon - put squares on the six
sides and connect the outer corners to form a 12-sided figure. Is this
figure equilateral? Is it equiangular? (What's the proper name for a
12-sided figure?)
Be sure to explain why your answer is correct!
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This is a problem that requires you to be detailed and keep track of a
few things carefully, and to test what you know about some basic shapes.
39 people got it right, and 13 got it wrong.
Six new schools this week: St. George's School in Middletown, Rhode
Island, Bokenskolen Gymnasium, Jokknokk, Sweden, West Leyden High School
in Melrose Park, Illinois, Grissom High School in Huntsville, Alabama,
Amity Junior High in Orange, Connecticut, and William Penn Charter School
in Philadelphia, Pennsylvania. Welcome!
It's a good idea in this situation to not just say, "It's obvious that
the other side is longer," (as in the first part of the question), but to
say exactly what all the angles are and how that affects the side
length. You could even take it a step further and explain what the side
lengths actually ARE! The answers I've highlighted are all very
thorough, and carefully explain each part of the answer.
Also, while a good way to start is to draw a fairly accurate picture,
it's probably not a good idea to measure the parts of the actual picture
to figure out the answer. In a situation like this, it's not too tough
to draw a reasonable picture and then fill in what you KNOW about the
pictures. As I have said before, drawings are nice, but often it's a
better idea to use your knowledge of the shapes involved to reach your
final conclusion.
It ws also tough to be brief and concise this week - there is a lot to
talk about. The trick is to say everything you need to say, but to lay
it out so that it is as easy to read as possible. That means using
separate paragraphs, for example, and maybe even numbering parts of your
answer to refer back to later.
Brent Tworetsky of JP Taravella High School drew some nice ascii pictures
to keep everything clear. Jenny Kaplan from Castilleja Middle School and
Berno Wiegel of Werner Heisenberg Gymnasium both wrote very good,
thorough explanations...Berno despite his plea that he is still learning
English!
Thomas Kuo of Burroughs High School sent in a superb explanation. Not
only is it clear, but it's well laid out and very thoughtful.
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The following students submitted correct solutions this week.
Highlighted solutions are included below. For a full list of solutions,
please check out
http://forum.swarthmore.edu/geopow/fullsolutions/022197.fullsolution.html.
Brent Tworetzky, Grade 10, JP Taravella High School, Coral Springs, Florida
Mr. Lewis' Geometry Class, Grade 9,10, St. George's School, Middletown,
Rhode Island
Neil Scully, Grade 10, Cheshire High School, Cheshire, Connecticut
Michael Brasher, Grade 11, Grissom High School, Huntsville, Alabama
Ken Duisenberg, Grade Hewlett Packard Engineer, Stanford University '93
Justin Lam, Grade 8, Sequoia Middle School, Pleasant Hill, California
Jon Sheffi, Grade 10, Commonwealth School, Boston, Massachusetts
Jonathan Goldberg, Grade 9, Germantown Academy, Fort Washington, Pennsylvania
Ernie Weiss, Grade 10, William Penn Charter School, Philadelphia, Pennsylvania
Heidi Taylor, Grade 10, Smoky Hill High School, Aurora, Colorado
Kelly Richburg and Amy Mousaw, Grade 10, Cheshire High School, Cheshire,
Connecticut
Lauren Shea and Tiffany Forbes, Grade 10, Cheshire High School, Cheshire,
Connecticut
Kate Gosselin and Ashley Hackett, Grade 10, Cheshire High School,
Cheshire, Connecticut
Alanna Yee and Chris Wilby, Grade 9, Cheshire High School, Cheshire, Connecticut
Erik Day and Mike Savenelli, Grade 10, Cheshire High School, Cheshire,
Connecticut
Chris Goodwin, Grade 9, Cheshire High School, Cheshire, Connecticut
Jake DeGennaro, Grade 9, Cheshire High School, Cheshire, Connecticut
Jessica Carnevale and Lauren Orlowski, Grade 9, Cheshire High School,
Cheshire, Connecticut
Catherine Nichols, Grade , Lakeside School, Seattle, Washington
Jess Gilburne and Eric Faden, Grade 8, Georgetown Day School, Washington, DC
Richard Twogood, Grade 9, Granada High School, Livermore, California
Jenny Kaplan, Grade 6, Castilleja Middle School, Palo Alto, California
Keah Kendall, Grade 9, Smoky Hill High School, Aurora, Colorado
Roger Mong, Grade 8, Zion Heights Junior High School, Richmond Hill,
Ontario, Canada
Berno Wiegel, Grade 8, Werner Heisenberg Gymnasium, Germany
Sheridan Waller, Grade 10, Smoky Hill High School, Aurora, Colorado
Josh Parker, Grade 10, Smoky Hill High School, Aurora, Colorado
Michael Chin, Grade 10, Smoky Hill High School, Aurora, Colorado
Nicole Heidelberg, Grade 9, Granada High School, Livermore, California
Kirk Vissat, Grade 10, Smoky Hill High School, Aurora, Colorado
Janet Eckart, Grade 9, Granada High School, Livermore, California
Lincoln Simon Cheung, Grade , North Sydney, Australia
7th grade class, Grade 7, Scuola Media Salvo D'Acquisto, Bologna, Italy
Suzanne Davis, Grade 10, Newport High School, Bellevue, Washington
Thomas Kuo, Grade 9, Burroughs High School, Ridgecrest, California
Giscard Pongnon, Grade 12, George Wingate High School, Brooklyn, New York
Christine Nguyen, Grade 9, Granada High School, Livermore, California
Shameica Edwards, Grade 11, George Wingate High School, Brooklyn, New York
Tim Ayers, Grade 9, Rivers School, Weston, Massachusetts
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From: Brent Tworetzky
bla...@ix.netcom.com
Grade: 10
School: JP Taravella High School, Coral Springs, Florida
shown are the angles and the diagram of the hexagon.
* * * * * * * * *
* * * *
* * * *
* * 120 * *
* 90 * 90 *
* * * *
C * 60* *
* * * *
* * * * 90 * *
* 90*60 60* *
* 120 A * * * * * *120 *
*90 90*
* * * *
30 * *
* *90 * *
B * * * * * *
for this to be equilateral, squares cannot be adjacent. in fact,
only one square can be inscribed in a regular hexagon with a common
side without intersecting with another square. (this is because the
maximum height/width of a regular hexagon with side s is s(sqrt3)
which is less than 2s- 2s is what is needed for two non-intersecting
squares to be inscribed in a regular hexagon.) however, this is
equiangular, because all sides of the hexagon are 120. this is because
CAB is 120 (360 minus 90-90-60) and because the shape is symmetric,
the remaining angles, ACB and ABC, are 30. each angle of the hexagon
is 30 + 90 = 120.
now for the dodecagon- 12 sided polygon.
* * * * * * *
* * * *
* *
* * * *
* * * * * *
* * * * * *
* * * * * * * * *
* * * *
B 90 * * *
60* 90 * * *
* * * * * *
60 A 120 * *
* * 90* * * *
* * * *
C * * *
* * * *
* * * * * * * * *
* * * * * *
* * * * * *
* * * *
* *
* * * *
* * * * * * *
this is a regular dodecagon because it is equilateral and equiangular.
it is equiangular because all angles are 150. a regular hexagon's angle is
120. right angles (from the square) are 90. the remaining angle (BAC) is
60 (360-120-90-90=60). the shape we have is symmetrical, so ACB and ABC
are both 60. each angle of the dodecagon is now 150, so it is equiangular,
and all the sides are equal, because the sides of a square are all equal,
so are the sides of an equilateral triangle. because the squares and
equilateral triangles share sides, there sides are all equal and thus all
sides of the dodecagon are equal.
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From: Jenny Kaplan
a...@pacbell.net
Grade: 6
School: Castilleja Middle School, Palo Alto, California
The figure made with the equilateral triangle and the squares is not
equilateral, but it is equiangular and this is why. For the figure to be
equilateral the triangles made the sides of two of the squares and the
line that
connects them must be equilateral. If each side of the center equilateral
triangle is x, than each side of the square will be x. When you draw the
triangle between the squares, the outside side (the one that is not part of a
square) must also be x. If those triangles are equilateral all of their angles
will be 60 degrees. To find out the angle you would add the two 90 degree
angles
from the square and the 60 degree angle from the triangle, you would then
subtract that number (which happens to be 240) from 360, that = 120. That means
that it is not equilateral because you can not have a equilateral triangle with
one angle 120! The figure is equiangular because each angle of the square
is 90,
and for every 90 degree angle there is a 30 degree angle from one of the out
side triangle. Therefore each on of the figures angles is 120 degrees!
The figure made with the regular hexagon and the squares is both
equilateral and
equiangular. For the figure to be equilateral each one of the out side
triangles
must be equilateral. Each one of the inside angles of a hexagon is 120 degrees
if you add that to 180 (from the two right angles of the triangles) and
subtract
it from 360, it would = 60 degrees, making each of the outside triangles
equilateral, making the outside figure equilateral! The figure is also
equiangular (for the same reason the figure around the equilateral triangle
was), for each 90 degree angle of the square there was a 60 degree angle to
add to it, making each angle of the outside figure 150!
The proper name for a 12 sided figure is a dodecagon!
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From: Berno Wiegel
berno....@rhein-neckar.de
Grade: 8
School: Werner Heisenberg Gymnasium, Germany
Hi Annnie
I1ve learned English about 3 1/2 years at school, but it1s very difficult for me
to give you a good explanation in this language. Also the dictionary doesn1t
contain all the mathematical terms. But I1m going to try it.
The hexagon with the triangle at the center is equiangular, but it1s not
equilateral.
Explanation: You need three equilateral outer triangles between the three
squares around the triangle at the center to get a equilateral hexagon. - Each
angle of the triangle at the center is 60°. The angles in the squares are 90°.
Add together one angle of the triangle in the center and two angles of the
squares (you find these angles at each corner of the triangle at the center)
than subtract the result from 360°. The difference is 120°. From this it
follows that the angles of the three outer triangles are 120°, 30°, 30°.
Therefore the third side of the outer triangle can1t be equal to the sides of
the squares.
Each of the six angles in the hexagon consists of an angle of a triangle (30°)
and an angle of a square (90°). Therefore the hexagon is equiangular.
The 12-sided-figure is equilateral and equiangular.
Explanation: A regular hexagon has got six sides with the same length and
six angles with 120°. If you put squares at the six sides and connect the outer
corners you get six triangles between the squares. These triangles are
equilateral and equiangular, because two of their three sides are also two
sides of the squares, and the angles between the squares are 60°
(360°-{90°+90°+120°} ,
you find these angles at each corner of the regular hexagon
at the center).
Bye for now,
Berno
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From: Thomas Kuo
wk...@ridgecrest.ca.us
Grade: 9
School: Burroughs High School, Ridgecrest, California
Subject: Solution for POW 2/17-21/97
Nice new format for the new POW page :^|~
-Thomas
From: Thomas Kuo
Email: wk...@ridgecrest.ca.us
School: Sherman E. Burroughs High School, Ridgecrest, California
Grade: 9th
1) The resulting hexagon for the first problem is only equiangular.
a) The equilateral triangle and the squares have an equal side length
meaning the triangle formed by the sides of the square is isosceles.
b) Find the value of the vertex angle of the formed isosceles triangle.
360 - 60 - 90 - 90 = 120 degrees
c) Find the value of the base angles of the formed isosceles triangle.
(180 - 120)/2 = 30 degrees
d) Find the value of the angles of the formed hexagon.
Each base angle is bordered by the square meaning each angle of the
formed hexagon is 90 + 30 degrees = 120 degrees.
e) The sides of the formed hexagon can easily be seen not to be
equilateral because the formed isosceles triangle is not
equilateral. But for more proof...
ALL VALUES IN DEGREES
x = length of side of square
y = length of the base of the formed isosceles triangle
sin(30) / x = sin(120) / y by the Law of the Sine
x * sqrt(3)/2 = y * 0.5
y = sqrt(3) * x which means that x is not equal to y
2) The resulting "DODECAGON" for problem 2 is a regular polygon.
a) The regular hexagon and the squares have an equal side length
meaning the triangle formed by the sides of the square is isosceles.
b) Find the value of the vertex angle of the formed isosceles triangle.
360 - 120 - 90 - 90 = 60 degrees
c) Find the value of the base angles of the formed isosceles triangle.
(180 - 60)/2 = 60 degrees
d) Find the value of the angles of the formed dodecagon.
Each base angle is bordered by the square meaning each angle of the
formed hexagon is 90 + 60 degrees = 150 degrees.
e) The sides of the formed dodecagon can easily be seen to be
equilateral because the formed isosceles triangle is equilateral.
3) Just in case you didn't notice...
A 12-sided figure is a dodecagon.
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