1. Assume that an algorithm to solve a problem takes f(n) microseconds for some function f of the input size n. For each time t labeled across the top, determine the exact largest value of n which can be solved in time f(n) where f(n) ≤ t. Use a calculator! You will find it helpful to convert the t values to microseconds, and you may find it helpful to insert a row for n. Note that “lg n” is the log2 n. Note that the only row you can’t write out the values for fully is the “lg n” row—only there may you write 2x for the appropriate value of x. Use the Windows built-in scientific calculator (under Accessories menu) as necessary. A couple values are filled in to get you started. Important: “exact values” means precisely that. Check your answers with values above and below! Time t = f(n) = 1 second 1 hour 1 day 1 month =30 days n2 1,609,968 lg n n3 2n n lg n 2,755,147,513 2. Use loop counting to give a O( ) characterization of each of the following loops basing each upon the size of its input: a. Algorithm Loop1(n): s ← 0 for i ← 1 to n do s ← s + i b. Algorithm Loop2(p): p ← 1 for i ← 1 to 2n do p ← p * i c. Algorithm Loop3(n): p ← 1 for j ← 1 to n2 do p ← p * i d. Algorithm Loop4(n): s ← 0 for j ← 1 to 2n do for k ← 1 to j do s ← s + j e. Algorithm Loop5(n): k ← 0 for r ← 1 to n2 do for s ← 1 to r do k ← k + r 3. Order the following functions from smallest to largest by their big-O notation—you can use the letters in your answer rather than copying each formula. Be clear which is smallest and which is largest, and which functions are asymptotically equivalent. For example, if g, h, and m are all O(n lg n), you would write g = h = m = O(n lg n). a. 562 log3 108 b. n3 c. 2n lg n d. lg nn e. n3 lg n f. (n3 lg n3)/2 g. nn h. 56n i. log5 (n!) j. ncos n k. n / lg n l. lg* n m. 4. a. Which of these equations is true, and why? b. Which of these is smaller for very large n? Trisecting the Circle: A Case for Euclidean Geometry Author(s): Alfred S. Posamentier Source: The Mathematics Teacher, Vol. 99, No. 6 (FEBRUARY 2006), pp. 414-418 Published by: National Council of Teachers of Mathematics Stable URL: http://www.jstor.org/stable/27972006 Accessed: 09-02-2018 18:19 UTC JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected] Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at http://about.jstor.org/terms National Council of Teachers of Mathematics is collaborating with JSTOR to digitize, preserve and extend access to The Mathematics Teacher This content downloaded ...

1. 1. Assume that an algorithm to solve a problem takes f(n)
microseconds for some function f of the input size n. For each
time t labeled across the top, determine the exact largest value
of n which can be solved in time f(n) where f(n) ≤ t. Use a
calculator! You will find it helpful to convert the t values to
microseconds, and you may find it helpful to insert a row for n.
Note that “lg n” is the log2 n. Note that the only row you can’t
write out the values for fully is the “lg n” row—only there may
you write 2x for the appropriate value of x. Use the Windows
built-in scientific calculator (under Accessories menu) as
necessary. A couple values are filled in to get you started.
Important: “exact values” means precisely that. Check your
answers with values above and below!
Time t =
f(n) =
1 second
1 hour
1 day
1 month
=30 days
n2
1,609,968
lg n
n3

2. 2n
n lg n
2,755,147,513
2. Use loop counting to give a O( ) characterization of each of
the following loops basing each upon the size of its input:
a. Algorithm Loop1(n):
s ← 0
for i ← 1 to n do
s ← s + i
b. Algorithm Loop2(p):
p ← 1
for i ← 1 to 2n do
p ← p * i
c. Algorithm Loop3(n):
p ← 1
for j ← 1 to n2 do
p ← p * i
d. Algorithm Loop4(n):
s ← 0
for j ← 1 to 2n do
for k ← 1 to j do

3. s ← s + j
e. Algorithm Loop5(n):
k ← 0
for r ← 1 to n2 do
for s ← 1 to r do
k ← k + r
3. Order the following functions from smallest to largest by
their big-O notation—you can use the letters in your answer
rather than copying each formula. Be clear which is smallest
and which is largest, and which functions are asymptotically
equivalent. For example, if g, h, and m are all O(n lg n), you
would write g = h = m = O(n lg n).
a. 562 log3 108
b. n3
c. 2n lg n
d. lg nn
e. n3 lg n
f. (n3 lg n3)/2
g. nn
h. 56n
i. log5 (n!)
j. ncos n
k. n / lg n
l. lg* n
m.
4. a. Which of these equations is true, and why?
b. Which of these is smaller for very large n?

4. Trisecting the Circle: A Case for Euclidean Geometry
Author(s): Alfred S. Posamentier
Source: The Mathematics Teacher, Vol. 99, No. 6 (FEBRUARY
2006), pp. 414-418
Published by: National Council of Teachers of Mathematics
Stable URL: http://www.jstor.org/stable/27972006
Accessed: 09-02-2018 18:19 UTC
JSTOR is a not-for-profit service that helps scholars,
researchers, and students discover, use, and build upon a wide
range of content in a trusted digital archive. We use information
technology and tools to increase productivity and
facilitate new forms of scholarship. For more information about
JSTOR, please contact [email protected]
Your use of the JSTOR archive indicates your acceptance of the
Terms & Conditions of Use, available at
http://about.jstor.org/terms
National Council of Teachers of Mathematics is collaborating
with JSTOR to digitize,
preserve and extend access to The Mathematics Teacher
This content downloaded from 128.103.149.52 on Fri, 09 Feb
2018 18:19:28 UTC
All use subject to http://about.jstor.org/terms

5. DEL m er Alfred S. Posamentier
Trisecting the Circle:
A Case for Euclidean
Geometry
Editors' note, part 1: The paper that Alfred Posa
mentier submitted to "Delving Deeper" was simply
entitled "A Case for Euclidean Geometry" and con
tained a number of lovely examples of problems
and ideas illustrating the variety of rich mathemat
ics one can encounter from a start that is "just" Eu
clidean geometry. We shamelessly chopped the
paper into pieces and are publishing just one of
them?coincidentally, roughly a trisection of his
paper. We are holding the other parts for later and
are hoping, also, for contributions by other authors
who find Euclidean geometry a fertile source of
This department focuses on mathematics content that appeals to
secondary school
teachers. It provides a forum that allows classroom teachers to
share their mathematics
from their work with students, their classroom investigations
and projects, and their
other experiences. We encourage submissions that pose and
solve a novel or interesting
mathematics problem, expand on connections among different
mathematical topics,
present a general method for describing a mathematical notion
or solving a class of prob
lems, elaborate on new insights into familiar secondary school
mathematics, or leave the
reader with a mathematical idea to expand. Send submissions to
"Delving Deeper" by

6. accessing mt.msubmit.net.
Edited by Al Cuoco, acuoco?edcorg
Center for Mathematics Education, Newton, MA 02458
E. Paul Goldenberg, pgokSenberg^edcorg
Education Development Center, Newton, MA 02458
mathematical ideas that go beyond here's-yet-an
other-odd-theorem.
As an undergraduate mathematics major, a
prospective teacher usually takes at least one
geometry course. Typically, these courses
focus on non-Euclidean geometry (sometimes pre
sented as Modern Geometry), or vectors, transforma
tions, or topology. Instead, we at the City College of
New York offer a course on more advanced Euclid
ean geometry in which prospective teachers investi
gate a plethora of geometric theorems (or relation
ships) that enrich their understanding of Euclidean
geometry and, consequently, their teaching of it.
Encountering such a wide range of geometric theo
rems provides more than breadth; the connections
among them give teachers an opportunity to delve
more deeply into questions that the typical high school
student would ask. Our students are pretty adept at
dividing a circle into two equal parts, but to divide a
circle into three equal parts might be a bit more chal
lenging?beyond the obvious method: the pizza cut.
Let's take a look at a few of these (see fig. 1). As
we consider the task of trisecting a circle, here are

7. just four possible methods. Each one uses some
nice geometric relationships.
414 MATHEMATICS TEACHER | Vol. 99, No. 6 ? February
2006
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2018 18:19:28 UTC
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Fig. 1 Four ways to cut a pizza in thirds
FOUR WAYS TO CUT A PIZZA INTO THIRDS
Traditional pieces: Figure 1a
Somebody who would actually take a protractor to a
pizza restaurant would draw laughs from others but
would be able to draw the radii that would divide the
circle into three equal parts (fig. 2), each of which is
360?
a = ?= 120?
3
?if the cheese cooperates. Wrapping a piece of
string around the circumference of the pizza would
also help. Dividing the string into three parts would
let us mark the circumference in thirds:
h_C _2nr ~3~ 3
Another way would be to take a stick and measure
the radius, mark that radius off six times around
the circle, and note every other marked-off point to

8. get the three equally spaced division markers.
Everybody gets an equally big slice (sector).
But there are intriguing alternatives to the
traditional.
Using concentric circles to get three equal
areas: Figure 1b
Of course, this concentric division is hardly suit
able in a restaurant; yet from a mathematical point
of view, this version is quite interesting.
Without any loss of generality we can assume the
radius of the initial circle to be 1. We must deter
mine the two radii rx and r2 so that the three colored
regions shown in figure 3 will be equal in area.
Because the outermost circle's radius is 1, its
area is . The problem requires us to make the area
of each colored region (the innermost yellow disk
and the blue and red annuii) a third of that. The
radius rx of the circle that bounds the yellow disk
must be
Fig. 2 Partitioning a circle into three equal sections
Given: radius of the largest circle = 1
Sought after: and r2 to make the
three areas equal

9. Solution
: r1
Fig. 3 Concentric circles method
and yellow) must then be 2/3 of the area of the out
ermost circle, or
2/r
3 '
so r2 itself must be
I
The combined area of the inner two regions (blue
Another way to express the three radii is
Vi V2 ,S
V3 V3 V3
Vol. 99, No. 6 ? February 2006 | MATHEMATICS TEACHER

10. 415
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Or one can rationalize the denominators, but that
makes this relationship somewhat less obvious:
S /6
3 ' 3 '
Using the teardrop shape: Figure 1c
This curious trisection of the area begins with
semicircles built on a trisection of the diameter.
Figure 4 shows how this will be divided into
(semi-) circular arcs.
We derive the solution by noting first that the
semicircles described by rx and r2 fit perfectly
within the outer circle, and therefore the sum of
their radii must equal the outer circle's radius: rx +

11. r2 = 1. We want values for and r2 that make the
area of the green teardrop the now-familiar /3.
The equation we want is built from four facts:
by , we get
Area of the outer circle:
Area of the circle built on r?
Area of the circle built on r2:
Area of the green teardrop:
Art =-A t green ? outer 2
Dividing both sides of
^outer ~~ 7 ?
Aj = /TTj2
A2 = 7cr22
1 1
+ -A, = 2 2 3
1 1 2 - ?7zt +

12. 2 2 1 2
1 2 1
2 3
Given: radius of the outer circle = 1
Sought after: rx and r2, so that the
congruent green and yellow teardrop
regions are each one-third of the area
of the circle
2 1