FEBRUARY
Two points are chosen on the parabola
defined by y = x2, one with a positive
x-coordinate and the other with a
negative x-coordinate. If the points are
(a, b) and (c, d), where a < 0 and c > 0,
find the y-intercept of the line joining the
two points in terms of a and c.
Flip a fair coin 5 times. Which is more
likely to occur—all 5 flips land heads up,
or the flips land in the order heads-tailsheads-tails-heads?
1
A box containing a block of aluminum
weighs 8 lbs. When a steel block weighing
3 times the aluminum block replaces the
aluminum block inside the box, the weight
is now 20 lbs. What is the weight of the
box?
Two people agree to meet between 6 p.m.
and 7 p.m. They also agree to wait only
15 minutes for the other person to arrive.
What is the probability that they meet
each other? (Assume that arrival times are
uniformly distributed over the one-hour
interval.)
4
The difference between the cube of the
sum of two positive integers and the sum
of the cubes of the same two integers is
210. Compute the sum and the product of
the two integers.
A square-wheeled tricycle can be pedaled
smoothly if the road consists of appropriately spaced bumps. What is the name of
the mathematical shape that would allow
for a smooth ride?
Which event is more likely to occur—
exactly 1 head in 2 flips of a coin or exactly
5 heads in 10 flips of a coin?
9
In ABC, AB = 16, AC = 15, and BC = 9.
Find the ratio
m∠B
.
m∠A
3
The average of 5 quiz grades is 10. When
the lowest grade is dropped and the new
average is calculated, it turns out to be 11.
What was the score of the dropped grade?
6
10 # 5 = 9
11 # 6 = 7
13 # 8 = 3
8
Four semicircles are drawn in the interior
of a square using each side of the square as
a diameter. If the area of the square is 64,
find the area of the shaded region.
2
5
The # symbol is defined as an arithmetic
operation. Use the following three examples to figure out the value of 6 # 3:
The numbers 1–9 are written on nine index cards, one number per card. Arrange
the cards into three piles so that the sum of
the numbers in each pile is 15.
10
In ABC, AB = BC = 12. Side AC is extended through C, a length equal to itself to
a point D. Point E is on AB; DE intersects
BC at F; and BF = 8. Find AE.
7
Sixty-five candy bars were to be shared
by a group of children. Each Pascal bar
was cut in half, and one piece was given
to each child. Every Fermat bar was cut in
thirds, and each Mandelbrot bar was cut
in quarters, and these pieces were equally
distributed to the children. How many
children were in the group if all 65 candy
bars were shared?
11
Which design will allow a marble to roll
from A to B in the least amount of time?
m∠B
.
m∠A
AB;
AB;
DE
12
The numbers 1, 2, 3, 5, and 6 are written
on index cards, one number per card. The
cards are then placed in a hat. Three cards
are chosen from the five cards. The numbers on the three cards are added together.
How many different sums are possible?
13
BC
A rectangle with sides in the ratio of 3 to 5
is inscribed in a circle.
What
x − 12
12 ratio
+ 3xis−the
+ 6x of
− 12
the area of the rectangle to the area of the
circle?
11
1
3
+
−
=
log2 x 2 log25 x log8 x
 1  x
x x 3  −
=0
x
 x 
In how many three-digit numbers are at
least two of the digits the same?
17
When a cube is sliced with a single cut,
which of the following cross-sectional
polygons are not possible?
20
The number of days in a non–leap year
is not divisible by either 2 or 3. If the day
exactly in the middle of a year is a Sunday,
on what day of the week will the 1/3 day
and the 2/3 day fall?
24
Circle O, with radius OA = 5, and circle P,
24 radius PB = 7, have a common inter>with
nal tangent AB. If OP = 20, find AB.
1
log b x
A
scalene triangle, isosceles triangle,
equilateral triangle, nonsquare rectangle, isosceles trapezoid, nonisosceles
trapezoid, parallelogram, square,
nonsquare rhombus,
pentagon, hexagon, octagon
21
Factor
224 – 1
into its prime factors.
m∠B
.
m∠A
B
15
DE
BCall real values of x:
Solve for
m∠B
x − 12. + 3x − 12 + 6x − 12 > 24
m∠A
1
3
1
; +
AB11
−
=
log2 x 2 log25 x log8 x log b x
P
O
x −2
16
14
18
What is the set
; integer values for a
ABof
for which sin x • sin ax = 1 will have
a solution?
DE
DE
 1  x x −2
x x 3  −
=0
x
BC x 
19
Find the
b 12
that+ satisfies
the
12 + 3ofx −
6x − 12 >
24
x − value
following:
11
1
3
1
+
−
=
log2 x 2 log25 x log8 x log b x
BC
x − 12 + 3x − 12 + 6x − 12 > 24
 1  x x −2
=0
x x 3  −
x
 x 
22
23
26
27
11
1
3
1
+
−
=
log2 x 2 log25 x log8 x log b x
Stan’s nine-digit Social Security number
Find all x that satisfy this equation:
begins and ends with 5. It also has the
 1  x x −2
strange property that the sum of any three
x x 3  −
=0
x
 x 
consecutive digits is 12. Find the middle
digit.
25
What is the smallest natural number
divisible by each of the integers 1–10,
inclusive?
28
© National Council of Teachers of Mathematics, 1906 Association Drive, Reston, VA 20191-1502
Copyright © 2014 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.
This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
Solutions
to calendar
Looking for more
Calendar problems?
Visit www.nctm.org/publications/
calendar/default.aspx?journal_id=2
for a collection of previously published
problems—sortable by topic—from
Mathematics Teacher.
Edited by Margaret Coffey, Margaret
[email protected], Thomas Jefferson High
School for Science and Technology, Alexandria, VA 22312, and Art Kalish, artkalish@
verizon.net, Syosset High School (retired),
Syosset, NY 11791
The Editorial Panel of Mathematics Teacher
is considering sets of problems submitted by
individuals, classes of prospective teachers,
and mathematics clubs for publication in the
monthly Calendar. Send problems to the Calendar editors. Remember to include a complete solution for each problem submitted.
Problems 2, 6, 15, and 21 are based on exhibits at MoMath, the National Museum of
Mathematics, in New York City.
Other sources of problems in calendar form
available from NCTM include Calendar Problems from the “Mathematics Teacher” (a
book featuring more than 400 problems,
organized by topic; stock number 12509,
$22.95) and the 100 Problem Poster (stock
number 13207, $9.00). Individual members
receive a 20 percent discount off this price.
A catalog of educational materials is available at www.nctm.org.—Eds.
1. Equally likely. The probability of
a head is 1/2 on each flip of the coin.
Therefore, the probability of any particular sequence of heads and tails is 1/2
raised to the power of the number of
flips. In this case, each of the two events
has a probability of 1/32.
2. –ac. The slope of the line through
(a, a2) and (c, c2) is m = (c2 – a2)/(c – a) =
c + a. Therefore, the equation of this line
is y – c2 = (c + a)(x – c). The y-intercept
occurs when x = 0, so y – c2 = –c2 – ac,
and y = –ac. We can generalize as follows: A line that intersects the parabola
y = x2 at two points, one on either side of
the y-axis, will intersect the y-axis at the
point whose value is the negative of the
product of the x-coordinates of the intersection points.
3. {1, 5, 9; 3, 4, 8; 2, 6, 7} or {2, 4, 9; 1,
6, 8; 3, 5, 7}. The sum of the numbers in
each pile must be 15, so the 7, 8, and 9
must be in separate piles. The card with
the 6 cannot be in the pile with the 9. If
the 6 is with the 7, then that pile must
also contain the 2. The 5 cannot be with
the 8, since that pile would also need
the 2; thus, the 5 is with the 9 and the 1.
Placing the 6 with the 8 and then applying similar reasoning will result in a second solution. These two possibilities can
be found by completing a standard 3 × 3
magic square in which rows, columns,
and diagonals each add to 15.
4. 2 lbs. Let b = the weight of the box,
and let a = the weight of the aluminum
block. Then a + b = 8, and 3a + b = 20.
Subtract the first equation from the second to get 2a = 12 → a = 6. Therefore,
b = 2.
442 Mathematics Teacher | Vol. 107, No. 6 • February 2014
Copyright © 2014 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.
This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
Alternate solution: Let b = the weight
of the box. Then 3(8 – b) = 20 – b →
b = 2.
5. 7/16. Sketch a graph of the overlapping of the arrival times for the two people. The graph indicates that if the first
person to arrive comes at 6:00 p.m., he
will wait until 6:15 for the second person. The square represents the potential
times for the two people to arrive. The
probability that they meet is the area of
the shaded region divided by the area
of the square. If we convert units into
minutes, the area of the square is 602 =
3600. The area of the shaded portion
can be found by subtracting the areas of
the two unshaded right triangles from
the area of the square: 3600 – 2(0.5 • 45 •
45) = 3600 – 2025 = 1575. The probability is 1575/3600 = 7/16.
6. Catenary. A catenary is a curve
formed by a free-hanging chain or cable
of uniform weight supported at the two
ends. The hyperbolic cosine function
can be used to produce a graph of a catenary. (To see a square-wheeled tricycle
move along a catenary, go to https://
www.youtube.com/watch?v=LgbWu8z
Jubo. Visitors can ride on a square-
wheeled tricycle at MoMath, the
National Museum of Mathematics, in
New York City. See also Douglas Wilcock, “A Road for Every Wheel,” Mathematical Lens, MT Nov. 2012, vol. 106,
no. 4, pp. 254–59.)
7. 6. The average of the 5 scores is 10,
so the sum of the scores is 5 • 10 = 50.
Similarly, the sum of the 4 scores is 4 •
11 = 44. Therefore, the dropped score
was 50 – 44 = 6.
8. The sum = 7; the product = 10. Let
a and b be integers. Then (a + b)3 –
(a3 + b3) = (a3 + 3a2b + 3ab2 + b3) –
(a3 + b3) = 3ab(a + b). It follows that
3ab(a + b) = 210 = 2 • 3 • 5 • 7. Divide
both sides of the equation by 3. Then
ab(a + b) = 2 • 5 • 7. Since all the factors
on the right are prime, the three factors
on the left individually must equal 2, 5,
or 7. Therefore, a and b are 2 or 5 (in
either order), making the sum 2 + 5 = 7
and the product 2 • 5 = 10.
9. 15. Notice that, for any of the three
examples, a # b = c, a + b + c = 24.
Therefore, 6 # 3 = 24 – (6 + 3) = 15.
10. Send student solutions to the Problem of the Month editors, Seán Madden
([email protected]) or
Ricardo Diaz ([email protected]).
11. 60. Groups of 2, 3, and 4 children
were to share the candy, so the number
of children present had to be divisible by
all three numbers. The smallest number
of children would be 12. Under this condition, there would be 6 Pascal bars,
4 Fermat bars, and 3 Mandelbrot bars,
for a total of 13 bars shared. Since
65 candy bars were shared, the number
of children had to be 5 times as many:
5 • 12 = 60.
12. 8(4p – 8) = 32p – 64. Each petal of
the four-leafed shape can be cut in half
by a diagonal of the square. Construct
a radius from the midpoint of a side to
the center of the square. Then each half
petal is a segment of a circle of radius
4. The area of the segment is the area
of the quarter circle minus the area of
the isosceles right triangle: (1/4)(16p) –
(1/2)(4)(4) = 4p – 8. Since there are
8 such pieces that form the shaded
region, the total area of the shaded portion is 8(4p – 8).
B
8
12 E
Alternate solution: Two of the half
petals have an area equal to the area of
a semicircle minus one-fourth the area
of the square, or 8p – 16. Thus, 4 such
pairs would be 4(8p – 16) = 32p – 64.
x
A
F
4
4
G q C
p
D
15. Cycloid. The cycloid (brachistochrone) is the curve of quickest descent.
Tracing a point on the circumference of
a circle as it rolls without slippage along
a straight line will form a cycloid.
16. 8. There are 5C3 = 10 possible ways to
choose 3 cards from 10. However,
1 + 6 + X = 2 + 5 + X. Therefore, the
only possible sums are 6, 8, 9 (2 ways),
10 (2 ways), 11, 12, 13, and 14.
13. 2. Apply the law of cosines twice:
162 + 92 − 152 7
=
• •
1622 +99216
− 152 187
cos B =
=
2
2
• 16
2 • 916
182 7
and
2
2 +9
2 − 15
15B+=16 − 9
5 =
cos
cos A =
2 • 9 •=16 .
18
•
•2
1522 +151616
− 92 65
cos A =
= .
2 • 15
15• 216+ 162 −692 5
−1 7
cos A
=
= .
B cos
18
6
=
, 2 • 15 • 16
−−11 5
7
Therefore,
AB cos
cos 618
,−1
=
−1 5
7
A cos
B cos
6
18
,
=
−
1
AB
A cos 5
cos B =
( )
( ))
() ( )
()
17. 30/17p. Let the sides of the rectangle
be 3 and 5; its area is thus 15. The diagonal of the rectangle is 632 + 52 = 334.
Since the radius of the circle is one-half
the diagonal, the area of the circle is
6
AB
and a calculator
is needed to obtain 2.
AD
AB
(Note: The required
2
2 ratio2 will always
AD= 16 + 9 − 15 = 7
equal cos
2 ifBthe
sides
of
the triangle are
p + p3 2p2 • 9 •416
18
AD
formed using
integers
m and
where
=
→ px
= 8pn→
x=6
p
4
3
x
0<n<m
<
2n
and
the
sides
are
AB =
p + 3 22p
4
=15 +→
px
2
162 −BC
92== 8np
52 → x = 6
m2 – ncos
, AC
and
4 == pmn,
A
=4 .)
.
+x p3 23p
2 • 15
6 = 8p → x = 6
= • 16→ px
3
x
14. 6. Construct4the diagram
as
1 7
−draw
described
and
a
line
through
F parB cos 18
= that−1 intersects
,
allel to AB
AD
at
G.
Let
A cos 5
x = AE; p = AC =6CD; and q = GC. Use
the fact that GFC ~ ABC and that
FDGAB
~ EDA to set up the following
proportions: FG/q = 12/p and (p + q)/FG
= 2p/x. Replace FG with 4 to get q = p/3
AD
and substitute into the second equation:
I♥
Fibonacci
numbers.
MATHEMATICS
IS ALL AROUND US.
( )
()
p + p3 2p
4
=
→ px = 8p → x = 6
4
3
x
Vol. 107, No. 6 • February 2014 | MatheMatics teacher 443
32 + 52
2
2
34. 3 + 5
2
 3434.

17π
π
.
 =
2
 2 
2
 34 
17π
.
π
 of the
 =rectangle
The ratio of the area
to
2
2
30 
15
that of the circle=is then
.
17 π
17π
2
15
30
=
.
17 π
π
17
PC 2
18. 16. Construct PC parallel to AB and
PC
AB
extend OA to
intersect PC at D. Since
∠OAB is a right angle, ∠ODP is also a
AB
right angle, and
is a right trianOA ODP
gle. Apply the Pythagorean theorem
2
OD2 + DP 2 = OPOA
along with the knowlPC
edge that OD = OA + AD = 5 + 7 = 12
and DP = AB. We then find that AB2 =
PCAB = 16.
202 – 122 = 256, so
C
D
A
P
O
B
19. x < 1.2 or x > 4.5. First, discover all the
critical values in the equation. The critical
values are the numbers that make each
absolute value expression 0. They are 2, 4,
and 12. Place these on the domain number
line, dividing it into 2 segments and 2 rays.
Set up and solve four inequalities based on
the restricted domain:
When x < 2: (–x + 12) + (–3x + 12) +
(–6x + 12) > 24 → –10x + 36 > 24 →
x < 1.2
When 2 < x < 4: (–x + 12) + (–3x + 12) +
(6x – 12) > 24 → 2x + 12 > 24 →
x > 6 (no solutions)
When 4 < x < 12: (–x + 12) + (3x – 12) +
(6x – 12) > 24 → 8x – 12 > 24 →
x > 4.5
When x > 12: (x – 12) + (3x – 12) +
(6x – 12) > 24 → 10x – 36 > 24 → x > 6
Putting this together, we have solutions
when x < 1.2 or when x > 4.5.
20. 252. There are 900 three-digit
numbers. There are 9 • 9 • 8 = 648 threedigit numbers whose three digits are all
Solution 24
different. Therefore, 900 – 648 = 252
three-digit numbers that have at least
two of the same digit.
21. Nonsquare rhombus and octagon.
The ten different types of cross sections
that are possible are the scalene triangle,
isosceles triangle, equilateral triangle,
square, nonsquare rectangle, nonisosceles trapezoid, isosceles trapezoid, parallelogram, pentagon, and hexagon.
22. {. . . , 1, 5, 9, . . .} or numbers in
the form 1 + 4n, n ∈ integers. Since sin x
is bounded by –1 and 1, we have sin x •
sin ax = 1 only if both factors are 1 or –1.
When x = p/2, we get sin x = 1. For that
choice of x, sin ax = 1 requires a = 1 + 4n.
Thus, when x = p/2 and a = 4n + 1, each
of the factors equals 1, so the product
is 1. Similarly, the two original factors
will be –1 when x = –p/2 and again a =
1 + 4n. It follows that the product, sin x •
sin ax is also 1. (Noninteger solutions for
a are possible. For example, choosing x =
5p/2 and requiring that sin ax = 1 results
in additional noninteger values of a of the
form a = (1 + 4n)/5.)
23. 20. Apply a reciprocal relationship
for logs, loga b = 1/logb a, to get:
1
11log x 2 + log x 25 − 3 log x 8 = log x b
2

25 

→ log x 211 3  = log x (20)
8 

It follows that b = 20.
24. The 1/3 day is a Tuesday, and the
2/3 day is a Friday. Remove 5 days from
the 365 days in the year. The days to
remove are the first and last days, the day
in the middle, and the 1/3 and 2/3 days.
Doing so leaves 360 days to be divided
into 3 equal sections of 120 each. The middle section must be then divided into 2 sections of 60 days each, as shown. Since the
444 Mathematics Teacher | Vol. 107, No. 6 • February 2014
middle day is a Sunday and 61/7 leaves a
remainder of 5, the 1/3 day will be 5 days
before Sunday, which is a Tuesday; the
2/3 day will be 5 days after Sunday, which
is a Friday. (See solution 24 above.)
25. 32 • 5 • 7 • 13 • 17 • 241. Use the
methods for factoring the difference of
two squares and the sum of two cubes as
follows:
224 – 1
= (212 – 1)(212 + 1)
= (26 – 1) • (26 + 1)(24 + 1)(28 – 24 + 1)
= 63 • 65 • 17 • 241
= 32 • 5 • 7 • 13 • 17 • 241
26. x = 1 or x = 3. Applying the rules of
exponents and dividing both sides of the
equation by x, we obtain the following:
1
x
x3
− x x−4 = 0
−3
x x = x x−4
We can now equate exponents, so –3/x
= x – 4, or –3 = x2 – 4x → x2 – 4x + 3 = 0
→ (x – 3)(x – 1) = 0. The two solutions
are x = 1 or x = 3.
27. 2. Let the number be 5 a b c d e f g 5.
The sum of three consecutive numbers is
always 12, so a + b = 7, making c = 5. Similarly, f + g = 7, so e = 5. Since c and e are
both 5, the middle number, d, must be 2.
28. 2520. Start with 10 and count
down to 1 to find new factors. We need
to include 10, whose factors are 2 and
5. We also need to include 9, which
has two factors of 3. We do not need 8
since we already have a factor of 2, but
we must include the factor 4. We must
include 7 but can omit 6, 5, 4, 3, 2, and 1
because those are already factors of the
number we seek. Multiply to get 10 • 9 •
4 • 7 = 2520.