Supplement for 11-6
http://illustrativemathematics.org/standards/hs:
G-GMD, G-MG Tennis Balls in a Can
Alignments to Content Standards

Alignment: G-GMG.1.2
HSG
Geometry
Domain
HSG-GMD: Geometric Measure and Dimension
Cluster
Visualize relationships between two-dimensional and three-dimensional objects
Standard
Apply concepts of density based on area and volume in modeling situations (e.g., persons
per square miles, BTUs per cubic feet)

Alignment: G-MG.1.1
HSG
Geometry
Domain
HSG-MG: Modeling with Geometry
Cluster
Apply geometric concepts in modeling situations
Standard
Use geometric shapes, their measures, and their properties to describe objects (e.g.,
modeling a tree trunk or a human torso as a cylinder).
Tags
Tags: MP 4
The official diameter of a tennis ball, as defined by the International Tennis Federation, is at least
2.575 inches and at most 2.700 inches. Tennis balls are sold in cylindrical containers that contain
three balls each. To model the container and the balls in it, we will assume that the balls are 2.7
inches in diameter and that the container is a cylinder the interior of which measures 2.7 inches
in diameter and 3×2.7=8.1 inches high.
1. Lying on its side, the container passes through an X-ray scanner in an airport. If the
material of the container is opaque to X-rays, what outline will appear? With what
dimensinos?
2. If the material of the container is partially opaque to X-rays and the material of the balls
is completely opaque to X-rays, what will the outline look like (still assuming the can is
lying on its side)?
3. The central axis of the container is a line that passes through the centers of the top and
bottom. If one cuts the container and balls by a plane passing through the central axis,
what does the intersection of the plane with the container and balls look like? (The
intersection is also called a cross section. Imagine putting the cut surface on an ink pad
and then stamping a piece of paper. The stamped image is a picture of the intersection.)
4. If the can is cut by a plane parallel to the central axis, but at a distance of 1 inch from the
axis, what will the intersection of this plane with the container and balls look like?
5. If the can is cut by a plane parallel to one end of the can—a horizontal plane—what are
the possible appearances of the intersections?
6. A cross-section by a horizontal plane at a height of 1.35+w inches from the bottom is
made, with 0<w<1.35 (so the bottom ball is cut). What is the area of the portion of the
cross section inside the container but outside the tennis ball?
7. Suppose the can is cut by a plane parallel to the central axis but at a distance of w inches
from the axis (0<w<1.35 ). What fractional part of the cross section of the container is
inside of a tennis ball?
Commentary
This task is inspired by the derivation of the volume formula for the sphere. If a sphere of radius
1 is enclosed in a cylinder of radius 1 and height 2, then the volume not occupied by the sphere is
equal to the volume of a “double-naped cone” with vertex at the center of the sphere and bases
equal to the bases of the cylinder. This can be seen by slicing the figure parallel to the base of the
cylinder and noting the areas of the annular slices consisting of portions of the volume that are
inside the cylinder but outside the sphere are the same as the areas of the slices of the doublenaped cone (and applying Cavalieri’s Principle). This almost magical fact about slices is a
manifestation of Pythagorean Theorem. We see it at work in Part 6 of this task. The other parts
of the task are exercises in 3D-visualization, which build up the spatial sense necessary to work
on Part 6 with understanding. The visualization required here is used in calculus, in connection
with procedures for calculating volumes by various slicing procedures.
Submitted by James Madden on behalf of the participants in the Louisiana Math and Science
Teacher Institute On-Ramp.
Solutions
Solution: Solution
1. The shadow is a rectangle measuring 2.7 inches by 8.1 inches.
2. The shadow is a light rectangle (2.7 × 8.1 inches) with three disks inside. It looks like a
traffic light:
3. The image is similar to the previous one, but now only the outlines are seen:
4. The intersection with the container is a narrower rectangle. The intersections with the
balls are smaller circles. Because each ball touches the container along its whole
“equator,” the circles must touch the long sides of the rectangle:
5. The intersections are two concentric circles, except when w=0,2.7,5.4,8.1 and when
w=1.35,4.05,6.75 . In the former case, we see a circle (from the container) and a point
(where the plane touches a sphere). In the latter case, we see a single circle corresponding
to a place where the equator of a ball touches the container.
6. The intersection of the plane with the interior of the container is a disk of radius 1.35
inches. Its area is π(1.35) 2 in 2 . The intersection with the ball is a smaller disk that is
contained in the first disk. The radius r of the smaller disk is the square root of
(1.35) 2 −w 2 , as we see from the diagram below depicting the intersection of a plane
through the central axis of the container with the bottom ball. Thus, the area of the
smaller disk is π((1.35) 2 −w 2 ) . Accordingly, the area inside the larger disk but
outside the smaller is πw 2 , provided that 0≤w≤1.35 . (It is notable that the radius of
the ball does not appear explicitly in the expression for this annular area.)
7. Referring to Problem d), we see that we wish to find the ratio of the total area of three
congruent disks to the area of a rectangle, one of whose dimensions is equal to the
diameter of the disks. The same picture used in the previous problem, but interpreted as a
view from one end of the container, gives us the radius of the small disks — namely,
(1.35) 2 −w 2 − − − − − − − − − − √ , so the total area of the disks is
3π((1.35) 2 −w 2 ) . The area of the rectangle is
(8.1)2(1.35) 2 −w 2 − − − − − − − − − − √ . So, the ratio is
w2
3π((1.35) 2 −w 2 )(8.1)2(1.35) 2 –
− − − − − − − − − √ =π(1.35) 2 −w 2 − − − − − − − − − − √ 5.4
G-MG The Lighthouse Problem
Alignments to Content Standards

Alignment: G-MG.A.1
HSG
Geometry
Domain
HSG-MG: Modeling with Geometry
Cluster
Apply geometric concepts in modeling situations
Standard
Use geometric shapes, their measures, and their properties to describe objects (e.g.,
modeling a tree trunk or a human torso as a cylinder).
Tags
• This task is not yet tagged.
Suppose that even in perfect visibility conditions, the lamp at the top of a lighthouse is visible
from a boat on the sea at a distance of up to 32 km (if it is farther than that, then it is obscured by
the horizon).
1. If the "distance" in question is the straight-line distance from the lamp itself to the boat,
what is the height above sea level of the lamp on top of the lighthouse?
2. What are two other interpretations of the distance being investigated in this problem?
Describe how to solve the alternate versions.
Commentary
1. In addition to the purely geometric and trigonometric aspects of the task, this problem
asks students to model phenomena on the surface of the earth. This first step, observing
that the bound of 32km on the visibility of the lighthouse is due to the curvature of the
earth, will likely be a difficult part of the modeling process. Instructors should expect to
provide guidance for this part, and possibly also the diagram in the solution.
2. The instructor can also choose whether or not to provide the radius of the earth, or to
have students discover its relevance and then look it up in an available resource. Students
may discover that since the earth is not a perfect sphere, the "radius" of the earth varies
from one location to the next. As such, students may end up with one of many possible
such values, with which the calculations in the solution could be repeated verbatim.
3. This problem is freely translated from the math and physics problem column in La Presse
de Tunisie (October, 2006). It asked only the first question. La Presse is a general
circulation newspaper in Tunis and this was the easiest of the three math problems.
Presumably the first interpretation of ‘distance’ was intended in the original problem. But
the variants give an opportunity for students to see the differences in the three values and
use several trigonometric approaches.
4. It is tempting in b.ii to actually solve for θ and divide by cosθ . Not only is this more
work but it might introduce round off error.
5. One should note that the small differences in the three answers are dwarfed by the
approximations implicit in estimating the radius of the earth and the rounding off to even
kilometers in stating the problem. In particular, the calculated differences between part a
and b.i of less than 1/1000th of a meter and between b.i and b.ii of 1/100th of meter are
negligible.
Solutions
Solution: Original Solution
1. Consider the triangle with vertices C , the center of the earth, A , the lamp, and B , the
boat. The radius of the earth can be found on the internet as 6,371 km.
(Note that the diagram is drawn to a highly
disproportionate scale to aid with visibility.)
With notation as above, we have AC=6371+h , BC=6371 and AB=32 . Since we are
told that the lighthouse can be seen from up to 32 km, CBA is a right-angle. By the
Pythagorean theorem,
6371 2 +32 2 =(6371+h) 2 =6371 2 +2(6371)h+h 2 ,
which simplifies to h 2 +2(6371)h=32 2 , which in turn can be solved by the quadratic
formula:
h=−12742±12742 2 −4(1)(−(32 2 ) − − − − − − − − − − − − − − − − − √ 2 =.08036
3643 km or 80.36 meters.
2.
For the sake of measuring the discrepancy in our answer coming from different
interpretations of "distance," we maintain a grossly unreasonable level of precision for
this part of the problem.
Let D be the base of the lighthouse, R the radius of the earth and continue the notation
of part a). Let θ be the angle between ADC and BC . Then RR+h =cosθ so
h=Rcosθ −R .
The other two interpretations are given below.
1. Take the distance to be the length of the circular arc from D to B . Then
θ360 =322π6371 or θ=360⋅322π6371 =.2877829139 degrees.
Now the height of the tower is 6371cos(.2877829139) −6371=.0803649948 km , or
approximately 80 meters.
2. Take the distance to be the straight-line distance from the boat to the base of the
lighthouse, so BD=32 . We calculate the angle θ using the law of cosines
applied to the triangle BCD . We have BD 2 +DC 2 =2(BC)(BD)cosθ . So,
cosθ=BD 2 −(BC 2 +DC 2 )−2(BC)(BD)
or in our case
cosθ=32 2 −(2R 2 )−2R 2 =.9999873859
The height of the lighthouse is 6371.9999873859 −6371=.0803654448 km or 80.37
meters.
For comparison purposes, we continue to find θ . Substituting,
cos −1 (θ)=.9999873859 we see θ=.2877832164 degrees which agrees to 4
places with the arcsin calculation.
G-MG Toilet Roll
Alignments to Content Standards

Alignment: G-MG.A.1
HSG
Geometry
Domain
HSG-MG: Modeling with Geometry
Cluster
Apply geometric concepts in modeling situations
Standard
Use geometric shapes, their measures, and their properties to describe objects (e.g.,
modeling a tree trunk or a human torso as a cylinder).
Tags
• This task is not yet tagged.
Picture a roll of toilet paper; assume that the paper in the roll is very tightly rolled. Assuming
that the paper in the roll is very thin, find a relationship between the thickness of the paper, the
inner and outer radii of the roll, and the length of the paper in the roll. Express your answer as an
algebraic formula involving the four listed variables.
Commentary
The purpose of this task is to engage students in geometric modeling, and in particular to deduce
algebraic relationships between variables stemming from geometric constraints. The modelling
process is a challenging one, and will likely elicit a variety of attempts from the students.
Teachers should expect to spend time guiding students away from overly complicated models.
Similarly, the task presents one solution, but alternatives abound: For example, students could
imagine slicing the roll along a radius, unraveling the cross-section into a sequence of trapezoids
whose area can be computed.
Solutions
Solution: A geometric model
We begin by labeling the variables, for which the above diagrams may be useful. Let t denote
the thickness of the paper, let r denote the inner radius, let R denote the outer radius and let
L denote the length of the paper, all measured in inches. We now consider the area A ,
measured in square inches, of the annular cross-section displayed at the top of the first image,
consisting of concentric circles. Namely, we see that this area can be expressed in two ways:
First, since this area is the area of the circle of radius R minus the area of the circle of radius r ,
we learn that A=π(R 2 −r 2 ) .
Second,iIf the paper were unrolled, laid on a (very long) table and viewed from the side, we
would see a very long thin rectangle. When the paper is rolled up, this rectangle is distorted, but - assuming r is large in comparison to t -- the area of the distorted rectangle is nearly identical
to that of the flat one. As in the second figure, the formula for the area of a rectangle now gives
A=t⋅L .
Comparing the two formulas for A , we find that the four variables are related by:
t⋅L=π(R 2 −r 2 ).