POW
The vertices of quadrilateral ABCD are at the points A:(1,2), B:(2,-2), C:(-5,-4), and D(-6,0). What's the most
specific name you can give to this quadrilateral?
This grid might be helpful:
Extra: What's the area of quadrilateral ABCD?
810: That's A Lot of Rock!
Last weekend I drove through a tunnel and realized that a lot of rock would need to be removed from the inside
of a mountain to make a tunnel.
The tunnel I drove through was approximately parabolic in shape, and was about 5 meters wide and 10 meters
high. My car's odometer was broken, so I didn't know the length of the tunnel; however, I do know that I was
driving at the posted speed limit of 50 miles per hour, and it took me 3 minutes to get through the tunnel.
Based on this information, what is the volume of the tunnel?
Bonus: What is the longest tunnel in the world?
Comments
Although this was a traditional calculus problem, only about half of the submissions involved taking an integral
'by hand'. Other methods used were 4/3 the area of the largest inscribed triangle, 2/3 the product of the base and
POW
height, approximating areas by using triangles and trapezoids, and using the graphing calculator to find the
integral of the parabola function. Many of the highlighted solutions below show these methods.
The most common mistakes made were mixing metric and English units, and using cross-sections of rectangles
or ellipses rather than cross-sections of a parabola. Another mistake made by several submitters was converting
from cubic meters to cubic km. Remember that a cubic km is really a cube 1000 meters on a side, so to convert
from cubic meters to cubic km, you need to divide by 1000^3, not 1000.
I received several different answers to the bonus, and there was some confusion as to what would be classified
as a tunnel. Many people thought that the longest tunnel was the English Channel Tunnel or Chunnel; however,
it is about 2 miles shorter than the Sei-kan Tunnel in Japan. This tunnel was completed in 1988 and is about
53.6 km or 33.3 miles long. To read more, visit the Web page The Seikan Tunnel from PBS.
Thanks to everyone who submitted a solution to this problem.
Highlighted solutions:
From: Carlos Adriel Salmerón Arroyo, age 18
School: Facultad de Ciencias, U.N.A.M., México, Distrito Federal, México
Volume=134 083.333m^3
I resolved this problem using Analytic Geometry and Calculus. I´ve
taken the point on the top of the parabolic segment as H(0,10) and
the points in both sides on the flor P'(-5/2,0) and P(5/2,0) (the
extreme equidistant points to the origin (0,0); on the axis x).
First, I determinated the equation of the parabola to get a function
of x; then I solved the integral:
5/2
(integral)f(x)dx=Area
-5/2
The general form of this kind of parabola is:
y-y'=-4p(x-x')^2
(I'll take k=-4p)
Where (x',y') is the superior vertex (0,10).Then: y-10=kx^2
Using P(5/2,0) or P'(-5/2,0):
-10=k(5/2)^2=25k/4
k=-10(4)/25=-40/25=-8/5
Now: y-10=(-8/5)x^2
y=(-8/5)x^2+10=f(x)
To get the area:
5/2
5/2
Area=(integral)f(x) dx=(integral)((-8/5)x^2+10)dx
POW
-5/2
-5/2
0
5/2
But: (integral)f(x)dx=(integral)f(x)dx because the parabola is
-5/2
0
symmetric to axis y. So:
5/2
5/2
Area=2*(integral)((-8/5)x^2+10)dx=2*(-(8/5)(1/3)x^3+10x)]=
0
0
5/2
=2(2x)(5-(4/15)x^2)] = 4(5/2)(5-(4/15)(25/4))=10(5-(25/15))=
0
= 10(5-(5/3))=10(15-5)/3=10(10)/3=100/3m^2
Driving 50milles/hour:
50milles/1hour=50mill/60min=5mill/6min=(5/6)mill/1min=
=(15/6)mill/3min=(5/2)mill/3min
So, te length of the tunnel in meters is:
1mill=1609m, (5/2)mill=1609(5/2)m
The volume will be:
Volume=(100/3)(5/2)1609m^3=134 083m^3
BONUS:the longest tunnel in the world is the "Eurotunnel" (Eurotúnel
in spanish) wich connect France with Great Britain.
From: Keith Chan, age 17
School: James Ruse Agricultural High School, Carlingford, Australia
Volume of tunnel is 134112 cubic meters.
Firstly we can use Simpson's Rule to find out the area of the
parabolic cross-section. We simply place this parabola with the road
on the x-axis (thus f(a) and f(b) = 0) and f((a+b)/2) = height
Area = (b-a)/6 * [ f(a) + 4f((a+b)/2) + f(b) ]
= 5/6 * [ 0 + 4*10 + 0 ]
= 100/3 m^2
Now, for the length of the tunnel, Distance = Speed * Time
Length = 50 * 3/60
= 2.5 mi
= 4023.36 m
POW
Volume = 4023.36 * 100/3
= 134112 m^3
The world's longest tunnel is Japan's Sei-kan tunnel
POW
1182: The Supermarket Shelf
You are the manager of a supermarket, and you are working on rearranging some of the products in your store. Your
main concern is where to place the most expensive items. You decide that putting the items on a middle shelf is
probably a good idea. That way customers won't have to stretch to reach the items off a top shelf or strain their backs
when they bend to reach an item from a bottom shelf. However, what is the optimal height for a middle shelf? Is it 3
feet off the ground? Or 4 feet off the ground? Or somewhere in between?
Luckily, one of your new employees is a math major just starting his summer break. You enlist him to help
figure out the answer to your question. You have the following data on customers, collected by a previous
employee who was a statistics major on summer break.

A typical customer stands about 3 feet away from the shelf when looking at items.

When a customer looks at an item, his or her head is bent at an angle 15 degrees below the horizontal.

A female customer's eyes are about 59 inches above the ground, and a male customer's eyes are about 64
inches off the ground.
Based on this information, your math major employee determines the optimal shelf height. What is the optimal
shelf height? Round your answer to the nearest inch and be sure to show work to support your answer.
Comments
The most common solution methods were to use the tangent ratio or to use the Law of Sines. The tangent ratio can be
seen in solutions by Stephen, Mai, and Giacomo. Giacomo's solution also included an uploaded diagram. Bob's solution
shows how the Law of Sines can be used and includes a diagram of his solution.
The most common mistakes made were with having calculators in the wrong mode, and not rounding the
answer to the nearest inch.
Thanks again to everyone who submitted a solution.
Highlighted solutions:
From: Giacomo Boccardo, age 17
School: Liceo Scientifico Righi, Bologna, Italy
The optimal height is 52 inches.
Look at the picture!
First I convert the customer's distance from the shelf in inches:
AM=BF=HK= 3 feet = 36 inches
I consider the triangles AMN and BFG:
they are obviously egual
AM=BF=36 inches
POW
AMN=BFG=15°
AN
BG
Tan(AMN)=--=Tan(BFG)=-AM
BF
AN=BG=Tan(AMN)*AM=Tan(15°)*36=9.65 inches
NK=MH-AN=54.35 inches
GK=FH-BG=49.35 inches
The optimal shelf height is between the average of the two height:
Height=(54.35+49.35)/2=52 inches
The optimal height is 52 inches.
From: Stephen Emeneau, age 17
School: Parkview Education Centre, Bridgewater, Canada
The shelf should be 52 inches above the ground
The person stands 3 feet from the shelf, that's 36 inches, and looks
down at 15 degrees. if you let x represent the number of inches down
the shelf the person looks then:
tan15 = x/36
x = 36 * tan15
x = 9.6 inches
since a males eyes are 64 inches above the ground we can calculate
the distance from the ground his eyes will spot the shelf.
64 - 9.6 = 54.4inches
since a females eyes are 59 inches above the ground we can calsulate
the distance from the ground her eyes will spot the shelf.
59 - 9.6 = 49.4inches
assuming the same number of males atend the supermarket as females we
take the average of the two previous numbers to get the best shelf
POW
height:
(54.4 + 49.4)/2 = 51.9 inches or approximately 52 inches
POW
906: Cup Capacity
The Plastic Container Supply Company has hired you to help in the design of its plastic cups.
You have developed a large plastic cup that is 3.75 inches tall. The bottom of the cup is a circle with a diameter
of 1.75 inches, and the top of the cup is a circle with a diameter of 2.75 inches.
Using calculus, determine the volume of the cup.
Bonus: What United States company is credited with creating the first paper cup?
Comments
This was a very challenging problem for many people. Many correct answers were given by working with volumes of
cones; however, calculus was required in order to receive credit for a correct solution.
The answers shown below are essentially the same, but each emphasizes a different aspect of the solution.
Reading them carefully can help you to understand why the disk method works in solving this problem.
Thanks to everyone who submitted a solution to this problem.
Highlighted solutions:
From: Shyam Sitler, age 17
School: DeLand High School, DeLand, FL
15.156 inches^3
First we must find a line so that when it is rotated around the x-axis
it produces a cone. The cup must be able to fit exactly in one section
of the cone. The volume of this section (equal to the volume of the
cone) can then be found be using the disk method.
First convert diameters to radii.
1.75in.=2r
2.75in.=2r
r=.875in.
r=1.375in.
Then fit to line.
y=ax
(.875)=a(x)
x=(.875/a)
(1.375)=a(3.75+x)
1.375=3.75a+xa
1.375=3.75a+(.875/a)a
1.375=3.75a+.875
.5=3.75a
a=(.5/3.75)
a=(1/7.5)
;this is the first x vlue
;add height to the first x-value to get the
second x value
POW
y=(1/7.5)x
Now we must find the points to integrate.
x=(.875/a)
x=.875/(1/7.5)
x=6.5625
;first value
second x=(first x+3.75)
second x=(6.5625+3.75)
second x=10.3125
;second value
Now we must use the disk method
Evaluate y=(((1/7.5)x)^2*pi)dx from 6.5625 to 10.3125
The volume equals 15.156 inches^3
From: David Saunders, age 18
School: Greenwich High School, Greenwich, CT
The first paper cup is credited to Hugh Moore's Dixie Cups company
and the volume of the cup is 15.154 inches cubed.
We begin by placing the cup on a set of axis. We do this such that
the bottom of the cup lies on the y-axis, and the center of the
circle which is on the bottom has the x-axis passing through it, thus
placing half of the cup in the first quadrant, and half of the cup in
the fourth quadrant of the axis. To find the volume of this cup, we
must first determine the linear equation which represents the side of
the cup. We can do this by using the diameters of the top and bottom
of the cup. Since we know the x-axis passes through the center of the
cup, if we were to find the radius of the top and bottom, we could
determine two points on the graph so we may determine a linear
equation. We first must conclude that if the bottom of the cup is at
x=0 that the top of the cup is at x=3.75 since the cup is 3.75 inches
tall.
Next, we find the radius of the bottom of the circle using:
R = (.5)(D) Where R = radius and D = Diameter
We then plug in our values
R = (.5)(1.75in)
R = .875in
Since we know the bottom of the cup lies on the y-axis, we know that
its x coordinate is 0, therefore our first point is (0, .875) since
the radius represents the maximum y-value in the first coordinate for
our graph.
Next, we do the same to find the point using the top of the cup
R = (.5)(2.75in)
R = 1.375in
So we can make our second point (3.75, 1.375) using the x coordinate
which is the top of the cup (as prieviously explained) and using the
POW
radius as the y value. We now must find the linear equation for the
side of the cup. We begin by finding the slope which is:
(Y(1) - Y(2))/(X(1) - X(2))= slope
Where X(1), Y(1) = (3.75, 1.375) and
X(2), Y(2) = (0,.875)
Thus:
(1.375in - .875in)/(3.75in - 0in) = Slope
.5in/3.75in = slope
slope = .1333
We also know the y value where x = 0, is .875, the radius of the
bottom of the cup, and this value becomes the y-intercept. So, the
equation for the side of the cup using our computed slope and yintercept is given by:
y = .1333Xin + .875in
We now apply calculus to our algebra and geometry above. We know that
a cup is essentially a series of connecting circles one upon another
to form a cylinder. The volume for this cylinder may be found using
the formula for a solid of revolution - or a part which, rotated
around the x-axis creates a three dimensional shape, such as our cup.
the formula for this is:
V = the integral between the original x value and ending x value of
the formula to compute the area of part of the solid.
Although sounding confusing, simply put, the area of a circle is
given by:
A = (pi)(r^2), this then becomes our formula to compute the area of
part of the solid. Yet, further, we must replace r with our linear
equation for the side of the cup because between the top and bottom
of the cup, the radiuses are changing. so we get the formula:
A = (pi)(.1333Xin +.875in)^2
We then find the integral of this function within the range between
0, the bottom of the cup, and 3.75, the top of the cup. So our
formula for the volume of the cup is:
V = integral between 0 and 3.75 of (pi)(.1333Xin +.875in)^2
Thus
V = 15.154 in^3
And so, the volume of the cup is 15.154 inches cubed. While the first
paper cup is credited to Hugh Moore's Dixie Cup Company which we
remember best for those paper cups we loved as children which
featured sesame street characters or little riddles.
POW
Circles and Tangents
This is a problem that a student sent me a few years ago. You'll need to draw a good picture!
AOD is a diameter of circle O. B is any point on the circle. At B, a tangent is drawn to the circle. From the
center, O, a line is drawn parallel to AB, meeting the tangent at P. Prove that PD is tangent to the circle.
Comments
Now this was a tough one! I think the word "proof" scared a lot of you off, as we only got 317 responses - and only 85 of
those folks got this one right. You have all written a lot of proofs this year; we just didn't call them that. Don't be afraid
of the word "proof," and don't think that it means you have to write one of those two-column things or you won't get
credit. When we say "prove," we mean "write a convincing argument as to why something is true."
The basic strategy in solving this problem is to show that PD is perpendicular to OD, since we know that
tangents are perpendicular to the radius at the point of tangency. To do this, you probably want to find
congruent triangles. Let's look at a few ways you might do that.
Claire Ripsteen of Piedmont High School and Marc Azar of Sagesse High School both used a very handy circle
theorem to head towards those congruent triangles. If two parallel lines cut through a circle, the two arcs they
make are congruent. From there they used vertical angles to get SAS, and off they went. Both of their proofs are
what I would call "modified two-column." They don't have the two columns, but they give a step and state a
reason pretty clearly.
Al Phelps of Algonquin Regional High School, Daya Alderfer of the American Embassy School, and Pete Stein
of Akiba Hebrew Academy all used the isosceles triangle, alternate interior angles, and corresponding angles to
get their congruent triangles. Al and Daya wrote your basic two-column proof, while Pete wrote his in
"paragraph form." That means that instead of outlining the separate steps, he wrote complete sentences that
explained how to show that the triangles are congruent. I like Pete's picture as well.
Alex Mittal of Greenwich High School got the necessary congruent angles by using angle addition theorems.
While others did this as well, it wasn't nearly as popular as the first two methods. He provided a nice picture,
too - I like that it isn't exactly to scale, which forces you to concentrate on what you know, not what you see.
There were a few common mistakes that people made. A number of people tried drawing a picture and stated,
from that, that it was obvious that PD only hit the circle at D, so it was a tangent. You need to remember that
you can't ever measure or study a drawing to get the right answer. You can start that way to gain some insight
into the problem, but then you must use what you know about the figures involved to find the relation. Still
others stated that BD is perpendicular to OP. We don't know that right off. You can prove it, but you can't just
state it. Similarly, you can't say that BPDO is a kite, which more than a few people did.
One thing that quite a few people did was to set the drawing up so that BO was perpendicular to AD. That sure
makes everything easy, but it isn't what the problem says. B is any point on the circle. That meansthat before
you go too far with your explanation, you should try drawing it in a few different places to make sure that
everything still works.
POW
A lot of people had problems knowing what to put in the "Answer" box, where we ask for a complete sentence.
You don't want to put the whole proof in there, obviously. A good thing to try is something like what Daya
wrote: "I proved that PD was a tangent by doing a two column proof. I proved that it was a tangent by using
concepts of parallel lines, Isoceles triangles, circles, congruent triangles and tangency." That gives the reader an
idea of what they're going to read in your explanation, so they're well prepared.
One student asked the question of whether or not these types of tangents are related to the ones that we learn
about in trig. They are indeed! The "tangent" of an angle is the actual length of a segment that is tangent to a
"unit circle" (a circle with a radius of 1). You can read more about this in Tan 90, an answer in our Dr. Math
archives.
- Annie
Highlighted solutions:
From: Claire Ripsteen, age 15
School: Piedmont High School, Piedmont, CA
PD IS tangent to the circle.
Here are the steps that I used in order to prove that PD truly is
tangent to the circle:
In order to prove that PD is tangent to the circle, I went through a
couple of steps described below.
1. Because that line that goes through points O and P is parallel to
chord AB, arcs BG and AK are congruent.
2. Therefore, angles BOG and AOK are congruent.
3. Angle AOK is congruent to angle GOD because they are vertical
angles.
4. OB is congruent to OD because they are both radii of circle O.
5. OP is congruent to OP because of the reflexive property of
equality.
6. Therefore, triangles BOP and DOP are congruent based on SAS.
7. Therefore, angle ODP is congruent to angle OBP because
corresponding parts of congruent triangles are congruent.
8. Because angle OBP is a right angle and because BP is tangent to
the circle, PD is also tangent to the circle.
NOTE: See my attached picture.
POW
POW
Making Money at the Movies
You are the business manager at a local movie theater. Based on past ticket sales, you know that the theater averages
5,000 customers per day. The current ticket price for all show times is $5 per person. There are no discounted prices for
children, students, or senior citizens.
The general manager of the theater wants to increase prices in order to increase profits. However, he is afraid
that if the price is increased too much, he could end up losing money. He has asked you to determine how much
he should raise the price in order to maximize his profit.
The current profit is $2 per ticket. Any increase in ticket price would be a direct increase in profit.
You survey customers randomly over a period of two weeks. Your results show that you would lose 800
customers for every 50-cent increase in ticket price. For example, if you increased the ticket price to $6, you
would lose 1600 customers.
What new ticket price would lead to a maximum profit?
Be sure to include a complete and detailed explanation of what you did to solve this problem.
Comments
Thanks to everyone who submitted a solution. There are several ways to solve this problem. Some of the solutions can
be seen below.
The most common solution method was to use the first derivative of a profit equation. This can be seen in
solutions from Jenny, David, Gordon, Jennifer, Le'Mark, and James. In addition, Jennifer used the second
derivative to test for a maximum, and James evaluated the function on each side of the maximum. Le'Mark's
solution also shows a graph of the function.
The most common mistake made by people using this method was to assume that for a decrease in ticket cost,
there would be a corresponding increase in the number of customers. However, the survey only asked people
what they would do if the price increased; it did not ask what they would do if the price decreased.
Others realized that the function was a parabola, and found the vertex of the parabola. This can be seen in Olga's
solution. Garret and Burt also used the idea of zeros of a parabola to solve this problem.
Jenny's and Carolyn's solutions involve developing a table to look at how the profit changes as the ticket price
changes.
Thanks again to everyone who submitted a solution to this problem.
Highlighted solutions:
From: Jenny Taylor, age 16
POW
School: Fontainebleau High School, Mandville, LA
The new ticket price that would lead to maximum profit would be $5.56.
To begin with, I read through the problem and set up a general word
equation for finding the answer:
Profit = (ticket price - 3) x (number of customers)
I developed this equation because the current profit is $2 on a
$5-ticket per customer. This means that the other $3 is used towards
the necessary costs of the movie theater. The total profit equals the
ticket price minus the $3 to run the theater, multiplied by the number
of customers that attend the movie theater.
Then I came up with an equation for the number of customers that
attend the movie theater:
number of customers = 5000 - 1600(ticket price - 5)
I wrote this equation using the information about losing 1600
customers per $1 increase of ticket price after the original ticket
price of $5 with the 5000 being the typical number of customers at the
movie theater.
I substituted the second equation into the first:
Profit = (P - 3) x [5000 - 1600(P - 5)]
P = ticket price
I simplified the equation to:
Profit = (P - 3) x (5000 - 1600P + 8000)
Profit = (P - 3) x (13000 - 1600P)
I divided both sides of the equation by 1000 to get the profit of the
movie theater in thousands of dollars.
Profit = (P - 3) x 1000 x (13 - 1.6P)
Profit / 1000 = (P - 3) x (13 - 1.6P)
I multiplied the equation out to get:
Profit / 1000 = -1.6(P^2) + 17.8P - 39
I rewrote the equation to:
y = -1.6(x^2) + 17.8x - 39
I took the derivative of the equation:
dy / dx = -3.2x + 17.8
Then I plugged 0 in for y (because to get the maximum point of the
parabola, which is the highest profit, the slope is 0) to get:
0 = -3.2x + 17.8
I simplified the equation and solved it:
-17.8 = -3.2x
POW
x = 5.5625
x = P = ticket price
Therefore the ticket price that would yield the most profits would be
$5.56.
The movie theater profit would be:
Profit = (P - 3) x (13000 - 1600P)
= (5.5625 - 3) x [13000 - (1600 x 5.5625)]
= (2.5625) x (13000 - 8900)
= (2.5625) x (4100)
Profit = $10506.25
POW
The Joke - posted July 21, 2008
Recently a friend forwarded an email to me that contained a joke he thought I would appreciate. You probably
get the same sort of thing from time to time, too. It got me thinking about the whole idea of forwarding jokes on
the internet.
Suppose someone writes a really funny joke and emails it to 50 people. Assume that everyone who receives the
joke then forwards it to 8 of their friends, who each forward it to 8 of their friends and so on. Answer the
following questions:
1. If the first forwarding is considered to be the first stage (400 emails are sent), how many people will
receive the joke on the fourth stage?
2. Write a formula that will allow you to easily determine how many people will receive the joke on the
nth stage.
3. Use your formula to calculate how many people will receive the joke on the sixth stage.
4. The world population is currently estimated at 6,370,000,000 people. On what stage would the number
of messages being forwarded exceed the population of the world?
Extra: In reality, not everyone forwards a joke when they receive one. They may think it's not funny, or they
may object to clogging the Internet. Also, a person may receive a joke again as it circulates, and even if they
forwarded it the first time they won't do it again because they have already sent it to their friends.
Suppose you write a new joke that you think is really funny, and you send it to 50 friends. The joke then begins
to get forwarded. In stage one, 70% of your 50 friends think the joke is funny and each of them forwards it to 12
people. However, 1/5 of the people they forward it to also saw it when you first sent it, so they don't forward it
again. Assume that this continues for each stage of forwarding, with one-fifth of the recipients having already
seen it and 70% of those for whom it is new thinking it's funny enough to forward to 12 people each.
After 9 such stages of forwarding, how many distinct people will have seen your joke at least once?
POW
Print This Problem | Teacher Support Page
Summer: Train Travel - posted July 28, 2008
[French version] [Spanish version]
A train one mile long travels at 30 mph. It enters a tunnel one mile
long at 4 pm. At what time does the end of the train emerge from the
tunnel?
Extra: A train a mile and a half long takes a minute and a half to go through a tunnel a
mile and a half long. How fast is it going?
Filling Glasses - posted September 1, 2008
If we were to fill a glass with water at a constant rate (for example, 1 cubic inch per minute), we could graph the
height of the water in the glass as time goes by.
Suppose we fill the three glasses below in such a manner. Match up each glass with the graph that best
describes the height of the water in the glass over time.
Glass A
Glass B
Glass C
POW
Graph 1
Graph 2
Graph 3
Graph 4
This problem may seem different from the types of problems that you're used to, but don't assume that that
means you can't do it. Think about how the shape of the glass changes the rate at which the water rises. If you're
stuck, no problem. Tell us what you are thinking and make sure to leave a comment about the help you want.
Extra: Describe the glass that would correspond to the leftover graph.
POW
Summer: Seeing Stars - posted July 28, 2008
Below are some "regular stars". They're formed by extending the edges of regular polygons.
We could show that the angles at the star points of the regular hexagonal star (indicated by the *) have a
measure of exactly 60 degrees.
1. What's the angle measure of the star points of the regular five-pointed star?
2. What's the angle measure of the star points of a regular eight-pointed star?
Extra: What's the angle measure of the star points of a regular n-pointed star?