Problem Of The Month

Problem March/April 2015

Proposed by student Linh Nguyen

Given a parallelogram with the following characteristics:

1) The points E and F are the midpoints of AD and CD respectively,

2) G is the point of intersection of BE and AC,

3) H is the point of intersection of BF and AC,

4) The area of the parallelogram is 24.

Find the area of the region EFHG.  See picture below.

PoMMarchApril

Solutions to this problem are due by 11:59pm on April 10 to Dr. Heidi Hulsizer (hhulsizer@hsc.edu).

Problem #33, April 2014

Find two distinct sets of 3 non-zero real numbers that have the same sum
and the same product. That is, find sets {a, b, c} and {d, e, f } such that
{a, b, c} does not equal {d, e, f }, but a + b + c = d + e + f, and abc = def .

Due Date: Wednesday, April 30, 11:59 PM

Problem of the Month rules:

1. Submissions must include the answer along with a valid mathematical justification.
2. Solutions must be submitted by individuals only, not groups.
3. Only one solution per solver.
4. The use of calculators and computers is allowed.
5. Solutions must be your own work.

The Problem of the Month is sponsored by the Hampden-Sydney Math/CS Club. There are many benefits to solving the Problem of the Month, including

• immortality: as a correct solver, your name will be enshrined for e-eternity on this web page!
• riches: as a correct student solver, you have a chance to win a valuable prize awarded at the end of the year to the student solver with the most correct solutions!

Solutions should be submitted in writing or by email to Dr. Pendergrass or Dr. Lins.

Previous Problems of the Month

Problem #32, March 2014

March 2014 Problem

Problem #31, February 2014

The hands of a clock coincide at exactly 12:00:00.  When is the next time that they will coincide?

Problem #30, December 2013

The date November 29, 2013 is a prime date, because when it is expressed in the format (MMDDYY) you get 112913 which is a prime number.  Which dates are prime this month?

Problem #29, November 2013

You are on a row boat in the middle of a large circular lake.  On the shore a hungry monster is watching you row.  The monster can move four times faster than you can row, but you know that you can outrun it on dry land.  If you can get to a place on the shore before the monster does, then you can run away safely.  But the monster is smart and he is following your boat.  He will try to catch you wherever you land before you can run away.  Are you doomed?

Problem 29 was correctly solved by Linh Nguyen

Problem #27, September 2013
September 2013 Problem

Problem #26, May 2013
May 2013 Problem

Problem #25, April 2013
A person took an amazing journey: starting at a certain point on the Earth’s surface, this person traveled 50 miles due south, then 50 miles due east, then finally 50 miles due north.  At the end, he was right back at the starting point!  Describe the set of all points on the Earth’s surface from which this journey is possible.

Problem #24, February 2013

February 2013 Problem

Problem #23, January 2013

Suppose that you have just opened a brand new 1000 piece jigsaw puzzle. When assembled, the puzzle pieces will fit together so that there are 25 rows with 40 pieces in each row.

If you select two pieces at random from the box, what is the probability that they will fit together?

Problem #22, December 2012.
Suppose that I have a dark gray circle with radius 1, surrounded by n other light gray circles all with radius r. Each circle just barely touches its neighbors. Find a formula for the radius r as a function of the number of light gray circles.

Problem #21, September 2012.

The following problem was taken from Terrence Tao’s blog.

Suppose there is a very long terminal in an airport. You must walk from one end to the other in order to catch your flight. There are several moving walkways to help you along. Unfortunately, you need to stop and tie your shoe. Would it be faster to tie your shoe while you are on one of the walkways, or to tie your shoe off of the walkways?
You may assume that you can walk while you are on the walkways, and that no one is blocking your way.

Due Date: Sunday, September 30, 11:59.99 PM

Problem #20, April 2012.

13 pirates find a treasure chest filled with gold coins.  When they try to divide the coins evenly amongst themselves, there is one coin left over.  In the ensuing fight over the last coin, one pirate is slain.  When the 12 remaining pirates try to divide the coins evenly, there is still one coin left over.  Once again the pirates fight and a pirate is slain.  When the 11 surviving pirates try to divide the coins, they are relieved to discover that  the coins can now be evenly divided.  What is the smallest number of coins that could have been in the treasure chest?

Due Date: Monday, April 30, 11:59.99 PM

Correct solutions to problem #20 were submitted by Logan Spraker, Tanner Clement, Watt Mountcastle, Stuart Welch, Clifton Hudson, Shawn Stum, Luke Driscoll, Kyle Davis, Mitch Cavallarin, and Arley Morelock.

The Problem of the Month is sponsored by the Hampden-Sydney Math/CS Club. There are many benefits to solving the Problem of the Month, including

• immortality: as a correct solver, your name will be enshrined for e-eternity on this web page!
• riches: as a correct student solver, you have a chance to win a valuable prize awarded at the end of the year to the student solver with the most correct solutions!

Solutions should be submitted in writing or by email to Dr. Pendergrass or Dr. Lins.

Problem of the Month rules:

1. Submissions must include the answer along with its mathematical justification.
2. Solutions must be submitted by individuals only, not groups.
3. Only one solution per solver.
4. The use of calculators and computers is allowed.
5. Solutions must be your own work.

Problem #19, December 2011.

Two holiday-spirited students are building a Christmas tree out of stacked soda cans.  The base level of the tree is an equilateral triangle, and each subsequent level is also an equilateral triangle, where the centers of the cans are positioned over the gaps between the cans in the previous level.  The picture below shows a tree with three levels.  There are six cans in the base level, and a total of ten cans in all.  How many cans are in a Christmas tree with n levels, where n is an arbitrary positive integer?

Problem #18, November 2011.

There are 100 envelopes in a box.  Each envelope contains a different number.  The numbers could be large or small, rational or irrational – anything goes as long as they are all different.  Among all the envelopes in the box, the one with the largest number is the winner.  If you turn it in, you will receive $100. For any other number, you get nothing. The rules for the game are as follows. You can draw an envelope out of the box and open it to read the number. If you don’t like it, you can select another envelope, but you must tear up the previous number first. You can repeat this process until you find a number you like, or you reach the last envelope in the box. Describe a strategy for this game that guarantees a 25% or better chance of winning the$100.   If you can, find the expected value of your strategy.  The person who submits the best strategy gets bonus points.

Problem 18 was correctly solved by Andrew Arnold, Ke Shang, Ryan Sheridan, and Bryan Talbert.  Bonus points to Ke Shang, who found the absolute best strategy, and computed the expected winnings under that strategy.

Problem #17, October 2011.

In the diagram below, point A’ is the midpoint of segment CC’, point B’ is the midpoint of segment AA’, and point C’ is the midpoint of segment BB’.  For an arbitrary triangle ABC, describe a procedure for finding the points A’, B’, and C’, given the points A, B, and C.

Problem 17 was correctly solved by Professor Paul Hemler.

Problem #16, September 2011.

Alice and Bob enjoy meeting for lunch under the Old Oak Tree. They each live exactly 10 miles from the Old Oak Tree. Being young and healthy, they like to ride their bikes from their homes to the tree. They each ride at exactly 20 miles per hour. They each leave their homes at precisely the same moment. Now it happens that at the moment they begin riding, a fly, which has until now been sitting on Alice’s nose, begins flying at 30 miles per hour directly towards Bob’s nose.  When it gets there, the fly immediately reverses direction and flies directly back to Alice’s nose.  Once there, it again reverses direction and flies back to Bob’s nose in the same manner. This goes on, back and forth, until Alice and Bob meet at the Old Oak Tree, nose to nose. When they finally meet, how far has the fly flown?

Problem 16 was correctly solved by Franklin Bowers, G. W. Butler, Tanner Clements, Jack Devine, Thomas Duhamel, Samuel Haden, Professor Paul Hemler, Clifton Hudson, Professor William Porterfield, Ke Shang, Shawn Stum, and Professor Patrick Wilson.

Problem #15, April 2011.

A baker makes 60 cents profit on each fresh loaf he sells.  Day-old loaves must be sold at a discount, and the baker loses 20 cents on each of these that he sells.  The number of fresh loaves in demand per day is equally likely to be anywhere between 0 and 100.  How many loaves should the baker make per day in order to maximize his profit?  (Assume that the baker can sell all his day-old loaves at the discounted price.)

There were no correct solutions to Problem #15.

A 12cm by 12cm paper square is folded so that the bottom right corner touches the exact middle of the top side as shown below.

Find the dimensions of the three right triangles A, B, and C.

Problem #14 was correctly solved by Ke Shang, Shannon Web, and Professor Patrick Wilson.

Problem #13, February 2011.

John had three true loves, but he couldn’t settle on the one he would give his heart to completely.  So he sent all three a Valentine’s card, each of a different color.  But, poor John, he didn’t sign the cards, and each of his loves mistakenly thought their card came from a different person.  So John was left out in the cold.

John’s true love’s names were Mary, Cindy, and Laura.  Their last names were Hunter, Gardner, and Carpenter, but not necessarily in that order!  The boys they thought sent them the cards were David, Stephen, and Brian, but not necessarily in that order!  From the clues below, work out the names of John’s sweethearts (first and last), the color of the card that each received (red, pink, or blue), and the name of the boy that each recipient mistook for John.

1. When Miss Gardner received her card, she thought it was from David.
2. When Laura received her blue-colored card, she told Miss Hunter and together they figured out who the card was from.  They were mistaken, of course!
3. The girl who received the red card was convinced it came from Brian.
4. Neither Cindy nor Miss Carpenter received a pink card.

Put all aright in this Valentine Day’s mess by filling in the following table, the first column of which we’ve helpfully completed for you:

(Sweetheart’s Name)
First Last Card Color Boy
Mary

Cindy
Laura

Problem 13 was correctly solved by Glen Bowman, Ke Shang, Malik Springer, Katie Stratton, and Shannon Webb.

Problem #12, December 2010.

The model Christmas tree above is going to be placed inside a clear plastic sphere. What is the (inner) radius of the smallest sphere that will contain the tree?

Students Eric Schafer and Ke Shang correctly solved Problem 12.

Problem #11, October/November 2010.

Two flies meet their demise on a one foot by one foot piece of flypaper.  On average, what is the horizontal separation between these unfortunate creatures?  For bonus points: what is the average distance between the two flies?

Students Eric Schafer and Ke Shang solved Problem 11, as well as Professor Jonathan Keohane.  Student Eric Schafer gets bonus points for his simulation, which correctly estimated the distance between the two flies.

Problem #10, September 2010.

If you shuffle a deck of cards and then begin drawing cards off the top, without replacement, which card in the deck is most likely to be the first red king?

Professor Jonathan Keohane submitted a correct solution to problem 10.

Problem #9, April 2010.

There are two coins in a hat.  One coin is fair – its probability of heads is 1/2.  The other coin is not fair, but you’re not sure in which direction:  it’s either heads with probability 2/3 or tails with probability 2/3.  You select a coin at random from the hat and flip it four times.  All four flips result in tails.  What is the probability that the coin you selected is the fair coin?

Michael Collins submitted a correct solution to problem #9.

Due Date: Friday, April 30, 11:59.99 PM

Problem #8, March 2010.

Two ships are sailing in the middle of the ocean with constant speeds and headings.  They both pass near a small island around the same time.  In general, how many different times can the two ships and the island all be aligned (i.e. all lie on the same line)?  Assume that the ships’ headings are not parallel and that the ocean is flat.

No correct solutions to problem #8 were submitted.

Problem #7, February 2010.

This problem was inspired by Japanese temple problems called sangaku.

The figure above depicts a unit square (side length equals one). Find the total area enclosed by the blue and green circles.

Correct solutions to problem #7 were submitted by Stuart Callahan, Tony Carilli, Andy Collins, Michael Collins, Richard McClintock, Katie Stratton, Shihao Tian, Patrick Wilson, and Hendrik Ziller.  Thanks to everyone who submitted solutions!

Problem #6, January 2010

We call this a “cross-number” puzzle, because it’s like a cross-word, except that the answers are numbers, and the clues can refer to other answers.

So for instance the first clue says that 1-across is equal to 1-down plus twenty.  Bonus points to solvers who can spot mathematically notable (or otherwise notable) numbers in the solution.

Correct solutions to problem #6 were submitted by Tony Carilli, Michael Collins, and Jake Reynolds.  Thanks to everyone who submitted solutions.  Bonus points to Professors Carilli and Collins, who spotted several notable numbers in the cross-number!

Problem #5, December 2009.

Imagine if Christmas lasted 365 days!

On the first day of Christmas,
my true love gave to me,
a partridge in a pear tree.
(1 Present on Day 1)

On the second day of Christmas,
my true love gave to me,
two turtledoves
and a partridge in a pear tree.
(3 Presents on Day 2)

On the third day of Christmas,
my true love sent to me
three French hens,
two turtle doves,
and a partridge in a pear tree.
(6 Presents on Day 3)

So at the end of the third day you will have received 1+3+6=10 presents. If this pattern of gift giving continues, how many presents will you have received by the end of the 365th day?

Correct solutions to Problem #5 were submitted by Ben Brawley, Andy Collins, Michael CollinsPaul Cottrell, Jonathan Keohane, Ke Shang, Herb Sipe, Alex Smith, and Katherine Stratton.

Problem #4, November 2009.

Angie, Bill, Caleb, and Darlene each have some money in their pockets.  Together Angie and Bill have exactly as much as Caleb and Darlene have together.  Together Angie and Caleb have twice what Bill and Darlene have together.  Together Angie and Darlene have three times what Bill and Caleb have together.  Rank Angie, Bill, Caleb and Darlene from poorest to richest.

Correct solutions for Problem #4 were submitted by Ben Brawley, Andy Collins, Michael Collins, Paul Cottrell, Susie Hemler, Roberto Ibarra, Shiva Patel, Robert Pickus, Ke Shang, Herb Sipe, Katie Stratton, Shihao Tian, Patrick Wilson, and Roger Yamada.  One incorrect solution was submitted.

Problem #3, October 2009.

George and Lennie are walking towards each other beside a railroad track.  A train going in the same direction as George takes 20 seconds to pass him as he walks.  Ten minutes later the train is at Lennie’s position, and it takes 18 seconds to pass him as he walks.  How much further time elapses before George and Lennie meet?

Correct solutions for Problem #3 were submitted by Shihao Tian (student), Professor Michael Collins, and Professor Patrick Wilson.  One incorrect solution was submitted.

Problem #2, Spring 2009.

The ancient Babylonians had a method for finding the square root of a positive number $a$. They started with a guess, call it $x_0$, that was close to $a$. Then they would calculate $a/x_0$. If $x_0$ was too big (bigger than the square root), then $a/x_0$ would be too small. On the other hand, if $x_0$ was too small, then the fraction $a/x_0$ would be bigger than $\sqrt{a}$. Taking the two numbers $x_0$ and $a/x_0$ and averaging them gives a better approximation call it $x_1$.

Repeating this process, the Babylonians got a recursive formula for better and better approximations of the square root $\sqrt{a}$:

$x_{n+1} = \textrm{average}(x_n,a/x_n) = \frac{1}{2}(x_n + a/x_n)$

Now, there is more than one way to find the “average” of two numbers. The Babylonian algorithm simply averages $x_0$ and $a/x_0$ by adding them together and dividing by two. This is the usual method and is known as the arithmetic mean. There is also the geometric mean of two numbers $x$ and $y$ given by $\sqrt{xy}$. Another mean is the harmonic mean of $x$ and $y$ which is given by the formula $\dfrac{2xy}{x+y}$.

To solve this problem of the month, you have to answer the following two questions.

1. Explain why replacing the arithmetic mean used in the Babylonian algorithm with the geometric mean would defeat the purpose of an algorithm to compute square roots.
2. Would replacing the arithmetic mean in the Babylonian algorithm with a harmonic mean cause the algorithm to converge to the square root faster or slower? Explain.

Correct solutions submitted by:

Problem #1.

A rectangular steel plate of constant density has a rectangular hole cut out of it, as shown in the figure below:

Draw a single straight line the cuts the plate into two pieces of equal mass.

Correct solutions submitted by:  Alex Smith