students walk on University of Idaho campus

Visit U of I

Learn about the many reasons the University of Idaho could be a perfect fit for you. Schedule Your Visit

Parents on campus during orientation

Homecoming Oct. 1-7

Join other Vandal families for a week of celebration and Vandal traditions. View Calendar

campus full of students

U of I Retirees Association

UIRA has a membership of nearly 500 from every part of the University. Learn More

Vandal Science News Puzzlers

We love our Puzzlers.

We hope you do too.


The puzzler for this issue comes from Physics, and is appropriate to summer recreational pursuits.

Suppose you are floating on a rubber raft in a swimming pool. For some reason, you have a bowling ball sized rock with you on the raft. You decide you've had enough of holding the rock, so you put it overboard and watch it sink to the bottom of the pool. What happens to the water level of the pool?

A correct answer must include a short explanation. Have fun thinking this one through while you're poolside this summer!


As Archimedes would have known, the solution is all about displacement. The water level of the pool will drop a very small amount when the rock is thrown overboard.
The reason is this:

  • While the rock is floating with the raft, its weight is displacing an amount of water with weight equal to that of the rock.
  • Once it sinks, however, it displaces an amount of water with volume equal to that of the rock.

Now the rock is more dense that water (after all, we're told that it sinks), so the amount of water weighing the same as the rock will have a volume greater than the volume of the rock. This means more water was being displaced when the rock was in the raft, so the water level will be higher at that time. Eureka!

Correct Solvers

  • Tim Householder (BS Mathematics, 2002)
  • Mike Nickerson
  • Lee Ogren (Chemistry, 1974)
  • Laura Podratz (Geology, 2007)
  • Greg Stenback (Geological Engineering, 1985; M.S. Statistics, 1987)
  • John Stutz (MS Physics, 1972)


Take a square piece of paper measuring 12 inches on each side. Call the corners (in cyclic order) A, B, C, and D. Make four folds in the paper as follows:

  • Fold from the midpoint of AB to the midpoint of CD.
  • Fold from the midpoint of BC to the midpoint of DA.
  • Fold diagonally from A to C.
  • Fold from D to the midpoint of AB.

These folds should divide the square into nine regions. Find the area of the smallest of these regions.


The four folds and nine resulting regions are shown in the diagram. The area of the (pink shaded) smallest region is 3.
You can find that area lots of different ways – probably the most common approach would be to put the square on the coordinate plane with the origin at point A, find the equations of the various lines, and work it out in coordinate geometry. Here’s a different way that’s fairly simple and uses only a little geometric reasoning and elementary algebra.

  • Label all of the points as in the figure, and let x represent the area of the small triangle PQR.
  • Since the square is 12-by-12, the area of the whole is 144. The first two folds (between midpoints of sides) divide the square into four squares, each with area 36.
  • It’s clear that Q must be the midpoint of HP, so HQD is a right triangle with legs of length 3 and 6. That makes its area 9.
  • But triangle PQE is congruent to HQD, so its area must also be 9.
  • Now notice that the triangle AER is similar to PQR, but double the scale. That means its area will be four times as large as the area of PQR, so area(AER) = 4x.
  • The areas of AER and PRE must add to be 18 – half of the square AEPH. So:
    area(AER) + area(PRE) =
    4x + (9 – x) =
    3x =
    x =
solution to February 2014 Vandal Science News Puzzler

Correct Solvers

  • Kelli Anderson (BS Mathematics, BS Biochemistry, 2011)
  • Craig Beisel (BS Mathematics, 2001; MS Bioinformatics and Computational Biology, 2007)
  • Jason Evans (BS Computer Science 1996; PhD Bioinformatics and Computational Biology, 2009)
  • Tim Householder (BS Mathematics, 2002)
  • John Spence (MS Mathematics, 2001)


This puzzler is an intriguing geography challenge. As you know, a country that has no ocean shoreline is said to be "landlocked". You need to cross at least one international border to get from a landlocked country to the Ocean. We'll say that a country is "doubly landlocked" if you have to cross at least two international borders to get from that country to an ocean. Another way to say this would be to say that a doubly landlocked country is landlocked and all of its neighbors are landlocked as well.

There are two doubly landlocked countries in the world. Name at least one of them.


The two doubly-landlocked countries are Uzbekistan and Liechtenstein. The map here shows Uzbekistan in red, and all of its neighbors in other colors. A quick scan verifies that no neighbor touches an ocean. Liechtenstein, on the other hand, is a tiny country (just 62 square miles) having only two neighbors, Austria and Switzerland, both of which are landlocked

Correct Solutions:

  • Laura Baldwin (Geology, 2007)
  • Zephyr Bizeau (Zoology, 2000)
  • Fred Eberle (MS Geography, 1984)
  • Jessica Edwards (Lincoln Middle School, Pullman WA)
  • Jason Evans (Computer Science, 1996; PhD Bioinformatics and Computational Biology, 2009)
  • Brian Hill (Chemistry, 1965)
  • Tim Householder (Mathematics, 2002)
  • Dorah Mtui (Forney Lab, Department of Biological Sciences)
  • Lee Ogren (Chemistry, 1974)
  • Elizabeth Shawkey Owosina (MS Geology, 1995)
  • Rick Potter (Geology, 2004)
  • Kim Salisbury
  • Trace Yates


diagram of the 30 square grid for the june 2013 Vandal Science News PuzzlerConsider the five-by-six grid of squares shown here. We'll say that the length of each square side in this grid is one. There are lots of length 11 paths in the grid that start at the green dot in the lower left corner and end at the red dot in the upper right corner – each of these paths will consist of a sequence of eleven "up" or "right" moves. What is the probability that such a path, generated at random, will pass through the purple dot?


The probability of a path going through the purple dot will be the fraction

[# of paths passing through the purple dot] / [total number of paths from green dot to red dot]

The blue path shown above is a typical corner-to-corner path. Note that we can associate it with a sequence of “R” (right) or “U” (up) moves – 11 moves in all, of which six must be “R” and five must be “U”. So the total number of such paths is the number of ways of arranging six R’s and five U’s into a sequence. You might remember from algebra class that the answer to this is the “binomial coefficient” C(11,6). (It’s called a binomial coefficient because it would be the number in front of the x6y5 term in (x+y)11 – the 11th power of the binomial x+y.) You can find this number from the famous “Pascal’s Triangle”, or you can simply use the formula C(n,r) = n!/[r!(n-r)!]. In our case, C(11,6) = 11!/[6!5!] which comes out to be 462.

A typical path through the purple dot is shown in orange. Note that it really consists of two paths – one from green dot to purple dot and the other from purple dot to red dot. Using the same logic as we did above, the number of paths from green to purple is C(6,4) (since it corresponds to a sequence of six moves, four of which must be “R”) and the number of paths from purple to red is C(5,2). Computing these, we have C(6,4) = 6!/[4!2!] = 15 and C(5,2) = 5!/[2!3!] = 10. Each of the 15 paths from green to purple could be continued in any of the 10 ways from purple to red – so the total number of paths through the purple dot will be 15*10 = 150.

So now we can conclude that the probability of a path going through the purple dot is 150/462 = 25/77, or just slightly less than one-third.

Correct Solvers

  • Luke Edwards (Biology and English, 2006)
  • Paul Hohenlohe
  • Timothy Householder (Mathematics, 2002)
  • John Stutz (MS Physics, 1973)


The students in Dr. Bitterwolf’s chemistry class compare notes and find that all of them have a math, biology, or geography class in addition to chemistry.

  • Everyone with a geography class also has a math class, but nobody has both a geography class and a biology class.
  • One out of every three students with a math class also has a geography class, and half of those with a biology class also have a math class.
  • There are 14 students with a geography class, and 26 students with a biology class.

How many students are in Dr. Bitterwolf’s chemistry class?


There are 55 students in Dr. Bitterwolf’s Chemistry class. Getting that answer is a matter of organizing the information, and probably the easiest and best way to do that is with a Venn diagram such as the one shown here. From the clues tell us:

  • The set of Geography students lies entirely inside the set of Math students.
  • The Biology and Geography sets are disjoint.
  • The 26 Biology students are split with 13 inside the Math set and 13 outside.
  • The Geography set contains 14 students.
  • The portion of the Math set outside of Geography has to contain 28 (two times as many as Geography), and 13 of them are already known to be inside the Biology set – so the remaining region must contain 15 students.

Adding it all up gives us our total of 55.

Correct Solvers

  • Mike Andrews (current UI Computer Engineering major)
  • Jenny Baker (current UI Political Science major)
  • Zephyr Bizeau (Zoology, 2000)
  • Katarina Brownell (current UI Creative Writing major)
  • Fred Eberle (MS Geography, 1984)
  • Carol Hansen Sokel (Mathematics, 1973)
  • Timothy Householder (Mathematics, 2002)
  • Levi Kilian (Math Club officer at Lewis Clark State College)
  • Kaitlyn Miller (current UI Biology major)
  • Amy Pendegraft (current UI History major)
  • Noah Qualls (current UI Mathematics and Finance major)
  • Tom Rice (Cartography 1988 and MS Geography 1993)
  • Daniel Schmalz (current UI Electrical Engineering major)
  • Teresa Silveus  (Education 2008, MEd 2012)


Solution to July 2012 Vandal Science News PuzzlerSuppose we start with a square of side length one. We then create a new larger figure by attaching four squares, each of side length 1/3, to the middle thirds of each of the four sides.  This new figure now has 20 segments making up its boundary. We’ll create our third iteration by attaching small squares (side length 1/9) to the middle thirds of each of those 20 segments. The new figure is rather complicated and has 100 segments in its boundary. We could create an even more complicated figure by attaching tiny (side length 1/27) squares to each of those segments. If we continue this forever, what is the limiting area of the resulting figure? [Warning: there’s both a hard way and an easy way to do this!]


As we said, there is both an easy way and a hard way to do this one – we’ll give both here.

Easy way :Notice that as we add smaller and smaller squares in the various steps to create this shape, it seems to be filling in a new square, rotated 45 degrees from the original square. Also, the new square has a side length equal to the diameter of the original square. Since the original square’s diagonal is the square root of 2, the area of the new square will be 2.

Solution to the October 2012 puzzler

Hard way :The mathematical purists among us may prefer to show analytically that the area is two. We can do this by writing the total area as an infinite series:

  • The original square has area 1, so at stage 1 the area is 1.
  • We create the stage two figure by adding four small squares, each with area 1/9 – thus: 1 + 4/9.
  • At stage 3 we add 20 small squares, each with area (1/9)2 – thus: 1 + 4/9 + 20/92.
  • At stage 4 we add 100 smaller squares, each with area (1/27)2 = (1/9)3. Thus: 1 + 4/9 + 20/92 + 100/93.
  • There’s an obvious pattern developing that becomes even more apparent with a little rewriting:
    1 + 4/9 + 20/92 + 100/93  + . . .  =  1 + (4/9)[1 + 5/9 + (5/9)2 + (5/9)3 + . . . ]
  • If you remember infinite series from calculus, you might recognize the sum inside those brackets as a geometric series. Then remembering that the sum of 1 + r + r2 + r3 + . . .  is 1/(1-r), we can see that the sum in the brackets is 1/(1-(5/9)) = 1/(4/9) = 9/4.
  • So the area of the limiting figure is 1 + (4/9)[9/4] = 2.

Correct Solvers

  • Tim Householder (Mathematics, 2002)
  • Zephyr Bizeau (Zoology, 2000)
  • Fred Eberle (MS Geography, 1984)
  • Mark Daily (Physics, 1981)
  • Carey Edwards (Forest Products, 2002, GIS Certificate, 2010)
  • Sam Bacharach (Journalism, 1971, MS Geography 1980)


Consider an ordinary circular clock face with the numbers 1 through 12 marking the hour positions. It is possible to draw two straight lines through the face, neither touching one of the twelve numbers, so that each region created has the same sum of hour numbers. Describe how to do this.


Image of click with 2 diagonal lines for Vandal Science News PuzzlerThe answer is to draw one line from between the 10 and 11 to between the 2 and 3, and the other from between the 8 and 9 to between the 4 and 5, as shown here. You could try lots of “guess and check” work to find this, but the elegant method uses some mathematical detective skills:

  • First, do we want two lines that cross (determining four regions) or non-crossing lines (determining three regions)? Well, the sum of the clock numbers is 1+2+3+…+11+12=78, which is not a multiple of four, but is a multiple of three – so the non-crossing lines is what we need.
  • Then, since 78 divided by three is 26, we know we’re looking for sums of 26, and two of them will need to be with consecutive clock numbers. It isn’t hard to find that 11+12+1+2 and 5+6+7+8 work.

Correct Solvers:

  • LuAnn Scott (Lab Manager, UI Department of Biological Sciences)
  • Mark Borth (Technician, UI Environmental Health and Safety)
  • Gary Green (Mathematics, 1964)
  • Carey Edwards (BS Forest Products, 2002; GIS Certificate, 2010)
  • Laura Podratz Baldwin (Geology, 2007)
  • Kelly Rush Gerber (Mathematics, 1995)
  • Fred Eberle, (MS Geography, 1984)
  • Greg Stenback (Geological Engineering, 1985; M.S. Statistics, 1987)
  • Travis Nelson (Current Chemical Engineering student)
  • Sam Bacharach (BS Journalism, 1971; MS Geography, 1980)
  • Lee Ogren (Chemistry, 1974)


Suppose a cannon is mounted on a cart, pointed perpendicular to the cart’s bed. Now suppose the cannon shoots a ball upward while the cart is traveling at a constant velocity in a straight line on a horizontal surface, as in the left half of the figure. If we disregard friction, most people would guess (correctly) that the ball will eventually land back in the cart.

But now suppose the cart is actually rolling freely down a hill subject to the influence of gravity. If the cannon is again fired, will be ball land ahead of, behind, or exactly on the cart? This isn’t multiple choice: credit will be given only if the answer is accompanied by an explanation!

Computer drawing of cannon mounted on a cart.


The ball will land in the cart, the same as when the cart was moving on a level surface. Brian Hill (Chemistry, 1965) knew that the key is to look at the problem a bit askew – adopt a new axis system in which the “vertical” axis is perpendicular to the slope the cart is on. In this axis system, gravity now provides a force somewhat off of vertical, but it’s clear that this force will be acting on both the cart and the ball.


Suppose you have two fuses that each burn for exactly one hour. However, they don't burn at uniform rates, nor do they necessarily burn at the same rate. (That is, if you light the two fuses simultaneously they would both burn out exactly an hour later, but they may not always appear to be the same length at intermediate times, and would not necessarily be half their original length after a half-hour.) How can you use these two fuses (and a pack of matches) to time exactly 45 minutes?


You first light both ends of one fuse, and one end of the other. When the first fuse burns completely out you know that 30 minutes have passed and that the second fuse has 30 minutes of burn time left. At that moment you light the second end of that second fuse. With both ends burning, its remaining 30 minutes of burn time is shortened to 15 minutes, and when it burns completely out, a total of 45 minutes will have passed.

Correct solvers:

  • William R. Cordwell (Math and Physics, 1975; M.S. Math, 1975)
  • Jerry Fairley 
  • Claude Freaner (Math, 1966)
  • Mark Garber
  • Quinn MacPherson (current Physics and Materials Science student)
  • Alex Main (current Mathematics and Geography student)
  • Travis Nelson (current Chemical Engineering student)
  • Garrett Stauffer (current Electrical Engineering student)
  • Greg Stenback (Geological Engineering, 1985; M.S. Statistics, 1987)
  • Ian Tanimoto
  • Gary Troyer (Chemistry, 1968)
  • Mark Wilkins (Math and C.S., 1987)


There is a place in the world where traveling nine miles in any direction will require you to set your watch ahead one half hour, then back one half hour, then finally ahead again one half hour (in order to keep the watch set correctly at all times). In what country is this place located?


The June Vandal Science News featured the following tricky Geography puzzler:

There is a place in the world where traveling nine miles in any direction will require you to set your watch ahead one half hour, then back one half hour, then finally ahead again one half hour (in order to keep the watch set correctly at all times). In what country is this place located?

The simple answer is that the place is located in India. More specifically, it is a very small enclave of India called Dahala Khagrabari that lies within an enclave of Bangladesh called Upanchowki that itself lies within an enclave of India called Balapara Khagrabari (that lies within Bangladesh). Add to that confusion the fact that India and Bangladesh maintain time zones separated by one half-hour, and you get the situation described by the puzzler.

See Cooch Behar district (Wikipedia) for more information, or an annotated map of Cooch Behar of this messy piece of the India / Bangladesh border.


College of Science

Physical Address:
Mines 321

Mailing Address:
875 Perimeter Drive MS 3025
Moscow, ID 83844-3025

Phone: (208) 885-6195

Fax: (208) 885-6904


Web: College of Science