Mathematics- Physics High School SMILE Meeting
28 March 2000
Notes Prepared by Earl Zwicker

Announcement by Porter Johnson

Recent Academic year SMILE write-ups have been placed on the SMILE home page. Check them out at the URL http://www.iit.edu/~smile/. The write-ups are being put first on my home page, http://www.iit.edu/~johnsonp/ and then transferred over to the SMILE site.

`                     OUR NEXT MEETING...                ...will be April 11, 2000                           4:15 p.m.                           111 LS               AT OUR LAST MEETING (Mar 28)...`

Fred Schaal (Lane Tech HS)
first mentioned Weird Science (Lee Marek) on one of Letterman's shows, in which two cans of soda pop are shaken in an apparatus like a paint shaker. The cans are then simultaneously pierced, and streams of pop shoot out to the ceiling in opposite directions. Spectacular - but not done live because of the messiness. Also one may use an ultrasonic cleaner with high frequency vibrations in place of a paint shaker.

Next, Fred brought us back to something he had done with us earlier: The truncated cone. Start with a circle of radius R cut from paper. Draw two radii on the circle from its center, and so define an angle of 360o - ko between them. Cut out the sector between them, as shown:

`            s = 2 p r = p [k/180] R`

Tape the radii together, to form a truncated cone surface with the paper. Let r be the radius of base of the cone, and h its altitude. Then

r = R * [ko/360o]

and

h = (R2 - r2)1/2 = R * [ 1 - (ko/360o)2]1/2.

The conical volume is given by

V = pr2h/3

= p R3/3 * [ko/360o]2 * [ 1 - (ko/360o)2 ]1/2

"What value should the angle ko have in order for volume V to be a maximum?" was the question Fred asked us. But lacking for any volunteers, Fred left us with the unanswered question. Maybe next time somebody will have an a solution?

Ed Robinson (Collins HS)
posed the following challenge to us. We are given twelve coins, one of which is either heavier or lighter than the others. Using a balance, how - with no more than three weighings - can we determine which coin is the odd one? He noted that 1990s pennies have a zinc cladding, and therefore weigh different from a 1972 penny, which is made of solid copper. This would make it possible for us to do this experimentally on the balance he had placed on the table.

Ed proceeded to show us a theoretical strategy to do this, starting with just 6 coins, and 9 coins---namely, split the coins into 3 sets, and go from there. A guiding principle is that you can tell which coin of 3 is bad by making one weighing. One can expand the same strategy to 12 coins. Ed actually carried out the experiment with a set of coins, and it worked! He also handed out this paper:

A Strategy For Solving The Forged Coin Problem
Edward J Robinson [Collins HS]

There are twelve coins, one of which is forged. The forged coin is either lighter or heavier than the others. By using at most three weighings on a balance scale, find the forged coin and determine whether it is heavier or lighter than the other coins.

Divide the twelve coins into three groups of four, groups A,B,C.

First weighing

Place group A on the left and place group B on the right. Set group C aside. If the balance is equal then the forged coin is in group C. If the balance tends to the left, then the forged coin is either in group A or B. If the forged coin is in group A, then it is heavier; if the forged coin is in group B, then it is lighter. If the balance tends to the right, then the forged coin is either in group A or B. If the forged coin is in group A, then it is lighter; if the forged coin is in group B then it is heavier.

• Result of first weighing- Balance equal

Second weighing

Place any three coins from group A on the left and any three coins from group C on the right.

Result of Second weighing

• If the balance is equal, the coin set aside from group C is forged and you can weigh it against any of the other coins for the third weighing to determine whether it is heavier or lighter. Place the forged coin on the left. If the balance tends to the left then the forged coin is heavier. If the balance tends to the right then the forged coin is lighter.

• If the balance tends to the left. the forged heavier coin is on the left. Take off the three coins from the right side. Place on the right any one of the three coins that are on the left. Put aside the third coin from the left. If balance is equal then the third coin set aside is forged and heavier. If balance tends to the left then the coin on the left is forged and heavier. If the balance tends to the right then the coin on the right is forged and heavier.

• If the balance tends to the right, then the forged lighter coin is on the left. Take off the three coins from the right side. Place on the right any of the three coins that are on the left. Put aside the third coin from the left. If balance is equal then the third coin set aside is forged and lighter. If the balance tends to the left then the coin on the right is forged and lighter. If the balance tends to the right then the coin on the left is forged and lighter.

• Result of First weighing- Balance tends to the left

Second Weighing

Switch the positions of the two coins that are on the far left on both the left side and the right side. On the right side, replace the non- switched coins, (group U) with three coins from group C.

Results of Second weighing

• If the balance is equal, then the forged coin is either one of the three replaced coins (group U), which can only be a heavier coin. You can weigh any two of the coins. If the balance is equal then coin not weighed is heavier and forged. If the balance tends to the right then the forged heavier coin is on the right side. If balance tends to the left, then the forged heavier coin is on the left.

• If the balance tends to the left then the lighter forged coin is on the right side but not the farthest left coin. The farthest left coin on the right side cannot be a lighter coin because it was on the left when the balance tended to the left after the first weighing. The farthest left coin on the left side cannot be a heavier coin because it was on the right when the balance tended to the left after the first weighing. Take off the coins on the left side. Take off the farthest left coin on the right side. Move one coin from the left side to the right side. Place one coin aside from the right side. If the balance is equal, then the coin placed aside is lighter and forged. If the balance tends to the right then the forged lighter coin is on the left side. If balance tends to the left, then the forged lighter coin is on the right.

• If the balance tends to the right then the forged coin is either the farthest left coin on the left side or the farthest left coin on the right side. Leave these two farthest left coins on the scale and take the other coins off the scale. Weigh either coin against a known regular coin to determine the forged heavier or lighter coin. For example, replace the coin on the right side with a known regular coin. If the balance is equal, then the coin placed aside is forged and heavier. If the balance tends to the right, then the forged lighter coin is on the left side.

• Result of First weighing- Balance tends to the right

Second Weighing

Switch the positions of the two coins that are on the far left on both the left side and the right side. On the right side, replace the non- switched coins, (group U) with three coins from group C.

Results of Second weighing

• If the balance is equal, then the forged coin is either one of the three replaced coins(group U), which can only be a lighter coin. You can weigh any two of the coins in group U. If the balance is equal then coin not weighed is lighter and forged. If the balance tends to the right then the forged lighter coin is on the right side. If balance tends to the left, then the forged lighter coin is on the left.

• If the balance tends to the right, then the heavier forged coin is on the right side but not the farthest left coin. The farthest left coin on the right side cannot be a heavier coin because it was on the left when the balance tended to the right after the first weighing. The farthest left coin on the left side cannot be a lighter coin because it was on the right when the balance tended to the right after the first weighing. Take off the coins on the left side. Take off the farthest left coin on the right side. Move one coin from the left side to the right side. Place one coin aside from the right side. If the balance is equal, then the coin placed aside is heavier and forged. If the balance tends to the right then the forged heavier coin is on the right side. If the balance tends to the left, then the forged heavier coin is on the left side.

• If the balance tends to the left then the forged coin is either the farthest left coin on the left side or the farthest left coin on the right side. Leave these two farthest left coins on the scale and take the other coins off the scale. Weigh either coin against a known regular coin to determine the forged heavier or lighter coin. For example, replace the coin on the right side with a known regular coin. If the balance is equal, then the coin placed aside is forged and lighter. If the balance tendsto the left, then the forged heavier coin is on the left side.

For a JAVA applet to try your hand at solving this problem at a virtual level, see the website Find the Forged Coins; http://www.cut-the-knot.org/blue/OddCoinProblems.shtml.

PJ Comment: Also, check these websites:

You got us thinking, Ed! Thanks!

Sally Hill (Clemente HS)
put us through a paper-and-scissors exercise to construct a geodesic dome. A template, which consists of 21 equilateral triangles, is given on the website http://www.pitsco.com/Neatstuff/geodome.htm. We crowded around the table and carefully cut out section after section from the pattern copied onto many pieces of paper. Fitting and gluing (use stick glue to make it easy) 5 sections together will do it!

Sally, thanks for introducing us to this useful idea and website resource!

Porter Johnson (IIT Physics)
passed out copies of "Weighing Problem,"a write-up from five years ago, which is reproduced below: How does it compare with Ed's Approach?

PORTER JOHNSON
SMILE MATHEMATICS A
21 MARCH 1995

WEIGHING PROBLEM

One is given a set of 12 coins, one of which is counterfeit and either lighter or heavier than the other 11. We are allowed to use a [crude] balance, which can only tell us whether the coins on the left pan weigh less than, the same as, or more than the coins on the right pan. Devise a scheme which involves only three different weighings to determine which coin is counterfeit, and whether it is lighter or heavier than the others.

Remark: To solve this problem one must make full use of the three possible results of weighing, varying which and how many coins are on each side, to develop the scheme. I have found a scheme which works, which I will describe here. It may not be unique.

Let us begin by numbering the coins 1 through 12. [It is important that the coins be distinguishable, in spite of having the same weight--perhaps the dates on them are all different.

Weighing #1: Put coins 1, 2, 3, 4 on the left pan and coins 5, 6, 7, 8 on the right pan.

• Case I: [0] The pan balances. We conclude that coins 1-8 all have the same mass, and that the counterfeit coin is either 9, 10, 11, or 12.

Weighing #2: Put coins 1, 2, 3 on the left pan and coins 9, 10, 11 on the right pan.

• Case A: [00] The pan balances. We conclude that coins 9, 10, 11 are OK, and that 12 is the bad coin.

Weighing #3: Put coin 1 on the left pan and coin 12 on the right pan. If left side is heavier [001] 12 is light, whereas if the left side is lighter [002] 12 heavy. [if the pans balance we should properly annihilate ourselves, to avoid a cataclysmic meltdown of the logical framework of the universe!]

• Case B: [01] The left side is heavy. We conclude that either 9, 10, or 11 is light.

Weighing #3: Put coin 9 on the left pan and coin 10 on the right pan. If they balance [010] coin 11 is light. If the left pan is heavier [011] coin 10 is light. If the left pan is lighter [012] coin 9 is light.

• Case C: [02] The left side is light. We conclude that either 9, 10, or 11 is heavy.

Weighing #3: Put coin 9 on the left pan and coin 10 on the right pan. If they balance [020] coin 11 is heavy. If the left pan is heavier [021] coin 9 is heavy. If the left pan is lighter [022] coin 10 is heavy.

• Case II: [1] The left side of the pan is heavier than the right side. We conclude that coins 9-12 are OK; and either one of the coins 1-4 is heavy, or one of the coins 5-8 is light.

Weighing #2: Put coins 1, 9, 10, 11, 12 on the left pan and coins 2, 3, 4, 5, 6 on the right pan.

• Case A: [10] The pan balances. We conclude that coins 1, 2, 3, 4, 5, 6 are OK, and either 7 or 8 is light.

Weighing #3: Put coin 1 on the left pan and coin 7 on the right pan. If left side is heavier [101] 7 is light, whereas if the pan balances [100] 8 is light. [if the left side is lighter [102] we again should undergo annihilation!]

• Case B: [11] The left side of the pan is heavier. We conclude either that either 1 is heavy, or that 5 or 6 are light.

Weighing #3: Put coin 5 on the left pan and coin 6 on the right pan. If left side is heavier [111] 6 is light, whereas if left side is lighter [112] 5 is light. If the pan balances [110] 1 is heavy.

• Case C: [12] The left side of the pan is lighter. We conclude that one of the coins 2, 3, 4 is heavy.

Weighing #3: Put coin 2 on the left pan and coin 3 on the right pan. If the left side is heavier [121] 2 is heavy, whereas if the left side is lighter [122] 3 is heavy. If the pan balances [120] 4 is heavy.

• Case III: [2] The left side of the pan is lighter than the right side. We conclude that either [a] one of the coins 1-4 is light, or [b] one of the coins 5-8 is heavy.

Weighing #2: Put coins 1, 9, 10, 11, 12 on the left pan and coins 2, 3, 4, 5, 6 on the right pan.

• Case A: [20] The pan balances. We conclude that coins 1, 2, 3, 4, 5, 6 are OK, and either 7 or 8 is heavy.

Weighing #3: Put coin 1 on the left pan and coin 7 on the right pan. If the left side is lighter [202] 7 is heavy, whereas if the pan balances [200] 8 is heavy. [if the left side is heavier [201] we again should undergo annihilation!]

• Case B: [21] The left side of the pan is heavier. We conclude either that either 1 is light, or that 5 or 6 are heavy.

Weighing #3: Put coin 5 on the left pan and coin 6 on the right pan. If the left side is heavier [211] 5 is heavy, whereas if the left side is lighter [212] 6 is heavy. If the pan balances [210] 1 is light.

• Case C: [22] The left side of the pan is lighter. We conclude that one of the coins 2, 3, 4 is light.

Weighing #3: Put coin 2 on the left pan and coin 3 on the right pan. If the left side is heavier [221] 3 is light, whereas if the left side is lighter [222] 2 is light. If the pan balances [220] 4 is light.

SUMMARY

 First Weighing Second Weighing Third Weighing *[000] harikari [100] 7 light [200] 8 heavy [001] 12 light [101] 8 light *[201] harikari [002] 12 heavy *[102] harikari [202] 7 heavy [010] 11 light [110] 1 heavy [210] 1 light [011] 10 light [111] 6 light [211] 5 heavy [012] 9 light [112] 5 light [212] 6 heavy [020] 11 heavy [120] 4 heavy [220] 4 light [021] 12 light [121] 2 heavy [221] 3 light [022] 10 heavy [122] 3 heavy [222] 2 light

Note: In three weighings with three types of outcomes, there are 27 possible total outcomes. We have distinguished 24 different configurations [each of 12 coins light or heavy], and there are three logical inconsistencies.