dan's math@home - problem of the week - archives
 
 
Problem Archives page 18
with Answers and Winners. For Problems Only, click here.
 
1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+ . prob index
 
171 - Drowning Child!
172 - To Catch A Thief
173 Three Equal Areas
174 Pigs Pkeets Profits
175 Fruit Season Again
176 - Jack - and - Diane
177 - Cover That Table!
178 - - The Cat-Census
179 Around The World
180 Cover That Board?
 
 
Problem #171 - Posted Saturday, May 3, 2003
Help ! Drowning Child ! . (back to top)
At Newport Beach, CA, the long, straight shoreline separates the clear
water (blue) from the sand (tan). Mitch, the lifeguard, is at M, 80 meters
from the water, and sees a drowning child C, 120 meters from the sand.
The points A and B are 280 meters apart along the shoreline. Mitch can run
only 4 m/sec but he can swim 8 m/sec. At what spot on the shoreline AB
should Mitch aim in order to reach the child at C as soon as possible?
Compute the minimum time to the nearest tenth of a second, compare it to the times
for the paths: MC, MAC, MBC. Show steps and reasoning. One pt penalty for resub.
Can Mitch save the child?
 
Solution: Gaurav writes it out this way (one arithmetic (copying) error has been fixed):
Let a = dist from A at which Mitch hits the water . . . Equation of time becomes
t = (X/4) + (Y/8) = sqrt(80^2 + a^2)/4 + sqrt(120^2 + (280-a)^2)/8
8t = 2*sqrt(80^2 + a^2) + sqrt(120^2 + (280-a)^2) . . . Take differential :
8(dt/da) = (2*2a)/(2*sqrt(80^2 + a^2)) - 2*(280-a)/(2*sqrt(120^2 + (280-a)^2))
At maxima or minima, dt/da = 0. Also all the numerators are always positive (sums of squares).
2a*sqrt(120^2 + (280-a)^2) = (280-a)*sqrt(80^2 + a^2) . . . Square both sides :
4*a^2 (120^2 + (280-a)^2) = (280-a)^2 * (80^2 + a^2) . . . On solving , a = 40 m
Which means X = sqrt(80^2 + 40^2) = 89.44 ; Y = sqrt(120^2 + 240^2) = 268.328 m
time taken = 89.44/4 + 268.328/8 = 22.361 + 33.541 = 55.902 sec
time taken in MAC = 80/4 + sqrt(280^2 + 120^2)/8 = 20 + 38.078 = 58.078 sec
time taken in MBC = sqrt(80^2 + 280^2)/4 + 120/8 = 72.801 + 15 = 87.801 sec
Line MC will divide AB at a point Q such that AM:BC = AQ:QB => AQ = 112 m
time for MC = sqrt(80^2 + 112^2)/4 + sqrt(120^2 + 168^2)/8 = 34.409 + 25.807 = 60.216 sec
Path Taken Time on sand Time in water Total time
MOC 22.36 s 33.541 s 55.9 s <== Optimum
MAC 20.00 s 38.078 s 58.1 s
MBC 72.801 s 15.00 s 87.8 s
MC 34.409 s 25.807 s 60.2 s
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
A couple of you noticed that, oddly, lifeguard Mitch swims faster than he runs; this is of course because he's
filmed while running in slo-mo! (One of you would prefer the longest run possible if 'Pam' were subst. for 'Mitch'.)
 
WINNERS - Problem 171 . (back to top) . leader board . . . .
Dan's Note : One point deducted from some scores for various reasons: resubmissions,
lack of rounding, first three entries were largely right with one error for one of the times...
Jayavel Sounderpandian. 9 pts - Nice answer, but could use more steps ; miscalc on the MC time
Nick McGrath . . . . . . . . 8 pts - Your excellent answer was lost in transit for a while, there you go!
Gaurav Agrawal . . . . . . 7 pts - I like your text-picture of the path ; mis-step in ans for MPC time
Arthur Morris. . . . . . . . 6 pts - Avoided calculus "Snell did the calc for us"; a bit off in MPC time
Hermen Jacobs . . . . . . . 5 pts - Right to notice it's Snell's Law of refraction; BASIC beats algebra!
Vince LoCascio . . . . . . 4 pts - Your second entry was fine; thanks for the problem suggestion!
Steve Carlon . . . . . . . . . 3 pts - An EXCEL user admits there's a more elegant calculus solution!
Tim Poe . . . . . . . . . . . . . 3 pts - Good ans, nice spreadsheet design, including my picture!
Ed Wern. . . . . . . . . . . . . 3 pts - Right; 3x^4 - 1680x^3 + 286400x^2 + 3584000x - 501760000 = 0; x=40.
Jack Dostal . . . . . . . . . . 3 pts - Another combo of Pythagoras and Microsoft gets the answer!
Marcello Cammarata. . 3 pts - Nice exposition, correct ans; try to e-mail rather than attach if possible.
Trevor Wells (new). . . . . 2 pts - Welcome to my contest! Some ans were ok but you need interior pt.
Phil Sayre . . . . . . . . . . . 2 pts - Correct answers ; you needed to round to nearest tenth too.
Allen Druze . . . . . . . . . 2 pts - Couldn't find non-calc solution ; I agree the pure alg one is hard.
 
 
Problem #172 - Posted Monday, May 19, 2003
~ To Catch A Thief ~ . (back to top)
There are five men married to five women in a big five-bedroom house near me. One of the men
is a thief. The other men are: a poor man, a rich man, a baker, and a doctor. The women don't work.
One member of each couple always tells the truth and the other one always lies. In four of the five
couples, it's the husband who lies; the exception is the poor man, who tells the truth. All ten people
have names that can designate men or women. In these statements, nobody mentions the name of their own spouse:
 Bobby:  1. Pat is married to Terry.
 Elly:  2. Kim is married to the rich man.
 Freddie:  3. Elly is married to Ronnie.
 Jerry:  4. Elly is married to Willie. 5. Freddie is the baker.
 Kim:  6. Jerry is not married to Terry.
 Lou:  7. Pat is not the poor man. 8. Bobby is married to Jerry.
 Pat:  9. Freddie is married to Ronnie. 10. Willie is not the doctor.
 Ronnie:  11. Jerry is the rich man or is a woman. 12. Bobby is married to Kim.
 Terry:  13. Freddie is married to Lou. 14. Elly is a woman.
 Willie:  15. Kim is the thief. 16. Freddy tells the truth.
Decide who is married to whom, which are men, which are women, and
the profession of each man. Who was the thief?
 
Solution: Here's Vince's thought process, and a table pasted from Marcello. Thanks guys!
 
A. "Freddie is married to Ronnie" (9) is a lie since Freddie mention Ronnie in (3). Therefore, Pat is a liar.
B. Willie is the doctor. (10); C. Freddie is a liar (16); D. Lou tells truth teller (7). So Jerry is married to Bobby (8)
E. Kim is truth teller in (6) because (8) has been established as true. That makes Ronnie a liar (12).
F. From A, B, C, E we have 4 of 5 liars: Pat(A), Willie(B), Freddie(C), and Ronnie(E). Jerry or Bobby is the final liar.
G. If Jerry is the Liar, then, the 5 truth tellers would be Bobby, Ellie, Kim, Lou and Terry. But Willie can't be married
to any of them. Not Bobby (8), not Kim (2), not Lou (13), not Terry (1), and not Ellie (4). Thus, Bobby is final liar.
H. Couple can now be established as WE (4), BJ(8), FL(13), PK (1), RT (1). I. Husbands designations:
Willie/Doctor (10), Pat/Rich (2), Freddie/Baker (5), Jerry/Poor (11), Ronnie was the Thief (from all above).
 Profession  Men  Women
 Doctor  Willie  Elly
 Baker  Freddie  Lou
 Rich man  Pat  Kim
 Poor man  Jerry  Bobby
 Thief  Ronnie  Terry
 
WINNERS - Problem 172 . (back to top) . leader board . . . .
Steve Carlon . . . . . . . . 10 pts - Good series of clues: "Assume Willie told truth . . . contradiction" etc.
Arthur Morris. . . . . . . . 7 pts - Some of us did it without "Visio" drawing and linking software! Imagine!
Tim Poe . . . . . . . . . . . . . 5 pts - Hard-working solution eliminates all erroneous paths, good job!
Ed Wern. . . . . . . . . . . . . 4 pts - Uses a 10 x 10 matrix and elim's possib. of who can't be married to whom.
Vince LoCascio . . . . . . 4 pts - Your solution is immortalized above (as long as my site stays up!)
Marcello Cammarata. . 3 pts - Thanks again for a correct and nicely laid-out solution!
Allen Druze . . . . . . . . . 2 pts - Ok . . . right answers, no reasons given, there we go!
Hermen Jacobs . . . . . . . 2 pts - You got most of our ten married folks, thanks for your thinking!
Ravi Raja (new) . . . . . . . 2 pts - Welcome! Lots of intertwining reasons (mis)led you to Bobby as thief.
 
 
Problem #173 - Posted Monday, June 2, 2003
Three Equal Areas . (back to top)
A point P is placed inside a triangle so that segments
from this point to the vertices X, Y, and Z divide the
big triangle into three smaller triangles of equal area.
Decide which of these statements is true, and prove it.
a) P is the center of the inscribed circle of the triangle
b) P is the center of the triangle's circumscribed circle
c) The three angles meeting at P are each 120 degrees
d) P is the intersection of the altitudes of the triangle
e) P is the intersection of the medians of the triangle
f) P may not exist, or be unique, or be determinable
Bonus: Point a is the "incenter." What are geometric names
(if any) for the other four points described in b, c, d, and e?
Show steps and reasoning.
Three equal areas?
 
Solution: Most of you figured out the 'equal area point' was e), the intersection of the medians. Terms: (Dutch from Hermen in italics)
a) Incenter, equidistant from sides, inpunt - b) Circumcenter, equidistant from vertices, ompunt - c) (1st) Fermat point, Steiner point, isogonic
center, Kimberling center - d) Orthocenter, culminating point, hoogdepunt - e) Centroid, center of mass, geocenter, barycenter, zwaartepunt.
Some of you had especially good disproofs of the choices a, b, c, and d: Here's Nick, Tim, Tim, and Vince :
 
a) False. If P is the "incenter" and r is the radius of the incircle then r will be equal to the altitudes of triangles XPY,YPZ and ZPX.
The areas of the 3 triangles will be .5*XY*r, .5*YZ*r and .5*ZX*r which will be equal only if XY=YZ=ZX i.e. XYZ is equilateral.
b) False. The center of the circumscribed circle lies at the circumcenter, the intersection of the perpendicular bisectors of the sides.
If the triangle were obtuse, then the intersection of these lines would lie outside of the triangle.
c) The three angles meeting at P are each 120. False. A tall narrow isosceles triangle would have two larger angles and one smaller.
d) Using a simple 3,4,5 right triangle, I concluded that a, b, and d were not even true for this simple trianlge let alone all triangles.
 
e) Many of you proved that using medians gave equal areas, fortunately that proof is reversible. Here's Jay(avel)'s version:
"I shall use the figure you had given... Extend ZP to meet XY at its midpoint M. The triangles MXZ and MYZ have equal bases
MX and MY, and equal altitude, and therefore have equal areas. Write it as Area(MXZ) = Area(MYZ) . (1) Similarly, Area(MXP)
= Area(MYP). (2) Subtracting (2) from (1) we get Area(PXZ) = Area(PYZ)... (3) Similarly we prove Area(PXY) = Area(PYZ).
 
WINNERS - Problem 173 . (back to top) . leader board . . . .
Arthur Morris . . . . . . . .10 pts - Bonus pt included; good prf of e); no reasons why a - d don't work...
Jayavel Sounderpandian . 7 pts - Good answer, thanks for links to mathworld! Also show a-d are bad.
Nick McGrath . . . . . . . . 6 pts - Good 'disproofs': a) wrong if orig is skinny isosceles, etc. (got your 171)
Hermen Jacobs . . . . . . . 5 pts - Clever: f) is right bec. of 'or' and P being determined! Thanks 4 Dutch terms.
Tim Poe . . . . . . . . . . . . . 4 pts - Hard-working solution eliminates all erroneous paths, good job!
Vince LoCascio . . . . . . 4 pts - I like your idea of a specific triangle eliminating several of the choices.
Gaurav Agrawal . . . . . . 4 pts - Right; a and b are ruled out unless it's an equilateral triangle.
Steve Carlon . . . . . . . . . 3 pts - Glad you're enjoying 'my' problems, Steve! (I steal a lot of 'em.)
Akifumi Iwahashi (new) 3 pts - Welcome to my contest, a former student of a former student!
Marcello Cammarata. . 3 pts - Very hip to show all the Greek and Latin roots of the terms!
Allen Druze . . . . . . . . . 2 pts - Good solid area proof of e), no bonus terms or disproofs of a - d.
Ravi Raja . . . . . . . . . . . 1 pt - You got part-way on this problem ; thanks and keep entering!
 
 
Problem #174 - Posted Saturday, June 14, 2003
Pigs, Parakeets, and Profits . (back to top)
The Pet Shop Boys bought a certain number of guinea pigs, and half that many pairs of
parakeets, to sell in their shop. On every pet sold, there was a ten percent markup on what
the Boys paid for it. They paid twice as much for each pig as for each parakeet. The price
in dollars for a guinea pig was equal to half of the second perfect number (the one after six).
 
After all but seven of the pets had been sold, the Boys found they had taken in the same
amount of money they'd paid out for the pets. Their profit was the combined retail value
of the remaining pets. a) What was their profit? b) How many of each type of pet had been sold?
c) Hadn't been sold? Show steps and reasoning.
 
Solution: There was only one solution to this problem; remember there are "pairs" of parakeets. Here's Nick:
Let G and P be the number of Guinea pigs and parakeets respectively and let g and p be their respective costs.
The the total number of pets, N = G+P = 2G . . . Total Cost, C=Gg+Pp =Gg + G.g/2 = 3Gg/2
When there are 7 pets left to sell assume there are x pigs and 7-x parakeets.
Selling price of a pig = 1.1*g and selling price of a parakeet=1.1*g/2 so the total revenue is:
(G-x)*1.1*g + (G-(7-x))*1.1*g/2 which we are told is equal to the total cost C = 3Gg/2
Equating and simplifying we get 3G=11x+77. Since G must be an integer and x<=7 we have 2 possible
solutions: (1) x=2, G=33 or (2) x=5, G=44. We can reject (1) because we would have started with 33
parakeets which is not a whole number of pairs so we must have started with 44 pigs and 44 parakeets.
Profit is the revenue form the last 7 pets sold of which 5 are pigs and 2 are parakeets. The price of a pig is
half the second perfect number = $14 => price of parakeet = $7 . . . a) Profit = 5*14 + 2*7 = $84
(Note: it is not totally clear from the question whether $14 is the selling price or the boys' purchase price.
The above assumes the former. If the latter the profit will be 5*15.40+2*7.70 = $92.40)
b) 39 guinea pigs and 42 parakeets had been sold . . . c) 5 guinea pigs and 2 parakeets hadn't been sold.
 
WINNERS - Problem 174 . (back to top) . leader board . . . .
Nick McGrath . . . . . . . 10 pts - That's right, 39 pigs and 42 keets; I agree 'price' wording was a bit vague
Hermen Jacobs . . . . . . . 6 pts - Good answer(s), then "fixed up in resub" as we say in the trade.
Steve Carlon . . . . . . . . . 5 pts - You have a good 'proof style', I'll have to send you some harder problems!
Gaurav Agrawal . . . . . . 4 pts - Right; a and b are ruled out unless it's an equilateral triangle.
Vince LoCascio . . . . . . 4 pts - You did have the 2 answers but the early submission gained you pts
Jack Dostal . . . . . . . . . . 4 pts - Nice solution; right, you can't sell more animals than you got!
Allen Druze . . . . . . . . . 3 pts - Profit was $84 wholesale, but $92.40 after you sell the rest.
Ravi Raja . . . . . . . . . . . 3 pts - That's it; set up a nice sys of eqns with x unsold pigs and solve!
Jeff Limber (new) . . . . . . 3 pts - Welcome to the contest! Good steps and (pairs of) comments!
Martin Gritsch (new) . . . 3 pts - A new contestant all the way from Austria. Wilkommen! Good proof.
Tim Poe . . . . . . . . . . . . . 3 pts - Good orig ans except for the 16 1/2 pairakeets, resub fixed it right up.
Arthur Morris . . . . . . . .2 pts - Bonus pt included; good prf of e); no reasons why a - d don't work...
Akifumi . . . . . . . . . . . . 2 pts - Welcome to my contest, a former student of a former student!
Ed Wern. . . . . . . . . . . . . 2 pts - Partly ok but with 4 unsold pigs you have fractional birds I'm afraid...
 
 
Problem #175 - Posted Friday, June 27, 2003
Fruit Season Again ! . (back to top)
What could be better than fresh fruit season? Yumm! Doctors say to get plenty of
carbohydrates and protein each week, but not too much fat or too many calories !

Fruit

 Protein grams

Fat grams

Carbo grams

Calories

Apple

0.6

0.8

24.1

106

Banana

5.0

6.0

89.0

100

Cantaloupe

1.5

0.0

23.5

50

Peach

 0.5

0.1

7.6

33
You've been ordered to eat, each week, a total of at least 230g of protein, at least
4400g of carbos, and at most 260g fruit fat, while minimizing your total calories.
How much of each fruit should you have weekly, if: a) you're required to eat
whole numbers of fruits? b) you can cut up the fruit and eat any amounts?
 
Solution: Dan's note: When I made up this problem I found a solution in integers, and then used the W.O.P.
(Well-Ordering Principle) of the integers to guarantee that there would be a minimal calorie solution. Well,
the answer I got was nowhere near as lo-cal as you sharp contestants. I humbly list my total after these:
Nick says: . . . we have (where A,B,C,P are the numbers of apples, bananas, etc)
. . . . 0.6*A+5*B+1.5*C+0.5*P>=230 . . . . 0.8*A+6*B+0.1*P<=260
24.1*A+89*B+23.5*C+7.6*P>=4400 . . . . we want to minimize: 106*A+100*B+50*C+33*P
Jay writes: Per calorie, Banana and Cantaloupe have the most Protein and Carbohydrate.
Hence the diet must contain as many Bananas as the fat constraint will allow and enough Cantaloupes
to make up the remaining Protein and Carbohydrate requirements.
The solutions are: a) 43 Bananas and 25 Cantaloupes. b) 43 1/3 Bananas and 23.1206 Cantaloupes.
Dan's further notes: Gee, I never realized that apples and peaches were so unhealthy!
Nick agrees mathematically but mentally can't fathom eating over 3 canteloupes per day!
My orig ans was A=24, B=40, C=12, P=7 : 7375 cal ; 24.2, 39.9, 8.02, 9.57 : 7280 cal.
WINNERS - Problem 175 . (back to top) . leader board . . . .
Jayavel Sounderpandian 10 pts - Good; you and Steve both looked at calories per other things.
Steve Carlon . . . . . . . . . . . . 7 pts - Exact solution B=130/3, C=3260/141 to minimize cal, nice job.
Vince LoCascio . . . . . . . . . 5 pts - "Eat as many bananas as possible." Words for us primates to live by!
Nick McGrath . . . . . . . . . . 4 pts - first entry! a) 43 B and 25 C : 5550 ; 43.101 B and 24.001 C had 5510 cal.
Hermen Jacobs . . . . . . . . . . 3 pts - You meant 25 cants not 23, totals ok; your B=43,3, C=23,3 had 5495 cal
Allen Druze . . . . . . . . . . . . 3 pts - Your answers were clear, rounded to nearest calorie; more pts for reasons.
 
 
Problem #176 - Posted Tuesday, July 8, 2003
"Jack and Diane" . (back to top)
They've been immortalized in song, but how old are they? When Jack was twice as old as
Diane was when Jack was twice as old as Diane was when Jack was half as old as he is now,
Diane was half as old as Jack was when Diane was a year older than half as old as Jack is now.
Assuming their ages are whole numbers of full years, a) How old are Jack & Diane if they're both in their
twenties? b) What are all possible solutions if their ages are under 100?
 
Solution: (From energetic contestant Steve C.)
Define variables first: J = Jack's current age; D = Diane's current age; J-D = diff. in their ages (assume Jack is older)
Break the problem into two parts: 1) "When Jack was twice as old as Diane was when Jack was twice as old as
Diane was when Jack was half as old as he is now" . . . Working backwards:
D1 = Diane's age when Jack was half as old as he is now = J/2 - (J-D) = (D-J/2)
D2 = Diane's age when Jack was twice as old as D1 = 2D1 - (J-D) = 2D-J - (J-D) = 3D-2J
So Jack's age for part one is 2*D2, which is 6D-4J . . . 2) "Diane was half as old as Jack was when Diane was a
year older than half as old as Jack is now." . . . Again, working backwards:
J1 = Jack's age when Diane was a year older than half as old as he is now = (J/2+1) + (J-D) = 3J/2-D+1
So Diane's age for part two is 1/2*J1, which is 1/4(3J-2D+2) . . . Now we know that when Jack was 6D-4J years
old, Diane was 1/4(3J-2D+2) years old, which creates the equation:
(6D-4J) - (J-D) = 1/4(3J-2D+2), which simplifies to : 30D = 23J + 2
From this we can deduce that Jack's age must end in a 6, so we try J = 6, 16, 26,...,86, 96 looking for 23J+2 to be
divisible by 30. This yields the following solutions: {J,D} = {26,20}, {56,43}, {86,66}
A) Jack is 26 and Diane is 20. . . B) {J,D} = {26,20}, {56,43}, {86,66} . . . Thanks for the problem!
 
WINNERS - Problem 176 . (back to top) . leader board . . . .
Nick McGrath . . . . . . . . . .10 pts - Balanced combo of algebra & spreadsheet gives D = (23J+2)/30.
Vince LoCascio . . . . . . . . . 7 pts - Yes, divide the problem up into several points in time, you got it!
Steve Carlon . . . . . . . . . . . 5 pts - Managed to make a complicated sol'n understandable; see above.
Tim Poe . . . . . . . . . . . . . . . 4 pts - That's the key, age difference is constant (like I tell my older sister)
Jack Dostal . . . . . . . . . . . . 4 pts - Nice use of Jn & Dn to stand for the ages n years ago. Go Jack!
Jayavel Sounderpandian . 3 pts - That's the right equation, 23J+2=30D; just need to see those steps!
Hermen Jacobs . . . . . . . . . 2 pts - This was an intricate interdependence of "when" statements, it's true.
Gaurav Agrawal . . . . . . . . 2 pts - Good start; your answers don't work in the equation but nice try!
Ravi Raja . . . . . . . . . . . . . 2 pts - There's more to it than D = 3y - 2x but glad you enjoyed the attempt!
 
 
Problem #177 - Posted Wednesday, July 16, 2003
Cover That Table! . (back to top)
I was eating in a restaurant recently where the tables were squares
1.5 meters on a side. They were putting down paper tablecloths
for the next people, using four 1 meter squares for each table.
"That seems wasteful," I told them. "I bet you could cover a table
using only three uncut squares of paper." a) Was I right about three? Show/explain how to do it. b) What's the largest square table that
can be covered with three 1-meter squares? Answer is independent of a).
Show steps and reasoning. (Time's up on this one.)
graphics dan bach
 
Solution: a) No, you can't cover a table 1.5m x 1.5m. A question was e-mailed
to me about whether it was possible to cover a 1.3 x 1.3 table with three 1x1's,
I mistyped the problem statement (as well as forgetting to say three uncut sq's).
b) Well, I thought this would be easier, but my answer turned out to be the
worst of those tables over 1m x 1m. I thought a different angle would produce
an answer of about 1.3, but maximizing the side produced only 1.06m.
Art and Vince had different layouts but the same side of s = (1 + sqrt[2])/2
=1.207. Tim and Allen used trilateral symmetry; Tim cut out some squares
and estimated a side of 1.17, but Nick, Jay, and Steve won the day with a side
of sqrt[sigma] = 1.272, where sigma = the Golden Ratio = (1 + sqrt[5])/2 =
1.618 (all decimals approx.)
(Dan's note: My diagrams, at the right, were a lot prettier in Photoshop before
GIF animation compression. Click Reload or Refresh if not animating.)

WINNERS - Problem 177 . (back to top) . leader board . . . .
Nick McGrath . . . . . . . . . . 9 pts - First answer, best result (but -1 pt for missing paren and resub.)
Jayavel Sounderpandian . 7 pts - Nice algebra and explanation, co-biggest table, thanks 4 steps.
Steve Carlon . . . . . . . . . . . 5 pts - Good use of trig, getting s = sec[B] where sin[B] = (sqrt[5] - 1)/2
Tim Poe . . . . . . . . . . . . . . . 3 pts - Fell a bit short of optimal size, but I liked the hands-on approach.
Vince LoCascio . . . . . . . . . 3 pts - Your side was 1.207 w/diff arrgmt than Art (not 1.212 you claimed)
Arthur Morris . . . . . . . . . . 3 pts - That's another way to get 1.207, welcome back Art!
Hermen Jacobs . . . . . . . . . 2 pts - No penalty for resub, right about limit = sqrt[3] with tiny pieces.
Erika Brandner (new) . . . . 2 pts - Welcome to my contest! Attachments ok for this prob, see pic.
Phil Sayre . . . . . . . . . . . . . 2 pts - Good to see you back, I had some slow-routed e-mail (thru school)
Allen Druze. . . . . . . . . . . . 2 pts - Nice symmetry, as shown above, but side wasn't quite as big.
 
 
Problem #178 - Posted Thursday, July 24, 2003
The Cat-Census (This problem comes to us in limerick form!) . (back to top)
"They've counted the cats in Llanfair,
Which number a third of a square.
If a quarter were gone,
Just a cube would stay on.
How many, at least, must be there?"
Show steps & reasoning.
 
Solution:
I can't find too much wrong with Jay's answer, Allen D's was similar, as were lots using algebra :
"Let c be the number of cats. Then c = (1/3)n^2 for some integer n, which makes n divisible by 3.
Let n = 3m for some integer m. Then, we get c = 3m^2. When a quarter is gone we are left with
(3/4)c = (9/4)m^2. This number must be a cube, say, k^3 for some integer k. We then get
m^2 = (4/9)k^3. Or, m = (2/3)k^3, which makes k divisible by 3. The smallest such k is 3,
which makes m = 18 and c = 972. Thus, the smallest number of cats is 972." (Maybe k^(3/2)-Dan)

Some of you 'pounced' on this cat problem, evidently not one of my toughest. One bonus point
for the first person to kiss up and offer the next-smallest answers, point goes to returning
contestant Alan O'Donnell for the next ten! After 972 cats comes 62208, then 708588, etc.
 
Another grateful point to Ed Wern who offers his own limerick!
There once was a puzzler named Ed
Who tried to count cats in his head.
What with cubing and squaring
He got to despairing,
So used paper and pencil instead.
 
WINNERS - Problem 178 . (back to top) . leader board . . . .
Nick McGrath . . . . . . . . . . 9 pts - First answer, and correct but lacking some algebra cohesiveness
Quasi-C . . . . . . . . . . . . . . . 7 pts - Next answer sent, but just received! Summer e-mail didn't all fwd!
Jayavel Sounderpandian . 6 pts - Algebra proof was really good, but is sqrt of k^3 needed?
Mahbubul Hasan (new) . . . 5 pts - Welcome to my contest! (Also thanks for table idea, I mistyped it.)
Alan O'Donnell. . . . . . . . . 5 pts - I like your approach: "Simply check n^3, if 4/3 of this is a square.."
Jack Dostal . . . . . . . . . . . . 4 pts - That's the way, make Excel do roots, and pick out the integers.
Ed Wern . . . . . . . . . . . . . . 4 pts - Includes a bonus pt for the limerick, you're a both-brainer!
Arthur Morris . . . . . . . . . 3 pts - I like it: 'web search on Llanfair proved fruitless.' Rats, no cats.
Hermen Jacobs . . . . . . . . . 3 pts - That's good reasoning as usual, HJ. Smart to try p = 3^q.
Tim Poe . . . . . . . . . . . . . . . 3 pts - Any restrictions on what numbers you scan to see if x works?
Ravi Raja . . . . . . . . . . . . . 3 pts - Right, the min k that gives 3x/4 = k^3 is 12, gives 972 cats.
Phil Sayre . . . . . . . . . . . . . 3 pts - Happy 2CU! Thanks 4 extras beyond 972; Credit for last week too.
Marcello Cammarata . . . . 3 pts - That's what it boils down to ; 1/3 of 54 squared = 972.
Allen Druze. . . . . . . . . . . . 3 pts - Nice symmetry, as shown above, but side wasn't quite as big.
Zahi Teitelman . . . . . . . . . 3 pts - I got this and some other answers in another folder, good job.
Steve Carlon . . . . . . . . . . . 3 pts - Right, N = x^2/3, x is div by 6; then try 6, 12, ... until x^2 = 4 y^3.
Vince LoCascio . . . . . . . . . 2 pts - Good try got 2,304 cats which is 48^2, it's a big square cat army!
 
 
Problem #179 - Posted Wednesday, August 6, 2003
Around The World ! . (back to top)
A group of airplanes is based on a small island. Each plane holds just enough
fuel to make it halfway around the world. Any amount of fuel can be transferred
from one plane to another in flight but the only source of fuel is on the island.
What is the smallest number of planes that will ensure the flight of one plane all
the way around the world (on a great circle), and how is it done?
Assume the planes have the same constant ground speed and rate of fuel consumption, that
no time is lost refueling in the air or on the ground, that all planes return safely to base island,
and can land nowhere else. Show steps and reasoning. (Time's up on this one.)
One plane makes it
around the world.
Solution: From Art "All-Time" Morris:
"Start with some [more] assumptions. Any amount can be transfered only up to the original amount of fuel.
The same plane can be used more than once. They do not have to take off at the same time.
Each plane can receive and/or offload more than once.
The general approach is to empty one plane of fuel as soon as the remaining planes can receive it.
Obvious that there is no solution with two planes. [Here's the method with three planes, P1, P2, P3:]
P3 will fly around the world; times are given in units of P3 position. P1, P2, P3 take off together.
At 1/8 around, P1 gives 1/4 tank to P2, P1 gives 1/4 tank to P3, and P1 returns home.
At 1/4 around, P2 gives 1/4 tank to P3 and P2 returns home. At 1/2 around, P1 takes off.
At 3/4 around, P1 gives 1/4 tank to P3 and turns around. At 3/4 around, P2 takes off.
At 7/8 around, P2 gives 1/4 tank to P3, P2 gives 1/4 tank to P1 and turns around.
P1, P2 and P3 arrive home together. There are a few other solutions with ... different schedules."
 
 
WINNERS - Problem 179 . (back to top) . leader board . . . .
Arthur Morris . . . . . . . . . 10 pts - Good answer, and apparently not the only one! See above.
Tim Poe . . . . . . . . . . . . . . . 7 pts - Nice use of degrees and longitude (if they fly on equator?)
Jayavel Sounderpandian . 5 pts - Good idea to divide 'world' into eighths, similar to Art's.
Alan O'Donnell. . . . . . . . . 4 pts - First 4 planes, then re-sub was nice thorough way w/ three.
Ed Wern . . . . . . . . . . . . . . 4 pts - Thanks for sharing all the false starts and mental struggles!
Leora (new) . . . . . . . . . . . . . 4 pts - Welcome to my contest! Nice and well-explained, 3 planes.
Sam Wilson . . . . . . . . . . . 4 pts - Welcome back! Amelia makes it alive with Orville & Wilbur.
Jack Dostal . . . . . . . . . . . . 3 pts - Good explanation, solution works, but 4 planes used.
Hermen Jacobs . . . . . . . . . 3 pts - Your first (4 planes) sol'n was ok, no charge for re-sub.
Nick McGrath . . . . . . . . . . 3 pts - Thought 4 planes at first, then redid w/ 3 w/o resubmitting!
Phil Sayre . . . . . . . . . . . . . 2 pts - Used 4 planes, table was readable and correct but extra plane.
AZ Runner . . . . . . . . . . . . 2 pts - Three start out, refuel at 1/8, etc. Works but has extra plane.
Ravi Raja . . . . . . . . . . . . . 2 pts - I understood your text e-mail ok, no need for attchmt, but 4 planes
Bugz Podder (new) . . . . . . 2 pts - Welcome to the contest; not true that A needs to be full at 1/2 mark.
Quasi-C . . . . . . . . . . . . . . . 2 pts - Did it with 3 planes but illegally dumped one plane in the water
Steve Carlon . . . . . . . . . . . 2 pts - Used 7 or 8 planes and managed to send one around
Vince LoCascio . . . . . . . . 2 pts - "An inelegant blend of 18 planes" works but not so efficient
Allen Druze. . . . . . . . . . . . 1 pt - Not sure about your eqn; after refueling the one plane is empty
 
 
Problem #180 - Posted Sunday, August 17, 2003 . (back to top)
Cover That Board ?
Can you cover the board at the right
with just copies of the T-shaped tile?
Show how to
do it, or else
prove that it
can't be done.
No cutting, overlapping, or hanging T's off the edge!
(You can refer to columns A,B,... , rows 1, 2, ... if you want.
Show steps & reasoning; 1 pt penalty for resubmissions.)
Either show the board can
or can't be covered by T's.
 
Solution(s): Most of you realized that coloring the board chess-style facilitated an odd counting argument (literally).
This is a classic problem; in my Prealgebra textbook I offer the checkered board in the picture as a hint.
Here are a few short solutions from loyal contestants Art, Jay, and Nick, respectively and respectfully:
 
It cannot be done because an odd number of tiles (9) are required. Consider the 6 by 6 board as a checker board with
18 black and 18 white squares. Since any tile can only cover 3 black squares and 1 white square, or 3 white squares
and 1 black square, they must be used in pairs to cover the equal number of black and white squares (18 of each). Art
It is not possible to cover the board with T-shaped tiles.  Color the little squares on the board black and white alternatingly
as in a chessboard. Then each tile must cover 3 white and 1 black squares, or 3 black and 1 white squares.  Let x denote
the first type and y the second type of tiles.  Since there are 18 white squares and 18 black squares, we have .3x + y = 18
and . x + 3y = 18. Solving these, we get x = y = 4.5.  Since x and y are not integers, it is not possible to cover the board. Jay
 
It is NOT possible to cover the board. Paint the 6x6 board chessboard fashion so it has 18 black and 18 white squares.
Paint the tiles chessboard fashion so a tile has either 3 black and 1 white square (type A) or 3 white and 1 black square
(type B). (in each case the 'odd man out' color being in the center of the row of 3 squares) If it is possible to cover the board we clearly
need 9 tiles. Assume we need x Type A's and (9-x) Type B's. We then have a total of 3*x + 1*(9-x) = 2x+9 black squares
on the 9 tiles. However there is no integer value for x which results in a total of 18 black squares; the tiling is not possible. Nick
 
 
WINNERS - Problem 180 . (back to top) . leader board . . . .
Arthur Morris . . . . . . . . . 10 pts - Nice point; the 3w1b and 3b1w must be used in pairs, ok!
Alan O'Donnell. . . . . . . . . 7 pts - Covered all cases, 9wwwb,0bbbw --> 0wwwb,9bbbw, all odd.
Tim Poe . . . . . . . . . . . . . . . 5 pts - Good non-checkered proof, no charge or bonus for b/w re-sol.
Nick McGrath . . . . . . . . . . 4 pts - Cool, x bbbw, 9-x wwwb --> 2x+9 black sq's covered, not 18.
Ed Wern . . . . . . . . . . . . . . 4 pts - Slick coord proof showing that square F1 is uncoverable!
Quasi-C . . . . . . . . . . . . . . . 4 pts - Nasty alternating colors spoil the 2N+9 white squares theory.
Jayavel Sounderpandian 4 pts - I liked your pure algebra no-denying-it proof. See above.
Hermen Jacobs . . . . . . . . . 4 pts - Nice idea of T's 'covering up' the black and white counters.
-- These five answers were just as correct but sent in a day or three after the others...
Steve Carlon . . . . . . . . . . . 3 pts - Right, with 3b1w or 1b3w and an odd number of pieces...
Marcello Cammarata . . . 3 pts - That's it, the "odd*odd = odd" argument works the best.
Sam Wilson . . . . . . . . . . . 3 pts - Yes, w/ N wwwb and 9-M bbbw you get 2N+9 black sqs covered.
AZ Runner . . . . . . . . . . . . 3 pts - Rules out case by case: if 2 bases of T's on one side, etc.
Marc Wallace (new) . . . . . . 3 pts - Good ans and nice use of text for graphics * * X X * *
-- The rest were several days later or partially correct-- thanks for all your entries!
Lisa Schechner . . . . . . . . . 2 pts - Hi, welcome back, just under the wire! Did Euler find this solution?
Vince LoCascio . . . . . . . . 2 pts - I got excited that you'd tiled the impossible, but yours was 8x8!
Phil Sayre . . . . . . . . . . . . . 1 pt - Filling a 4x4 corner is not the only way you might tile the board
Ravi Raja . . . . . . . . . . . . . 1 pt - Again, yes you can fill 4n x 4n, but doesn't preclude 6x6.
Ken Duisenberg (new) . . . . . 1 pt - Welcome! A couple weeks after others but still under the wire!
 
 
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YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 2003 A.D.
 
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