dansmath.com - problem of the week - archives
 
 
Problem Archives page 19
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
 
181 - A Triangular Duel
182 - Two More Trains
183 WhichKindaCandy
184 Crank Out Power(s)
185 - Money-Go-Round
186- Sums of 2 Squares
187- Right Triang Spiral
188- Chess Tournament
189- - -The Three Trees
190 -- The Flower Farm
 
 
Problem #181 - Posted Friday, August 29, 2003 . (back to top)
The Triangular Duel (Next-to-last problem in the sixth annual contest. Seventh Season Starts with Prob 183!)
Three
men: Fermat,
Galois, and Hilbert,
decide to fight a pistol duel.
They'll stand at the corners of an
equilateral triangle, and each man, in
order, will aim and shoot wherever he pleases.
They choose randomly who will be shooting first,
second, and third, and will continue in order until two
of them are dead. All three know Fermat always hits his
target, Galois is 80% accurate, and Hilbert hits his mark half
the time. Assuming that all three adopt the best strategy and that
nobody is killed by a wild shot not intended for him, who has the best
chance to survive, and why? 1 bonus pt: find the survival probabilities for each man.
Explain your reasoning carefully. Bonus answer not req'd. One pt penalty for resubmissions.
 
Who is most likely to survive?
Solution: Most of you figured out that the worst shot Hilbert has the best chance of surviving, mostly because
the better shots would try to kill each other off first. Some of you figured out that Hilbert maximizes his chance of
survival by shooting into the air whenever it's his turn (and he's still alive). Those of you that had F sh G, G sh F,
and H sh in air, got credit for a correct answer, and most of you figured out the correct probabilities listed below.
If you had H shoot F, then the percentages still favored Hilbert but this was counted as a "suboptimal strategy".
 
The probability tree is complicated. If H shoots first he 'passes', so there are two branches : F sh first (1/2),
F kills G 100%, or G sh 1st (1/2), G kills F 80%. Then H sh survivor F or G, (1/2), if miss they sh H (4/5 or 1).
Continue in this way and add all products of probs along the way to a result F or G or H is last survivor.
On at least one perpetual series of G and H surviving shots, you need to use an infinite geometric series. Wow!
Tim was first, but Nick and Hermen get the model citizen awards for correctly getting and giving the survival
probabilities : Fermat: 3/10 = 30.00% ; Galois: 8/45 ~ 17.78% (G screwed again!) ; Hilbert: 47/90 ~ 52.22%.
Recently joining our contest, Ken D. sends in a link to a similar problem with different percentages,
you can download a PDF at http://www.contingencies.org/novdec00/puzzles.pdf
 
Hermen J of Holland points out that Hilbert died in 1943, Galois (in a duel) in 1832, and Fermat in 1665,
so this duel (or is it a truel") could only take place in heaven , but nobody would shoot each other there!
And Ed W cracked me up with, "Why each agrees to take turns when his life is at stake is beyond me.
I would shoot first, shoot fast, and shoot often."
 
WINNERS - Problem 181 . (back to top) . leader board . . . .(blue score includes probability bonus point.)
Tim Poe . . . . . . . . . . . . . . . 10 pts - Laying off the probability point but correct in your strategy.
Nick McGrath . . . . . . . . . . 8 pts - "One of my favorites" Not your 'favourites'? Seen it b4, eh?
Hermen Jacobs . . . . . . . . . 6 pts - That's it, have Hilbert shoot in the air, let the other two die off.
Quasi-C . . . . . . . . . . . . . . . 4 pts - You got the intentional miss idea right, but percents a bit off
Marcello Cammarata . . . . 3 pts - Your probabilities were right assuming H shoots F; good job.
Steve Carlon . . . . . . . . . . . 3 pts - Glad you liked this week's (stolen) problem, branching tree good.
Alan O'Donnell. . . . . . . . . 3 pts - Good to average results over six orders, FGH, FHG, ...
Ed Wern . . . . . . . . . . . . . . 3 pts - Great quote, notes G's best chance, H's worst chance is to shoot 1st!
Allen Druze. . . . . . . . . . . . 3 pts - Your Hilbert didn't shoot in the air but calculations seemed good.
Jayavel Sounderpandian . 2 pts - Nice pacifist solution; "nobody shoot and everybody wins!"
Vince LoCascio . . . . . . . . . 2 pts - All three submissions got gradually better, good job Vince!
Bob Seegmiller (new) . . . . . 2 pts - Welcome! Did an empirical simulation, (F,G,H) = (27.7, 27.0, 45.3)
Ken Duisenberg (also #180) 1 pt - Even though not a real submission, thanks for sending the link!
 
 
Problem #182 - Posted Friday, September 12, 2003
Two More Trains! . (back to top)
(Last problem in the sixth annual contest! Seventh Season Starts next week with Prob 183!)
Every algebra class is required by law to have a "two trains" problem.
But this one doesn't involve distance, rate, or time!
A short freight train, made up of an engine and five boxcars, stops at
a small station. The station has a small siding that can hold three cars,
or an engine and two cars. A long passenger train is approaching from
behind, going the same direction. How can they let the passenger train
through, and then put the freight train back just the way it was ?
 
 
Solution: Dan's note: I could have spelled out the rules, but all of you except Quasi made the assumption
that the Passenger train could Pull any number (ok, three) of the boxCars. This is in fact what I had in mind
(PPC) when I stole the problem, but Quasi figured that certainly a passenger train could push some cars, eh?
That's another assumption, I agree, but it made the problem harder and more correct as stated in the random
opinion of the panel of dansmath.com judge, so Q-C was awarded first place this week. But the rest of you,
beginning with Tim, were ranked with the normal number of points.
 
Now for the actual solutions... I had fun verifying these with a quarter, five pennies, and a pen.
 
Jack Dostal sends in the popular solution, assuming Passenger train can Pull Cars (PPC):
"The freight train backs into the siding, drops off three boxcars, and proceeds forward with its remaining
two boxcars. The passenger train follows the freight train forward until it passes the siding. The passenger
train then backs into the siding, picks up the three boxcars, and pulls forward onto the main track. Both
trains then back up until the passenger train has passed the siding and stops. The freight engine and its two
boxcars back into the siding. The passenger train releases its three boxcars and proceeds forward on its way.
After the passenger train has passed the siding, the freight engine pulls forward onto the main track, then
backs up until it reaches the three boxcars, reattaches those cars and proceeds forward on its way."
 
But Quasi-C managed the job assuming (not) only that the passenger train could push cars, and (but) the
freight engine could hook its nose into a car and back up with it (which I hear they can, as it turns out):
1. The freight train backs onto the siding, and drops F4 & F5 onto the siding.
2. E returns to main track, backs up on the main track and drops F1,F2 &F3.
3. E pulls forward, and backs onto the siding.
4. P passes the siding pushing the group F1,F2,F3 forward.
5. E emerges from the siding with F4 & F5, and backs them down the main track a bit. E then goes back onto the siding.
6. P backs up past the siding, pushing F4 & F 5 further back if that has any appeal.
7. E emerges from the siding, hooks onto the "back" of F3, disconnects between F1 & F2, and pulls F2 and F3 with it onto the siding.
8.P pulls ahead of the siding, taking care not to crash into F1 sitting up there ahead somewhere.
9.E pushes F2 and F3 onto the track, and returns onto the siding.
10.P backs up encountering F2&F3 and pushes them past the siding all the way back to join F4&F5.
11.E emerges, goes forward, engages with F1, and pulls it onto the siding.
12.P goes forward well past the siding.
13.E pushes F1 onto the track, and returns alone to the siding.
14.P rolls back, bumping F1 back past the siding, and then proceeds forward down the clear track toward its destination.
15.E emerges from the siding with cars F1, F2, F3,F4 amd F5 down the track in the original order. It's all over but the re-linking."
 
WINNERS - Problem 182 . (back to top) . leader board .
Quasi-C . . . . . . . . . . . . . . . 9 pts - The resubmission was a great workaround, quoted above.
Tim Poe . . . . . . . . . . . . . . . 7 pts - Your answer was the first one received of those assuming PPC.
Nick McGrath. . . . . . . . . . 5 pts - Broke the process down into eight bite-sized steps, assuming PPC.
Jack Dostal . . . . . . . . . . . . 4 pts - Right, the freight can back up, even though that's an assumption too!
Hermen Jacobs . . . . . . . . . 4 pts - Nice solution, and thanks for the good wishes for next year's contest!
Jayavel Sounderpandian . 3 pts - Ok good notation (123456, empty)-->(123P,456) but some unexplained
Marcello Cammarata . . . . 3 pts - Good expl about when each thing goes FW or BW, keep it up!
Alan O'Donnell. . . . . . . . . 3 pts - Right you are, an easier one this week indeed! But try next week's.
Art Morris . . . . . . . . . . . . . 3 pts - Yes, no 'flying switches', 'drop switches', or 'gravity switches'!
Richard Hillman . . . . . . . 3 pts - Good answer. Yep, it's been a while, thanks for coming back!
Phil Sayre . . . . . . . . . . . . . 3 pts - Correct to state you're assuming P can pull cars, thanks.
Joe Alvord. . . . . . . . . . . . . 3 pts - Welcome back, how are the trains runnung up in Alaska?
Greg McWhirterer (new) . . 2 pts - Welcome! Some problems are easier, try again for a harder one!
Ken Duisenberg . . . . . . . 2 pts - Your solution slightly different, P puts 3 cars on siding, good!
Sandy Thompson (new) . . 2 pts - Glad to have you, good luck in the new 7th year contest!
Allen Druze. . . . . . . . . . . 1 pt - Your passenger train won't all fit on the siding, otherwise it'd work.
 
 
 
Problem #183 - Posted Wednesday, September 24, 2003
Which Kinda Candy? . (back to top)
(First problem in my seventh annual contest!)
The labels on these boxes of candy got so mixed up that none of the
boxes is labeled correctly. What is the least number of candies you must
taste test, and from which box(es), to determine which box has what?
Explain your reasoning carefully. One point penalty for resubmissions. 

3 chocolates
 
3 cremes
 
2 chocolates
1 creme
Solution: I was surprised that you only have to test/taste one candy. Take one candy from the third box,
the one marked "2 chocolates,1 creme". a) If your test candy is chocolate, since the label is wrong, the test box has three chocolates.
The one (mis)marked "3 cremes" now can't have 3 choc either so it's 2 choc 1 creme. The one marked "3 choc" has 2 choc 1 creme.
b) If the test candy is creme then the box is 3 cremes, the one marked "3 choc" is wrong so it's 2 choc 1 creme. And the one marked
"3 choc" has 2 choc 1 creme. Either way you've figured it all out eating only one candy. ("Bummer," as a couple of you candy fans said.)
 
A large group of contestants (28) this week, to start the new contest year! Welcome to the (apparently) growing dansmath community!
This time we have entrants from (among other places): Italy, India, The Netherlands, Alaska, Bangladesh, England, Canada, and Ohio!
Special mention (and bonus pt) goes to Quasi-C(onscious) who points out the problem assumes that you can relabel and have it be correct.
WINNERS - Problem 183 . (back to top) . leader board .
Anirban Bhattacharyya . 10 pts - Welcome back, and good job 2 cases for 3 boxes, first entry in!
Nick McGrath. . . . . . . . . . 7 pts - But did you go out and buy 3 boxes of candy for this problem?
Nikita Kuznetsov . . . . . . . 5 pts - Been a while, betchur busy, good to see u back!
Marcello Cammarata . . . . 5 pts - That's right; taste just one from the mixed box. Bene!
Here's a listing of the next group of correct questioners, arranged and scored by earliness.
 Mahbubul Hasan . . 4 pts  Tim Poe . . . . . . . . . 4 pts  Quasi-C. . . . . . . . . . 4 pts
 Alan O'Donnell . . . 3 pts  Hermen Jacobs . . . 3 pts  Jeremy Galvagni (new) 3 pts
 Joe Alvord. . . . . . . . 3 pts  S. Khoo (new) . . . . . 2 pts  Ravi Subramanian (new) 2 pts
 Allen Druze. . . . . . . 2 pts  Ed Wern . . . . . . . . 2 pts  AZ Runner . . . . . . . . . 2 pts
 Lisa Schechner . . . . 2 pts  Akifumi Iwahashi. . 2 pts  Ravi Raja . . . . . . . . . . 2 pts
 Greg McWhirter . . 2 pts  Sandy Thompson . . 2 pts  Mark Moyer (new) . . . . 2 pts
 Ken Duisenberg . . . 2 pts  Nemo Dumple (new) . 2 pts
Farid Mark Watson (new) 1 pt
And some well-meaning testers who ate too much candy - keep on entering!
 Art Morris . . . . . . . . . . . . . 1 pt  Vince LoCascio . . . . . . 1 pt
 Jack Dostal . . . . . . . . . . . . 1 pt  John Friend (new) . . . . . 1 pt
 
 
Problem #184 - Posted Sunday, October 5, 2003
Crank Out Power(s)! . (back to top) (Time's up on this one)
Here are a couple of innocent polynomial questions 4U:
(answer both questions to be ranked as correct; n^p means n to the power p)
 i) If we know x + y = 1 ,
and also x^2 + y^2 = 2,
then what is x^3 + y^3 ?
ii) We have a + b + c = 1 ,
and a^2 + b^2 + c^2 = 2,
and a^3 + b^3 + c^3 = 3,
What is a^4 + b^4 + c^4 ?
Explain your reasoning carefully. One point penalty for resubmissions.
 
Solution: a) This is one trippy system of equations. You'd want x^3 + y^3 to come out to be 3, but it won't.
The direct approach is to say y = 1 - x, and x^2 + (1-x)^2 = 2, and solve for x = (1 +/- sqrt[3])/2, then
compute and expand x^3 + (1 - x)^3. But a slicker way would be to use pure algebra. As Allen Druze puts it:
(x+y)^2 = x^2 + y^2 + 2xy = 2 + 2xy = 1, xy = - 1/2.
x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 1^3 - 3(-1/2)(1) = 5/2
b) These equations were, according to Nick (if not Britney Spears), "not so innocent."
Again, the answer, for  a^4 + b^4 + c^4 is "not 4." We can employ Mathematica to solve for a, b, and c,
but the solutions are complex (in both senses of the word), and the real graphs of the three equations as
surfaces in (a, b, c) space, don't intersect at all. Here's Anirban's take on the issue:
 
"It is given that, a + b + c = 1 ---- (i) ; a^2 + b^2 + c^2 = 2 ---- (ii) ; a^3 + b^3 + c^3 = 3 ---- (iii)
Multiplying (i) and (iii) we get,
a^4 + b^4 + c^4 + a*b^3 + a^3*b + a^3*c + a*c^3 + b*c^3 + b^3*c = 3
or, a^4 + b^4 + c^4 + a*b*(a^2 + b^2) + b*c*(b^2 + c^2) + c*a*(c^2 + a^2)=3
or, a^4 + b^4 + c^4 + a*b*(2 - c^2) + b*c*(2 - a^2) + c*a*(2 - b^2)=3 [from (ii)]
or, a^4 + b^4 + c^4 + 2*(ab + bc + ca) - a*b*c*(a + b + c)=3
Now 2*(ab + bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2) = 1-2 = -1
Again (a + b + c)^3 = a^3 + b^3 + c^3 + 3*(a + b)*(b + c)*(c + a)
= a^3 + b^3 + c^3 + 3*(1 - c)*(1 - a)*(1 - b) = a^3 + b^3 + c^3 + 3 - 3*(a + b + c) + 3*(ab + bc + ca) - 3*abc
Solving we get abc = 1/6. Hence a^4 + b^4 + c^4 = 3 - 2*(ab + bc + ca) + abc*(a + b + c) = 3 + 1 + 1/6 = 25/6
 
WINNERS - Problem 184 . (back to top) . leader board .
Joe Alvord. . . . . . . . . . . . 10 pts - Nicely worked, with Mathematica to expand (a+b+c)^4.
Anirban Bhattacharyya . 7 pts - You have a nice balance of detail and ideas; see above sol'n.
Mahbubul Hasan . . . . . . 5 pts - That's it; (x+y)^2 = x^2 + y^2 + 2xy ==> xy = -1/2 etc.
Marcello Cammarata . . . 5 pts - Good; among other things you found that abc = 1/6...
Quasi-C. . . . . . . . . . . . . . 4 pts - ... and you found that ab+ac+bc = -1/2. Are these Bernoullis?
Ed Wern . . . . . . . . . . . . . 4 pts - Used good ole quadratic formula to find x,y; then improved!
S. Khoo . . . . . . . . . . . . . . 4 pts - Excellent solution, comparing (a+b+c)^3 with (a+b+c)(a^2+b^2+c^2)
Ken Duisenberg . . . . . . . 4 pts - Now you're gettin' the hang of the complete logical argument!
Ravi Raja . . . . . . . . . . . . 4 pts - Extra point included for cool bonus equations, good (a2+b2+c2)^2.
Nick McGrath. . . . . . . . . 3 pts - Right, the innocent second part wasn't... and yes, x3-x2-x/2-1/6=0
Farid Mark Watson (new) 3 pts - Actually you were 'new' last week but this answer came first!
Akifumi Iwahashi . . . . . 2 pts - Yes, your college compter worked fine. Go Bears... grrr!
Allen Druze. . . . . . . . . . . 2 pts - Good expl'n on first part (see above); second answer correct also.
Jeremy Galvagni . . . . . . 2 pts - Right; there are no real solutions for a, b, c, but no need 4 'em.
Hermen Jacobs . . . . . . . . 2 pts - Yes, the algebra gets messy on the second part; 25/6 not 23/6.
Zahi Teitelman . . . . . . . . 2 pts - Good solution on first part, no 2nd part, I rounded your score 'up'!
Alan O'Donnell . . . . . . . 2 pts - Nice Excel sheets suggest no real sol's; where is (3sqrt3)/2 from?
Phil Sayre . . . . . . . . . . . . 2 pts - Cartan and Tartaglia knew a lot about cubics, and yes, it's 4 1/6.
Dennis Molter (new) . . . . . 1 pt - Welcome! Good discussion, solutions aren't whole or rational...
 
 
Problem #185 - Posted Saturday, October 18, 2003
Money-Go-Round! . (back to top)
Some card players sat in a circle, so that each had two neighbors, and each had a certain
number of dollars. The 1st player had $1 more than the 2nd player, who had $1 more than
the third, and so on. The first player gave $1 to the second, who gave $2 to the third, and
so on, each giving $1 more than they received, around and around the table as long as possible.
There were then 2 neighbors, one having 4 times as much money as the other.
(a) How many players were there? (b) How much money did the first player start with?
Explain your reasoning carefully. One point penalty for resubmissions.
 
Solution: (From super-solver Nick McGrath)
Each player's fortune diminishes by 1 for each turn he plays. So clearly the last player is the first to run out of money.
Play continues through the next round until the last player receives his coins from the second-to-last player but then
cannot go - so the game stops. At this point the last player has the most coins. Since all the other player's totals differ
from their neighbor's by one coin, we conclude that the last player now has four times as many coins as the first player.
Let the first player start with X coins. Let their be N players; so the last player starts with X-N+1 coins.
The last player's total diminishes by 1 for each round played; so after X-N+1 rounds he is left with zero.
On the next round the first player's total is reduced to X-(X-N+1)-1= N-2 coins
The last player receives fron the second-to-last a number of coins equal to the total number of turns since the game
started which will be N(X-N+1)+(N-1) = NX - N^2 +2N - 1.
Since the last player now has four times the first player's total:
NX - N^2 +2N -1 = 4(N-2) => NX = N^2 + 2N - 7 => X = N + 2 - 7/N
Clearly 1 and 7 are the only N's to give integer X. N=1 gives X=-2 which we can reject
So the only solution is X=8 N=7 => a) Number of players = 7 b) First player starts with $ 8.
 
WINNERS - Problem 185 . (back to top) . leader board . . . Yikes! Lots of contestants! (Ok, I love it. - Dan)
Nick McGrath. . . . . . . . . 9 pts - Good answer; your second effort had correct formula
Marcello Cammarata . . . 8 pts - Extra second place point for preceding resubmission
Tim Poe . . . . . . . . . . . . . . 6 pts - Nice to clarify that the 7th player has $20 and 1st has $5
S. Khoo . . . . . . . . . . . . . . 5 pts - I like the image of a snowball being added to & passed around
Jeremy Galvagni . . . . . . 4 pts - Good; d+1 rounds if last player has d to start with
Quasi-C. . . . . . . . . . . . . . 4 pts - The 'simple gifs' are the best, I agree totally .doc!
Dennis Molter . . . . . . . . 4 pts - Thorough explanation, I like the notation 987, 897, etc.
Jack Dostal. . . . . . . . . . . 3 pts - You let C++ play the game, but without visors or cigars.
Yakov Macak (new) . . . . . 3 pts - Thanks for entering, glad you happened on my site!
Alan O'Donnell . . . . . . . 3 pts - I can say I have world-class puzzlers entering my contest!
Omar Gymboski . . . . . . 3 pts - Welcome back! Nice clear explanation powered by algebra
Ed Wern . . . . . . . . . . . . . 3 pts - Now that's a play-by-play description of the 7-player game
Jim No-Spam (new) . . . . . 3 pts - Welcome to the contest, good answer spam or no spam!
Anirban Bhattacharyya 3 pts - Pairing of last and non-adj 2nd player gave n=11 players
Art Morris . . . . . . . . . . . 2 pts - You had 4 times as much but money could still be passed
Hermen Jacobs . . . . . . . . 2 pts - Nice effort in getting part-way to the solution.
Zahi Teitelman . . . . . . . 2 pts - Good solutions but assumed the game ends after 1 round
Phil Sayre . . . . . . . . . . . . 2 pts - Python did its job but it may have been misdirected to stop
Mahbubul Hasan . . . . . .1 pt - Your solutions worked if you stop too soon; resub. cost a pt.
Ajit Athle (new) . . . . . . . . 1 pt - Welcome to my contest, Your 4-player solution was close
Ken Duisenberg . . . . . . . 1 pt - Right, we do keep playing when one has 0, the next won't.
Farid Mark Watson . . . 1 pt - I tried the n=3, k=4 but ended with1, 0, 8; not 4x as much
Math 012378 (new) . . . . . . 1 pt - Thanks for entering; answer was partly right; write again!
 
 
Problem #186 - Posted Thursday, October 30, 2003
Sums of Two Squares.(meaning the sum of the squares of two positive integers) (back to top)
a) Prove that the sum of two squares, multiplied by the sum of two squares, is the sum
of two squares. b) Prove the sum of two different squares, multiplied by the sum of
two different squares, can be written as the sum of two squares in two different ways.
c) Give the smallest three examples (each) of a) and b).
Explain algebra or reasoning carefully. One point penalty for resubmissions.
 
Solution: This problem took me a long time to score -- 1) Problem was worded inexactly/incorrectly ; 2) My
semester was ending and I was grading tons of homeworks and final exams ; 3) Other project deadline nearing.
 
First, the phrase about sum of squares of positive integers: The smallest example is (1^2 + 1^2)(1^2 + 1^2),
which is 2*2 = 4 ; but 4 = 0^2 + 2^2, not as specified. Quasi-C, Nick, and others sought to salvage the result.
A better wording might be "sum of two distinct squares" but then we eliminate 50 = 5^2 + 5^2 = 1^2 + 7^2.
 
The proof: (a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2
= a^2c^2 + 2abcd + b^2d^2 + a^2d^2 - 2abcd + b^2c^2 = (ac + bd)^2 + (ad - bc)^2
= a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2 = (ac - bd)^2 + (ad + bc)^2 
The last two bold expressions are different as long as ac+bd =/= ad+bc or ad-bc.
 
Jack D. wants us to know that Leonardo Pisano (Fibonacci) proved this product of sums of squares in 1202
in Liber Abaci (but not at the piano!) Gee, I thought it would have been in the Pascal-Fermat era!
 
Farid Mark reminds us that in a later generation, Euler (1738) proved if N = a^2 + b^2 then the prime
fac'z'n of N contains only even powers of primes of the form 4k+3. This property is clearly multiplicative!
 
For a) the smallest products of two sums of non-zero squares are 4, 8, 10, 16, 20, 25, 26 (these allow 0^2 in sum),
For b) the smallest are 25, 50, 65, 85, 100 (25 = 3^2 + 4^2 = 0^2 + 5^2; 50 = (5,5), (1,7), 65 = (4,7), (1,8) is 'best')
WINNERS - Problem 186 . (back to top) . leader board . . . Yikes! Lots of contestants! (Ok, I love it. - Dan)
Hermen Jacobs . . . . . . . . 9 pts - Woulda been 10 if you'd proved two different sums...
Nick McGrath . . . . . . . . . 8 pts - Good questioning of premise of problem... bonus pt!
Anirban Bhattacharyya . 7 pts - There were a couple of smaller answers but good job.
Marcello Cammarata . . . 6 pts - Good algebra and nice condition on 'difference' of sums.
Quasi-C. . . . . . . . . . . . . . 5 pts - This was a classic 'Ross salvage job', although 65 < 100.
Jeremy Galvagni. . . . . . . 4 pts - Correct algebra but did it prove ac+bd =/= ad+bc for ex?
Jack Dostal . . . . . . . . . . . 4 pts - Great historical note about Fibonacci - see above comment
Ken Duisenberg . . . . . . . 4 pts - One of the cleanest answers and from a fellow puzzle siter!
Mahbubul Hasan. . . . . . . 3 pts - Good proof/example of both facts, although (1^2 + 1^2)^2 =/= 2.
Vince LoCascio. . . . . . . . 3 pts - Interesting point: if ad=bc then one answer square will be 0.
Zahi Teitelman . . . . . . . . 3 pts - Separates into cases; some a,b,c,d equal. Smaller examples exist.
S. Khoo . . . . . . . . . . . . . . 3 pts - Expands (ac+bd)^2 + etc. Then recombines; it works!
Farid Mark Watson . . . . 3 pts - Unique approach; using Euler's 1738 result (comment above)
Ed Wern . . . . . . . . . . . . . . 3 pts - Seems to get all the small examples right, and a proof to boot!
Art Morris . . . . . . . . . . . . 2 pts - First entry, 'algebra-free' with examples only; need proof.
Allen Druze . . . . . . . . . . . 2 pts - Brevity is the soul of wit, not always of detailed proofs
Phil Sayre . . . . . . . . . . . . 2 pts - That's true about ac+bd etc; need to see the 'little algebra'
Yakov Macak . . . . . . . . . 2 pts - It does make a diiference (sum) if equal; 'smallest' is total.
Greg McWhirter . . . . . . . 2 pts - 'Late' entry ok as long as prob still up. Yes, zero is a confuser.
Ravi Raja . . . . . . . . . . . . . 2 pts - Right, add & subtract abcd to complete the two squares.
Alan O'Donnell . . . . . . . . 1 pt - 4 is indeed smallest except for the zero question. Salvage?
Barbara Garcia (new) . . . . .1 pt - Welcome Barb G! The squares could be any squares here.
Mario Roederer (new) . . . . 1 pt - Late memo; true that (1^2+1^2)(2^2+2^2)=16 needs 0^2, 0 not pos.
 
 
 
Problem #187 - Posted Tuesday, November 11, 2003
Right Triangle Spiral (back to top)
This problem still open! 1 point max while this sentence is up.
a) How many triangles will it take to go all
the way around and overlap the first?
b) Will the pattern go around and around forever,
or is there a limit to the number of rotations?
Prove your answers, explain carefully.

Solution: This picture has always intrigued me... First, it's a graphic demonstration that -/n is
"constructible" by ruler & compass, assuming the unit length is given. Second, it looks cool.
And third, I'd wondered if the triangles do go 'round 4ever. (Cheap square root symbol -/n)
 
a) Many (most) of you used Excel or s'mother program to decide on the first triangle that touches
(overlaps) the first triangle; a couple or 3 of you interpreted 'overlap' as covering the first triangle,
needing a sum of 405o rather than 360o. You weren't penalized for this reasonable approach.
 
Letting an = 'central' angle of nth triangle; we see a1 = 45o , and an = arctan(1 / -/n), It then takes
16 triangles to reach 351.1o, the 17th having an angle of 13.6o making a total of 364.7o.
So the intended answer was 17 triangles to go around once. (For 405o it takes 21 triangles.)
In fact, Vince and Ed say successsive revolutions are reached at triangles # 54, 110, 186, 281, ...
And Tim Poe figures that after a million triangles, we have 317 revolutions. (radius 1000)
 
b) The angle an = arctan(1 / -/n) so we want to show that S = Sum [n=1 to oo]{arctan(1 / -/n)} will
either converge on some finite limit or will grow without bound. This is a calculus-free contest,
so I should be clever about not using methods such as the integral test for divergent series (but
you folks can use calculus whenever you wish). In radians, we see that arctan(x) > x for small x,
so our sum S > Sum[n=1 to oo](1 / -/n). But as Zahi points out. the sum Sm of the first m terms,
1/-/1 + 1/-/2 + . . . + 1/-/m > 1/-/m + 1/-/m + . . . + 1/-/m = m / -/m = -/m , so taking the limit
as m -> oo, Sm grows without bound, and our sum S diverges.
 
I drew an animation in Mathematica of the first few hundred triangles, and to me it appears that
the radii of successive revolutions are about equally spaced. (1 bonus pt for first proof to Rick M.)
WINNERS - Problem 187 . (back to top) . leader board . . .
Quasi-C. . . . . . . . . . . . . . . 9 pts - First answer, largely right; tan x ~ x a bit loose (eg 1/n vs 1/n^2)
Nick McGrath . . . . . . . . . 7 pts - arctan x > x > -/x; Bernoulli 1st to show sum(1/n) diverges.
Art Morris . . . . . . . . . . . . 5 pts - Yes, 17 triangles and angle > 1/-/n but why is sum ~ 2-/n ?
Hermen Jacobs . . . . . . . . 4 pts - 17th triangle, yes; but needed 2 see more proof of divergence...
Tim Poe . . . . . . . . . . . . . . 4 pts - You were first to a million triangles, but convinced =/= proved.
Mahbubul Hasan. . . . . . . 4 pts - Ok, 12190 tri = 48 rev, but terms > 0 doesn't diverge sum
Zahi Teitelman . . . . . . . . 4 pts - Different interp. of 'cover' was ok; good elem proof of div.
Marcello Cammarata . . . 3 pts - You chose to use an = arcsin(1/-/n) starting n=2 to 18. Ok!
Jack Dostal . . . . . . . . . . . 3 pts - True that '17th tri gives 2 pi' the angles approach but don't reach 0.
Vince LoCascio. . . . . . . . 3 pts - Used cos(an) = -/(n/(n+1)) ok; sum(an) apparently goes on 4ever.
S. Khoo . . . . . . . . . . . . . . 3 pts - True that arcsin(x) > x just like sin(t) < t ; keep entering!
Phil Sayre . . . . . . . . . . . . 3 pts - Good ans; arctan(1/-/n) > 1/-/n not < ; fixed in proof.
Farid Mark Watson . . . . 3 pts - Nice non-attachment text table; good use of divergence p-test
Martin Gritsch . . . . . . . . 3 pts - Nice to hear from you again; we both have faith in Mathematica!
Jeremy Galvagni. . . . . . . 2 pts - Seemed valid but your graphics didn't show in attachmt.
Allen Druze . . . . . . . . . . . 2 pts - The pattern continues forever but do the rotations?
Alan O'Donnell . . . . . . . 2 pts - 4 is indeed smallest except for the zero question. Salvage?
Dennis Molter . . . . . . . . . 2 pts - Yes, the angle never reaches 0 in theory or in practice.
Ed Wern . . . . . . . . . . . . . 2 pts - Nice table graphic attch; is it enough to prove area -> inf?
Ajit Athle . . . . . . . . . . . . 2 pts - Good to see you again; thanks for liking my di(tri)agram.
Anirban Bhattacharyya. 2 pts - I'll go for the 405o but not that sum of tan-1(1/-/n)) converges.
Lisa Schechner . . . . . . . . 2 pts - True that 21 triangles will cover 405o, and arctan x ~ x + O(x^3)
Ken Duisenberg . . . . . . . 2 pts - I agree tan x ~ x but is that enough to prove divergence (1/n vs 1/n^2)?
Marisa Carcallas (new) . .1 pt - Welcome to my contest... the angles decrease but good try!
Mario Roederer . . . . . . . 1 pt - Got it in before these ans. posted; 17, diverges, good.
Rick Montgomery . . . . . 1 pt bonus for good numerical and algebraic proofs of equal spacing.
 
Problem #188 - Posted Friday, November 21, 2003
Chess Tournament! (back to top)
I was on my junior-high chess team! One time, another seventh grader and I entered
an eighth-grade tournament. Every player played every other player once (round-robin);
a win counted as one point and a draw was 1/2 point. My friend and I got a total of 8
points, while all the 8th-graders got the same number of points (as each other).
How many eighth-graders were in the tourney, and why? Prove your answers, explain carefully.
 
Solution: Yes I was on my junior high chess team (by high school the math got too interesting).
Let's let n = the number of eighth graders, each amassing m points. Counting the seventh graders,
there were a total of nm + 8 points earned, equal to the number of games played. There were n+2
players, each playing n+1 opponents; the total number of games being (n+1)(n+2)/2 = nm + 8 ;
we get n^2 + 3n + 2 = 2nm + 16 ; and so n(n + 3 - 2m) = 14. This means n is a divisor of 14;
we rule out n = 1 and 2 because 2 or 3 opponents isn't enough for two players to get 4 pts each.
If n = 7 then 7(7+3-2m) = 14 ; m = 4. If n = 14 then 14(14+3-2m) = 14 ; m = 8.
There are 2 answers: there could have been 7 or 14 eighth-graders.
 
For n=7, Ed points out that the only solutions are: a) all players got 4 pts, and the tourney is a
complete draw, or b) one seventh grader beat the other one, and won the tournament.
 
WINNERS - Problem 188 . (back to top) . leader board . . .
Art Morris . . . . . . . . . . . .10 pts - First all-time, first this week! x(x-1)/2 from round-robin format.
Nick McGrath . . . . . . . . . 8 pts - Your correct answer arrived just 2 min after Art's, so extra point!
Zahi Teitelman . . . . . . . . 6 pts - I liked your way of solving for y = x+3 - 14/x so x = 7 or 14
Tim Poe . . . . . . . . . . . . . . 5 pts - Good (half) point : n-2 div n(n-1)/2 - 8 w/rem 0 or .5
Jack Dostal . . . . . . . . . . . 4 pts - Thanks for including factorial formula for combo-nation.
Vince LoCascio. . . . . . . . 4 pts - Glad to have you back; good clear answer, not so cryptic!
Marcello Cammarata . . 4 pts - Right, n=7 or n=14, and all sol's n>300 not very round-robin-able
Quasi-C. . . . . . . . . . . . . . 4 pts - "AOL ate my answer" ;-} I placed you 12 hours in the past.
Ed Wern . . . . . . . . . . . . . 4 pts - Nice concise answer, bonus pt for further analysis (see above)
S. Khoo . . . . . . . . . . . . . . 3 pts - Right and good expl of n=7, but quadratic factors in another way.
Akifumi . . . . . . . . . . . . . 3 pts - Nice to see you again; Good expl. of n=7; 14 works too.
Hermen Jacobs . . . . . . . . 3 pts - Good try first time; better 2nd ans. Still 1 pt/gm played not 2.
Yakov Macak . . . . . . . . . 3 pts - Well-explained; thanks 4 continuing to enter my contest!
Jeremy Galvagni. . . . . . . 3 pts - Thanks for vote of support; good ans but n(n-1) not n(n+1).
Ken Duisenberg . . . . . . . 3 pts - Got 7 or 14 8th gr; yes (and if 8 pts each 7th gr, 6, 10, 15 8th)
Rick Montgomery (new) . 3 pts - Welcome! Good answer & glad you like the rest of my site too!
Jon Stearn . . . . . . . . . . . 3 pts - Good combo of intelligent algebra and brutal xlspreadsheet.
Mahbubul Hasan . . . . . 2 pts - Early entry, good try; I think you assumed each 7th gr got 8 pts?
Alan O'Donnell . . . . . . . 2 pts - First answer worth 2 pts, good 2nd ans was 3-worthy, resub chg
Ajit Athle . . . . . . . . . . . . 2 pts - Good expl for 7 eighth gr, but can't stop looking after one...
Allen Druze . . . . . . . . . . 2 pts - Reasonable expl but with n = 10 players there's somethg fishy.
Lisa Schechner . . . . . . . 2 pts - Gave plausible 7 or 14 player scenarios; other possibles?
Wanda Cahill . . . . . . . . 2 pts - Your solution looked good overall; couldn't read eqns in word attch.
Joe Alvord . . . . . . . . . . . 1 pt - Semi-trick question, it had 2 solutions; can't assume all draws...
Mario Roederer . . . . . . . 1 pt - If each 7th got 8 pts then n>=9 yes but 10 players not 9?
 
 
Problem #189 - Posted Tuesday, December 2, 2003
The Three Trees (back to top)
Driving out in the "Western Plains" states of the U.S. is like being on a flat, infinite plane.
Three trees are now growing at random points on the plane (or plain). What's the probability
they form an obtuse triangle? Prove your answers, explain carefully.
 
Solution: Man, this problem surprised me! I thought I had a good solution, but you guys connvinced me there
are many equally valid (or invalid) arguments, and the answer can be 0%, 100%, or several values in between!
The problem is different interpretations of three "random" points, and choosing from an infinite interval:
 
(a) Put A and B on the x-axis and draw vertical lines up from each, dividing the upper half plane into regions T1, T2, T3. Draw a semicircle above x-axis with diam. AB. Then assuming (WLOG) that C is above the x-axis, if C in T1, angle A is obtuse; if C in T2, angle B is obtuse, and if C is in the semicircle, angle C is obtuse. The only region for C to be acute is in the strip T2 above the circle; this area has finite width compared to the infinite width regions T1 and T3, therefore the ratio of the acute to the total is zero; Prob(obtuse) = 100%.
 
(b) Let the three trees be A, B, and C, and assume side AB is longest. Draw circles centered at A and B with radius AB. The lune shape must contain C as AB is longest. Now draw a circle with AB as diameter. If C is inside, on, or outside this circle then C is an obtuse, right, or acute angle respectively. The ratio of the areas of the circle to the lune is [pi r^2] / [(8 pi - 6\/3) / 3] ~ 0.63938... which is the prob that /_\ ABC is obtuse.

 
(c) Three points determine a circle, so let's assume three random points on the circumference. Say tree A is
at the 0 degree mark; then B and C will be somewhere between 0 and 360, say distributed uniformly.
If B and C are both less than or both more than 180, the triangle will be obtuse; prob 1/4 for each.
If B and C are more than 90 apart the triangle will be obtuse, prob 1/4. Total: 3/4 = 0.75 in this scenario.
 
(d) The angles add to 180: A+B+C = 180 is a plane in 3-space (A,B,C); use the slanted equilateral triangle
where all A, B, and C are >= 0. In this region /_\ ABC will be obtuse if A > 90 or B > 90 or (180-A-B) > 90.
This makes 3/4 of the region, so the prob ABC is obtuse is again 0.75.
 
(e) AB might be the long side, the middle side, or the short side. Compute the prob of obtuse for each and
average the results: (0.63938 + 0.82102 + 1) / 3 = 0.82013, very close to that middle value!
 
(f) Pick one point, put tree A there, draw two lines through it in a big X; tree B is on one leg, C on another.
The prob that angle BAC is obtuse is 1/2 ; if acute add the prob that angle B or C is obtuse; total ~ 80.7%.
 
(g) Nick wrote a simple program to generate six random numbers from 0 to 1 and pair them up as coords
of three points A, B, C. In 10,000,000 trials, about 72.5% of the triangles were obtuse!
 
(h) (My idea): A random triangle has two sides, x > y, and a third side z where x - y < z < x + y. In the
interval [x-y, x+y], length 2y, the obtuse triangles have z > sqrt(x^2 + y^2) or z < sqrt(x^2 - y^2), so the
prob. of obtuse is [sqrt(x^2 + y^2) - sqrt(x^2 - y^2)] / (2y). Scale this so y = 1 and we have
[sqrt(x^2 + 1) - sqrt(x^2 - 1)] / 2. This ratio can take on values from 0 to sqrt[2]/2 = 0.7071, dep. on x!
.
WINNERS - Problem 189 . (back to top) . leader board . . .
So many different good ideas meant more higher scores this week! (Mostly ranked by time) - Dan
Art Morris . . . . . . . . . . . 10 pts - Initially thought your answer (a) was "wrong" but it now makes sense!
Tim Poe . . . . . . . . . . . . . . 9 pts - Offered two different answers (b, c) and only one minute after Arthur!
Quasi-C. . . . . . . . . . . . . . 8 pts - Nice answer (b) and timely entry gained muchos puntos this week.
Vince LoCascio. . . . . . . . 7 pts - I like the X idea; 1/8 deg intervals might undersample the infinite part?
Marcello Cammarata . . . 6 pts - Nice idea (d) of x+y+z =180 in three-space, good randomness there.
Jeremy Galvagni. . . . . . . 5 pts - Yes, (a) gives 0% ; J threw 3 coins at a wall and got 15/20 obtuse.
Ed Wern . . . . . . . . . . . . . 5 pts - "In the Plains nothing's over 90 deg this time of year!" ;-} Thanx 4 text rescue
Rick Montgomery . . . . . 5 pts - Good proof of (b), proofs of p = 0 and 1 not so bogus! (1 pt for 187)
Jack Dostal . . . . . . . . . . . 5 pts - Beautiful word attachment with colored semicircles and triangles!
Joe Alvord . . . . . . . . . . . 4 pts - Under transf x -> 1/x regions 01 are equal; prob=50%.
Nick McGrath . . . . . . . . 4 pts - Yes, it was Lewis Carroll; I liked your empirical simulation!
S. Khoo . . . . . . . . . . . . . . 4 pts - Well-explained, essentially proof (a); keep on entering!
Ajit Athle . . . . . . . . . . . . 4 pts - I like the argument you gave for (c); and 3/4 is close to exper.
Hermen Jacobs . . . . . . . . 3 pts - Good answer, especially the second one. Thanks for persisting!
Mahbubul Hasan . . . . . . 3 pts - You got through most of arg.(a) except comparing the infinities.
Mohamed Omar . . . . . . . 3 pts - Yours was like a 2-dimensional version of (d). Nice job!
Phil Sayre . . . . . . . . . . . . 3 pts - Refined argument (a) inside expanding shell of rad r -> infinity
Mark Moyer (new) . . . . . . 3 pts - Welcome! Thanks for the link note, nice proofs of (b) and (e).
Alan O'Donnell . . . . . . . 3 pts - Reasonable sum 1/2 + 1/2 - 1/4 = 3/4 total consid angle from AB.
Ken Duisenberg . . . . . . . 3 pts - Managed to do proof (c) looking at 3/8 + 3/8 for obtuse angle.
Yakov Macak . . . . . . . . . 2 pts - Good discussion of infinity; needed some limit argument.
Joanne Dunlap (new) . . . 2 pts - Welcome to my contest! Essentially 50% chance of obtusity.
Abbey the Mathematician (new) 1 pt - Right; the prob of one angle at least 180 is 0; above 90?
Mario Roederer . . . . . . . 1 pt - Suggests mathworld.wolfram.com/ObtuseTriangle.html Thanks!
Allen Druze . . . . . . . . . . 1 pt - Most recently received, thanks for sending it in!
 

 

 
Problem #190 - Posted Tuesday, December 23, 2003
The Flower Farm (back to top)
In the comic strip 'Foxtrot' (Nov. 3, 2003), a math tutor solves a problem about planting
as many flowers as possible in an 8m by 10m garden,where each flower was at least
1m away from the closest other flower. Apparently, his solution (shown at right) had 99
flowers (planting at the edge is ok). Can you do better? If so, give the maximum number
of flowers you can plant and show how it's done; if not, try to prove 99 is the maximum.
Prove answer, explain carefully. 1m = 1 meter.
99 and counting...
 
Solution: Cool problem, eh? The answer is 'no,' 99 trees is not max. Although 36 trees is the most around the perimeter,
that blocks off too much space, leaving a 6 x 8, then 28, etc; but 36+28+20+12+3 = 99 only.
 
The next step is to see that offsetting the flowers by 1/2 in alternating rows saves space, the next row is d = sqrt[3]/2 ~ 0.8660
from the prevoius. So since 10 / d = 20/sqrt[3] = 11.54, we have room for 12 rows: 9+8+...+9+8 = 6*9 + 6*8 = 102 flowers.
We could also cram in 2 rows of 9 instead of the last 2 rows of 8, since making 9+8+...+8+9+9+9+9 = 104 flowers.
(A quick check shows 11 d = 9.526 < 10m for the 102, and 8 d + 3 = 9.928 < 10 for the 104.)
 
Turning things 90 degrees, put 11 flowers on the left edge, then 10 in the 2nd column, etc; 8/d = 9.24, room for 10 columns.
(Again, 9 d = 7.79 < 8 so we're ok.) Then 11+10+...+11+10 = 5*11 + 5*10 = 105 flowers, the most popular answer.
Nick was the first to see there was room to subst an 11 for the last 10 : (8 d + 1 = 7.928 < 8) ; making 106 flowers.
(Dan's note: Jeremy G. offered a triangle tiling proof that the theoretical maximum is 112, but agrees with
Nick, Tim, and me that 106 is the best. I'd have made a 5-frame animation with all these but I'm too busy!)
. . .Sorry to all of you for making you wait again! Is it Problem-of-the-Fortnight or what?
 

 

WINNERS - Problem 190 . (back to top) . leader board . . .
Nick McGrath . . . . . . . . 10 pts - Thanks for explaining the progression of planting improvements!
Tim Poe . . . . . . . . . . . . . . 6 pts - Also excellent expl; 105 flowers 'resubbed' to 106, worth the -1pt.
Jeremy Galvagni. . . . . . . 5 pts - I pretty much understood your n < 112.54 proof, and infinite case
Art Morris . . . . . . . . . . . 4 pts - I wasn't sure how you arranged 2" circles, you were first with 105.
Hermen Jacobs . . . . . . . . 4 pts - That's it; 99->102->105; room for one more row of 11 though.
Marcello Cammarata . . 3 pts - You got 105 ok, I wonder what arrangement of hexagons gives 104.
Joe Alvord . . . . . . . . . . . 3 pts - You were about to send 102 when you turned sideways just in time.
Jack Dostal . . . . . . . . . . . 3 pts - Pythagoras proof and attached picture were nice; thanks.
Phil Sayre . . . . . . . . . . . . 3 pts - Good series of steps made interesting reading, 105 flowers.
Yakov Macak . . . . . . . . . 3 pts - That's it; you showed 105 but didn't prove it's max (good thing!)
Jon Stearn . . . . . . . . . . . 2 pts - Good try, (108, 111) but there isn't room for that many rows/cols
Mario Roederer . . . . . . . 2 pts - I got excited at your 126 and 128 but space = sqrt[3]/2 not sqrt[2]/2.
Allen Druze . . . . . . . . . . 2 pts - "The answer is yes." (I forgot the question!) 102 is pretty good.
Vince LoCascio. . . . . . . . 2 pts - The border logic cuts down the available space, need closer dots.
Alan O'Donnell . . . . . . . 2 pts - Got 105; "max is in the 105-109 range" I agree; 107 might pe possible!
Ken Duisenberg . . . . . . . 2 pts - Another member of the 105-club. Good! (Could prove 10 rows fit...)
Quasi-C. . . . . . . . . . . . . . 2 pts - Good proof 105 flowers, a few extra days closer to New Years.
Ed Wern . . . . . . . . . . . . . 2 pts - Yes, the 102 proves 99 isn't max, then your 105 proves 102 isn't.
Ajit Athle . . . . . . . . . . . . 1 pt - Just under the wire back there in January, but 99 isn't best.
 
 
 THANKS to all of you who have entered, or even just clicked and looked.
My site is in its seventh season - OVER 45,000 HITS so far! (Not factorial.)
Help it grow by telling your friends, teachers, and family about it.
YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 2003 A.D.
 
Problem Archives Index
 
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Problems only . . . . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
 
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