263 -- Cosine My Loan? 264 Cubes&Sqrs 2gether 265 - Four Mini-Pollies! 266- Those Factorialoots 267- Squareful Numbers 268 Nine Tangled Wires 269 AnotherBrick inWall 270-SpheresFillingSpace
- problem #261 - posted wednesday, february 21, 2007
- Here's To Toast ! (back to top) . . .
- My old toaster can hold at most two slices of bread and toast them on one side at a time.
- I want my bread toasted on both sides and buttered on one side. It takes 3 seconds to put
- each slice into the toaster, 30 sec to toast one side of 1 or 2 slices, 3 sec to reverse a slice,
- and 12 sec to butter one (that side must already be toasted). Assume only one of these
- operations can be done at a time. In how short a time can I toast & butter three slices?
- Show steps & reasoning. Winners judged on shortest total time, then order received.
- Solution: Again, thank you for your extreme patience and concern during my summer absence!
- Dan's comment: In my secret source for this problem, as some of you asked, it did take 3 sec to remove
- a piece of toast; but the problem was valid (albeit unclear) as stated, such as whether you could butter
- a piece while another was toasting: yes;butter while turning, no. My bad; I'll do my best scoring these!
- I got a variety of solutions and times; I ranked them according to shortest time and validity. Here are some:
- (Alan O'Donnell's solution, total 108 seconds) . . . Assumptions:
1 - you have 2 hands, so can hold a slice of bread in each hand. ; 2 - you can butter toast while toaster is working.
3 - since not stated, it takes 0 time to remove toast for buttering ; 4 - you want a cup of tea with your toast.
- 0s - pick up 2 slices to insert into toaster (simultaneously)
3s - toast loaded. put on kettle.
33s - toast pops up. reverse one slice, and replace one slice with a fresh slice of bread (simultaneously)
36s - butter 1/2 toasted slice
48s - put teabag in cup.
66s - toast pops up. replace fully toasted slice with semi-toasted/buttered slice, and turn over semi-toasted/unbuttered slice.
69s - butter fully toasted slice
81s - pour hot water onto teabag in cup.
96s - butter remaining unbuttered slice
108s - toast done. Prepare cuppa to taste and have a rest.
- Then there was the question of whether putting two pieces in simultaneously meant two operations at once,
- so maybe the best we can do with the restriction is 111 sec? Philippe sent in an awesome bar graph
- Placing or turning (green), toasting (brown), and buttering (yellow). Yumm! (bonus point)
- But after months of deliberation I applaud Nick McG, who saw no reason not to take partially toasted bread out
- and do other things to save time. "What an interesting and tricky problem. Since the time required to remove a slice from the toaster
- is not specified I have assumed it to be zero. - First, since it takes less time to butter than to toast it seems that we should be able to find
- time for buttering in parallel with toasting. If this is true, we need to find the optimum time for just toasting.
- Second, if inserting and flipping slices took zero time we can see that 3 slices could be toasted in 90 seconds:
First 30 secs. -Toast Slice 1A (slice 1 side A) & 2A: Second 30 seconds - Toast 1B and 3A : Third 30 seconds - Toast 2B and 3B.
- However, since it takes 3 seconds to insert or flip a slice, and each slice has two sides, each slice needs a total of 66 seconds to toast.
The toaster holds 2 slices so, if we maximize the utilization of the toaster (i.e. both parts of the toaster being used at all times - either- toasting or having slices inserted or flipped) we have a minimum theoretical time of 66 * 3 / 2 = 99 sec. But we need to be careful
- because for the first 3 seconds we can utilize only one part of the toaster while we insert Slice 1. So, our minimum theoretical time, MTT,
- is actually given by 3 + 2*(MTT - 3) = 3 * 66 => MTT = 100.5 seconds. ( which we can round up to 101 seconds since all of the
- possible operations take an integer number of seconds).
- So, how might we achieve this. Consider one side of the toaster: If we start by inserting slice 1 (3 seconds): toasting 1A (30 seconds) :- removing slice 1 and inserting slice 2 (3 seconds): toasting 2A (30 seconds) : flipping slice 2 ( 3 seconds): toasting 2B ( 30 seconds)
- we have taken 96 seconds and we still have to butter slice 2 so we would take a minimum of 108 seconds.
- This is significantly greater than our theoretical minimum time but, at first, it seems we can do no better.
- It seems counterintuitive that we can actually do better by partially toasting one side of a slice since a) it seems to be wasting time to- remove a slice from the toaster before it is ready to be buttered and b) it adds 3 seconds to our insertion/flipping time each time we do it.
- - For example, if we toast one side of one of the slices in two stages we are increasing the total time to toast 3 slices from 66 * 3 = 198 sec
- to 66*3 + 3 = 201 seconds. Then our MTT expression becomes 3 + 2*(MTT - 3) = 3 * 66 + 3 => MTT = 102 seconds.
- This is still better than what we have achieved in practice so far (108 seconds) and we find with some experimentation that a total of 102- seconds is indeed possible as shown in the tables below which show the timeline from 1 to 102 seconds both for each slice of bread and
- for each half of the toaster. We can see the toaster is fully utilized ( a requirement to minimize the time) and also that our assumption
- was correct i.e. that if we minimize the toasting time then we can do the buttering in parallel. Shortest time is 102 seconds"
- ( No law against toasting a slice already buttered on the other side, except for the electrical damage... )
- ( Nick had a table with 102 rows, but I compressed it for your viewing convenience! - Dan )
- Mark R. sent in solutions from 105, 120, 138, and 144 sec. depending on severity of rule enforcement.
- And Yakov from Australia waxes nostalgic about these old toasters:
"You have brought on a wave of nostalgia for the old days: smoke rising from the toaster, charcoal being- scraped from charred bread, being told that charcoal is 'good for you', mum throwing smoking toast out
- the window so it doesn't smoke up the kitchen any more . . . . How many of your contestants would have
- had these experiences, and what have they missed??"
- A classic problem to say the least!
- WINNERS - Problem 261 . . listen to dansmathcast - my math podcast! - (back to top) . leader board
- Nick McGrath . . . . . . . 10 pts - Fabulous sol'n 102s and table, same colors as Philippe! Thanks for e-mail.
- Mark Rickert . . . . . . . . 9 pts - That 105s was pretty thorough, and first received, had to go with Nick's 102.
- Alan O'Donnell . . . . . . 7 pts - That was a nice easy-to-understand table for 108s, so I made it quotable!
- Nats Kroy . . . . . . . . . . . 6 pts - Good idea 105s, but the toaster can't do both sides w/o human intervention
- Denis Borris. . . . . . . . . 6 pts - Down to 99s if we can use a buttering brush on bread currently in toaster!
- Etienne Desclin . . . . . . 5 pts - Nice early entry, and 111s was the next best answer, albeit with two hands.
- David Stigant. . . . . . . . 4 pts - Nice list, total 114s; could make it 111 if place 2 slices at first second.
- Philippe Fondanaiche. 4 pts - Great expl. for 111s on attached doc; great bar graph for us visual learners!
- Marcello Cammarata. . 4 pts - Good to realize solutions 120s or 111s depended on the rules applied.
- Yakov Macak . . . . . . . . 3 pts - 120s = 2 min. plus great images in your toast history! Got toast racks?
- Claudio Baiocchi . . . . . 3 pts - 120s or 114s ; True that "no time to extract a slice" is strange. My mistake!
- Tim Poe . . . . . . . . . . . . . 3 pts - 117s - Yes, old type toaster (with rust and grime) flip 2nd while 1st toasts.
- K Sengupta . . . . . . . . . . 3 pts - Easy-to-read list of events, but 138s is slower bec. 3rd slice toasts alone.
- Hermen Jacobs . . . . . . . 2 pts - 192s ; Extra time for those "private rooms" with one slice at a time there.
- Allen Druze. . . . . . . . . . 2 pts - 144s ; some time needed to butter toast while next slice toasts; right.
- Oxana Urdaneta (new). . 2 pts - Welcome (finally) Oxana! 3 min = 180s ; could be faster with overlaps.
- time, in minutes? Show your steps and reasoning.
- Solution: This is Mark's winning entry, received six hours before any others!
Hi, Dan . . . > > C's pace was 7.8 minutes per mile. (given)
> Based on the clues (numbers of runners passed, etc.):
A finished 2, 3, or 4 . . . B finished 4 or 5 . . . D finished 3, 4, or 5 . . . E finished 2 or 3
> Therefore, C finished 1. . . Therefore, the possible orders of finish are:
C, A, E, B, D . . . C, A, E, D, B . . . C, E, A, B, D . . . C, E, A, D, B . . . C, E, D, A, B.
Since A reached the 3-mile mark 6 minutes after C, A's pace is 2 minutes slower per mile than C's;- so A's pace is 9.8 minutes per mile. . Since D passed C 42 minutes into the race, D's pace is
- 42 / (10 - 42 / 7.8) = 9.1 min. per mile. This means D is faster than A; so the order is: C, E, D, A, B
Since B got to the turnaround 7 minutes after D, B's pace is 1.4 minutes slower per mile than D's;- so B's pace is 10.5 minutes per mile. . Since E ran 50/9 miles when B ran 40/9 miles, E's pace is
- 5/4 times faster than B's pace; so E's pace is 8.4 minutes per mile. . Therefore, we have:
1 - Carla, 78 min . . 2 - Emo, 84 min . . 3 - Dan, 91 min . . 4 - Annie, 98 min . . 5 - Ben, 105 min
- WINNERS - Problem 262 . . subscribe to dansmathcast - my math podcast - (back to top) . leader board
- Mark Rickert . . . . . . . . 10 pts - I like your 'elimination of possible orders' approach. Nice job, and quotable.
- Etienne Desclin . . . . . . . 7 pts - The data do indeed 'fit together in only one way', like all good puzzles!
- Tim Poe . . . . . . . . . . . . .. 6 pts - Thanks for always including the problem stmt; yes Carla 7:48mpm, Dan 9:08.
- Nick McGrath . . . . . . . . 5 pts - I like your descriptive notation Ta, Sa, etc. They help 'run through' the logic!
- Nats Kroy . . . . . . . . . . . 5 pts - Right: Emo fin 2, 3, or 4; Dan 3,4,5; Annie 2,3,4; Ben 4,5; means Carla won!
- Denis Borris . . . . . . . . . 4 pts - Nice charts there; Carla 5/13 mi after turn, Dan 4 8/13 mi in C 5 5/13 time...
- Zahi Teitelman . . . . . . . 4 pts - Conds => Carla 1, Emo 2, Ben 5. Dan 42 min for 60/13 mi; Ve = (4/5)Vb, etc.
- Doug Babcock. . . . . . . . 4 pts - Carla won, pass Dan at 2520 sec (supercomposite!-D), Emo 2800 sec for 5 5/9 mi.
- Yakov Macak . . . . . . . . 3 pts - Good answer as well; these are over 48 hrs after Mark's at this point ...
- Philippe Fondanaiche . 3 pts - Nice notation S(n) stmt abt n; r(n) = rank; x/Vd = (10-x)/Vc for x=Dan's Dist.
- Wenjun Kang . . . . . . . . 3 pts - Good logic, fine 10step solution, well-explained! Dan 42 Sd = 10 - 42 Sc.
- Ken Duisenberg . . . . . . 3 pts - Right, only Carla is 'not behind' anyone so she wins; but why assume Dan is 3rd?
- K Sengupta . . . . . . . . . . 3 pts - You misinterp. Carla: ran 7 min 48 sec for each mile, not all ten. Early entry!
- Allen Druze. . . . . . . . . . 2 pts - Good solution except for Emo 89.6 min; shoulda been 84 min. Others ok.
- Aclaira. . . . . . . . . . . . . . 2 pts - Right order, and good logical flow in your argument; earlier for more pts!
- Frank Mullin . . . . . . . . 2 pts - Clues => Anne 2-4, Ben 4-5, Dan 3-5, Emo 2-3, so Carla 1 (one and won)
- Claudio Baiocchi . . . . . 1 pt - Sorry about the language barrier; Google translator wasn't enough help either...
- problem #263 - posted sunday, april 1, 2007
- Show your steps and reasoning.
- Solution: a) by Giridhar Prasannan
Consider x^7 + 1 = 0. [Letting p = pi] the roots of the equation are e^ip/7, e^i3p/7, ..., e^i13p/7
The sum of these roots should be zero since there is no x^6 term. Hence the real parts sum should also to zero.
cos p/7 + cos 3p/7 + cos 5p/7 + cos 7p/7 + cos 9p/7 + cos 11p/7 + cos 13p/7 = 0
Since cos(2p - x) = cos x, we have: cos 9p/7 = cos 5p/7 , cos 11p/7 = cos 3p/7 , and cos 13p/7 = cos p/7
Also cos7p/7 = - 1 , So we get 2(cos p/7 + cos 3p/7 + cos 5p/7 ) = 1 , i.e. cos p/7 + cos 3p/7 + cos 5p/7 = 1/2
Finally since cos(-x) = - cos x , cos 5p/7 = - cos 2p/7. Hence, cos p/7 - cos 2p/7 + cos 3p/7 = 1/2.- b) by Hermen Jacobs: "Hi Dan, I send you my solution of problem # 263 B. cos^2(x) = (1 + cos 2x)/2
Thus cos^2(x) + cos^2(2x) + cos^2(3x) = 1 + cos 2x + cos 4x + cos^2(3x) The equation is- (cos 2x + cos 4x)/2 + cos^2 (3x) = 0 ; cos 3x . cos x + cos^2(3x) = 0 ; cos 3x.(cos x + cos 3x ) = 0 ;
- 2.cos 3x. cos 2x. cos x = 0 . . . If cos p =0 then p = pi/2 + k.pi = pi*(2k +1)/2 [same for cos 2x and cos 3x]
All the solutions are pi*(2k+1)/2 and pi*(2k+1)/4 and pi*(2k+1)/6"- [Dan's note: odd mults of pi/4 and of pi/6; the odd mults of pi/2 are included; you don't need the first ones.]
- Another way using polynomials by Prachai K. "Well, this one is just plain bashing... Let cos x = a. Then
- cos 2x = 2a^2 - 1, cos 3x = 4a^3 - 3a. Plug those into the equation, you get: a^2 + (2a^2 - 1)^2 + (4a^3 - 3a)^2 = 1,
- which reduces to 16a^6 - 20a^4 + 6a^2 = 0, or a^2(2a^2 - 1)(4a^2 - 3) = 0. [Roots a = 0, |a| = sqrt[2]/2, sqrt[3]/2.]
- The solutions are a = pi/2, 3pi/2, pi/4, 3pi/4, 5pi/4, 7pi/4, pi/6, 5pi/6, 7pi/6, 11pi/6 and + or - 2kpi, for any integer k.
- WINNERS - Problem 263 . . subscribe to dansmathcast - my math podcast - (back to top) . leader board
- Nick McGrath . . . . . . . 10 pts - Nice going, even some English spellings this time. Gunning for season win.
- Mark Rickert . . . . . . . . 7 pts - Found evidence on www.mathcomp.com thanks for the source, Mark!
- Prachai K. . . . . . . . . . . . 6 pts - Right, roots of z^7 + 1 = 0 are odd powers of e^ip/7, for b see above.
- K Sengupta . . . . . . . . . . 5 pts - Thanks for finding deMoivre on Wikipedia, emerging as defacto source.
- Etienne Desclin. . . . . . . 5 pts - I like your polynomial factoring approach, no charge for clarif, thanx 4 puzzle!
- Denis Borris . . . . . . . . . 4 pts - Good remark x^7 + 1 = 0 has no x^6 term so sum of roots is 0; steps on b).
- Ravi Raja . . . . . . . . . . . 4 pts - Nice to hear from you again, keep on entering. Good to get cx c2x c3x = 0.
- Akifumi. . . . . . . . . . . . . 4 pts - Hard-working PhD student takes time out to do dansmathcontest; thanks!
- Giridhar Prasannan . . 4 pts - Very quotable solution; nice work on part b too; see above for part a ans!
- Jin Won Park . . . . . . . . 3 pts - Good c = cos x; c^2 + (2c^2-1)^2+ (4c^3 - 3c)^2 = 1 and similar for a.
- Marcello Cammarata . . 3 pts - Thanks for the excel spr, monotone func on subintervals. A 14-gon too!
- Philippe Fondanaiche . 3 pts - Using good ideas with Trigo; one part had pi/3 when we wanted pi/6.
- Zahi Teitelman . . . . . . . 3 pts - Right; one is cos 7pi/7 = -1, others add to 1, and pair up, so ans = 1/2.
- Adam Morgan. . . . . . . . 3 pts - Yes, pi/7 is root to cos x - cos 2x + cos 3x = 1/2, 8x^3 - 10x + 3 = 0.
- Arijit Bhattacharyya . . 3 pts - Welcome to my contest; interesting that a) can be 1/2 or -3 from your eqn.
- Phil Sayre . . . . . . . . . . . 3 pts - Invoked the name Chebyshev, and I appreciate it! T1^2 + T2^2 + T3^2 = 1.
- Radu Ionescu . . . . . . . . 3 pts - Welcome back; it's been a while, good answer esp working with cos pi/14.
- Claudio Baiocchi . . . . . 3 pts - Nice to see people using Derive (but not when you're Derunk;-). Good prf of b).
- Garry Malashkin . . . . . 3 pts - Good proof in there; I found some of the steps hard to fill in but result ok.
- Hermen Jacobs . . . . . . . 3 pts - Just part b) but an excellent method (see above) - Thanks for all these years!
- Al Nelson . . . . . . . . . . . 3 pts - I like the white board attach; are you still using it in class? Good content too.
- Nats Kroy . . . . . . . . . . . 2 pts - Part b only, but nice work in the polynomial realm with factoring & expands.
- Haley Buxton . . . . . . . . 2 pts - Good identity in part a) to get 8c^3 - 4c^2 - 4c + 1 = 0; and that works!
- Yakov Macak . . . . . . . . 2 pts - You seem to have enough "math tools" to do this job, good on ya Y-Mac!
- problem #264 - posted sunday, april 22, 2007
- Cubes & Squares Together (back to top)
- Solution: Several of you found this proof or a similar one online; like this:
- We have (m + 1)^3 - m^3 = 3m^2 + 3m + 1 = n^2, for some integers m and n.
Hence 12m^2 + 12m + 4 = 4n^2, from which 3(2m + 1)^2 = (2n - 1)(2n + 1).
Now, 2n - 1 and 2n + 1 are relatively prime.(By Euclid's algorithm, their greatest common divisor divides- their difference, namely 2. Since both are odd, their greatest common divisor must be 1.)
Therefore we must consider two possible cases, with a and b relatively prime:
(Case 1) 2n - 1 = 3a^2, 2n + 1 = b^2 . . . . (Case 2) 2n - 1 = a^2, 2n + 1 = 3b^2
Taking the first case, we have b^2 = 3a^2 + 2. Impossible, as any square is congruent to 0 or 1, modulo 3.
In the second case, notice a must be odd. Setting a = 2k + 1, we have 2n - 1 = (2k+1)^2 = 4k^2 + 4k + 1.
Hence 2n = 4k^2 + 4k + 2 = 2[k^2 + (k + 1)2]. So n = k^2 + (k + 1)^2.- I liked this proof from Garry, using Pell's Equation and continued fractions always gets my attention!
- a) (m+1)^3 - m^3 = n^2 (1) 3m^2 + 3m + 1 = n^2 12m^2 + 12m + 3 = 4n^2 - 1
3*(2m+1)^2 = (2n - 1)*(2n + 1) (2) integers 2n-1 and 2n+1 coprime so one must be perfect square
from (1) we have that n is odd so 2n+1 = 3(mod 4) and 2n+1 can't be a perfect square modulo 4
so 2n-1 is a perfect square and 2n-1 is odd we have 2n-1 = (2k+1)^2 = 4*k^2 + 4k + 1 and
n = 2*k^2 + 2k + 1 = k^2 + (k+1)^2 Q.E.D.
b) Let's suppose 2n=a and 2m+1=b From (2) => a^2 - 3*b^2 = 1 (3)
Note that if a is even and b is odd then from pair (a,b) satisfying (3) there is a pair (n,m) satisfying (1)
The equation (3) has infinite set of solution (a,b) where a,b are integer and a>0,b>0
Proof: Suppose that a pair (a1, b1) is solution (3). Define a pair (ak, bk) by equation :- . .(a1 + b1*sqrt(3))^k = ak + bk*sqrt(3) (4) obviously that (4) => (a1-b1*sqrt(3))^k=ak-bk*sqrt(3)
so (ak)^2 - 3*(bk)^2 = (ak - bk*sqrt(3))*(ak + bk*sqrt(3)) = ( (a1-b1*sqrt(3))*(a1+b1*sqrt(3)) )^k- . . . = ( (a1)^2 - 3*(b1)^2 )^k = 1^k = 1 and a pair (ak, bk) also a solution of (3). (2,1) is a solution (3) so
- . . . we can to build a infinite solution set (ak,bk) from (4) Q.E.D.
Easy to calculate first five pairs for (3) (2,1) and (2+sqrt(3))^2=7+4*sqrt(3) so (7,4)
and (2+sqrt(3))^3=26+15*sqrt(3) so (26,15) and (2+sqrt(3))^4=97+56*sqrt(3) so (97,56)
and (2+sqrt(3))^5=362+209*sqrt(3) so (362,209) from first pair (2,1) 2n=2 and 2m+1=1 ==>- . . n=1, m=0 is a trivial example from third pair (26,15) 2n=26 and 2m+1=15 => n=13, m=7
from fifth pair (362,209) 2n=362 and 2m+1=209 => n=181, m=104.- The actual numbers, the next few solutions, and a pattern are in this table from Etienne of Belgium:
7^3 + 13^2 = 343 + 169 = 512 = 8^3 (and 13= 2^2 + 3^2 )
104^3 + 181^2 = 1124864 + 32761 = 1157625 = 105^3 (and 181=9^2 + 10^2 )
1455^3 + 2521^2 = 1456^3 (and 2521= 35^2 + 36^2 )
20272^3 + 35113^2 = 20273^3 (and 35113= 132^2 + 133^2 )
282359^3 + 489061^2 = 282360^3 (and 489061= 494^2 + 495^2 )
3932760^3 + 6811741^2 = 3932761^3 (and 6811741= 1845^2 + 1846^2 )
n | cube square
0 | 0 1
1 | 7 13 c1 = 8*s0 - 1
2 | 104 181 c2 = 8*s1
3 | 1455 2521 c3 = 8*(s2 + s0) - 1
4 | 20272 35113 c4 = 8*(s3 + s1)
5 | 282359 489061 c5 = 8*(s4 + s2 + s0) - 1
6 | 3932760 6811741 c6 = 8*(s5 + s3 + s1)
- WINNERS - Problem 264 . . subscribe to dansmathcast - my math podcast - (back to top) . leader board
- Nick McGrath . . . . . . . 10 pts - First and formost again, Nick; good proof of cube-different squares!
- Mark Rickert . . . . . . . . 6 pts - Similar proof, valid algebra; second example ok first is 8^3 - 7^3 not 9^3 - 8^3
- K Sengupta . . . . . . . . . . 5 pts - I like the references to MathWorld and Pell's Equation (Mark too)
- Garry Malashkin . . . . . 5 pts - Good use of recursions: (2 + \/ 3)^3 = 26 + 15 \/ 3 ; 26/2 = 13; 13^2 = 169
- Denis Borris . . . . . . . . . 4 pts - I couldn't reconcile your integer sequences links but the followup helped
- Claudio Baiocchi . . . . . 4 pts - The two 'Lemmata' were more than enough to prove sum-of-squares property.
- Philippe Fondanaiche . 4 pts - Nice proof using rel prime factors and Pell's Equation. Keep on entering!
- Etienne Desclin. . . . . . . 3 pts - Good table of examples, and intuition (not proof) about patterns that emerge
- Ravi Raja . . . . . . . . . . . 3 pts - I like mult by 4 then factoring out a 3. Very clever use of quadratic residues
- Nats Kroy . . . . . . . . . . . 2 pts - Part b answer was great; good to list \/(3x^2+3x+1) to see if it's an integer
- Jean Moreau de Saint-Martin (new) 2 pts Bienvenue a mon concours; good proof! See examples of first sol's above.
- Zahi Teitelman. . . . . . . 2 pts - Right that 105^3 - 104^3 = 181^2 and 181 = 10^2 + 9^2, same with 8, 7, 13.
- Radu Ionescu . . . . . . . . 1 pt - You found the one solution; right that 3x(x+1) = (y+1)(y-1) this yields more
- Hermen Jacobs . . . . . . . 1 pt - Used trusty Basic to find solution with 13^2 = 8^3 - 7^3; let it run longer.
- Phil Sayre . . . . . . . . . . . 1 pt - "7 and 2. 104 and 9." Somewhat right; I sense there's a man in a hurry here ;-}
- problem #265 - posted wednesday, may 23, 2007
- Solution: Thanks to K Sengupta [my comments in brown]
- (a) By conditions of the problem: x = V7 + V 11 . . . Or, x^2 18 = 2V 77 [sq both sides; ok x not z.]
Or, x^4 36*x^2 + 324 = 308 . . Or, x^4 36*x^2 + 16 = 0. Consequently, f(x) = x^4 36*x^2 + 16- (b) By the problem: (x- 2V3)^2 = (6i)^2 . . . Or, x^2 4*V3*x + 12 = -36
Or, 4*V3*x = x^2 + 48 . . . Or, (4*V3*x)^2 = (x^2 + 48)^2 . . . Or, 48*x^2 = x^4+ 96*x^2 + 2304
Or, x^4 + 48*x^2 + 2304 = 0 . . . . Consequently, f(x) = x^4 + 48*x^2 + 2304- (c) By the problem: x V5 = cuberoot(2) . . . Or, (x - V5)^3 = 2 . . . Or, (x^3 + 15x) V5(3x^2 + 5) = 2
Or, x^3 + 15x 2 = V5(3*x^2 + 5) . . . Or, (x^3 + 15x 2)^2 = 5*(3*x^2 +5)^2
Or, x^6 + 30*x^4 - 4*x^3 [+ 225x^2] 60x + 4 = 45*x^4 + 150*x^2 + 125 [oops]- x^6 15x^4 4x^3 + 75x^2 60x 121 = 0 ; f(x) = x^6 15*x^4 4*x^3 + 75*x^2 60x 121
- (d) Put x = tan h, where h = 18 degrees ; Accordingly, 5h = 90 degrees, giving: 2h = (90 3h)
Or, tan 2h = tan (90-3h) = cot 3h . . . Or, tan 2h = 1/(tan 3h) [these are like Chebyshev quotients]
Or, 2*tan h/(1 - tan^2 h) = (1 - 3*tan^2 h)/(3*tanh - tan^3 h) (resubstituting the value x = tan h)
Or, 2x/(1-x^2) = (1- 3*x^2)/(3x- x^3) . . . Or, 6*x^2 2*x^4 = 1 - 4*x^2 + 3*x^4
Or, 5* x^4 - 10*x^2 + 1 = 0 . . . Consequently, f(x) = 5* x^4 - 10*x^2 + 1
- WINNERS - Problem 265 . . subscribe to my show - dansmathcast - (back to top) . leader board
- K Sengupta . . . . . . . . . . 9 pts - Very good answer except the one coefficient; thanks for entries!
- Claudio Baiocchi . . . . . 7 pts - I like yr correction to precision of stmt: real rel prime integer coeffs.
- Nick McGrath . . . . . . . 6 pts - Good that tan(36+54) = infinity, so denom = 0; tan36*tan54 = 1.
- Ed Wern . . . . . . . . . . . . 5 pts - Welcome back Ed; MathWorld says tan(pi/10) = \/(25 - 10\/5) / 5.
- Mark Rickert . . . . . . . . 5 pts - Good idea, expand powers of \/7 + \/11 etc and elim radical terms
- Hermen Jacobs. . . . . . . 4 pts - Early entry Hermen! Nice work on a, b, c; small coeff error on b.
- Al Nelson . . . . . . . . . . . 4 pts - Nice hand-writ pdf; d) is just 4th deg, u had extra factor x^4 - 10x + 5.
- Phil Sayre . . . . . . . . . . . 4 pts - First 3 parts were good and basic, a surd came along and helped on d.
- Radu Ionescu . . . . . . . . 3 pts - First entry; right polys no steps on a,b,d; resub nicely solved part c.
- Denis Borris . . . . . . . . . 3 pts - I was ok on a, b, c but tan(2*9) = 2tan9/(1-tan29) = \/2 (\/5 - 1) etc ?
- Zahi Teitelman. . . . . . . 3 pts - Yes the factor theorem works with a, b, c. And d) extra 5x^4 + 10x^2 + 1.
- Philippe Fondanaiche . 3 pts - Very nice work, mais you missed an x on yr 12i term x4 - 96x2 + 2448
- Nats Kroy . . . . . . . . . . . 3 pts - Wow, a vacation with no internet is like three weeks off, in the 1960s!
- James Laverty (new) . . . 2 pts - Welcome to my contest; nice answers on a,b; good intuition on degrees c, d.
- Bengt Blomster (new). . . 2 pts - Hello Bengt; I liked that cos72 = 1/(2phi) = (\/5 - 1)/4. Keep on entering!
- Wenjun Kang . . . . . . . . 2 pts - Nice answer; right amount of detail, well done (received before 266 up)
- Ajit Athle . . . . . . . . . . . 2 pts - Your a, b, c are ok; in d you had tan36 so got its poly z^4 - 10z^2 + 5.
(oo is infinity . . . Again,
try for a numerical value and an exact proof)
- Solution: Welcome new contestant and dansmath.com website fan Joe Fendel, who writes:
- A) I'd say T is larger. If we compare S^(100) and T^(100), we get 99!*S and 100!. Divide by 99! and we
- are comparing S and 100. Clearly S < 100, since 100^99 is term-wise more than 99!. Thus T > S.
- Excel bears me out: S is about 37.6231 and T is about 37.9927.
B) Here I'd say S/99 is bigger than T/100. Let's start with 99! / 99^99, just because that's such a fun number.- If we multiply this by (99/100)^99, we get a smaller number, which is equal to 99! / 100^99, and this is the
- same as 100! / 100^100. The 99th root of the larger number is S/99, and the 100th root of the smaller
number is T/100. And so S/99 must be smaller than T/100 (since they're each larger than 1).- Again, Excel confirms: S/99 is about 0.380031 and T/100 is about 0.379927.
C) This sure looks like it's headed toward 1/e, which is about 0.3679. Makes sense in the context of limits- involving exponents, too. How can we prove it? Well, 1/e is the limit as n -> oo of (n / n+1)^n.
- That looks a lot like that "if we multiply this by (99/100)^99" step above! So for example, let x be the limit
- of [(n!)^(1/n)]/n as n -> oo. Then since for all n, we have x = x^(n+1)/x^n, it's true in the limit, and thus
- x is the limit of the ratio of (n+1)!/(n+1)^(n+1) and n!/n^n, which is (n / n+1)^n, and that leads to 1/e!
- Dan's note: Joe was one of the few people not to use Stirling's formula, which as you may know, says:
- n! ~ [\/(2 pi n)] * [(n/e)^n] ; but did you know that there's a whole series making this more accurate:
- n! ~ [\/(2 pi n)] * [(n/e)^n] * [1 + 1/(12n) + 1/(288n^2) + . . . ] : Cool!
- WINNERS - Problem 266 . . subscribe to my show - dansmathcast - (back to top) . leader board
- Philippe Fondanaiche 10 pts - Nice attachmt; is this your first weekly win? Fe'licitations Philippe!
- Nick McGrath . . . . . . . 7 pts - Shows logT - logS > 0; good proof of monotonicity (long word!)
- K Sengupta . . . . . . . . . . 5 pts - Very interesting & original steps; a bit tricky, good to Wikipedia Stirling!
- Moreau deSaint-Martin 4 pts - I liked the second proof you sent in, and your use of Stirling's f!
- Joe Fendel (new) . . . . . . . 4 pts - Thanks for finding my site (and liking it!) Quotable answer too. (above)
- Mark Rickert . . . . . . . . 4 pts - Not all proofs included but first entry received and numericals were ok!
- Radu Ionescu . . . . . . . . 3 pts - Early entry gains point; major steps left out of b, c but right idea Radu.
- Art Morris . . . . . . . . . . 3 pts - Welcome back Art! Good numericals and nice prf on a) and Stirling too!
- Marcello Cammarata. . 3 pts - Nice answer but I think logT = log1+log2+...+log100)/100 not times...
- Etienne Desclin. . . . . . . 3 pts - Very pretty PDF layout 8-} Yes the sequence in c) does indeed decr.
- Zahi Teitelman. . . . . . . 3 pts - Stirling yes but in the limit, not alw. exact. I liked Excel LN, FACT, EXP.
- Claudio Baiocchi . . . . . 3 pts - Thanks for the support and the great proof of part c). Radius of conv!
- Nats Kroy . . . . . . . . . . . 3 pts - In c) limit is 0.37 to 2 places, cool use gamma function to smooth out n!
- Steve Tonnesen (new) . . 3 pts - Welcome to my contest; nice answers on a,b; good intuition on degrees c, d.
- Hermen Jacobs. . . . . . . 2 pts - Good to program it but I think you used a z=n! in place of n in expon.
- Denis Borris . . . . . . . . . 2 pts - Thanks for the resub to suggest Stirling; not sure abt ratio argt in b)...
- James Laverty . . . . . . . 2 pts - Interesting approach in a); thanks for trying for prf of b); Stirling rules c)
- Phil Sayre . . . . . . . . . . . 2 pts - A bit after the Labor Day rush, good Stirling; pretty close call on b).
- problem #267 - posted thursday, september 27, 2007
- Next-to-last problem of this season! 11th season starts with Problem 269.
- Solution: a) Etienne (de Belgique) nous donne les 28 solutions des chiffres "pleins de pouvoir":
| 01) 1 02) 4 = 2^2 03) 8 = 2^3 04) 9 = 3^2 05) 16 = 2^4 06) 25 = 5^2 07) 27 = 3^3 |
08) 32 = 2^5 09) 36 = 2^2 * 3^2 10) 49 = 7^2 11) 64 = 2^6 12) 72 = 2^3 * 3^2 13) 81 = 3^4 14) 100 = 2^2 * 5^2 |
15) 108 = 2^2 * 3^3 16) 121 = 11^2 17) 125 = 5^3 18) 128 = 2^7 19) 144 = 2^4 * 3^2 20) 169 = 13^2 21) 196 = 2^2 * 7^2 |
22) 200 = 2^3 * 5^2 23) 216 = 2^3 * 3^3 24) 225 = 3^2 * 5^2 25) 243 = 3^4 26) 256 = 2^8 27) 288 = 2^5 * 3^2 28) 289 = 17^2 |
- WINNERS - Problem 267 . . try dansmathcast - (back to top) . leader board
- Joe Fendel . . . . . . . . . . . 10 pts - Wow, a weekly winner in the second entry. Fast track to partner ;-}
- Tim Poe . . . . . . . . . . . . . 7 pts - Nice how you elim a certain amount by each condition imposed.
- Nats Kroy. . . . . . . . . . . . 6 pts - Good mult table pitting squares against cubes - clever Nats!
- Nick McGrath . . . . . . . . 5 pts - True, these are widely-er known (Sloane) as powerful numbers.
- Philippe Fondanaiche . 5 pts - Nice recursion for Pell's eqn x2 - 8y2 = 1: x' = 3x + 8y; y' = x + 3y
- K Sengupta . . . . . . . . . . 4 pts - Small typo double expon for triple, good proof (1 is ok) and Pell's.
- Garry Malashkin . . . . . 4 pts - That's a pretty complete proof; good that ak + bk \/2 = (1 + \/2)^k.
- Claudio Baiocchi . . . . . 3 pts - Right; you prove a stronger result, the infin. consec. sqfl and sqs!
- David Stigant . . . . . . . . 3 pts - Welcome back; thanks for showing the nice 'Haskell' program!
- Moreau deSaint-Martin 3 pts - Good recursion, nice 12167=23^3, 12168=2^3 39^2 both non-sq.
- Ed Wern . . . . . . . . . . . . 3 pts - List > 1 with 27 elts and 27 as an elt! I like any symmetry I can get.
- Marcello Cammarata . . 3 pts - Powers of primes 2..17 ; new recursion (a,b)->(4a(a+1),(2a+1)^2)
- Mark Rickert . . . . . . . . 3 pts - Movin up that all-time list; yet another recursion an+1 = 6 an - an-1.
- Denis Borris . . . . . . . . . 2 pts - That was the right link, rather short answer; next prob more interesting!
- James Laverty . . . . . . . 2 pts - Numbers like 12 aren't squareful/powerful bec 3 goes in but 9 doesn't.
- Etienne Desclin. . . . . . . 2 pts - Nice proof especially in part b) read others' ans here for info part c.
- Steve Tonnesen . . . . . . . 2 pts - Keep on entering, good thinking and correct on those first 4 pairs.
- Zahi Teitelman. . . . . . . 1 pt - Same as James; 63 has the same expon structure as 12 and isn't sqfl.
- problem #268 - posted wednesday, october 10, 2007
- Last problem of this season! Eleventh annual contest starts with Problem 269.
- Nine Tangled Wires /\|\|/\|\ (back to top)
- I was working on the circuits in my house, and found a bundle of nine identical wires,
- going into the wall and up to the attic. In the attic were the nine wires going down to
- the main floor. Suppose I can twist any number of wires together at either end, and I
- have a circuit tester that lights when a loop is complete. How can I match up all nine
- pairs of wire ends, using the smallest number of trips up to (and down from) the attic?
- Assume the wires are insulated except for a few bare cm at the ends. Wires cannot be pulled thru
- the wall or pipe. Show your steps and reasoning.
- Solution: Here's a detailed solution by longtime solver Tim Poe; his first weekly win since Problem 233:
- "First, I note that the problem asks for the minimum number of round trips to the attic, not the number of tests.
- I assume that your circuit tester is portable (more or less a standard multimeter) and you can take it with you
- and test from either end. I also assume you begin downstairs.
- Downstairs: connect a pair of wires, a trio of wires, and a quartet of wires.
- Attic (.5 round trips): Using the loop tester, find the ends of each group. (A wire that completes a loop with
- only one other is one of the pair; one that makes a loop with two others is one of the trio, and one that makes
- a loop with three others is part of the quartet.) Once this is done, you*ll have the same bundles as you have
- downstairs, but within each bundle you won*t know which wire is which. Label the pair (in any order) A and B;
- the trio (again in any order) C, D and E; and the quartet (still in any order) F, G, H and I. Connect them in the
following groups: {A,C,F}, {B,G}, {D,H}, {E} and {I}
- WINNERS - Problem 268 . . try dansmathcast - (back to top) . leader board
- Tim Poe. . . . . . . . . . . . . 10 pts - I like the permutation cycle notation. First weekly win since 2005!
- Claudio Baiocchi . . . . . 7 pts - Adapted a 10-wire solution with and w/o a "ground wire" (cheating)
- Nick McGrath . . . . . . . . 5 pts - Another yearly win to go along with 2004 and 2005; jolly good!
- Marcello Cammarata . . 4 pts - I wonder if 'Marcello Cammarata' is Italian for Martin Gardner?
- Ken Duisenberg . . . . . . 4 pts - Nicely worded solution; great to have a fellow puzzler along!
- Joe Fendel . . . . . . . . . . . 4 pts - I'm proud to show my face in your google search results! ;-}
- David Stigant . . . . . . . . 4 pts - Nice thorough explan; bonus pt for proof it can't be done in one.
- Denis Borris . . . . . . . . . 3 pts - I managed to patch yr correct 15-wire sol'n to your resub instruc!
- Etienne Desclin. . . . . . . 3 pts - Cool notation 1111,222,33 as twisty groups, then go upstairs...
- Philippe Fondanaiche . 3 pts - I see you are part of the "four pairs plus one" camp. Ca marchera!
- Mark Moyer (new) . . . . . 3 pts - Great to have you along for the dansmath ride... keep entering!
- Steve Tonnesen. . . . . . . 3 pts - Reduced the number of trips in resub, solved while on a driving trip!
- Ed Wern . . . . . . . . . . . . 3 pts - I like yr fake sol'ns: call electrician, buy a really long wire, yell w/partner.
- Mark Rickert . . . . . . . . 2 pts - Early entry with update; not sure why you corrected 3 trips with 4.
- James Laverty . . . . . . . 2 pts - You answered yr own ques abt interior wires; fewer trips needed
- Nats Kroy. . . . . . . . . . . . 2 pts - A loop has to involve more than one wire, still solved in 3 trips.
- Mary Noggin (new) . . . . 1 pt - "I'd hold a lightbulb between my wet feet, climb a ladder, wet my fingers..."
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- Solution: Wow, there were a lot of you with entries this time! Yes, it was a particularly easy one to start off YearEleven,
- and it did draw in some new contestants, so I have done my job %;-} The numerical answers for the bricks were:
- . . . . . . . . . . O: 165
- . . .. . . . . M: 80 , N: 85
- . . . . .J: 35 , K: 45 , L: 40
- . . F: 14 , G: 21 , H: 24 , I: 16
- A: 2 , B: 12 , C: 9 , D: 15 , E: 1
- Here are a couple of GIFs from JinWon's PDF (I couldn't grab the text)
- And now Garry's solution! . . .
- "We have the system : O = M + 85 ; M = J + K ; 85 = K + L ; J = 14 + G ; K = G + 24 ; L = 24 + I ; G = B + C ;
- I = 25 - C and L = J + 5 . . . nine equations, nine variables ( B,C,G,I,J,K,L,M,O )
I use the program which can solve a system of linear equations with exactly nine equations.- It get O=165, M=80, J=35, K=45, L=40, G=21, I=16, B=12, C=9. Easy to calculate A=14-B so A=2, and D=24-C so D=15."
- Many of the rest of you showed complete equation-solving details, and I thank you (and usually reward you) for that! - Dan
- WINNERS - Problem 269 . . try dansmathcast - (back to top) . leader board
- Nick McGrath . . . . . . . .10 pts - Last year's winner, & off to a fast start again! Go Nick McG!
- Tim Poe . . . . . . . . . . . . . 7 pts - Thanks, it was nice to be back, or bach in my case, back then.
- Mark Rickert . . . . . . . . 5 pts - Hey Mark; good to notice the key bricks were G and I (not Joe)
- Denis Borris . . . . . . . . . 5 pts - This "one of my top cuties" was photographed in a bookstore.
- Claudio Baiocchi . . . . . 4 pts - Good method, solve for variables "floor-by-floor"-- it worked!
- Etienne Desclin. . . . . . . 4 pts - Liked yr method of gray variables, filled in by black numbers!
- Art Morris . . . . . . . . . . . 3 pts - Art "one-full-decade-of-entries" Morris 'Excel-s' yet again ;-}
- Ondrej Calda (new) . . . . 3 pts - A new math person 'Czechs in' from a new country for my contest!
- Garry Malashkin . . . . . 3 pts - Nine equations in 9 variables sounds do-able, by hand or 'solver'.
- Alan O'Donnell. . . . . . . 3 pts - Yes it's true this one was easier, but tougher challenges lie ahead!
- Phil Sayre. . . . . . . . . . 3 pts - Picked the top-down method, eh?
- Joe Fendel . . . . . . . . . . 3 pts - B=14-A, yes, etc. You got em all!
- Jin Won Park . . . . . . . 3 pts - Nice quoted ans (no attachmt needed)
- Nats Kroy. . . . . . . . . . . 3 pts - Brick pic didn't arrive but ans ok!
- Steve Tonnesen . . . . . . 3 pts - K+L=85, subst in, lots of eqns!!
- Hermen Jacobs . . . . . . 3 pts - Good sys of 11 eqns, sorry abt 268
- Yakov Macek . . . . . . . . 3 pts - Yer welcome 4 ez starter problem*
- *James Laverty . . . . . . . 3 pts - I liked "pyramid of values" & sys.
- Marcello Cammarata . 3 pts - Nice mix of eqns, thoughts, sol'ns.
- Allen Druze . . . . . . . . . 3 pts - Very clear expl'n, thnks! All good.
- David Stigant . . . . . . . 3 pts - Good eqns to sort things out, ok!
- Dawn Haught (new) . . . 3 pts - 85=K+L, L=G+19, G=K-24, go Dawn!
- Ravi Raja . . . . . . . . . . . 3 pts - J=x=F+G etc. Ok, who needs x?
- Mark Moyer . . . . . . . . . 3 pts - From diagram K+L=N & so on!<
- <Jae Kon Lee (new) . . . . 3 pts - Hello JK from fall semester Trig!
- Ed Wern . . . . . . . . . . . . 3 pts - 1."My sol'n" 2."How I got there"
- David Madfes . . . . . . . 3 pts - G=B+C etc and nice pyramid#s
- Frank Mullin. . . . . . . . 3 pts - Go SLO bikers! And good eqns.
- Philippe Fondanaiche . 2 pts - Ok but '5 more than'=/='5 times'
- Radu Ionescu . . . . . . . 2 pts - I see soln for J; steps 4 other 14?
- Ajit Athle . . . . . . . . . . 2 pts - Yay, max 15 eqns in 15 unknowns!^
- ^Riley Wern (new) . . . . . . 2 pts - Good job, & solved in the car ;-)
- Ken Duisenberg . . . . . . 2 pts - Some equations, all the answers!
- Kirk Bresniker . . . . . . . 2 pts - Spreadsheet applied to bricks, yes.
- Alex Kim (new) . . . . . . . 2 pts - Nice job, by a DVC college-mate!
- Brady Wern . . . . . . . . . 2 pts - Great, you're using your algebra!
- Art Wolfskill (new) . . . . 2 pts - Glad u like my site, good answer!
- Morandi Maurizio . . . . 1 pt - Welcome back! Yr first 5 letters ok.
- Kyle Misiaszek . . . . . . . 1 pt - Good reasons but multiple answers
- Bradley Wild (new) . . . . 1 pt - A few days later, but good answer!
- problem #270 - posted sunday, november 11, 2007 - Second problem of my eleventh contest!
- Spheres Filling Space (and vice-versa) (back to top)
- There are three common unit-sphere packings in 3D space, listed in increasing density:
- i) Stacked square-grid layers, ii) Stacked hexagonal layers, iii) Indented hexagonal layers.
- a) Find the exact density of each packing, then round to 0.1%.
- b) Fill in the main spaces for each packing i), ii), iii), with small spheres that are twice
- as dense as the unit spheres.Find the size of the small spheres in each case, and then
- rerank the total mass densities from lowest to highest, giving both the exact values
- and the approximations to 0.1%. Show steps and reasoning;
- Solution: Dan's note: Wow, I was ready to post this a couple of days ago but I realized there is more than one size hole
- in the third packing, and the "main spaces" might either be the largest, or the largest and the more numerous next-largest.
- None of this problem's responses took this into account, so I made the max score 8, then 6, then 4, 3, 2. Details below.
- Original note: Ok, not so many contestants this time, but you begged me to make up more interesting problems!
- There was a split on this one: a few of you didn't take into account the relative numbers of types of small spheres.
- Here's Jean Moreau de Saint-Martin with a great detailed report, followed by my calculation (which I believe to be correct).
- i) Stacked square-grid layers - Each sphere can be packed in a cubic box, with side 2.
Volume of a unit sphere 4pi/3, volume of the cube 8. Density (4pi/3)/8 = pi/6 = 0.5235987... rounded 52.4%
ii) Stacked hexagonal layers - The package of a sphere is now a prism with height 2, hexagonal base circumscribed to a unit circle
The hexagon has side 2/sqrt(3)., perimeter 6x2/sqrt(3) = 4sqrt(3), area (1/2)4sqrt(3) = 2sqrt(3).- The prism has volume 2x2sqrt(3) = 4sqrt(3). Density (4pi/3)/(4sqrt(3)) = pi/(3sqrt(3)) = 0.604599... rounded 60.5%
iii) Indented hexagonal layers - Let A, B the centers of 2 spheres in contact in 2 different layers. AB = 2.
Consider the prism of case ii), centered in A. B is located on a vertical side of this prism, so AB has a horizontal component of 2/sqrt(3).
The vertical distance between layers is the vertical component of AB, which is, by Pythagoras' theorem, sqrt(8/3).
Each sphere can be associated with a hexagonal prism having height sqrt(8/3) and (as in the stacked case) base area 2sqrt(3).
The volume of this prism is 4sqrt(2). Density (4pi/3)/(4sqrt(2))=pi/sqrt(18) = 0.740480... rounded 74.0%*** that was part a) *** now part b) ***
- i) The small spheres have their center at the apices of the cubes packaging the unit spheres, with distance sqrt(3) from the centers of the cubes.
They have radius (sqrt(3)-1), their volume is (sqrt(3)-1)^3 = 6sqrt(3) - 10 times the volume of a unit sphere.
The square grid of small spheres has the same mesh width as the square grid of unit spheres. So there is one small sphere for every- unit sphere, but they are twice as dense as the unit spheres, so the mass density is magnified by the factor
1 + 2(6sqrt(3)-10) = 12sqrt(3) - 19. Finally the density is (12sqrt(3) - 19)pi/6 = 0.934419... rounded 93.4%
ii) The small spheres have their center at the apices of the prisms packaging the unit spheres. From the center A of a prism, the small- sphere in contact has center B, and AB has projections 1 vertical, 2/sqrt(3) horizontal, so AB = sqrt(7/3) by Pythagoras' theorem.
So the small spheres have radius (sqrt(7/3)-1), their volume is (sqrt(7/3)-1)^3 = (16/9)sqrt(21) - 8 times the volume of a unit sphere.
Each small sphere lies on 3 unit spheres. The centers of these 3 spheres form a triangle with area half of the prism base area.- So there are 2 small spheres for every unit sphere; moreover they are twice as dense as the unit spheres, so the mass density is magnified
- by the factor 1 + 2x2((16/9)sqrt(21) - 8) = (64/9)sqrt(21) - 31.
Finally the density is ( (64/9)sqrt(21) - 31)pi/sqrt(27) = (64sqrt(7) - 31sqrt(27))pi/27 = 0.959623... rounded 96.0%
iii) Each unit sphere lies on 3 other unit spheres. Their centers form a regular tetrahedron with side 2.- Each tetrahedron center is the center of a small sphere. The distance from this center to the tetrahedron apices is sqrt(3/2).
So the small spheres have radius (sqrt(3/2)-1), their volume is (sqrt(3/2)-1)^3 = (9sqrt(6) - 22)/4 times the volume of a unit sphere.
Let A the center of a unit sphere, in contact with (in the upper layer) spheres with centers B,C,D. A is an apex for 4 tetrahedra:- 3 determined by A, one of points B,C,D, and 2 neighbours of A in its layer; and one tetrahedon ABCD. A is an apex for 4 other tetrahedra
- with spheres in the lower layer. As each center A belongs to 8 tetrahedra, and each tetrahedron has 4 apices, the tetrahedra are twice as
- many as the unit spheres, and so are the small spheres with density 2.
Thus the mass density is magnified by the factor 1 + 2x2(9sqrt(6)-22)/4 = 9sqrt(6)-21.
Finally the density is ( 9sqrt(6)-21)pi/sqrt(18) = (sqrt(27) - sqrt(24.5))pi = 0.774103... rounded 77.4%- From lowest density to highest density, the new ranking is
iii) Indented hexagonal layers, 77.4% i) Stacked square-grid layers, 93.4% ii) Stacked hexagonal layers, 96.0%.
- Dan's note: There was a variety of opinion on the number of small spheres (radius \/(3/2) - 1 = 0.2247) per unit sphere: 1, 2, or 3.
- But as I notice, there are larger (octahedral) holes between the spaces of layers, with small spheres of radius \/ 2 - 1 = 0.414,
- which have volume (4 pi/3)(\/2 - 1)^3 = (4 pi/3)(5\/2 - 7) and these are in a 1:1 ratio with the unit spheres, so the new density is
- (pi/6)(\/2) + (pi/6)(5\/2 - 7) = 0.7405 + 0.0372 = 0.7777 = 77.8%. If the small tetrahedral holes (2 to a unit sphere) are also filled
- ('main' spaces but not the mainest), the density becomes (pi/6)(6\/2 - 7) + (pi/3)(9\/6 - 22) = 0.7777 + 0.0119 = 0.7896 = 79.0%
- Either way this formerly densest packing finishes last, and the winner is the stacked hexagonal at 96.0% total (mass) density.
- Another French contestant Philippe F attached a beautiful word document from which I stole these pictures:
- cubical
- stacked hexagonal
(i'm leaving out all his nice equations!)
- indented square / hex
- WINNERS - Problem 270 . . try dansmathcast - (back to top) . leader board
- Moreau deSaint-Martin . . . 8 pts - This early answer came the closest; see quoted solution above!
- Nick McGrath . . . . . . . . . . . 6 pts - I believed you until I found some of those spheres were bigger.
- Marcello Cammarata . . . . . 4 pts - First answer, claimimg the reverse order, good expl of sphere sizes
- Philippe Fondanaiche . . . . 4 pts - That was an awesome document you produced, with embedded picts!
- Claudio Baiocchi . . . . . . . . 3 pts - Correct original densities, smaller spheres are twice as heavy per unit vol
- Nats Kroy . . . . . . . . . . . . . . 3 pts - Great argmt with boxes LMN and limit of number of spheres inside
- Hermen Jacobs . . . . . . . . . . 2 pts - I believe you on the first cubical part, the tetrahedral is good locally
- K Sengupta . . . . . . . . . . . . . 2 pts - Very good job on part a) all three parts, enjoy the space ball debate.
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