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Topic: Birthday twins (Read 5178 times) |
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johnP
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Birthday Twins
The twins in question were probably born in a "normal" 365 day year. The oldest one on 28th of February at 23.59 (as close as possible to 00.00), the youngest one on the first of March (at 00.00). Whenever the day has 366 days (argh, what's the English name for the 29th of February every fourth year?), the young one will celebrate birthday two days after his/her older sibling.
After doing some googling, it seems that the English word for these 366 years is "leap year" and the 29th of Feb is "leap day".
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| « Last Edit: Sep 1st, 2003, 6:57pm by Icarus » |
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PKo
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Add to this that He-man is a girl instead of a boy and Shela is a boy (nothing say otherwise), then when the riddle speaks about "her" birthday, we're talkink about He_man's one, which is two day later on a leap-year !
Bingo
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PKo
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Forget the previous message 
My english is poor and I misunderstood the sentence.
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drdedos
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JohnP,
Your answer is what i thought originally. But read the question more carefully:
"Sheila and He-Man are twins; Sheila is the OLDER twin. Assume they were born immediately after each other, an infinitesimally small - but nonzero - amount of time apart. During one year in the course of their lives, Sheila celebrates her birthday two days AFTER He-Man. How is this possible?"
If Sheila is the OLDER twin, that means she was born BEFORE He-Man. How could she have a birthday two days AFTER him? I'm confused.
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Gamer555
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Re: Birthday twins
« Reply #4 on: Jul 30th, 2002, 9:13am » |
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I bet so many people fell into that trap. I did. I don't know! Does it have anything to do with boys versus girls? *sigh*
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drdedos
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Maybe they were in a plane flying across the prime meridian on February 28 / March 1.
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Everend
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They are born while crossing the International Dateline (not prime meridian) in the easterly direction. With Sheila being born on March 1 and He-man being born on Feb 28 (after crossing the Dateline). Normally Sheila celebrates the day after He-man except every leapyear when it falls two days after his.
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johnP
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Damned! I should have known there was a trap in there. You're right... and I was arrogant 
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Ron Spiegelhalter
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The question states that she celebrated her birthday on the actual date; that it wasn't a rescheduled celebration. However, it says nothing about HIS birthday celebration. So he celebrated his birthday two days early, and she celebrated hers two days later on the actual date.
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Cathy
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Originally, I was focusing on the leap year idea,too. But then I figured this approch would never work because Sheila is the OLDER one.
Then I realize, it could work with the help of the time zone difference. Say, on a leap year, the twins were given birth on an airplane (or a boat) as it was crossing the pacific ocean travelling from East to West. Sheila was born on one side of the time zone, on March 1st (before 1am). And right after the transportation device had crossed to the next time zone, He-Man was born. Because the western time come after eastern time, so the clock was set back for one hour. Which means He-Man was actually born on Feb 28th (right before 00:00am).
Therefore, even though Sheila is biologically OLDER, He-Man actually has the birthday first in term of callendar.
So, He-Man has the birthday 2 days before Sheila every leap year.
p.s. Feel free to point out the errors I might have in explaining the time zone concepts, for geography has always been a subject I willfully neglected. But I believe the idea in my explanation is correct.
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prince
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Another variant solution depends on the definition of OLDER (does it depend on order of birth, or conception). Since Sheila and He-man are not the same sex, they must be dizygotic twins (2 different sperm fertilizing 2 different eggs, not one fertilized zygote splitting in 2). If Sheila is conceived before He-man, she could be considered older, and still be delivered (born) after him.
This solution still requires the leap-year stuff, but not the crossing of the International Dateline.
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zarathustra
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Re: Birthday twins
« Reply #11 on: Aug 24th, 2002, 9:39pm » |
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If they did cross the international date line or any time zone when being born they would never know it and therefore wouldn’t be able to celebrate their birthday on different days. Since they were born an infinitesimal time apart then it would be impossible for them tell exactly the split second they crossed the line. The more likely solution I can think of if they were born by two different mothers (one is a surrogate) in two different countries on opposite sides of the date line on feb 28 and march 1 (even though its a different calendar date, and in different places they could still conceivably be born an infinitesimal time apart). Also this could take place any time during the day, not necessarily midnight.
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william wu
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Re: Birthday twins
« Reply #12 on: Aug 30th, 2002, 12:01pm » |
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So the standard solution has the twins born on the dateline while traveling opposite to the earth's rotation, on February 28 23:59:59 and March 1 00:00:00. Then a leap year comes along. Their birthdays are normally separated by just one day, but now February 29 inserts an extra day of separation, and their celebrations are separated by two days -- even if the twins are in the same time zone.
The problem as I phrased it says "Sheila celebrates her birthday two days AFTER He-Man does." Recently I was wonderng if someone could interpret this sentence in two different ways. One way is to just look at the calendar dates on which the two celebrate, and take the difference. Then I think we cannot do any better than 2 days (March 1 - February 28 = 2 days during a leap year). However, what if we interpret the words "2 days" as the amount of time separates Sheila and He-man's celebrations?
To make this point more clear, what if Sheila and He-man are placed on opposite sides of the globe during that leap year? Then I think we can get 3 days of separation.
What if during the leap year, one of the twins is in a country where daylight savings time is not recognized? Then can we get 4 days of separation?
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Yuvy Elly
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The maximum amount of time I can think of is as follows:
Assume TwinA is born before TwinB, but you still need TwinB's birthday a few years later, to occur the maximum amount of time before TwinA
If that is the case, crossing the meridian line before and after midnight will just hinder you (TwinA will be born at 23:59:59 on February 28th, TwinB will be born -after crossing east to west the meridian line - at 00:00:01 on February 28th).
(In other words, ignore the midnight crossover to get an extra day, if you need TwinA to celebrate his birthday AFTER TwinB. If you're trying to find the maximum amount of time 2 twins can celebrate their birthdays, then you would probably use the midnight crossover, going west to east across the meridian line to get an extra day)
Now for my solution: They were on a plane, crossing the meridian line during a leap year - from east to west.
So: TwinA gets born at (let's say) 1:00 pm March 1st on the east side of the meridian (Australia).
A few seconds later, TwinB gets born on the west side of the meridian at 1:00 on February 28th (USA)
That means, during any non-leap year, if they are in the same time-zone, they celebrate their birthdays a maximum of 48 hours one after each other.
Now, if TwinB happens to be in Australia/New Zealand one year, and TwinA is in Hawaii at the same time, they will celebrate an actual (almost) 72 hours after each other, with the first-born celebrating second.
Daylight savings:
I have no clue about daylight savings time in february for Hawaii and New Zealand, but assuming there is - at most it will give you an extra 2 hours difference during the birth, which you could leverage into an extra 2 days saying TwinA was born on 00:59 am on march 2nd (which at that exact instance across the world is 22:59 on february 28th) so a few seconds later TwinB could possibly still be born on february 28th.
Now assume TwinB celebrates at 00:01 am on February 28th (same birth-day day) and TwinA celebrate at 23:59 pm on March 2nd (same birth-day day), once again, on both sides of globe as before, then you have: Feb28th (24 hours), Feb 29th (24 hours), March 1st (24 hours), March 2nd (24 hours) and then the actual time for it to be 23:59 on March 2nd in Hawaii and you get almost (save a few seconds) 5 days!
phew
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GRAND_ADMRL_THUORN
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THE DARK TOWER IS NEAR!
  
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Re: Birthday twins
« Reply #14 on: Oct 30th, 2002, 5:11pm » |
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YOU ARE ALL WRONG, ALL WRONG!!!!
I SAW HE-MAN LAST NIGHT, HE DOESNT HAVE A SISTER, THE QUESTION IS A FARCE, A FARCE, WATCH FOR THE BLACK CHOPPERS, LOOK OUT FOR THE MEN IN BLACK, KEEP YOUR EYES OPEN FOR THE LITTLE BALD DOCTORS.
WILLIAM WU'S SITE HAD BEEN TAKEN OVER BY "THEM"
TRUST NO ONE, TRUST NO ONE!!!!!!!!!
ITS ALL A LIE!!!
HELP ME PUZZLE MASTER, HELP ME!!!
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Ronald A. Lau
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The problem I have with this riddle is the phrase...
"During one year in the course of their lives..."
One year only? Leap year occurs every 4 years. If its not one and only one year why the phrase "course of their lives.."
I know the wordings purpose is to hide any reference to leap year, but I think the "course of their lives" needs to be removed.
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william wu
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Re: Birthday twins
« Reply #16 on: Nov 7th, 2002, 10:04am » |
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Technically, I didn't say "During only one year in the course of their lives", so I don't think the phrasing is necessarily flawed as you suggest. Saying "during one year" does not imply that the statement "during only one year" is false. However I see how people could misinterpret the statement and think that I'm saying "during only one year", because of the number "one". I don't think removing "course of their lives" makes it much clearer though -- that phrase is just there for contextual clarity, to say that this phenomenon happens in a year during which Sheila and He-Man are alive (obviously). If any one has more ideas on how to rephrase the riddle without giving away the leap year periodicity, feel free to let me know. Thanks.
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| « Last Edit: Nov 7th, 2002, 10:05am by william wu » |
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Chronos
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Re: Birthday twins
« Reply #17 on: Nov 9th, 2002, 1:40pm » |
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I dunno... I think it's fairly obvious from the problem itself that the answer must somehow involve leap years. What's less immediately clear is how it involves leap years. So I wouldn't worry too much about giving that away.
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Penelope Pitstop
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Quote:| Sheila and He-Man are twins; Sheila is the OLDER twin. Assume they were born immediately after each other, an infinitesimally small - but nonzero - amount of time apart. |
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Quote:| What is the maximum amount of time by which Sheila and He-Man can be apart in their birthday celebrations during the same year? |
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Sheila is born shortly before midnight on 31 December 1900. He-Man is born shortly after midnight on 1 January 1901. So Sheila is the older twin, as desired. In the year 1904, Sheila celebrates her (fourth) birthday 365 days after He-Man celebrates his (third) birthday. 
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organica
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Re: Birthday twins
« Reply #19 on: Jan 6th, 2003, 11:14pm » |
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Heh... OK, Sheila & He-Man's mother Hella works for Clonaid. She is killed by Skeletor, but not before two different biomechanical spacecraft *capable of close-to-light-speed-travel* have each been bologically implanted with one clone twin. One of them crosses the dateline, and the othere one goes on holiday to Hawaii, but during the crucial infinitesimal/2 time juncture, Superman flies round the earth really quickly backwards, thus causing Time Itself to reset. But Skeletor isn't going to give in so easily, so he <author killed by humanitarian> <(from the future)>
g k
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D_a_v_e
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Boy, you people.
Nobody said they were twins to each other!
There birthdays could be up to 182 (3?) days apart.....
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redPEPPER
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Re: Birthday twins
« Reply #21 on: Jan 21st, 2003, 2:10am » |
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Even if they are not twins to each other, the ridde does say "Assume they were born immediately after each other, an infinitesimally small - but nonzero - amount of time apart".
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eviljed
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Re: Birthday twins
« Reply #22 on: Feb 25th, 2003, 9:20am » |
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What if it happened back in the middle ages (I don't remember the exact year), when a number of days was jumped in the calander to get it back in sync? That would give about 2 weeks in seperation.
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poseur
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Re: Birthday twins
« Reply #23 on: Feb 25th, 2003, 11:13am » |
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Wu asked if there's a better way to phrase the question. What if you said something like, On his 21st birthday He-Man mailed a birthday card to Sheila, which she received on her birthday 2 days later.
And if they're Jewish or Chinese, their birthdays could be a month apart. (They use lunar calendars and add a leap month about every 3 years).
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Anthony Gibson JR
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This is too easy
The real answer is:
This is probably not the answer you are looking for, but here goes. Their birthdays are both on March 1. The year is 2800, and He-Man is in Greece for his Birthday. 2800 would normally be a leap year, everywhere except for Greece, which will instead celebrate its leap year in 2900. Since Greece is on the other side of the international date line, when it is March 1 in Greece it is Feb 28 in the USA, because Greece skipped Feb. 29. Sheila, who is back in the US of A, then has to wait 2 days to celebrate her birthday, on March 1.
The scenario I outlined also works in Russia pre-1917 and in China pre-1949
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