• Log InLog In
  • Register
Liquid`
Team Liquid Liquipedia
EDT 16:24
CEST 22:24
KST 05:24
  • Home
  • Forum
  • Calendar
  • Streams
  • Liquipedia
  • Features
  • Store
  • EPT
  • TL+
  • StarCraft 2
  • Brood War
  • Smash
  • Heroes
  • Counter-Strike
  • Overwatch
  • Liquibet
  • Fantasy StarCraft
  • TLPD
  • StarCraft 2
  • Brood War
  • Blogs
Forum Sidebar
Events/Features
News
Featured News
[ASL19] Finals Recap: Standing Tall5HomeStory Cup 27 - Info & Preview18Classic wins Code S Season 2 (2025)16Code S RO4 & Finals Preview: herO, Rogue, Classic, GuMiho0TL Team Map Contest #5: Presented by Monster Energy6
Community News
Flash Announces Hiatus From ASL33Weekly Cups (June 23-29): Reynor in world title form?12FEL Cracov 2025 (July 27) - $8000 live event16Esports World Cup 2025 - Final Player Roster14Weekly Cups (June 16-22): Clem strikes back1
StarCraft 2
General
The SCII GOAT: A statistical Evaluation Weekly Cups (June 23-29): Reynor in world title form? StarCraft Mass Recall: SC1 campaigns on SC2 thread How does the number of casters affect your enjoyment of esports? Esports World Cup 2025 - Final Player Roster
Tourneys
FEL Cracov 2025 (July 27) - $8000 live event HomeStory Cup 27 (June 27-29) WardiTV Mondays SOOPer7s Showmatches 2025 $200 Biweekly - StarCraft Evolution League #1
Strategy
How did i lose this ZvP, whats the proper response Simple Questions Simple Answers
Custom Maps
[UMS] Zillion Zerglings
External Content
Mutation # 480 Moths to the Flame Mutation # 479 Worn Out Welcome Mutation # 478 Instant Karma Mutation # 477 Slow and Steady
Brood War
General
[ASL19] Finals Recap: Standing Tall Flash Announces Hiatus From ASL Help: rep cant save Where did Hovz go? BW General Discussion
Tourneys
[Megathread] Daily Proleagues [BSL20] GosuLeague RO16 - Tue & Wed 20:00+CET The Casual Games of the Week Thread [BSL20] ProLeague LB Final - Saturday 20:00 CET
Strategy
Simple Questions, Simple Answers I am doing this better than progamers do.
Other Games
General Games
Stormgate/Frost Giant Megathread Nintendo Switch Thread Path of Exile What do you want from future RTS games? Beyond All Reason
Dota 2
Official 'what is Dota anymore' discussion
League of Legends
Heroes of the Storm
Simple Questions, Simple Answers Heroes of the Storm 2.0
Hearthstone
Heroes of StarCraft mini-set
TL Mafia
TL Mafia Community Thread Vanilla Mini Mafia
Community
General
US Politics Mega-thread Trading/Investing Thread Things Aren’t Peaceful in Palestine Stop Killing Games - European Citizens Initiative Russo-Ukrainian War Thread
Fan Clubs
SKT1 Classic Fan Club! Maru Fan Club
Media & Entertainment
Anime Discussion Thread [Manga] One Piece [\m/] Heavy Metal Thread Korean Music Discussion
Sports
2024 - 2025 Football Thread NBA General Discussion Formula 1 Discussion TeamLiquid Health and Fitness Initiative For 2023 NHL Playoffs 2024
World Cup 2022
Tech Support
Computer Build, Upgrade & Buying Resource Thread
TL Community
The Automated Ban List
Blogs
from making sc maps to makin…
Husyelt
Blog #2
tankgirl
Game Sound vs. Music: The Im…
TrAiDoS
StarCraft improvement
iopq
Heero Yuy & the Tax…
KrillinFromwales
Trip to the Zoo
micronesia
Customize Sidebar...

Website Feedback

Closed Threads



Active: 603 users

Math Puzzle [num 17]

Blogs > evanthebouncy!
Post a Reply
1 2 Next All
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
Last Edited: 2010-08-18 19:58:37
August 18 2010 19:51 GMT
#1
It's been awhile huh!! Here goes!

Again, hide answer in spoilers. And good luck!

You have a broken calculator with only these buttons working:
sin
cos
tan
cot
sin^-1
cos^-1
tan^-1
cot^-1

where when I say sin^-1 I mean the inverse of sine, not one over sine.
i.e. sine(sine-1(x)) = x for x between -1 and 1.

Your calculator now show 0. How do you, after a finite number of pressing buttons, make ANY positive rational number to appear on your screen?

Ahaha maybe there's some confusion, by ANY positive rational number, I mean this:
If I want to make 15/23 appear, I can do it.
If I want to make any p/q appear, where p, q in natural number, and p, q has no common divisors, I can make it happen after finite number of moves.

sorry! xD

*****
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
KumquatExpress
Profile Joined October 2009
United States344 Posts
Last Edited: 2010-08-18 20:03:35
August 18 2010 19:54 GMT
#2
+ Show Spoiler +
...Wouldn't you just press cosine? Unless there's something I'm not getting here..
Edit: I see your edit. Will ponder now.
Speedythinggoesin, speedythingcomesout.
blabber
Profile Blog Joined June 2007
United States4448 Posts
August 18 2010 19:56 GMT
#3
+ Show Spoiler +
maybe I don't understand the question correctly, but wouldn't just pressing cos give you 1?
blabberrrrr
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2010-08-18 20:29:19
August 18 2010 20:01 GMT
#4
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress
Translator:3
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
August 18 2010 20:10 GMT
#5
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress


we're getting somewhere
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
BajaBlood
Profile Joined August 2009
United States205 Posts
August 18 2010 20:33 GMT
#6
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress


+ Show Spoiler +

I can't vouch for the truth/falsehood of the above. But if you can create a system for 1/a, then simply typing 'b' before the operation should yield the desired fraction.

barccot(tan(a)) = b/a

Unless it's cheating to use the implied multiplication


Empyrean
Profile Blog Joined September 2004
16978 Posts
August 18 2010 20:35 GMT
#7
The calculator doesn't have number buttons.
Moderator
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2010-08-18 20:51:55
August 18 2010 20:35 GMT
#8
On August 19 2010 05:33 BajaBlood wrote:
Show nested quote +
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress


+ Show Spoiler +

I can't vouch for the truth/falsehood of the above. But if you can create a system for 1/a, then simply typing 'b' before the operation should yield the desired fraction.

barccot(tan(a)) = b/a

Unless it's cheating to use the implied multiplication




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

As I stated above, sqrt(a/1) is possible.
But sqrt(a/2) is also possible.
If a is divisible by 2, then it just reduces to sqrt((a/2)/1), this gives us sqrt(2/2), sqrt(4/2), forever
If a is not divisible by 2, well
apply operation 1 to sqrt(2/1) to get sqrt(1/2)
apply operation 5 to that repeatedly to get sqrt(3/2), sqrt(5/2), forever

Now let us induct on n, using the same process.
Say we can get sqrt(a/1), sqrt(a/2), sqrt(a/3), ... ,sqrt(a/n) for any n
then, take sqrt((n+1)/1), sqrt((n+1)/2), ..., sqrt((n+1)/n) and apply operation 1 to them all
this gives sqrt(1/(n+1)), sqrt(2/(n+1)), ..., sqrt(n/(n+1)).
If we apply operation 5 to all of these, we can get sqrt(a/(n+1)) for any n.

But this gives us the square root of any rational number, so we can obtain any m/n by obtaining sqrt(m^2/n^2).
Translator:3
Roggles
Profile Joined December 2009
United States38 Posts
August 18 2010 21:42 GMT
#9
On August 19 2010 05:35 infinitestory wrote:
Show nested quote +
On August 19 2010 05:33 BajaBlood wrote:
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress


+ Show Spoiler +

I can't vouch for the truth/falsehood of the above. But if you can create a system for 1/a, then simply typing 'b' before the operation should yield the desired fraction.

barccot(tan(a)) = b/a

Unless it's cheating to use the implied multiplication




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

As I stated above, sqrt(a/1) is possible.
But sqrt(a/2) is also possible.
If a is divisible by 2, then it just reduces to sqrt((a/2)/1), this gives us sqrt(2/2), sqrt(4/2), forever
If a is not divisible by 2, well
apply operation 1 to sqrt(2/1) to get sqrt(1/2)
apply operation 5 to that repeatedly to get sqrt(3/2), sqrt(5/2), forever

Now let us induct on n, using the same process.
Say we can get sqrt(a/1), sqrt(a/2), sqrt(a/3), ... ,sqrt(a/n) for any n
then, take sqrt((n+1)/1), sqrt((n+1)/2), ..., sqrt((n+1)/n) and apply operation 1 to them all
this gives sqrt(1/(n+1)), sqrt(2/(n+1)), ..., sqrt(n/(n+1)).
If we apply operation 5 to all of these, we can get sqrt(a/(n+1)) for any n.

But this gives us the square root of any rational number, so we can obtain any m/n by obtaining sqrt(m^2/n^2).


how can you assume that you can get sqrt(a/n) for any n?

applying 1+5 only gets you sqrt(1+x^2). if you apply it again, you will get sqrt(2+x^2) etc. your claim is that by taking sqrt(2) and applying 1/x to make it sqrt(1/2), you can obtain sqrt(3/2), sqrt(5/2) etc as well, which is perfectly fine. However, if you tried it with sqrt(3), it would become sqrt(1/3), and then you would have a sequence of sqrt(1/3), sqrt (4/3), sqrt(7/3). combined with the natural numbers, you would get sqrts of 1/3, 3/3, 4/3, 6/3, etc. but we're missing the 2/3, 5/3, etc.

the 1+5 algorithm, when applied recursively, will take a fraction sqrt(a/b) and turn it into sqrt(1+a/b). therefore you will get the sqrt of a/b, (b+a)/b, (2b+a)/b, etc. which does not properly fill in the fractions between a/b and (b+a)/b.

tell me if I'm missing something stupid though...
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
August 18 2010 21:53 GMT
#10
On August 19 2010 06:42 Roggles wrote:
Show nested quote +
On August 19 2010 05:35 infinitestory wrote:
On August 19 2010 05:33 BajaBlood wrote:
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress


+ Show Spoiler +

I can't vouch for the truth/falsehood of the above. But if you can create a system for 1/a, then simply typing 'b' before the operation should yield the desired fraction.

barccot(tan(a)) = b/a

Unless it's cheating to use the implied multiplication




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

As I stated above, sqrt(a/1) is possible.
But sqrt(a/2) is also possible.
If a is divisible by 2, then it just reduces to sqrt((a/2)/1), this gives us sqrt(2/2), sqrt(4/2), forever
If a is not divisible by 2, well
apply operation 1 to sqrt(2/1) to get sqrt(1/2)
apply operation 5 to that repeatedly to get sqrt(3/2), sqrt(5/2), forever

Now let us induct on n, using the same process.
Say we can get sqrt(a/1), sqrt(a/2), sqrt(a/3), ... ,sqrt(a/n) for any n
then, take sqrt((n+1)/1), sqrt((n+1)/2), ..., sqrt((n+1)/n) and apply operation 1 to them all
this gives sqrt(1/(n+1)), sqrt(2/(n+1)), ..., sqrt(n/(n+1)).
If we apply operation 5 to all of these, we can get sqrt(a/(n+1)) for any n.

But this gives us the square root of any rational number, so we can obtain any m/n by obtaining sqrt(m^2/n^2).

+ Show Spoiler +


how can you assume that you can get sqrt(a/n) for any n?

applying 1+5 only gets you sqrt(1+x^2). if you apply it again, you will get sqrt(2+x^2) etc. your claim is that by taking sqrt(2) and applying 1/x to make it sqrt(1/2), you can obtain sqrt(3/2), sqrt(5/2) etc as well, which is perfectly fine. However, if you tried it with sqrt(3), it would become sqrt(1/3), and then you would have a sequence of sqrt(1/3), sqrt (4/3), sqrt(7/3). combined with the natural numbers, you would get sqrts of 1/3, 3/3, 4/3, 6/3, etc. but we're missing the 2/3, 5/3, etc.

the 1+5 algorithm, when applied recursively, will take a fraction sqrt(a/b) and turn it into sqrt(1+a/b). therefore you will get the sqrt of a/b, (b+a)/b, (2b+a)/b, etc. which does not properly fill in the fractions between a/b and (b+a)/b.

tell me if I'm missing something stupid though...


+ Show Spoiler +

You get sqrt(2/2), sqrt(4/2), sqrt(6/2), etc. because they are equal to sqrt(1), sqrt(2), etc.

You get sqrt(2/3) by flipping sqrt(3/2)
Translator:3
jalstar
Profile Blog Joined September 2009
United States8198 Posts
August 18 2010 21:58 GMT
#11
+ Show Spoiler +
You can't. You don't have an "=" button.
Hypnosis
Profile Blog Joined October 2007
United States2061 Posts
August 18 2010 22:01 GMT
#12
where do you get these problems? i feel like i missed something in calc or something because im done with calc 3 already and idk those identities for shit! Maybe i should review my trig proofs haha
Science without religion is lame, Religion without science is blind
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2010-08-18 22:03:17
August 18 2010 22:01 GMT
#13
On August 19 2010 06:58 jalstar wrote:
+ Show Spoiler +
You can't. You don't have an "=" button.

+ Show Spoiler +
Pressing the operator also shows the answer. Old school way.


EDIT: @hypnosis: this is a classic contest math algebra problem (I'm kinda cheating since I solved it a couple years ago, then forgot the solution)
Translator:3
Roggles
Profile Joined December 2009
United States38 Posts
August 18 2010 22:04 GMT
#14
On August 19 2010 06:53 infinitestory wrote:
Show nested quote +
On August 19 2010 06:42 Roggles wrote:
On August 19 2010 05:35 infinitestory wrote:
On August 19 2010 05:33 BajaBlood wrote:
On August 19 2010 05:01 infinitestory wrote:
a start
+ Show Spoiler +
Things we can do:
1. arctan(cot(x)) = arccot(tan(x)) = 1/x
2. arcsin(cos(x)) = arccos(sin(x)) = sqrt(1-x^2)
3. arctan(sin(x)) = arccot(cos(x)) = x/sqrt(1-x^2)
4. arcsin(tan(x)) = arccos(cot(x)) = x/sqrt(1+x^2)
5. arccos(tan(x)) = arcsin(cot(x)) = 1/sqrt(1+x^2)
(there is a sixth non-identity operation, but it's equivalent to 1 applied to 3.)

EDIT: redefinition of 1-5 for x = a/b (just need to put this down somewhere)
1. b/a
2. sqrt(b^2-a^2)/b
3. a/sqrt(b^2-a^2)
4. a/sqrt(b^2+a^2)
5. 1/sqrt(b^2+a^2)

will work on it

EDIT: You can get any fraction of the form a/1 or 1/a.
Obviously, 1 applied to 5 is sqrt(1+x^2).
Repeatedly applying this gives sqrt(2), ..... to infinity.
Contained in this sequence are sqrt(4) = 2, sqrt(9) = 3, etc.
We can apply 1 to those to get 1/2, 1/3, etc.


Edited, made progress


+ Show Spoiler +

I can't vouch for the truth/falsehood of the above. But if you can create a system for 1/a, then simply typing 'b' before the operation should yield the desired fraction.

barccot(tan(a)) = b/a

Unless it's cheating to use the implied multiplication




You can't type them, only obtain them by result of an operation

EDIT: solved
+ Show Spoiler +

As I stated above, sqrt(a/1) is possible.
But sqrt(a/2) is also possible.
If a is divisible by 2, then it just reduces to sqrt((a/2)/1), this gives us sqrt(2/2), sqrt(4/2), forever
If a is not divisible by 2, well
apply operation 1 to sqrt(2/1) to get sqrt(1/2)
apply operation 5 to that repeatedly to get sqrt(3/2), sqrt(5/2), forever

Now let us induct on n, using the same process.
Say we can get sqrt(a/1), sqrt(a/2), sqrt(a/3), ... ,sqrt(a/n) for any n
then, take sqrt((n+1)/1), sqrt((n+1)/2), ..., sqrt((n+1)/n) and apply operation 1 to them all
this gives sqrt(1/(n+1)), sqrt(2/(n+1)), ..., sqrt(n/(n+1)).
If we apply operation 5 to all of these, we can get sqrt(a/(n+1)) for any n.

But this gives us the square root of any rational number, so we can obtain any m/n by obtaining sqrt(m^2/n^2).

+ Show Spoiler +


how can you assume that you can get sqrt(a/n) for any n?

applying 1+5 only gets you sqrt(1+x^2). if you apply it again, you will get sqrt(2+x^2) etc. your claim is that by taking sqrt(2) and applying 1/x to make it sqrt(1/2), you can obtain sqrt(3/2), sqrt(5/2) etc as well, which is perfectly fine. However, if you tried it with sqrt(3), it would become sqrt(1/3), and then you would have a sequence of sqrt(1/3), sqrt (4/3), sqrt(7/3). combined with the natural numbers, you would get sqrts of 1/3, 3/3, 4/3, 6/3, etc. but we're missing the 2/3, 5/3, etc.

the 1+5 algorithm, when applied recursively, will take a fraction sqrt(a/b) and turn it into sqrt(1+a/b). therefore you will get the sqrt of a/b, (b+a)/b, (2b+a)/b, etc. which does not properly fill in the fractions between a/b and (b+a)/b.

tell me if I'm missing something stupid though...


+ Show Spoiler +

You get sqrt(2/2), sqrt(4/2), sqrt(6/2), etc. because they are equal to sqrt(1), sqrt(2), etc.

You get sqrt(2/3) by flipping sqrt(3/2)


+ Show Spoiler +

oh yeah that's right...we've filled in a/1 and a/2, and the different values for (a mod 3)/3 can be obtained by flipping previous values. so 1/3 can be gotten from flipping 3/1, and 2/3 can be gotten from flipping 3/2, and therefore a/3 is filled. and this pattern stacks on top of itself recursively.

just wanted a little clarification, sorry. I missed that little jump in the logic.
icystorage
Profile Blog Joined November 2008
Jollibee19343 Posts
August 18 2010 22:11 GMT
#15
On August 19 2010 05:35 Empyrean wrote:
The calculator doesn't have number buttons.

i think that's the challenge of the problem
LiquidDota StaffAre you ready for a Miracle-? We are! The International 2017 Champions!
FiBsTeR
Profile Blog Joined February 2008
United States415 Posts
August 18 2010 22:13 GMT
#16
I remember solving this when preparing for USAMO... I think this was on a past one.
]343[
Profile Blog Joined May 2008
United States10328 Posts
Last Edited: 2010-08-18 22:19:03
August 18 2010 22:18 GMT
#17
this is an old usamo problem

and VERY ANNOYING

+ Show Spoiler +

0: already exists

begin with cos 0 = 1.

now, notice that

f(x) = cot(arctan(x)) = 1/x
g_1(x) = cos(arctan(x)) = 1/sqrt(x^2+1)
g_2(x) = sin(arctan(x)) = x/sqrt(x^2+1)
h_1(x) = tan(arccos(x)) = sqrt(1-x^2)/x
h_2(x) = tan(arcsin(x)) = x/sqrt(1-x^2)
h_3(x) sin(arccos(x)) = sqrt(1-x^2)

so we can get any composition of these functions, applied to 1 or 0. call a number that can be reached by a sequence of keypresses "good." If n is good, so is 1/n, since 1/n = f(n). We will show that all numbers x such that x^2 is rational are good; this clearly implies the problem statement.

we only need to show that all rationals q less than 1 (and 1 is good) are good, since every rational greater than 1 is the reciprocal of a rational less than 1.

starting at 1, we find g_2(1) = 1/sqrt(2), g_2(g_2(1)) = 1/sqrt(3), and by induction, if g_2^(k)(1) = 1/sqrt(k+1) [where ^(k) means the function is composed k times], then g_2^(k+1)(1) = g_2(1/sqrt(k+1)) = (1/sqrt(k+1)) / (sqrt( (k+2)/(k+1) ) ) = 1/sqrt(k+2).

so all 1/sqrt(n), n a positive integer, can be reached.

furthermore, we find that all g_1(1/sqrt(n)) = sqrt(n/(n+1)) can be achieved.

Lemma. If sqrt(a/b) is good, so is sqrt(a/(ka+b)) for all nonnegative integers k.

Proof. Notice that g_2(sqrt(a/b)) = sqrt(a/(a+b)); by composing this repeatedly, by induction, we can achieve all sqrt(a/(ka+b)) for nonnegative integer k.

next, we use strong induction on the numerator of sqrt(a/b) to show that all square roots of rationals are good.

We can already construct all sqrt(1/b), so the base case is true.

Assume that for k = 1, 2, ..., n-1, sqrt(k/j) is good for all integers j.

Now, we need only show that all sqrt(n/j) are good, which would complete the induction. If gcd(j,n) = g > 1, then we can cancel g from the top and bottom and get a numerator between 1 and n inclusive, so the statement is true. Hence, we only need to consider cases where gcd(j,n)=1.

Otherwise, let j' be the smallest positive integer so that (j-j' ) is divisible by n; clearly j'<= n, and j' is not n because we don't need to consider when gcd(j,n)>1. Then sqrt(j'/k) is good by the induction hypothesis, so f(sqrt(j'/k)) = sqrt(k/j' ) is also good. Then by the lemma, we can construct all sqrt(k/(j'+k)), so we get all denominators that are j' mod k.

Since we have considered all residues mod k, the denominator can therefore be any positive integer, and the induction is complete.

Writer
Surrealz
Profile Blog Joined May 2010
United States449 Posts
August 18 2010 22:37 GMT
#18
On August 19 2010 07:18 ]343[ wrote:
this is an old usamo problem

and VERY ANNOYING

+ Show Spoiler +

0: already exists

begin with cos 0 = 1.

now, notice that

f(x) = cot(arctan(x)) = 1/x
g_1(x) = cos(arctan(x)) = 1/sqrt(x^2+1)
g_2(x) = sin(arctan(x)) = x/sqrt(x^2+1)
h_1(x) = tan(arccos(x)) = sqrt(1-x^2)/x
h_2(x) = tan(arcsin(x)) = x/sqrt(1-x^2)
h_3(x) sin(arccos(x)) = sqrt(1-x^2)

so we can get any composition of these functions, applied to 1 or 0. call a number that can be reached by a sequence of keypresses "good." If n is good, so is 1/n, since 1/n = f(n). We will show that all numbers x such that x^2 is rational are good; this clearly implies the problem statement.

we only need to show that all rationals q less than 1 (and 1 is good) are good, since every rational greater than 1 is the reciprocal of a rational less than 1.

starting at 1, we find g_2(1) = 1/sqrt(2), g_2(g_2(1)) = 1/sqrt(3), and by induction, if g_2^(k)(1) = 1/sqrt(k+1) [where ^(k) means the function is composed k times], then g_2^(k+1)(1) = g_2(1/sqrt(k+1)) = (1/sqrt(k+1)) / (sqrt( (k+2)/(k+1) ) ) = 1/sqrt(k+2).

so all 1/sqrt(n), n a positive integer, can be reached.

furthermore, we find that all g_1(1/sqrt(n)) = sqrt(n/(n+1)) can be achieved.

Lemma. If sqrt(a/b) is good, so is sqrt(a/(ka+b)) for all nonnegative integers k.

Proof. Notice that g_2(sqrt(a/b)) = sqrt(a/(a+b)); by composing this repeatedly, by induction, we can achieve all sqrt(a/(ka+b)) for nonnegative integer k.

next, we use strong induction on the numerator of sqrt(a/b) to show that all square roots of rationals are good.

We can already construct all sqrt(1/b), so the base case is true.

Assume that for k = 1, 2, ..., n-1, sqrt(k/j) is good for all integers j.

Now, we need only show that all sqrt(n/j) are good, which would complete the induction. If gcd(j,n) = g > 1, then we can cancel g from the top and bottom and get a numerator between 1 and n inclusive, so the statement is true. Hence, we only need to consider cases where gcd(j,n)=1.

Otherwise, let j' be the smallest positive integer so that (j-j' ) is divisible by n; clearly j'<= n, and j' is not n because we don't need to consider when gcd(j,n)>1. Then sqrt(j'/k) is good by the induction hypothesis, so f(sqrt(j'/k)) = sqrt(k/j' ) is also good. Then by the lemma, we can construct all sqrt(k/(j'+k)), so we get all denominators that are j' mod k.

Since we have considered all residues mod k, the denominator can therefore be any positive integer, and the induction is complete.



thank you sir
1a2a3a
palanq
Profile Blog Joined December 2004
United States761 Posts
August 18 2010 22:45 GMT
#19
since you have all the inverse functions, the problem is equivalent to taking any arbitrary rational number ( = a/b for some integers a and b) and turning it into zero. might be a useful approach going at it from the other direction
time flies like an arrow; fruit flies like a banana
TanGeng
Profile Blog Joined January 2009
Sanya12364 Posts
August 18 2010 22:59 GMT
#20
Incomplete
+ Show Spoiler +

method 1:
tan(cot-1(n)) = 1/n
so for any n=1/(a/b) you can also get to n=a/b

method 2:
cos(tan-1(n))) = 1/sq(n^2+1)
combined with method 1
tan(cot-1(cos(tan-1(sq(a)) = sq(a+1)
starting with 0, this pattern allows for all positive integers - but only for integers for now

now for some modulus:
some x where
x mod b is equivalent to a mod b
(x/b can be expressed as x/b+C where C is an integer)
x^2 mod b^2 is equivalent to x^2 mod b^2
also where
x mod b is equivalent to b - (x mod b)
x^2 mod b^2 is equivalent to x^2 mod b^2
this is less important though

in these cases n=x/b can be express as sq(x^2/b^2 + D) where D is some integer
combined with method 2, the conclusion is that given a/b you can get to any n=a/b+C where C is a positive integer

anyways the final piece of the puzzle is continued fractions where we want to express all rational numbers as
1/1/(1/ (a +C)+D) +E)....

I haven't gotten there yet.





Moderator我们是个踏实的赞助商模式俱乐部
1 2 Next All
Please log in or register to reply.
Live Events Refresh
OSC
20:00
Mid Season Playoffs
SHIN vs Bunny
Cham vs MaNa
SKillous vs TBD
PAPI vs Jumy
Gerald vs Moja
ArT vs TBD
SteadfastSC105
Liquipedia
[ Submit Event ]
Live Streams
Refresh
StarCraft 2
IndyStarCraft 232
UpATreeSC 161
goblin 116
SteadfastSC 105
ProTech85
Livibee 79
StarCraft: Brood War
Aegong 73
scan(afreeca) 31
yabsab 17
Dota 2
Gorgc7004
Pyrionflax228
League of Legends
Grubby4140
Dendi1193
Counter-Strike
fl0m2131
pashabiceps645
Foxcn275
sgares46
Heroes of the Storm
Liquid`Hasu576
Khaldor186
Other Games
summit1g5936
FrodaN1912
Beastyqt661
mouzStarbuck416
Sick104
Trikslyr87
Mew2King70
ZombieGrub61
Organizations
StarCraft 2
Blizzard YouTube
StarCraft: Brood War
BSLTrovo
sctven
[ Show 21 non-featured ]
StarCraft 2
• HeavenSC 41
• davetesta28
• sooper7s
• Migwel
• LaughNgamezSOOP
• IndyKCrew
• Kozan
• AfreecaTV YouTube
• intothetv
StarCraft: Brood War
• 80smullet 25
• ZZZeroYoutube
• STPLYoutube
• BSLYoutube
Dota 2
• WagamamaTV644
League of Legends
• Jankos2106
• Doublelift1875
• TFBlade1389
• masondota2493
Other Games
• Scarra1011
• imaqtpie1001
• Shiphtur211
Upcoming Events
Replay Cast
3h 37m
The PondCast
13h 37m
RSL Revival
13h 37m
ByuN vs Classic
Clem vs Cham
WardiTV European League
19h 37m
Replay Cast
1d 3h
RSL Revival
1d 13h
herO vs SHIN
Reynor vs Cure
WardiTV European League
1d 19h
FEL
1d 19h
Korean StarCraft League
2 days
CranKy Ducklings
2 days
[ Show More ]
RSL Revival
2 days
FEL
2 days
Sparkling Tuna Cup
3 days
RSL Revival
3 days
FEL
3 days
BSL: ProLeague
3 days
Dewalt vs Bonyth
Replay Cast
5 days
Replay Cast
5 days
The PondCast
6 days
Liquipedia Results

Completed

Proleague 2025-06-28
HSC XXVII
Heroes 10 EU

Ongoing

JPL Season 2
BSL 2v2 Season 3
BSL Season 20
Acropolis #3
KCM Race Survival 2025 Season 2
CSL 17: 2025 SUMMER
Copa Latinoamericana 4
Championship of Russia 2025
RSL Revival: Season 1
Murky Cup #2
BLAST.tv Austin Major 2025
ESL Impact League Season 7
IEM Dallas 2025
PGL Astana 2025
Asian Champions League '25
BLAST Rivals Spring 2025
MESA Nomadic Masters
CCT Season 2 Global Finals
IEM Melbourne 2025
YaLLa Compass Qatar 2025

Upcoming

CSLPRO Last Chance 2025
CSLPRO Chat StarLAN 3
K-Championship
uThermal 2v2 Main Event
SEL Season 2 Championship
FEL Cracov 2025
Esports World Cup 2025
StarSeries Fall 2025
FISSURE Playground #2
BLAST Open Fall 2025
BLAST Open Fall Qual
Esports World Cup 2025
BLAST Bounty Fall 2025
BLAST Bounty Fall Qual
IEM Cologne 2025
FISSURE Playground #1
TLPD

1. ByuN
2. TY
3. Dark
4. Solar
5. Stats
6. Nerchio
7. sOs
8. soO
9. INnoVation
10. Elazer
1. Rain
2. Flash
3. EffOrt
4. Last
5. Bisu
6. Soulkey
7. Mini
8. Sharp
Sidebar Settings...

Advertising | Privacy Policy | Terms Of Use | Contact Us

Original banner artwork: Jim Warren
The contents of this webpage are copyright © 2025 TLnet. All Rights Reserved.