Give me your math problems

[danex]

This isn’t a real problem exactly, but it’s something I was thinking about. I could probably find the answer pretty easily if I found my old trig notes and/or thought about it for more than 2 seconds, but ima share it anyway. 

You have 2 lines, one with a slope of 1/2, one with a slope of -1/2. They intersect. What is the degree of the angle that the intersection makes?

edit:

Spoiler

Thought about it for more than 2 seconds, pretty sure it’s 135 degrees. 

 

Edited October 28, 2020 by Danex

[FriarFritz]

6 hours ago, Danex said:

You have 2 lines, one with a slope of 1/2, one with a slope of -1/2. They intersect. What is the degree of the angle that the intersection makes?

edit:

  Reveal hidden contents

Thought about it for more than 2 seconds, pretty sure it’s 135 degrees. 

 

You are correct.

[Condensation]

If I have two books and Vapor gives me two more then takes away one, what do I have?

[Experience]

3 hours ago, Condensation said:

If I have two books and Vapor gives me two more then takes away one, what do I have?

An annoying sister.  

[Condensation]

1 minute ago, Experience said:

An annoying sister.  

Nope. Four books and some dispersed Vapor.

[theTruthshaper]

Ooh ooh Math problems!

On 10/28/2020 at 0:59 PM, Danex said:

This isn’t a real problem exactly, but it’s something I was thinking about. I could probably find the answer pretty easily if I found my old trig notes and/or thought about it for more than 2 seconds, but ima share it anyway. 

You have 2 lines, one with a slope of 1/2, one with a slope of -1/2. They intersect. What is the degree of the angle that the intersection makes?

edit:

  Reveal hidden contents

Thought about it for more than 2 seconds, pretty sure it’s 135 degrees. 

 

Spoiler

I am pretty sure that you are wrong:

arctan(1/2) ~ 26.5 degrees

which implies that the angle between the lines will be 57 degrees or 123 degrees.

 

[Experience]

1 hour ago, The_Truthwatcher said:

Ooh ooh Math problems!

  Hide contents

I am pretty sure that you are wrong:

arctan(1/2) ~ 26.5 degrees

which implies that the angle between the lines will be 57 degrees or 123 degrees.

 

Ewwwww. Degrees.  

[theTruthshaper]

4 minutes ago, Experience said:

Ewwwww. Degrees.  

I only used degrees 'cause the original question was in degrees

[FriarFritz]

4 hours ago, The_Truthwatcher said:

Ooh ooh Math problems!

  Reveal hidden contents

I am pretty sure that you are wrong:

arctan(1/2) ~ 26.5 degrees

which implies that the angle between the lines will be 57 degrees or 123 degrees.

 

I stand corrected.

Argh, it's been too long since I've done trig.

[Condensation]

On 11/4/2020 at 5:01 AM, The_Truthwatcher said:

Ooh ooh Math problems!

  Reveal hidden contents

I am pretty sure that you are wrong:

arctan(1/2) ~ 26.5 degrees

which implies that the angle between the lines will be 57 degrees or 123 degrees.

 

*clap clap clap clap clap*

[Narcoleptic Axolotl]

On 10/20/2020 at 3:53 PM, revelryintheart said:

aw dang you don't do statistics

that's the math im in right now hehe

Oh, I hate stats. I tend to like and understand most math, but stats . . . it frightens me.

 

On 10/25/2020 at 8:48 PM, Zelly said:

Welp, I'm fresh out of math problems, but here's some links to a few of my favorite math channels.

Do you like numberphile? Numberphile - Monty Hall

Or vsauce? How to count past infinity

Or vsauce2? The Secret of Snakes and Ladders

Or this series of Ted-Ed animated riddles (some are more mathy than others)? https://www.youtube.com/watch?v=z-ZEfxAL9SI

 

Don't forget 3blue1brown! One of the best calculus explanations available. Khan Academy is generally pretty good too.

[G2F4E6E7E8]

Here's an excellent source of fun problems: https://kskedlaya.org/putnam-archive/.

There are very substantial jumps in difficulty between the different numbers 1 through 6 given to the problems. Therefore don't try go through every problem for a single year and instead look at those across all years in the appropriate range. Some may require a little college-level math knowledge, so feel free to ignore those if you don't have the background. Finally, if you're not used to these kinds of problems, even 1's may be tricky at the beginning---don't be discouraged if it takes you 1 or 2 hours to figure out how to solve them.

2009 A1 is a good place to start and has a really cute solution:

Let f be a real-valued function on the plane such that for every square ABCD in the plane, f(A) + f(B) + f(C) + f(D) = 0. Does it follow that f(P) = 0 for all points P in the plane?

[Condensation]

I enjoy the TedEd riddles, or sometimes I just try to make my own logic puzzles and solve them.

My first year of middle school, I figured out everyone's schedules based on what classes I had with them and which directions they went.

[BloomAgeOne]

In a similar vein to Putnam, I really enjoy doing Olympiad mathematics, mainly because I feel that it really demonstrates the inherent beauty of mathematics, and it just gets really challenging.

Here's the archive of all the IMO (International Mathematics Olympiad) papers and their respective shortlists: https://www.imo-official.org/problems.aspx

The IMO can definitely require some prior knowledge, but there are definitely some accessible problems in there (particularly from earlier years). There's also many other Olympiads at different difficulties, and they're fun to try out sometimes.  

[Condensation]

Oh, the Olympiad! I totally forgot about that.

You can also look up State Math Competitions from previous years.

[BloomAgeOne]

43 minutes ago, Condensation said:

You can also look up State Math Competitions from previous years.

Nice! I haven't seen any of the State Math Competitions, probably since I don't live in the US myself. I have done a few problems from the USAMO, USAJMO, and various TSTSTs and TSTs though. Some of them get really hard. The US seems to have a limitless number of mathematics competitions going on.

[Condensation]

1 minute ago, BloomAgeOne said:

Nice! I haven't seen any of the State Math Competitions, probably since I don't live in the US myself. I have done a few problems from the USAMO, USAJMO, and various TSTSTs and TSTs though. Some of them get really hard. The US seems to have a limitless number of mathematics competitions going on.

Oh, cool! Yes, we have many math competitions. My school does a lot of them.

[Zelly]

I don't know if this counts as a math or a riddle or a bit of both, but.....

 

What's the pattern/rule between these 2 chains:

43, 10, 3, 5, 4, 4...

611, 16, 7, 5, 4, 4...

[Dunkum]

On 12/29/2020 at 7:42 PM, Zelly said:

I don't know if this counts as a math or a riddle or a bit of both, but.....

 

What's the pattern/rule between these 2 chains:

43, 10, 3, 5, 4, 4...

611, 16, 7, 5, 4, 4...

Number of letters in the english spelling of the prior number

[AquaRegia]

Card probability problem from real life: my wife and I are playing a variant of Pinochle.  4 players, 48 card deck, 9-10-J-Q-K-A only in 4 suits, so two of each card.  Each player holds 12 cards and bidding team pass 3 cards to each other before play begins.

If I need one specific card from my partner (say, an ace of clubs), what is the probability they will have it?

My thoughts: I already know the card I need is NOT in the 12 I have, so it must be in one of the other 3 hands.  P=1/3 that it is in my partner's hand, yes?  That means P=2/3 that it is in one of my opponents'.  But there are TWO aces of clubs, and I only need one.  So the only BAD outcome is if BOTH of them are in my opponents' hands.

Pbad = (2/3)(2/3) = 4/9 = 44.4%, therefore there is a 1 - 4/9 = 55.6% chance my partner has at least one of the aces I need.

Is this the correct solution?

And how would I extend this work to cover the case where I need TWO cards (say, K and Q of clubs)?  Is it just (5/9)(5/9) = 25/81?

[Condensation]

On 06/02/2021 at 5:17 PM, AquaRegia said:

Card probability problem from real life: my wife and I are playing a variant of Pinochle.  4 players, 48 card deck, 9-10-J-Q-K-A only in 4 suits, so two of each card.  Each player holds 12 cards and bidding team pass 3 cards to each other before play begins.

If I need one specific card from my partner (say, an ace of clubs), what is the probability they will have it?

My thoughts: I already know the card I need is NOT in the 12 I have, so it must be in one of the other 3 hands.  P=1/3 that it is in my partner's hand, yes?  That means P=2/3 that it is in one of my opponents'.  But there are TWO aces of clubs, and I only need one.  So the only BAD outcome is if BOTH of them are in my opponents' hands.

Pbad = (2/3)(2/3) = 4/9 = 44.4%, therefore there is a 1 - 4/9 = 55.6% chance my partner has at least one of the aces I need.

Is this the correct solution?

And how would I extend this work to cover the case where I need TWO cards (say, K and Q of clubs)?  Is it just (5/9)(5/9) = 25/81?

I think this is the right solution. You'd have to make sure by asking someone else, but I believe this is it. And yes, it is just (5/9)(5/9) = 25/81.

[Dunkum]

it seem reasonable, but i'm flipping back an forth on the two card case whether the fact that for the second card the size of the remaining hand is smaller matters/  eg for the 1 card case it is 12 cards each for opponents and partner.  but in t 2 card case, once you've got1 card its now 12 cards each for opponents vs 11 remaining cards in the partners hand, and I don't now if that affects anything or not.

[Condensation]

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15           May 16           May 19

June 17           June 18

July 14            July 16

August 14       August 15       August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard doesn't not know too.

Bernard: At first I don't know when Cheryl's birthday is, but I know now.

Albert: Then I also know when Cheryl's birthday is.

So when is Cheryl's birthday?

[Dunkum]

2 hours ago, Condensation said:

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15           May 16           May 19

June 17           June 18

July 14            July 16

August 14       August 15       August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard doesn't not know too.

Bernard: At first I don't know when Cheryl's birthday is, but I know now.

Albert: Then I also know when Cheryl's birthday is.

So when is Cheryl's birthday?

Spoiler

July 16th:

since Albert knows the month, and he knows that Bernard, knowing the day, doesn't know the full date, then that rules out the month of May, since the 19th is unique, and Bernard could know the date; and it rules out June, since the 18th is likewise unique.
Bernard, on hearing that is able to determine that the month must be either July or August, and he now knows the answer.  that rules out the 14th as the day, since that is shared between them and he wouldn't be able to know for sure which one it was if the 14th were the day
since this information is enough for Albert to determine the answer, that means he birthday must b July 16, since the other 2 remaining options a both in the same month and so Albert, knowing only the month, wouldn't be able to pick between them and so wouldn't know the answer

 

[Condensation]

*applause*

That is correct.