Math Problems and Teasers

[Cipher X]

This whole section is kinda getting boring so i thought hey why not have some math problems that are really freaky and somewhat hard.

1) If x > ½, find the simplest radical form expression for (1+sqrt(2x-1)) / (sqrt( x + sqrt (2x-1)))

IT might be easier to write it out on paper seeing that i couldnt put in the squareroot symbol oh btw sqrt = squareroot

Good Luck

[Cipher X]

The Length of each side of a certain right triangle is teh reciprocal of a different integer. What is the least possible sum of these three integers?

This one has got me stumped so Good Luck!

[Cipher X]

Together Al, Barb, Cal, Di, and Ed earned a total of $150, but in unequal amounts. In order to equalize their earnings exactly, Barb gave half (1/2) her earnings to Al. Next Cal gave 1/3 of his earnings to Barb. Then Di gave 1/4 of her earnings to Cal. Finally Ed gave 1/6 of his earnings to Di. How many dollars did Al earn before equalization?

This one just a little bit of thinking and you shoudl be able to get it

Have Fun

Cipher X

[rbayer]

Originally posted by Cipher X
The Length of each side of a certain right triangle is teh reciprocal of a different integer. What is the least possible sum of these three integers?

This one has got me stumped so Good Luck!

15, 20, and 12 will give you 47(:D). I'm not sure how you would do this with normal math stuff, but it's a great computational problem. If anyone has some kind of cool discrete math trick for stuff like this, I'd definately be interested in seeing it.


The Al, Barb, Cal, Di, Ed one isn't too hard:
Al had $11.


The first one is a little too much thinking for me at this time of night, and I assume the point would be that symbolic calculators are strictly forbidden. Maybe tomorrow during math I'll crunch through it...

[Richard Wallace]

sqrt(2)

[Gadget470]

Ok, first off. I'm stupid. It's late and I've done too much math and physics today. I was about to refute "Richard" .. as I typed it I realised he was right and that sqrt() IS radical. .. I've been sitting here working out a RATIONAL..

I'm not going to bother reasoning out the other two tonight. I think I'm gonna sleep tonight.

w00t for sl33p. b00 for th1n|<

[Caleb Fulton]

I'm tired/lazy, so I wrote a program on my calculator and I'll check it in the morning :)

[Cipher X]

Originally posted by Richard
sqrt(2)

Yeah the answer to that one is Sqrt(2). the only reason i know is because i put into my Ti-83 and had an epiphany that all the answers were sqrt(2). But if anybody knows how to solve it using algebra or some wierd rule in math that nobody knows about. I wanna know how to solve it !! lol



And as far as i can see the Right triangle problem. the only equation that i can come up with that might be worthwhile is (1/a)^2 + (1/b)^2 = (1/c)^2 where a,b and c are integers. Since it says that the numbers have to be reciprocal of an integer

Cipher X

[Richard Wallace]

square the original expression

this gives 1 + 2*sqrt(2x-1) + 2x - 1 in the numerator

and x + sqrt(2x-1) in the denominator

(note the ones cancel in the numerator)

so the square is 2

so the original expression is sqrt(2)

qed

[Cipher X]

Hey thanks richard,
i can finally rest. omg i feel like an idiot i squared it but i didnt see that teh ones cancel and yeah......
but anyways here is another one

What are all the ordered pairs of real numbers (x,y) for which

y^(x^2 - 7x + 12) = 1 and x + y = 6


pretty easy

[Yan Wang]

Take a weird shaped field 100'x26'. At one end is a goal 10' wide centered in the field's width (8' on each side). If you were only allowed to shoot at the goal from the sideline, you would want an optimal angle... hence, what would be that distance down the sideline and what would be the corresponding pheta degree value to 2 decimal places?

It's a challenge to do this without the first derivative of arctan but with a geometric method.

Note that the field is not to scale.... and that a geometric answer does not mean going into CAD and constraining a field of the following dimensions.

[AJ Quick]

Hey! Why not do my math homework!?!

I know I could do it, but it would really cut into my sitting around time.

[Richard Wallace]

Monsieurcoffee, thanks for a very engaging problem.

I changed ISPs this weekend, so my connection was down until now. This gave me some time to think of several solutions.

The obvious one is to express theta in terms of x using the transcendental arctan function, differentiate and solve for the maximum theta.

A little less obvious is to use the pythagorean theorem and the law of sines to express sin(theta) as an algebraic function of x, differentiate and solve for the maximum sin(theta).

More elegant is the following:

Let the near corner of the field be called O, the desired point on the sideline A, the near goalpost B, and the far goalpost C.

Let the angle ACB be called alpha, and the angle BAO be called beta. Note that (in degrees) the desired angle theta = 90 - alpha - beta.

Further note that as x increases, alpha increases and beta decreases.

Conclusion: theta is largest when alpha = beta.

Now express alpha and beta as:

alpha = artan(x/18), beta = arctan(8/x)

So theta is largest when x/18 = 8/x.

The solution is x = 12.

The corresponding value of theta is 22.62 degrees.

[Richard Wallace]

OK, it is my turn to pose a problem --

A regular polygon having N sides is circumscribed on a circle having unit radius. [Note: all sides of a regular polygon are equal, and all of its angles are equal. 'Circumscribed' means that the midpoints of each of the polygon's sides are tangent to the circle.]

As a function of N, find an expression for the fraction of the polygon's area that lies outside the circle. In other words, what fraction of the polygon would have to be removed to leave the circle?

Added challenge (ala Monsieurcoffee's previous problem): show a derivation of your expression that does not make use of transcendental functions such as sine, cosine, tangent, etc.

[Yan Wang]

Originally posted by Richard
More elegant is the following:

Let the near corner of the field be called O, the desired point on the sideline A, the near goalpost B, and the far goalpost C.

Let the angle ACB be called alpha, and the angle BAO be called beta. Note that (in degrees) the desired angle theta = 90 - alpha - beta.

Further note that as x increases, alpha increases and beta decreases.

Conclusion: theta is largest when alpha = beta.

Now express alpha and beta as:

alpha = artan(x/18), beta = arctan(8/x)

So theta is largest when x/18 = 8/x.

The solution is x = 12.

The corresponding value of theta is 22.62 degrees.

That is one solution... however, if you want elegant, check this out:

Draw a circle radius 13 within the rectangular field.
Translate it to the right so that the goalposts lie on the circle.
The goal = 10, the section below = 8
The missing part = x

From a theorum about secants/tangents, the part*(part+whole)=other part*(other part + whole)

so... 8(8+18)=x*x or 8(18)=x^2 which simplifies to 12.

[Richard Wallace]

Monsieurcoffee, I like your big circle solution. Can you provide a reference to the 'theorem about secants and tangents' on which it is based?

Revising my previous post to remove all reference to transcendental functions yields a solution that is more purely geometric:

Let the near corner of the field be called O, the desired point on the sideline A, the near goalpost B, and the far goalpost C.

Let the angle ACB be called alpha, and the angle BAO be called beta. Note that (in degrees) the desired angle theta = 90 - alpha - beta.

Further note that as x increases, alpha increases and beta decreases.

Conclusion: theta is largest when alpha = beta.

So theta is largest when triangles ACO and BAO are similar.

So theta is largest when x/18 = 8/x.

The solution is x = 12.

[Yan Wang]

Um... looking around. It's in the Course III math textbook here at Ithaca High School in the circles section. "McDougal, Littell Integrated Mathematics Course 3". It's Canadian.

Rather than using that theorum, after drawing that circle, you could have drawn radii of 13 to each goal post, forming a triangle, 13,10,13... half of which is 13,5,y... y=12. You can justify that y=x because x is greatest at the point of tangency of the circle and also that x and y are parallel and some other things which make them equal. I like your proof very much because it justifies things clearly and easily.

[Yan Wang]

Um, decided to draw it up from the book:

[jon virgi]

wow, great problems, TOO BAD THEY ARE STRAIGHT FROM THE PA MATH LEAGUE. wow, at least give them credit.

[Yan Wang]

I don't know what the PA Math League is (though now I guess I do since the title's pretty self explanatory)...

Anyway, I do the Mandelbrot math competition (national thing) and the NY Math League (NYML). That problem I had was from extra credit problems in math...

Anyway, I'll find some more :)

[Kevin Watson]

This is one of my favorite enigmas:

Three guys walk into a hotel. They pay ten dollars apiece for their room, which costs thirty dollars a night. After the three guys get to their room, the manager realizes that he over charged them five dollars, so he sends the bellhop up with the extra five dollars to repay them. On his way, the bellhop decides that he will give a dollar to each of the men and keep two dollars as a tip because he can't figure out how to divide five dollars between the three guys.

Here's the real problem: Since each guy got one dollar back, each only paid nine dollars apiece for the room or twenty-seven dollars altogether. When you add the two dollars the bellhop kept, you get twenty-nine dollars. Where'd the last dollar go?

-Kevin

[srjjs]

It didn't go anywhere. The hotel has 25 dollars, the bellhop has 2, and each of the 3 friends has 1. Your phrasing is just a little off.

[Kevin Watson]

Originally posted by srjjs
It didn't go anywhere. The hotel has 25 dollars, the bellhop has 2, and each of the 3 friends has 1. Your phrasing is just a little off.

Not much wrong with the phrasing (grammer, yes :)). It's designed to be deceptive. It's one of those puzzles that you either get immediately (wondering what the fuss is all about), or spend the next couple of hours wondering if you need non-Euclidean geometry to solve it.

This problem and many others can be found on a great web site devoted to "hardcore tech-interview style riddles":

http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml

-Kevin

[JJG13]

Here's my all time favorite math problem. It's really easy if you know the trick.

What is the sum of the first two numbers that are perfect squares, cubes, and fourth powers.

[Greg Ross]

Originally posted by JJG13
What is the sum of the first two numbers that are perfect squares, cubes, and fourth powers.
1?

[JJG13]

Originally posted by gwross
1?

That's the first perfect square, cube, and fourth, but whats the second?

[Richard Wallace]

4097 = 2^12 + 1

[JJG13]

Correct.

[Greg Ross]

Originally posted by JJG13
That's the first perfect square, cube, and fourth, but whats the second?
No! 1 is the second, and 0 is the first. So 1 is also the sum!:D

[tatsak42]

I didn't think 0 counted.

Here's a problem from a competition i just did:
abcd is a 4 digit number between 1000 and 9999. When you multiply a 3 digit number (efg) and a 2 digit number (hi) and a 1 digit number (j) you get abcd. a,b,c,d,e,f,g,h,i, and j use the numbers 0-9 once and only once. What are the two combinations that work?
abcd=efg*hi*j

[Gobiner]

The NSA sponsors a great math competition that ISN'T your basic sit down do math in x minutes type contest. You get a whole month to brood over 5 excellent problems, 4 times a year. I suggest you just go there are read all 345 posted problems. http://www.nsa.gov/programs/mepp/usamts.html

My own problem: What is the area of land that can be seen from a watch tower if it is 50 feet high? Assume a spherical Earth.
Answers that do not use calculus:yikes:deserve a pat on the back. Yes, the answers, not the person.

[Katy]

there is a cylinder that is a gas tank in a car that lays on its side (so if it wasn't held down it would roll back and forth). The meter is broken so you can't tell how much is in there but you do know if you stick a stick in through a hole in the top where it is usually filled and the tank is half way full the stick will be coated a quarter of the way.

How high is the stick coated when you only have one fourth of a tank of gas?

[Hailfire]

Here's one, what are the four magic cubes? Figure this out without using a computer program. Hint: If you don't know what a magic cube is, 153 is one of them because 1^3=1, 5^3=125, and 3^3=27. So 1+125+27=153. Well, there's one for you, find the other three!

[JasonStern]

Originally posted by Gobiner
The NSA sponsors a great math competition that ISN'T your basic sit down do math in x minutes type contest. You get a whole month to brood over 5 excellent problems, 4 times a year. I suggest you just go there are read all 345 posted problems. http://www.nsa.gov/programs/mepp/usamts.html

My own problem: What is the area of land that can be seen from a watch tower if it is 50 feet high? Assume a spherical Earth.
Answers that do not use calculus:yikes:deserve a pat on the back. Yes, the answers, not the person.

Nice problem, but not too bad. The maximum distance a person can see from the watch tower is a line tangent to the earth, which means this line is PI/2 rads in relation to the center of the earth. This gives a right triangle with a hypotenuse of R+50 (R= earth's radius), and one side of length R .

To get the angle between these to lines take the arc-cosine of (50+R)/R = arccos(1 + 50/R)= theta.

Now, the length of an arc equals theta times radius, so the length of the arc visible by the person in the watch tower is R*arccos(1+50/R).

Of course, this isn't the area, as requested. to get that, we imagine this is a straight line and use the area of a circle = PI* r^2. Therefore the total area is PI*R^2*arccos(1 + 50/R) ^2!!

I'm feeling lazy right now. can someone please plug in the value of R and see what that comes out to? Thanks!

[rbayer]

Ummm.... I don't think that works. First, I believe you actually mean theta=acos[R/(R+50)], since cos(theta)=a/h.

Second, I'm not sure where that area formula came from, but I don't believe that pretending like it's a straight line is valid without the use of some kind of fudge factor (my professor's words, not mine) like a Jacobian.

Anyways, using a bit of Calc (sorry, I don't know the formula for area of a partial sphere of the top of my head :D), I came up with:

A = 2*pi*r^2*(1-cos(phi)), where phi is the angle in spherical coordinates that defines where the line of sight and the sphere intersect.

Plugging in r=R and phi=acos[R/(R+50)], I get:

A=2*pi*R^2*[50/(R+50)]

I know you said a calc-free answer was best, but what really defines calc-free? After all, our good old friend A=4*pi*r^2 comes from calc, as does just about every other area/volume formula. If anyone does come up with a solution that doesn't somehow use any form of calc, they (and the solution!) definately deserve a pat on the back.

--Rob

[Gene F]

Here's one our team has a lot of fun with.

A, B and C are digits. Find the solution to the following

A+B^C = ABC

:)

[David Hoff]

Originally posted by Gene F

A+B^C = ABC

:)

This works

1+1^2 = 2

[tatsak42]

I'm thinking A, B, C, are different digits, and ABC is a 3 digit number...

And you want 7+3^6=736

[JasonStern]

Thanks rob! you are right, it should be arcos[R/(R+50)]. Also, looks like the "straighten it out" part doesnt work either. sorry about that everyone!!! I have got to stop doing math at 3am........

[Gene F]

Originally posted by tatsak42
I'm thinking A, B, C, are different digits, and ABC is a 3 digit number...

And you want 7+3^6=736

tatsak42 you are correct! Well done.
How about the form (A+B)^C=ABC.

:)

[odin892]

Originally posted by Cipher X
Hey thanks richard,
i can finally rest. omg i feel like an idiot i squared it but i didnt see that teh ones cancel and yeah......
but anyways here is another one

What are all the ordered pairs of real numbers (x,y) for which

y^(x^2 - 7x + 12) = 1 and x + y = 6


pretty easy

for (x,y)
(2,4) and (3,3)

simply factor (x^2 - 7x + 12) = 1 so that x must = 2 or 3 to make the expression always 0 so what ever y = the whole thing becomes 1. Fill in 3 and 4 for y.

[rbayer]

Originally posted by odin892
for (x,y)
(2,4) and (3,3)

simply factor (x^2 - 7x + 12) = 1 so that x must = 2 or 3 to make the expression always 0 so what ever y = the whole thing becomes 1. Fill in 3 and 4 for y.

Right idea, but i think you may want to double-check your factoring. (x^2 - 7x + 12) = 0 will factor to (x-3)(x-4)=0, giving x=3 or x=4 as solutions. Also, don't forget that if y=1, then the whole thing is 1 regardless of the exponent. Therefore, the solutions I come up with are:

(3, 3), (4, 2), (5, 1)

--Rob

[RBrandy]

Not sure if this is a good problem or not...

1, 2, 6, 42... what is the nex number in the pattern?

[Gene F]

Originally posted by RBrandy
Not sure if this is a good problem or not...

1, 2, 6, 42... what is the nex number in the pattern?

1806

3263442

After that my calculator can't keep track of the digits.

[doy]

tatsak42 you are correct! Well done.
How about the form (A+B)^C=ABC.

:)
(3+4)^3 = 343

[Gene F]

Thanks's doy, you are correct.

Make sure to watch our web page, www.metalinmotion.com, for our upcoming question of the day. It's a competition our team runs on our site with prizes awarded at Nationals. :yikes:

[Mike Rush]

OK, it is my turn to pose a problem --

A regular polygon having N sides is circumscribed on a circle having unit radius. [Note: all sides of a regular polygon are equal, and all of its angles are equal. 'Circumscribed' means that the midpoints of each of the polygon's sides are tangent to the circle.]

As a function of N, find an expression for the fraction of the polygon's area that lies outside the circle. In other words, what fraction of the polygon would have to be removed to leave the circle?

Added challenge (ala Monsieurcoffee's previous problem): show a derivation of your expression that does not make use of transcendental functions such as sine, cosine, tangent, etc.

A= nTan(360/2n) - Pi