Mathematics marathon

[Rinfiyks]

I have another problem!
There are two poles in the ground. They are both vertical. The height of both poles (from the ground) is 100 metres. A 50 metre rope is attatched to the top of both poles at either end. It is allowed to hang freely. The rope's closest point to the ground is 75 metres from the ground. What is the distance between the poles?

 

[qqwref]

Hehe.

Spoiler

In general this would be a pretty hard problem. But we see that the rope's lowest point (75m elevated) is 25m from its highest point (100m elevated), and its length is only twice that or 50m, so it must be completely doubled up. The poles are touching.

Here's one: If we took a solid regular dodecahedron, and cut it (using only planar cuts that go all the way through the dodecahedron) like a Megaminx, how many pieces would there be? How about a Gigaminx? Pyraminx Crystal? Starminx? Pentultimate?

 

[Sa967St]

2.
A 75 year old has a 50% probability of living for another 10 years.
A 75 year old has a 20% probability of living for another 15 years.
An 80 year old has a 10% probability of living for another 10 years.
What is the probability that an 80 year old will live for another 5 years?

Bump.

I got the numbers in the question wrong. I just did it from memory, which is why people's answers weren't making any sense. :/

The original question was:
A 75 year old person has a 50% chance of living at least another 10 years.
A 75 year old person has a 20% chance of living at least another 15 years.
An 80 year old person has a 25% chance of living at least another 10 years.
What is the probability that an 80 year old person will live at least another 5
years? (question 7a)

And the answer: 62.5,clicky

 

[Rinfiyks]

The diagram shows a square with side length 1. It's divided into 4 rectangles which all have the same area. Find x.

 

[MaeLSTRoM]

Got It!

Spoiler

Area of each rectangle = 0.25
height of bottom rectangle = 0.25 (1*0.25)
there fore remaining area = 0.75
Left rectangle is full height, so width is 1/3 (0.25/0.75)
Width 2 rectangles on right is 2/3
x=2/3 (same as width of rectangles on right)

 

[Igora]

Interestingly I got the same thing using a different method:

Spoiler

 

[Rinfiyks]

Well done now for a more difficult one.

The diagram shows two semicircles inside a square of side length 1. The common centre of both semicircles lies on a diagonal of the square.
What is the shaded area?

 

[qqwref]

I don't get it, what is stopping me from constructing a circle (cut in half) inside the square and saying that circle satisfies the problem? Are there any added constraints you didn't mention?

 

[Rinfiyks]

I don't get it, what is stopping me from constructing a circle (cut in half) inside the square and saying that circle satisfies the problem? Are there any added constraints you didn't mention?

The large semicircle must have its corners touching two adjacent edges of the square, and its arc must also touch two adjacent edges of the square. There's only one way to acheive this (that I can see.) The smaller one fits snugly in the remaining gap. Sorry for not making it clear.

 

[samehsameh]

Spoiler

Area =3Pi/2 * ( 3 - 2sqrt(2) )
or 0.8085 if u prefer numerical values

 

[qqwref]

The diagram shows two semicircles inside a square of side length 1. The common centre of both semicircles lies on a diagonal of the square.
What is the shaded area?

Spoiler

Call the radii of the two circles R (bigger) and r (smaller). Also, let A be the common center, B be the tangent point between the smaller semicircle and top line, and C be the intersection point of the larger semicircle's top corner with the top edge. Let D be the tangent point between the larger semicircle and the bottom line.

Now, a circle tangent to a horizontal line has its center directly above its point of tangency. This means that B, A, and D are on the same vertical line (and thus BD=1); since AB=r and AD=R, we have R+r=1. This also means that the angle ABC is a right angle. Now, by symmetry across the top-left/bottom-right diagonal, angle BCA must be a 45 degree angle. So triangle ABC is a right isosceles triangle. But AC=R (and again AB=r); this means R = r * sqrt(2).

So we have R+r=1 and R = r * sqrt(2). We can solve to find that r = 1/(sqrt(2)+1) = sqrt(2)-1 and R = 1-r = 2-sqrt(2). The shaded area is 1/2 pi r^2 + 1/2 pi R^2 = 1/2 pi (9-6sqrt(2)) ~= 0.80852.

 

[uberCuber]

Let \( P(a, a^2) \) be a point on the parabola \( y = x^2, a>0 \). Let \( O \) be the origin and \( (0,b) \) the y-intercept of the perpendicular bisector of line segment \( OP \). Find \( lim_{P \to O} b. \)

 

[Akash Rupela]

Let \( P(a, a^2) \) be a point on the parabola \( y = x^2, a>0 \). Let \( O \) be the origin and \( (0,b) \) the y-intercept of the perpendicular bisector of line segment \( OP \). Find \( lim_{P \to O} b. \)

Spoiler

Slope of OP= (a^2-0)/(a-0)=a
Thus slope of perpendicular bisector = -1/a
Also perpendicular bisector passes through midpoint of OP , i.e (a/2 , a^2/2)
so equation of perpendicular bisector is (y-a^2/2)/(x-a/2)=-1/a
Put x=0 in this to get y intercept b=(1+a^2)/2
as P tends to O, a tends to 0
Thus limit= 1/2

 

[qqwref]

Bump. I've been having trouble with this one, maybe someone can solve it... It doesn't originally exist in this wording, but I rewrote it to make it feel more like a math word problem.

There are C cabinets, each with D initially empty drawers in them. You place T things (0<T<C*D) into the drawers, randomly, but such that no drawer has more than one thing in it. What is the chance that exactly one cabinet is completely empty of things? Two cabinets? N cabinets?

 

[Akash Rupela]

Well, i dont really understand the question completely, is it like C and D are constants while T is variable which takes all values from 1 to CD-1, and due to random placing, all are equally likely . Or is T also fixed(then i dont make sense of this)?

Also is the answer 1/(CD-1) for all parts ? or maybe (CDcn)/2^(CD) ? where CDcn=CD!/(n! * (CD-n)! )
(i can see this problem in multiple ways, so trying to put all possible answers i can think of, if any is correct, i will elaborate)

 

[kprox1994]

A math thread? This isn't school! I have a question, if the answer to life, the universe and everything is 42; then what is the answer to death, the antiuniverse, and nothing?

 

[Akash Rupela]

24? or maybe 1/42?

 

[qqwref]

Well, i dont really understand the question completely, is it like C and D are constants while T is variable which takes all values from 1 to CD-1, and due to random placing, all are equally likely . Or is T also fixed(then i dont make sense of this)?

C, D, and T are all constants, although you don't know what they are. All possible placements of the T things in the drawers are equally likely.

Also is the answer 1/(CD-1) for all parts ? or maybe (CDcn)/2^(CD) ? where CDcn=CD!/(n! * (CD-n)! )

I don't have the answer, so what's your logic behind these guesses?

 

[Akash Rupela]

Oh i m so sorry i misread it , i read empty cabinets as drawers
so now i have T drawers full while cd-T drawers empty (i wil use lowercase c here to avoid confusion)
in 1st case i want 1 cabinet empty, means i have to distribute T objects among (cd-d) drawers.
for number of ways in which this can happen, i solve the integer equation, x1+x2+x3......x(cd-d)=t where each of the cd-d drawers have value 0 or 1
this can happen if i give a value of 1 to any random t drawers, and 0 to cd-d-t drawers
So i select those drawers with value=1 in (cd-d)Ct ways .
But we can also select the cabinet which will be empty in c ways
Also the total number of ways of selecting t drawers from cd drawers is equal to (cd)Ct
So answer of the 1st part should be c* (cd-d)Ct / (cd)Ct

Now for the general expression, first i select which n cabinets will be empty in cCn ways( in 1st part cC1= c)
then from the remaining (cd-nd) drawers, i select any t drawers which will be full while others empty, in (cd-nd)Ct ways
and total number of ways will still remain the same
so answer of N cabinets part should be cCn (cd-nd)Ct/ (cd)Ct = (if i simplify this , it will become too dirty without equation editor)

Put n=2 to get answer of 2nd part

I hope this is right

 

[qqwref]

So i select those drawers with value=1 in (cd-d)Ct ways .
But we can also select the cabinet which will be empty in c ways
Also the total number of ways of selecting t drawers from cd drawers is equal to (cd)Ct
So answer of the 1st part should be c* (cd-d)Ct / (cd)Ct

You can't just do this though, because you are not taking into account whether other cabinets will be empty or not. If exactly one cabinet is empty, you need to make sure the other ones are NOT empty.

As an example, let's have C=5, D=5, T=5. Your equation gives 5*(20 choose 5)/(25 choose 5) = 1.46, which is clearly wrong because probabilities can't be more than 1.