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Mathematics marathon

Rinfiyks

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That looks right to me! I can't remember the answer exactly but that sounds familiar. I think I see where I went wrong in this now.

If you know how to expand (a+b)^3, you can substitute to expand (a+b+c)^3:
\( (a+b+c)^3 \)
Let \( b + c = d \)
\( (a+d)^3 \)
\( = a^3 + d^3 + 3(a^2d + d^2a) \)
Then substitute d out and simplify.
\( = a^3 + (b+c)^3 + 3(a^2(b+c) + (b+c)^2a \)
\( = a^3 + (b^3 + c^3 + 3(b^2c + c^2b)) + 3(a^2b+a^2c + (b^2+c^2 + 2bc)a) \)
\( = a^3 + b^3 + c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
Lots of places to make a mistake though (especially when typing it up, which I did, and took me a while to find it :p)
 
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Rinfiyks

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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

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Hehe.

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

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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
 
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Rinfiyks

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The diagram shows a square with side length 1. It's divided into 4 rectangles which all have the same area. Find x.
6f9o2v.png
 

MaeLSTRoM

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Got It!
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

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Interestingly I got the same thing using a different method:
x1hgnt.png
 

Rinfiyks

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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?
fuc3kg.png
 

Rinfiyks

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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.
 

qqwref

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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?
fuc3kg.png

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

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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

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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 \( lim_{P \to O} b. \)

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

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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

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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)
 

qqwref

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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?
 
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