SUPER-FUN-HAPPY-MATHS-TIME

[Flaming_Arrow]

i have no idea how to any of these questions lol

[/0]

9.) If you break a stick into 3 pieces what is the probability that the 3 pieces can form a triangle?
(Source: Neobeo)

Let the stick of length L be broken into fragments of length x, y and z.
Since the problem is continuous probability, it's best to use a graph or something like that.
Hence, all possible breaks lie on the plane , with .



When the above graph is drawn, the area is represented by an equilateral triangle, and it is

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Now adding the restrictions

Since , this is the same as saying , or , and by symmetry, and .

So the allowable area is , with . This area is a mini-triangle within the larger triangle of the previous graph. The vertices (open points) of this triangle are , .



So the triangle is equilateral, and
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So the probability is

fun problem

[hard]

oh my god! i can't even draw 1/squareroot of 3-x

[TrueTears]

Had to do a bit of chinese text book reading and learning some new things for this question but anyways here goes:

First just some derivations:

Thus

Back to the question:

Treating the x in the question as a constant



For the definite integral with lower limit q and upper limit p:

To get the definite integral for this question let p = 1 and q = 0 , this yields:



BUT

subbing this in leads to a double integral namely:

after changing the order of integration we have:



So now we are integrating with respect to x first thus we treat t as a constant



Good learning experience for me this question.

[Damo17]

10.) Show (without calculus) that the minimum value of is .

Using the theorem:
If n positive functions have a fixed product, there sum is a minimum if it can be arranged that the functions are equal.
 

let

Although the three terms of the sum have a constant product, we run into trouble if we apply the aforementioned theorem as there is no value of x that satisfies these equations.

But if we set aside the constant, , and minimise the sum to we are able to solve the question.

The minimum comes by equating:

 



We take the positive solution as per the restriction of the question.

the minimum of is

(using the fact that if then the min value of the sum is therefore and adapting that to  gives .)

[dcc]

8.) Evaluate
(Source: Wikipedia)

.

[Over9000]

q12

Prove is irrational
let x be a rational number
let , where a and b are relatively prime integers
let



A rational result for this is impossible, therefore is irrational

[dcc]

13.) Show that .
(Source: Damo17)

[Over9000]

prove




Divide both sides by 9

[Over9000]

so N=1111.......1111 and so on with 2008 recurrences of the digit 1
We must find the 1005th digit after the decimal point in the expansion





so we begin to see a trend where an even amount of digits of 1 are of the form 3.31..... and each time two more digits of 1 are added there is a 3 added to left side of decimal point and a 3 added to right side of decimal point.
so for example, a term with 8 digits of 1 square rooted i.e has 4 3's before the decimal place and 4 3's after it followed by a 1 then 6 and so on.
so for an integer with 2008 digits of 1, there are 1004 3's before the decimal point and 1004 3's after it. We are looking for the 1005th term and as we can see in the trend, the last 3 is always followed up by a 1, therefore the 1005th term after the decimal point in an integer with 2008 digits of 1 using expansion , is 1

This isnt a very alegbraic way of doing this question but I just used patterns

[kamil9876]

Here is one i came up with in church today:

9.) If you break a stick into 3 pieces what is the probability that the 3 pieces can form a triangle?

If the stick has length L. And the three pieces have length a,b and c and c is the longest one then:






Say L=1 unit  for simplicity.

So basically we have the segment [0,1] and we must choose 2 points such that no segment has a length greater than 0.5. Say the first point we choose is in the first half. That means that the next point must be in the second half because if they would be both in the same half then we would get a piece greater than or equal to 0.5.

Example: say the first point is at point 0.2. that means the next point must be in (0.5,0.7).

Generally: if the first point is at x (and in the first half) then the next point must be in (0.5,x+0.5)

Let's turn this into a discrete probability problem

We will do this by splitting the stick up into n segments. In other words say the only possible points we can pick are 1/n, 2/n, 3/n.... n/n=1

Say we pick the kth point, in other words, x=k/n. using an analog of the general principle shown earlier, we can show that the next point must be in between the (n/2 +1)th and  (n/2 + k)th point (inclusive). (assume k is in the first half)
So that means that if we the kth point, the probability that we pick a valid point is k/n. The probability that we pick the kth point is 1/n. That means that the probability that we pick the kth point followed by a valid point is . Now we want the sum of the probabilities. So we want to add up k=1,k=2,k=3... all the way up to k=n/2 (since we are only considering the first half).

That means the probabilty of picking a point in the first half, followed by a valid point is:







Now we want a continous situation so we let n approach infinity.

This gives a probability of 1/8. However now we have to double that to take the second half into account so it is 1/4.

And well done \0 on ur shorter solution

[/0]

so N=1111.......1111 and so on with 2008 recurrences of the digit 1
We must find the 1005th digit after the decimal point in the expansion





so we begin to see a trend where an even amount of digits of 1 are of the form 3.31..... and each time two more digits of 1 are added there is a 3 added to left side of decimal point and a 3 added to right side of decimal point.
so for example, a term with 8 digits of 1 square rooted i.e has 4 3's before the decimal place and 4 3's after it followed by a 1 then 6 and so on.
so for an integer with 2008 digits of 1, there are 1004 3's before the decimal point and 1004 3's after it. We are looking for the 1005th term and as we can see in the trend, the last 3 is always followed up by a 1, therefore the 1005th term after the decimal point in an integer with 2008 digits of 1 using expansion , is 1

This isnt a very alegbraic way of doing this question but I just used patterns

correct, nice observation

[TonyHem]

i have no idea how to any of these questions lol

pretty much the same here

[humph]

11.) Find the maximum value of given that .

We have that

By the Cauchy-Schwarz inequality,

So is bounded above by .
In fact, equality holds in the Cauchy-Schwarz inequality when one vector is a multiple of the other; that is, if
for some .
As

and hence

we clearly have that equality holds when

and so achieves the upper bound for these values of .

[dcc]

new questions posted!