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September 23[edit]

Advanced geometry homewrok[edit]

Yes I know you dont do homework, but just point me in the right direction. I have been given a pig of a problem i dont have any idea how to start: Prove that in any planar triangle, the centroid, orthocenter, circumcenter and center of the 9 point circle all lie on a straight line??? Any pointers welcome.

Euler line may be a good start.

Consider where the 9 point center must lie between the altitude and perpendicular bisector from each side. See if you can show that it must be exactly halfway between the circumcenter and orthocenter. Use the fact that the centroid falls 1/3 of the way along each median to make a similar argument about where the centroid falls between the circumcenter and orthocenter. Rckrone (talk) 04:43, 23 September 2009 (UTC)[reply]

If getting started at all is a problem, start by deriving formula for each type of point mentioned above for a triangle (x1,y1) (x2,y2) etc..

Once you've got 2 points you have a line - the other points (in terms of x1,x2,y1,y2 etc) should satisfy the eqaution of this line (if it's true)83.100.251.196 (talk) 19:32, 23 September 2009 (UTC)[reply]

Ignoring the nine point centre for the moment, one nice way of proving that O, G and H are collinear is to consider a dilation of the large triangle by a factor of -1/2 centred at the centroid (i.e. shrink it to half its size and reflect it about the centroid). Then see what happens to the point O of the larger triangle. AMorris (talk)●(contribs) 11:01, 24 September 2009 (UTC)[reply]

Recruiting math exam questions...[edit]

Someone confirm that either I'm crazy and have forgotten basic math, or the answer key for this recruiting exam has errors:

1. In a 100m race, A can beat B by 25m and B can beat C by 4m. In the same race, A can beat C by:
Me: 29m || Key: 28m

2. A man bought fruit at the rate of 16 for $24 and sold them at the rate of 8 for $18. What is the percent profit?
Me: 50% || Key: 40%

3. Jack has two children, and at least one of them is a girl. What is the probability that both children are girls?
Me: 1/2 || Key: 1/3

4. How many times are the hands of a clock at right angles in a day?
Me: 48 || Key: 44

Help please? 218.25.32.210 (talk) 06:07, 23 September 2009 (UTC)[reply]

I think I understand number 3 now -- based on the Punnet square, Jack could have BB | BG | BG | GG, but because at least one is a girl, you discard the BB result and are left with a 1/3rd chance that he has 2 girls.
I'm still clueless about the rest of them though. 218.25.32.210 (talk) 06:32, 23 September 2009 (UTC)[reply]

(ec) 1. Key is right. B runs at 0.75x A's speed and C runs at 0.96x B's speed => C runs at 0.72x A's speed so A beats C by 28m.

2. (edited) hmm i'm confused.

3 Key is right. Possibilities are boy-boy, boy-girl, girl-boy, or girl-girl. Boy-boy is eliminated leaving three equal other possibilities. girl-girl is one of these 3.

4. (edited) Key is right. Hour hand turns at 1/12 rph (revolutions per hour). Minute hand turns at 1 rph. The first time the hands are at right angles (i.e. 1/4 revolution apart) is at t hours, where t/12=t-0.25 => t=12t-3 => t=3/11 hours = 16.36 minutes after 12. Your error is in thinking it was at 15 minutes after 12 which would only be right if the hour hand didn't move continuously as the minute hand turns. These right-angles occur every 6/11th hours so in 24h it happens 44 times. 70.90.174.101 (talk) 06:50, 23 September 2009 (UTC)[reply]

The key is right for 1, 3 & 4. Let me give you some clues for 4. Think of two times when the hands are obviously perpendicular. How long from one of these times to the next time? It's not going to be just an hour is it, because in one hour when the minute hand is in its original position, the hour hand will have moved forward 30°. In °/hr or revolutions/hr, what is the speed of the hour hand and the minute hand? Subtract the smaller of these from the larger. What is the significance of this number? It might be easier to work it out for the situation when there is no angle between the hands, eg. 12 noon, and then extrapolate from then. RupertMillard (Talk) 07:49, 23 September 2009 (UTC)[reply]

The key is wrong for question 2. Depending on which value you choose for base you either arrive at 50% or 33.3%, there seems no way to get 40% as an answer. Taemyr (talk) 11:24, 23 September 2009 (UTC)[reply]

To understand the clock problem in an obvious way: what is the angle between the hands at 12:15? The big hand is on the 3 and the small hand is 1/4 of the way between the 12 and the 1, so it's not 90 degrees. That should make the situation clear. 70.90.174.101 (talk) 23:44, 23 September 2009 (UTC)[reply]

40% would be if you assume your base was (24+36)/2, i.e. the average, but that seems a dumb thing to use. Dragons flight (talk) 11:40, 23 September 2009 (UTC)[reply]

For number 3, the article boy or girl paradox may be of interest. Pallida Mors 14:16, 23 September 2009 (UTC)[reply]

The key is right in #3. The three equally probable possibilities are:

girl, boy
boy, girl
girl, girl

The probability of the third one is 1/3. Michael Hardy (talk) 15:57, 23 September 2009 (UTC)[reply]

The key is wrong in #2. The simplest way to see that may be to imagine that 16 is the exact number he bought and sold. He paid $24; he got $36. The other way is to say that in the first case the price per item was $24/16 = $1.5, and in the second case it was $18/8 = $2.25. Michael Hardy (talk) 16:01, 23 September 2009 (UTC)[reply]

Problem 2 is not just wrong, it is poorly stated. In the real world, you usually don't sell more than you buy, but you certainly can sell less than you buy. In fact, it's common. The problem doesn't state that you sold all you bought. It is arguably an implicit assumption, but I would argue that implicit assumptions are acceptable if they match likely outcomes. For example, it is an implicit assumption that the value of the dollar hasn't changed between the date of the purchase and the date of the sale. Were I to challenge the problem on the basis that the implicit assumption isn't spelled out, I would be properly upbraided for nit-picking. However, it is not only common, it is usual for fruit vendors to sell fewer than they buy, and spoilage should be included in the calculation. The wording always makes a point of avoiding the issue - I think in this case it was incompetence, but if it was a finance class, not an arithmetic class, I'd suspect that the teacher wanted you to identify the missing assumption. If we accept that purchase count equal sales count the right answer is 33.3...%. 50% is the markup, not the profit.--SPhilbrick T 21:20, 24 September 2009 (UTC)[reply]

Introduction to number theory[edit]

Does anyone know of any good introductions to number theory (am I misusing the term)? I am trying to understand the Number Field Sieve but I keep realizing I just don't have the background to fully understand it. I would really like it if someone knew of an online resource. My math education stopped at calculus. Thank you. PvsKllKsVp (talk) 21:48, 23 September 2009 (UTC)[reply]

Tom Apostol's Analytic Number Theory is a nice introduction.--pma (talk) 21:58, 23 September 2009 (UTC)[reply]

Elementary Number Theory by Kenneth Rosen StatisticsMan (talk) 22:14, 23 September 2009 (UTC)[reply]

Thank you. PvsKllKsVp (talk) 22:34, 23 September 2009 (UTC)[reply]

I suggest "A Course in Number Theory and Cryptography" by Neal Koblitz. It starts at an undergraduate level and gets up through the quadratic sieve. I don't know whether the new edition includes the number field sieve, but it at least gets most of the way there. 70.90.174.101 (talk) 23:39, 23 September 2009 (UTC)[reply]

Calculating integrals (for use in probability theory)[edit]

I'm stuck on some integrals here. One problem is to find all the moments for

f(x)={\frac {1}{{\sqrt {2\pi }}x}}e^{-(\log x)^{2}/2},\qquad 0\leq x<\infty ,

or, more accurately, verify that they are $EX^{r}=e^{r^{2}/2}$ for r = 0, 1, ... .

I have two main options, as far as I can see. One is to just calculate the integral

\int _{0}^{\infty }x^{r}f(x)\,dx

but I don't know how to do that. I know how for r = 0. That is an easy u-substitution. But, for r > 0, I am not seeing it. I thought, for r = 1, I could use integration by parts with u = x and dv = f(x) dx. But, I don't really know how to do that integration by parts because it would end up with v as the integral of $e^{-x^{2}/2}\,dx$ , which has no closed-form antiderivative. The other way would be to use the moment generating function but that just adds in an $e^{tx}$ which doesn't seem to make things any easier as I think I'd end up with the same v. Any ideas? Thanks. StatisticsMan (talk) 22:23, 23 September 2009 (UTC)[reply]

Follow the first option: the integral is not that difficult. Make the substitution

t=\log(x)

, and complete the square in the argument of the exponential that results as integrand; finally make a translation. Is it OK?--pma (talk) 22:36, 23 September 2009 (UTC)[reply]

I see what you mean. That's pretty simple. But, I'm off by a factor of 2. Ah, I think I just figured it out. As x goes to 0, log x goes to -infinity but I just left the limits of integration as 0 to infinity. Thanks. As always, you are very helpful. This will help on another problem as well probably. StatisticsMan (talk) 23:01, 23 September 2009 (UTC)[reply]

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