Wikipedia:Reference desk/Archives/Mathematics/2007 January 25

=January 25=

Optimization Problem
I'm studying for a math midterm tomorrow and I have no idea how to approach this problem: A rectangle is inscribed in an acute triangle of base 15 cm and altitude 6 cm. If one of the sides of the rectangle lies on the base of the triangle, find the dimensions of the rectangle of maximum area. Any help would be much appreciated.

Thanks, anon.

P.S. I might post further doubts later tonight. —The preceding unsigned comment was added by 162.83.224.27 (talk) 23:50, 24 January 2007 (UTC).


 * Draw the triangle with the base on the x-axis, pointing upwards, with the lower-left vertex at the origin. Draw the rectangle inside the triangle. Let x be the distance between the origin and the left side of the rectangle. You should be able to find the length of the bottom of the rectangle in terms of x. The left side of the triangle is a line coming out of the origin. You can figure out the slope of the line by using the base and altitude lengths of the triangle. Once you have the slope of this line, find an equation for the line. Use this equation to determine the height of the rectangle - the upper left corner of the rectangle is on this line, and its x coordinate is x. Now you should be able to find the area of the rectangle in terms of x. If you need any more help, feel free to ask. - ST 134.10.40.183 02:14, 25 January 2007 (UTC)

Circle Geometry Problem
This is the same poster from above ^. Geometry hasn't been treating me very kindly lately. I also don't know how to go about this problem: A rectangle is inscribed in a circle of radius 7. The area of the rectangle as a function of its width w is...? Thanks again for any help. —The preceding unsigned comment was added by 162.83.224.27 (talk) 01:54, 25 January 2007 (UTC).


 * Draw both the rectangle and the circle centered at the origin. You only need to worry about the area in the first quadrant (the area where both x and y are positive), which has a quarter of a circle and a quarter of the rectangle in it. Can you find an equation for this part of the circle? (Or rather, an equation that gives you the top half of the circle?). If you can find this equation, f(x), then you've found an equation that gives you the height of the rectangle with a width of x. From there you can find the area of the rectangle (remember to multiply it by 4 to get the whole rectangle, rather than just the part in the first quadrant). -ST 134.10.40.183 02:20, 25 January 2007 (UTC)

Domain Question
What does it mean when it a graph's domain has a square bracket "[" on one side and a round parenthesis ")" on the other side? --24.76.228.161 03:10, 25 January 2007 (UTC)
 * That's called inclusive notation. The bracket means that the set starts with that number (greater/less than or equal to). A parenthesis denotes it is uninclusive, (greater/less than). X [' Mac Davis '] ( How's my driving? ) 04:38, 25 January 2007 (UTC)


 * It's as Mac Davis says, but the notation is also described at Interval (mathematics). —David Eppstein 05:37, 25 January 2007 (UTC)

Thank you Mac and David! --24.76.228.161 06:43, 25 January 2007 (UTC)

Statistical analysis of edit sessions
In an arbitration case involving suspected sock-puppetry some parties are using the fact that there are no overlapping edit sessions of two users. I though it might help if some statistics were applied to the data.

The data we have is two series with the time and dates of the two contributors. To start with I analysed my on edits to see what constitutes a continuous edit session. Anyone got any ideas as to how we could determin some statistically significant test? What would an appropriate null hypothesis be? —The preceding unsigned comment was added by Salix alba (talk • contribs) 13:13, 25 January 2007 (UTC).


 * My friend who is mathematician says: "Compare two distributions. All time moments t are lying between two edits as in time(edit_1) <= t < time(edit_2). Where both edits are made by user 1. Then define gap(t) = time(edit_2) - time(edit_1). First distribution is of gap(t) for random time t, Second distribution is of gap(t) for t = time(edit) of random edit of other user 2. Null hypothesis is two distributions are each others' same. Its explanation is that time(edit) of user 2 is random time to user 1 if two users 1 and 2 edit independent each from the other. But if it is sock-puppetry then Second distribution has its gaps on average larger." I hope is clear enough description, sorry for our English. Hevesli 17:21, 27 January 2007 (UTC)


 * I am afraid that in this form this is not a good test. If the two users are in antipodal time zones, then almost automatically most edits of one will be in the larger gaps of the other. Conversely, if they are in (almost) the same time zone, most edits of one will be in the relatively smaller gaps of the other. Looking at the users in question, M and T, we see that both have the same diurnal rhythm in editing. During the period from about 15:00 to 00:59, the edit intensity is almost constantly high for user M, which we can take to be "user 1". I think the idea of Hevesli & friend can be saved by only looking at edits in the window from 15:00 to 00:59, cutting out edits between 01:00 and 14:59. --Lambiam Talk  12:22, 29 January 2007 (UTC)

Gödel and computer algebra systems
Does Gödel's incompleteness theorems imply that computer algebra systems like Mathematica or Matlab are somehow limited? Or are these systems not more limited than any human mathematician? Mr.K. 17:55, 25 January 2007 (UTC)


 * The limitations have nothing to do with the nature of any agent attempting a proof, whether human or otherwise; they are intrinsic limitations of formal proof systems. --KSmrqT 22:58, 25 January 2007 (UTC)


 * Some contest this Penrose in Shadows of the mind comes up with some bunkum about humans being capable of a bit more. Apparently the process of thinking is not limited to formal logic. --Salix alba (talk) 02:33, 28 January 2007 (UTC)

Geometry puzzle
My aunt got posed this geometry puzzle, so I expected it to have a fairly clean solution, but I only managed to solve it by computing the root of a quartic. ¬.¬

Basically, you have a 1m cube box in the corner between the wall and the floor. You have an 8m ladder that you lean against the wall, i.e., the top of the ladder touches the wall, the bottom touches the floor, and somewhere in the middle it also rest on the corner of the cube. The task is to find how high up the wall the ladder rests (twol solutions, as it's symmetrical).

As I said, the only way I managed to solve this was to construct a quartic equation for the distance of the tip of the ladder above the cube and then solve that and add the 1m back. quartic was x4 + 2x3 - 62x2 + 2x + 1 = 0. I got 7.9somethinghorrible, no idea what surds it's composed of.

Anyone got any idea how to solve this in a nicer fashion?

Rawling 4851 20:54, 25 January 2007 (UTC)


 * The best I can come up with off hand is to consider this problem as a triangle inscribed with a sqaure, where the base of the triangle is length (x+1) and the height of the triangle is (y+1). There are two smaller triangles formed within this triangle that have congruent angles.  The first has a base of length x and a height of 1, and the second has a base of 1 and a height of y. Since the two smaller triangles and the larger triangle all have the same angles, their sides must be in the same ratios.  Therefore $$\frac{x}{1} = \frac{1}{y} = \frac{x+1}{y+1}$$.


 * Also, by the pythagorean theorem, $$ (x+1)^2 + (y+1)^2 = 8^2 = 64$$.


 * Therefore $$(\frac{1}{y} + 1)^2 + (y+1)^2 = 64$$.


 * However, when you solve that equation you'll still end up having to solve the same quartic. If anybody has any better ideas, feel free. :/ Dugwiki 21:34, 25 January 2007 (UTC)


 * Yup, that's the way I did it. Not at all pretty ;) Rawling 4851 21:38, 25 January 2007 (UTC)


 * According to Mathematica, the two positive solutions are $$\frac{\sqrt{65} - 1}{2} \pm \sqrt{ \frac{31 - \sqrt{65}}{2} }$$ --Spoon! 21:59, 25 January 2007 (UTC)


 * Wow, that's... really complicated. Thanks for trying :) Rawling 4851 22:35, 25 January 2007 (UTC)


 * You can write an equation for the angle the ladder makes with the floor: $$\cos\theta+\sin\theta=4\sin2\theta$$, but I don't see how that's any nicer. --Tardis 17:48, 29 January 2007 (UTC)

Probability
Suppose I have two friends, Alice and Bob. I give them a bag with 3 balls, 1 black, 1 red, and 1 white. Alice pulls one out, and carefully looks at the colour with out reveaing it. Bob says, 'The odds are 1/3 that Alice has drawn the red ball' Alice says, 'The odds are 1/1 that I have the red ball.' They can't both be right, can they? What's going on? Duomillia 23:43, 25 January 2007 (UTC)


 * Alice has information that Bob doesn't have. There is no "the odds"; there is only the odds given what each of them knows. So that's different for both of them. --Spoon! 00:12, 26 January 2007 (UTC)


 * I assume that Bob cannot see what ball Alice pulls out. You must get rid of the idea that there is an absolute probability. The probability depends on who you ask. Different entities knows different things and so must make the best use of what little knowledge they have obtained to compute a "true" probability for themselves. 219.105.39.135 10:01, 26 January 2007 (UTC)


 * "Randomness is ignorance." Things are only best described as random when we have a lack of information. If there was a football game and you missed it and didn't hear anything about it all day, the outcome is random to you. If a nuke hit the city, or the star player of your favorite team died recently then your favorite team is probably (no pun here!) less likely to have won, as an outcome. The more information you have, randomness collapses. X [' Mαc Δαvιs '] ( How's my driving? ) ❖ 11:49, 26 January 2007 (UTC)
 * So anytime someone says, 'the odds of that are X', they really mean, 'the odds are X according to the best possible information available to me' Duomillia 15:54, 26 January 2007 (UTC)


 * Yes! That's right! The odds are calculate by taking account of all the infomation available to the entity. In a maths exam this is not a problem because all the students have access to exactly the same infomation on the test papers. So you would expect all the students to calculate the exactly same "true" probability of an event. In real life, different people have access to different amount of infomation, sometimes conflicting infomation. 219.105.39.135 00:04, 27 January 2007 (UTC)