Wikipedia:Reference desk/Archives/Mathematics/2012 February 3

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February 3

Isosceles right triangle

I want to make an isosceles right triangle of a certain area. Let us say I want to make an isosceles right triangle with an area of 100 square units. How can I determine the lengths of the legs of that triangle? Thank you. Bus stop (talk) 01:54, 3 February 2012 (UTC)

An isosceles right triangle is half of a square. So what square would have area 2 x 100 = 200? The side of the square would be square root of 200 = 14.14..., and that would be the length of the legs of the triangle. Bubba73 ^{You talkin' to me?} 02:00, 3 February 2012 (UTC)

Wow—you are right. That is great. Thank you very much. Bus stop (talk) 03:05, 3 February 2012 (UTC)

Can I ask another question? I believe this is a Trapezoid of some sort. I want it to have an area of 100 square units. The base must be 10 units in length. The two vertical lines form right angles with the base. The other two interior angles are 45 degrees and 135 degrees. How can I determine the lengths of the two vertical lines? Thank you. Bus stop (talk) 17:50, 3 February 2012 (UTC)

Call the shorter of the two lines x, and draw the horizonal line across at the top of x. That is also 10 units long. The triangle above is a 45-45-90 and thus as 10 for the vertical side as well, which means that it is exactly half the square 10 on a side, which means an area of 50. This means the bottom rectangle also has an area 50 which means that x is 5 (and the longer one is 15). Essentially your diagram is that of a square 10 on a side with the triangle cut off gotten from connecting center top with the center left and then rotating that triangle around the center top.Naraht (talk) 18:27, 3 February 2012 (UTC)

Thank you. I think I understand you. Just to be sure, I've plugged my own numbers in, using numbers actually applicable to a project I'm working on. I've replaced 100 sq. units with 52.5 sq. units, and I've replaced a base length of 10 units with a base length of 7 units. Using these numbers I conclude that the shorter vertical (the one you designated x) will be 4, and that the longer vertical will be 11. Would that be correct? Thanks. Bus stop (talk) 00:32, 5 February 2012 (UTC)

That is correct. You can always check your answer by calculating the area of the shape using the lengths you found. Widener (talk) 03:45, 5 February 2012 (UTC)

Thanks. I actually did check it. I found 28 square units for the lower rectangle and 24.5 square units for the upper triangle. Added together that gets me to my intended 52.5 square unit area. I guess I still wasn't sure I did it right. I hope you won't mind if I ask another question. Here goes:

In reference to this diagram, how can I determine the placement of line A-B, which separates two trapezoids which each have an area of 52.5 square units? I know that the top trapezoid has a "roof" of 8 units. I also know that the bottom trapezoid has a "floor" of 22 units. And I also know that the distance from "roof" to "floor" is 7 units. Triangle X-Y-Z has two 45 degree angles and one 90 degree angle—and again—I know that the distance from Y to Z is 7 units. What I want to know is where to place line A-B so that my two resulting trapezoids will each contain 52.5 square units of area. Thanks for any help that can be offered. Bus stop (talk) 04:38, 5 February 2012 (UTC)

The length of AB is a function of its height h above the floor of the trapezoid. Since the angle is 45 degrees, the length of AB is equal to $22-2h$ (Can you see why this is the case?). The area of the lower trapezoid is therefore $h{\frac {22-2h+22}{2}}=22h-h^{2}$. We want to find the value for h for which the area is 52.5. Solving $22h-h^{2}=52.5$ for h gives 2.72 as an approximate solution. Do you know how to solve quadratic equations? Widener (talk) 05:27, 5 February 2012 (UTC)

I don't know how to solve quadratic equations. I can understand that the length of AB is a function of its height above the floor. I don't really understand why the length of AB is equal to 22 - 2h, in which h is the height of AB above the floor. I see that there are Quadratic Equation calculators online, but I doubt that I would know how to use them. I'm glad that you gave me the answer! I will use 2.72. I guess I can assume that measuring from the top down would yield a distance of 4.28 to AB. That is certainly precise enough. Lives are not at stake. I am only making a painting (but an important painting). Thanks a lot for your assistance! Bus stop (talk) 08:15, 5 February 2012 (UTC)

The exact solution is $11-{\frac {\sqrt {274}}{2}}$; if you want a more precise answer then you can round that off to whatever precision you desire. There is a formula for solving quadratic equations like this one. If you draw a vertical line from the point A down to the floor, you form a new right angle triangle with side length h. Because the angle is 45 degrees, the other side of that triangle also has length h. You can do the same for point B. You will notice that the length of the floor is equal to the length of AB + the two lengths of length h. That's why 22 = AB + 2h i.e. AB = 22 – 2h. Widener (talk) 09:19, 5 February 2012 (UTC)

I do now grasp why AB = 22 - 2h. In my last post I said that I did not understand that, but I do understand that now.

I don't think I can solve this yet on my own. But let me ask you a question. In solving for h you arrived at an answer of 2.72. In my diagram I indicated that from the "roof" of the top trapezoid to the "floor" of the bottom trapezoid was a height of 7. That would imply that the height of the "roof" above line AB was 4.28. I am wondering: is it is possible to check that the area of the upper trapezoid is 52.5? Bus stop (talk) 12:13, 6 February 2012 (UTC)

Do you know how to calculate the area of a trapezoid? Widener (talk) 13:49, 6 February 2012 (UTC)

Wow—you are right. It works. The top trapezoid is also 52.5! The area of the rectangle is 4.28 x 8 = 34.24. The area of the two triangles is 4.28 x 4.28 = 18.32. Added together gets me to my 52.5. But how did you figure out that h = 2.72? That math still escapes me. Thanks for prodding me into thinking about calculating the area of the upper trapezoid. Bus stop (talk) 14:14, 6 February 2012 (UTC)

What part of my explanation did you not understand? Widener (talk) 22:37, 6 February 2012 (UTC)

The first thing that I encounter that I don't understand is this: $h{\frac {22-2h+22}{2}}=22h-h^{2}$. Bus stop (talk) 23:40, 6 February 2012 (UTC)

Do you mean, you don't understand why that equality holds, or you don't understand why it is equal to the area of the lower trapezoid? The formula for calculating the area of a trapezoid can be found in the article trapezoid (the formula is quite easy to derive using the method you described above, or by splitting the trapezoid into two triangles). Here, a=22-2h, b=22. $h{\frac {22-2h+22}{2}}=h{\frac {44-2h}{2}}=h(22-h)=22h-h^{2}$ Widener (talk) 00:46, 7 February 2012 (UTC)

I understand this: $h{\frac {22-2h+22}{2}}$. This is the formula for finding the area of this trapezoid. To find the area of a trapezoid we add together the lengths of the two parallel sides and divide the result by two. We then multiply that result by the perpendicular distance between those two parallel sides (h). It is the subsequent expressions that perplex me. Actually I even understand that: $h{\frac {22-2h+22}{2}}=h{\frac {44-2h}{2}}$. But I don't understand how one gets to the next two expressions in your post above. And finally I don't understand how one eventually computes that h = 2.72. Bus stop (talk) 23:02, 7 February 2012 (UTC)

Both steps use the distributive property of multiplication. ${\frac {44-2h}{2}}={\frac {44}{2}}-{\frac {2h}{2}}=22-h.$ $h(22-h)=h\times 22-h\times h=22h-h^{2}.$ Finding the solution involves solving the quadratic equation $22h-h^{2}=52.5$ (which is equivalent to $22h-h^{2}-52.5=0$) and there are several ways you can do that; see quadratic equation. In this quadratic equation, $a=-1,b=22,c=-52.5$ Widener (talk) 00:12, 8 February 2012 (UTC)

I'm sorry to get off-topic but let me ask you this: Is the ratio between 2.72 and 4.28 going to be the same in all arrangements that are like these two trapezoids? These are two trapezoids that share a base. The legs are at a 45 degree angle to the bases. I think the legs can be thought of as one continuous straight line. The areas of both trapezoids are equal. Would there be a ratio, of the heights of each trapezoid to one another, that applies in all such arrangements? Bus stop (talk) 00:44, 8 February 2012 (UTC)

No, it won't. Only if the top and bottom of the trapezoid are in the ratio 8 : 22. Widener (talk) 06:46, 8 February 2012 (UTC)

Computing values for the Riemann zeta function

How does one compute values of the Riemann zeta function? Widener (talk) 06:16, 3 February 2012 (UTC)

It depends. For Re(x)>1, you could use the standard series. In the critical strip, Euler-Maclaurin formula is the oldest and probably the simplest, though the convergence rate is too slow to make it useful in practice. Isolated values for large Im(x) may be computed using Riemann-Siegel formula. To compute blocks of values simultaneously, one uses Odlyzko–Schönhage algorithm.--Itinerant1 (talk) 11:36, 3 February 2012 (UTC)

PARTIALLY resolved?

According to Hilbert's problems, Hilbert's sixth problem is "partially resolved". If you look at one of the sources, the title is "There is no axiomatic system for the quantum theory". It can be found here: http://arxiv.org/abs/0711.3130 Since quantum theory is a subset of physics in general, doesn't this provide a complete disproof for there being a solution to Hilbert's sixth problem? Widener (talk) 06:42, 3 February 2012 (UTC)

I'm no expert on this topic, but at a brief glance, Nagata's remarks on whether there's an axiomatic system don't appear to make a lot of sense. He seems to be claiming the quantum theory is self-contradictory. I don't understand why he concludes from that that the theory has no axiomatic system, rather than more straightforwardly concluding that the quantum theory is false. Is anyone here familiar with Nagata's work, or at least with his reputation, and can comment on how well this work is received? --Trovatore (talk) 06:49, 3 February 2012 (UTC)

I'm a little concerned about whether Nagata's papers should be considered reliable sources. For one thing, the "International Journal of Theoretical Physics" isn't exactly a top-notch journal. But that aside, the same journal published an apparent refutation of Nagata's paper: Comments on "There is no axiomatic system for the quantum theory" [MR2565160]. Internat. J. Theoret. Phys. 50 (2011), no. 6, 1828–1830. I think that the Nagata references should be removed from our article on Hilbert's problems, and the table should be changed back to indicate that the problem is not resolved. Sławomir Biały (talk) 14:42, 5 February 2012 (UTC)

It sounds like you are right. Widener (talk) 07:06, 7 February 2012 (UTC)