Wikipedia:Reference desk/Archives/Mathematics/2009 January 20

= January 20 =

Geometry problem
Consider two circles of non-equal radius overlapping. The boundary is then a pair of circular arcs meeting at two cusps, looking a bit like a body and a head. The "neck" should be seen as indented, i.e. both the arcs are more than semicircles. Now the problem - it seems intuitively obvious to me that the longest line through one of the cusps and ending at two points on the boundary will be parallel to the line of centres of the circles, but when trying to prove this I'm getting lost in the geometry/trigonometry. Is my supposition true, and is there a quick way to show it?→84.64.253.197 (talk) 13:33, 20 January 2009 (UTC)
 * Yes, it's true. It wasn't obvious to me, but here's a proof. Let g and h be the vectors from the cusp to the centres of the first and second circles respectively. Now draw your line, and let e be the unit vector parallel to it and pointing into the first circle. The distances along your line from the cusp to the intersections with the first and second circles are 2g·e and -2h·e respectively, where · is the standard dot product. Thus you need to maximize 2g·e - 2h·e = 2(g - h)·e, for a choice of unit vector e. This occurs when you choose e to be in the same direction as g - h, which is the vector from the centre of the second circle to that of the first. Therefore, your chosen line is parallel to the line through the centres. Joeldl (talk) 14:06, 20 January 2009 (UTC)
 * Note: You might also wonder if a segment from the cusp into just one of the circles might be longer. (Perhaps with the other side not pointing into the other circle.) The longest segment you can get this way is the diameter of the corresponding circle (which is the length of 2g or 2h), whereas the the segment you got before had the same length as 2 (g - h), that is, twice the distance between the centres. But this distance is more than either radius so long as neither centre is inside the other circle. This amounts to the assumption you made about the way the circles are situated. Joeldl (talk) 14:29, 20 January 2009 (UTC)


 * (After edit conflict - essentially the same method, but with calculus instead of vector products) Suppose one circle has radius rA and centre at A and the other has radius rB and centre at B, with the cusp at C. Suppose a line through C intersects circle A again at D and intersects circle B at E, and that DCE makes an angle &alpha; with AC and an angle &beta; with BC. Then the length of DCE is
 * $$L=2r_A \cos \alpha + 2r_B \cos \beta$$
 * and you want to maximise this subject to the constraint that
 * $$\alpha + \beta = 180^o - \angle ACB = \text{constant}$$
 * because triangle ACB is fixed. Use some calculus to find an expression for dL/d&alpha; and show that this is 0 when the lengths of the two perpendiculars from A and B to DCE are equal - so DCE is indeed parallel to AB. You then just have to verify that you have found a maximum for L, not a minimum. Gandalf61 (talk) 14:21, 20 January 2009 (UTC)


 * (OP) Thanks to both.→86.132.164.26 (talk) 19:56, 20 January 2009 (UTC)

Ordinary Annuity
Hello. Would you please clarify the following question? Instead of paying $4000 at the end of 5 years and $3000 at the end of 10 years, Erica agrees to make equal monthly payments for 10 years. Find the monthly payment if the interest charged is 8% per annum, compounded monthly. Explain your method. This is not homework counting for marks since I have the answer key. The solution is $48.97. Thanks in advance. --Mayfare (talk) 20:46, 20 January 2009 (UTC)


 * Clarify it in what way? What part isn't clear to you? --Tango (talk) 20:52, 20 January 2009 (UTC)


 * There does seem to be missing info, like the amount of money lent to Erica. Is it $7000 ?  I can't tell. StuRat (talk) 20:58, 20 January 2009 (UTC)

There is a bit of a trick since they tell you the principle in a funny way. The principle is gotten by discounting the two payments to time 0 and then converting them back into an annutiy. So we have:
 * $$P \,=\, {4000 \over (1+{.08 \over 12})^{5*12}}+{3000 \over (1+{.08 \over 12})^{10*12}} \, = \, 4036.41$$

Which is the discounting of the two payments. To convert that into a monthly payment I'll use
 * $$P \,=\,R\left[\frac{1-\frac{1}{\left(1+i\right)^n}}{i}\right] = R\cdot a_{\overline{n}|i}$$

From Annuity_(finance_theory). Using i = .08/12 = .00667, n = 10*12 = 120 gives me the following
 * $$a_{\overline{120}|.00667} \,=\,\left[\frac{1-\frac{1}{\left(1.00667\right)^{120}}}{.00667}\right]\,=\,82.42$$

Which finally gives me
 * $$R\,=\,{P \over a_{\overline{120}|.00667} }= {4036.41 \over 82.42} \,=\, 48.97$$

Happy hunting. (By the way, when I punched these into my calculater I didn't do the rounding you see above. Also I used .0066666667.  The answer is very sensitive to rounding like this).Anythingapplied (talk) 22:02, 20 January 2009 (UTC)

Positive roots of subsystems
I am trying to convince myself of something that is probably obviously true, but having trouble writing down a clear proof. The short version is "Can every root sub-system of a root system be ordered so that a root of the sub-system is positive in the sub-system if and only if it is positive in the original root system?"

A crystallographic root system is a set of real vectors of the same dimension that have a few nice properties listed in the definitions section. They are nicely classified, and given names like A1, B23, etc. If Φ is a root system, and J is a subset of Φ, then Z[J] ∩ Φ is also a root system, I'll call it (J). Root systems can be ordered so that some of the roots are called positive and some are called negative. There are lots of choices for the set of positive roots, but it is not completely arbitrary. An easy way of specifying it is specifying a set of roots to be the simple roots, though again not all subsets are sets of simple roots.

If J consists of simple roots, then (J) is called a parabolic sub-system, and one can define the simple roots of (J) just to be J, and then the positive roots of (J) are just the positive roots of Φ that happen to lie in (J). All good.

What if J consists of positive roots? Can the positive roots of (J) still be defined as the positive roots of Φ that happen to lie in (J)? JackSchmidt (talk) 22:29, 20 January 2009 (UTC)