Wikipedia:Reference desk/Archives/Mathematics/2010 May 17

= May 17 =

ridiculous indefinite integral question
Evaluate $$\int \cot x \cdot \ln{|\sin x|}dx$$ --115.178.29.142 (talk) 00:59, 17 May 2010 (UTC)
 * Change $$\cot x$$ to $$\frac{\cos x}{\sin x}$$ and solve by substitution, would be my guess at a glance. I'll look back when I'm not in a hurry. 76.228.199.229 (talk) 01:05, 17 May 2010 (UTC)
 * In fact, you don't even have to go that far. Hint: the antiderivative of cotangent is $$\ln |\sin x|+C$$. 76.228.199.229 (talk) 01:18, 17 May 2010 (UTC)

Geometry problem
Hey guys. I came across a really difficult (for me) olympiad question and I'm clueless as to how to solve it. Here it is.

Put O as the circumcenter of triangle ABC, and take an arbitrary point D on BC. Put E as the intersection of two circles; a circle with AD as its diameter, and the nine-point circle of ΔABC. Prove that angle ADE = angle ODC (E is not on BC)

I've tried everything from looking for similar triangles, finding cyclic quadrilaterals, to assigning every length using the law of sines. But I keep getting stuck no matter what I try. Can anybody help me?? thanks. Johnnyboi7 (talk) 12:47, 17 May 2010 (UTC)
 * This gave me a lot of trouble, and the way I ended up doing it isn't too pretty but here goes. Let X, Y, Z be the midpoints of BC, AC, AB respectively, N be the nine-point center, F the midpoint of AD (and so the center of that first circle) and G the base of the altitude from A.  NX is the nine-point radius, and AO is the circumradius which is twice the nine-point radius.  Since N is halfway between the altitude and the perpendicular bisector to BC, NX and AO are parallel.  The two circles intersect at G and E, so angle NFG = angle NFE.  Call this angle θ.  Reflect N over the line YZ and call that point N' (and note that A is the reflection of G).  N' is the midpoint of AO.  So then AFN' and ADO are similar triangles and so angle ADO = angle AFN' = θ.  angle EDG is inscribed in circle F, and has corresponding central angle EFG, so angle EDG = 180 - θ, which means angle CDE is also θ.  Since angle CDE = angle ADO, angle ADE = angle ODC. Rckrone (talk) 18:38, 19 May 2010 (UTC)
 * I guess I should add that you need to choose which vertex is B and which is C in order for it to work out, otherwise you may end up with angle ADE being supplementary to angle ODC rather than equal. Also depending on how things work out, angle EDG may end up being θ rather than 180 - θ depending on whether D falls between E and G on the circle or not.  That will determine whether to assign the label C to the vertex on G side of DG or to the one on the D side. Rckrone (talk) 20:41, 19 May 2010 (UTC)
 * Rckrone, I love you!! thank you so much for all the effort you've put into this:) I'm sorry I didn't reply sooner.
 * PS. Yeah I agree, the problem's a little bit shaky considering that the angle depends on where we put D. If I were the person who created this problem, I would have said "Prove that angle ADE = angle ODX (X being the midpoint of BC" "Johnnyboi7 (talk) 09:17, 23 May 2010 (UTC)

SAS instruction books
Does anyone have any suggestions as to what SAS software instruction book/guide would be the best one to buy?--160.36.39.85 (talk) 13:41, 17 May 2010 (UTC)
 * Anything by Andy McNab. Sorry. Terrlible joke. feel free to remove me;) Vespine (talk) 02:37, 21 May 2010 (UTC)