tag:blogger.com,1999:blog-5724670537875099184.post2681502655235442262..comments2017-03-25T16:46:26.799-05:00Comments on ZimBlog: Points In A Circle Problem: Summary To DateJason Zimbahttps://plus.google.com/100902559988995037755noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-5724670537875099184.post-81032922562640004342016-12-14T08:42:54.507-05:002016-12-14T08:42:54.507-05:00Well, that didn't come out too well. In linear...Well, that didn't come out too well. In linear format:<br /><br />A = 2r^2 uv(u+v)/((1+u^2)(1+v^2))Bill McCallumhttp://www.blogger.com/profile/09647093331377576865noreply@blogger.comtag:blogger.com,1999:blog-5724670537875099184.post-34454375906768402202016-12-14T08:42:00.014-05:002016-12-14T08:42:00.014-05:00Here's another formula, which you can derive f...Here's another formula, which you can derive from that one. If u and v are the tangents of two of the vertex angles then <br /><br /> uv(u + v)<br />A = 2r^2 ----------------<br /> (1 + u^2)(1+v^2)<br /> <br />Here you have to allow u or v = ∞ by taking limits. It u = ∞ then you have a right angled triangle sitting on the diameter of the circumscribed circle, and the formula gives 2r^2 v/(1+v^2). Fun exercise to check that is half the product of the two legs. If u and v are both infinity then the formula gives zero, which makes sense because a triangle with two right angles had zero area.<br />Bill McCallumhttp://www.blogger.com/profile/09647093331377576865noreply@blogger.comtag:blogger.com,1999:blog-5724670537875099184.post-82772111317956297422016-12-13T08:23:18.295-05:002016-12-13T08:23:18.295-05:00That's cool! There is a lot to know.... (Remin...That's cool! There is a lot to know.... (Reminds me, I had also never seen the "angle-side-angle" area formula 1/2(cot(A) + cot(B))^(-1)c^2 that I used in the Halloween Challenge post earlier this year.) Now since every triangle is inscribed in some circle, the "r" in your area formula is what I believe is called the circumradius of the triangle. I have seen some nice formulas for this in terms of the sides. Jason Zimbahttp://www.blogger.com/profile/02616845376407530843noreply@blogger.comtag:blogger.com,1999:blog-5724670537875099184.post-36323192571866650922016-12-12T23:41:01.110-05:002016-12-12T23:41:01.110-05:00Seriously, this lead me to discover a formula for ...Seriously, this lead me to discover a formula for the area of a triangle inscribed in a circle of radius r, with vertex angles a, b, and c:<br /><br />A = (r^2/2)(sin(a + b - c) + sin(a - b + c) + (sin -a + b +c))<br /><br />I assume this is well-known to those who know it well, but I was not one of them.Bill McCallumhttp://www.blogger.com/profile/09647093331377576865noreply@blogger.comtag:blogger.com,1999:blog-5724670537875099184.post-43456295251590782152016-12-12T22:56:48.696-05:002016-12-12T22:56:48.696-05:00L. M. A. O.L. M. A. O.Bill McCallumhttp://www.blogger.com/profile/09647093331377576865noreply@blogger.comtag:blogger.com,1999:blog-5724670537875099184.post-2236010467106263682016-12-12T19:27:03.568-05:002016-12-12T19:27:03.568-05:00"Don't tempt me, Frodo!" https://dr..."Don't tempt me, Frodo!" https://drive.google.com/file/d/0B9YMTchBa3Y2RGQzYU5mNFVicDA/view?usp=sharingJason Zimbahttp://www.blogger.com/profile/02616845376407530843noreply@blogger.comtag:blogger.com,1999:blog-5724670537875099184.post-79968301611234626172016-12-12T17:36:39.657-05:002016-12-12T17:36:39.657-05:00Dammit Jason, I'm supposed to be working!Dammit Jason, I'm supposed to be working!Bill McCallumhttp://www.blogger.com/profile/09647093331377576865noreply@blogger.com