tag:blogger.com,1999:blog-2400513859305780710.post7258043487237257638..comments2023-04-04T07:53:53.789-07:00Comments on Mathematical Food For Thought: Perimeter To Angle? Topic: Geometry/Trigonometry. Level: Olympiad.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-2400513859305780710.post-40448066797785154002006-04-16T06:22:46.000-07:002006-04-16T06:22:46.000-07:00Inscribing a circle into triangle ABC, and using a...Inscribing a circle into triangle ABC, and using a = x + y, b = y + z, c = z + x, and r, we find that the identity is equivalent to<br><br>r^2 / xz + r^2 / xy + r^2 / yz = 1<br><br>By Heron's, the area of the triangle A = \sqrt{xyz(x+y+z)} = (x+y+z)r, so A^2 / A^2 = 1 = r^2 (x + y + z) / xyz = 1, which is equivalent to the above identity. QED.QCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-20089293437199335832006-04-16T09:58:23.000-07:002006-04-16T09:58:23.000-07:00Don't you think you should start to slow down ...Don't you think you should start to slow down a little. USAMO is in two days!!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-25606934769304528762006-04-16T11:00:38.000-07:002006-04-16T11:00:38.000-07:00Tomorrow is my cool down period for the year :).Tomorrow is my cool down period for the year :).paladin8noreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-22607310647340661222006-04-16T14:04:48.000-07:002006-04-16T14:04:48.000-07:00Haha, nice. Well, good luck on the USAMO. You ar...Haha, nice. Well, good luck on the USAMO. You are a great role model. You've worked hard and improved so much in one year.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-82033872842424750432006-04-28T13:31:31.000-07:002006-04-28T13:31:31.000-07:00I wonder if this proof is valid:From the equations...I wonder if this proof is valid:<br><br>From the equations AP + AQ + PQ = 2, AP = 1 - PB, and AQ = 1 - QD we get<br>PQ = PB + QD<br><br>This means that triangles PCB and QCD can be "flipped" inward along PC and QC respectively to fill up triangle PCQ. <br><br>Then we have 2(alpha + beta) = 90, or alpha + beta = 45 which means alpha + beta = angle PCQ = 45.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-34072047502161020082006-04-28T13:39:05.000-07:002006-04-28T13:39:05.000-07:00That's interesting. Intuitively, it makes sens...That's interesting. Intuitively, it makes sense, but I'm not completely sure if I would be able to rigorously prove it. A good idea, though!paladin8noreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-48943343979060195742006-04-28T14:11:04.000-07:002006-04-28T14:11:04.000-07:00Hmm, I guess "flipping triangles" inward...Hmm, I guess "flipping triangles" inward is not the best way to put it.<br>How about: <br>rotate triangle PCB so that BC lines up with DC. Then the newly formed triangle<br>"PCQDB" is SSS congruent with the original PCQ, so alpha + beta = angle PCQ, and alpha+beta + angle PCQ = 90, so PCQ = 45Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-62085315099316773772006-08-20T18:33:08.000-07:002006-08-20T18:33:08.000-07:00I think you can prove it... draw an altitude down ...I think you can prove it... draw an altitude down from c to pq and use algebra about the triangle areas and stuffanonymounoreply@blogger.com