tag:blogger.com,1999:blog-6500033298667240354.post6987568723062264083..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 3278foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger15125tag:blogger.com,1999:blog-6500033298667240354.post-85462752699250329202019-08-05T15:04:51.842-07:002019-08-05T15:04:51.842-07:00Do not use all of these Private Money Lender here....Do not use all of these Private Money Lender here.They are located in Nigeria, Ghana Turkey, France and Israel.My name is Mrs.Ramirez Cecilia, I am from Philippines. Have you been looking for a loan?Do you need an urgent personal or business loan?contact Fast Legitimate Loan Approval he help me with a loan of $78.000 some days ago after been scammed of $19,000 from a woman claiming to be a loan lender from Nigeria but i thank God today that i got my loan worth $78.000.Feel free to contact the company for a genuine financial call/whats-App Contact Number +918929509036 Email:(fastloanoffer34@gmail.com)loanhttps://www.blogger.com/profile/11403605777656746915noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-30860202206990081442011-02-06T22:49:09.720-08:002011-02-06T22:49:09.720-08:00A solution is a solution. It may be useful for tho...A solution is a solution. It may be useful for those who study calculus.<br />Thank you Bleaug.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-81538822108818878172011-02-05T05:48:06.162-08:002011-02-05T05:48:06.162-08:00well if you insist here it is:
http://bleaug.free....well if you insist here it is:<br />http://bleaug.free.fr/8foxes/8foxes327.pdf<br /><br />I apologize in advance for the overpowered, tedious and lengthy calculations contained in that document compared the one-line elegant geometric argument given by Anonymous above :-D<br /><br />BleaugAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-79815999579273271892011-02-05T00:02:07.589-08:002011-02-05T00:02:07.589-08:00Thank you Anonymous and Newzad for the geometric s...Thank you Anonymous and Newzad for the geometric solutions.<br />Bleaug, I really want to see the Taylor series expansion. Would you like to comment it out here?<br /><br />Although, this is solved, I see merits of generating a general case for this one. Later...8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-35021328303542355512011-02-04T19:11:38.556-08:002011-02-04T19:11:38.556-08:00Ha, hence gravity/weights. I don't know why I...Ha, hence gravity/weights. I don't know why I didn't see that earlier; nice problem.sixnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-41014677242584421772011-02-04T10:39:50.361-08:002011-02-04T10:39:50.361-08:00http://geometri-problemleri.blogspot.com/2011/02/p...http://geometri-problemleri.blogspot.com/2011/02/problem-97-ve-cozumu.htmlAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-18375461590797543932011-02-04T04:59:13.773-08:002011-02-04T04:59:13.773-08:00let be O the center of Pluto,r the radius,ABCD
th...let be O the center of Pluto,r the radius,ABCD <br />the rectangle<br />OA=10.00000173+r=a,OB=30+r,OC=20+r,OD=r<br />the triangles OAC and OBD have the same median<br />hence<br />(r+a)²+(r+20)²=r²+(r+30)²<br />r=1156Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-58638632265405444762011-02-04T02:19:30.026-08:002011-02-04T02:19:30.026-08:00Also, the arc length isn't exactly the same; h...Also, the arc length isn't exactly the same; however you get the picture.sixnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-84548265061559020892011-02-04T02:17:32.213-08:002011-02-04T02:17:32.213-08:00I think it's more like, consider the rightmost...I think it's more like, consider the rightmost weight to be the top of the circle. Now since the circle barely curves near the top, the distance between the two rightmost weights are negligible (up and down). Now as you move further left on the circle, you see the same situation again, except you have a weight and a fixed point on the rectangle. Notice that this is the same distance (arc length) as the two rightmost weights. However, the up and down ratio has changed a little. <br /><br />If you just make the rightmost weight (20) the top of the circle in your picture, things should be more clear.<br /><br />Like bleaug, I too am still looking for a geometric argument for it. I'm trying to imagine different angles the rectangle could be in, and something that wouldn't change given the radius of the circle. I feel like I'm close, yet I'm not quite there.sixnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-9517034062081575112011-02-04T01:25:28.075-08:002011-02-04T01:25:28.075-08:00http://i51.tinypic.com/29d3r5.gif
And maybe I didn...http://i51.tinypic.com/29d3r5.gif<br />And maybe I didnt understand question wellAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-45785804417798419222011-02-03T15:04:33.440-08:002011-02-03T15:04:33.440-08:00Nothing is missing. Answer is C, but I'm still...Nothing is missing. Answer is C, but I'm still looking for a geometric argument for it.<br /><br />Here is the calculus approach: express each measured height to center of Pluto as 2nd degree Taylor series in 1/R (R being the radius we are looking for). You will get something like the following:<br /> R ~ 10*20/0.00000173 cm ~ 1156 km<br />irrespective of brick dimensions or angle provided that R >> brick size.<br /><br />bleaugAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-84296269511803290432011-02-03T14:52:15.068-08:002011-02-03T14:52:15.068-08:00Repeated objections made us to check the answer, w...Repeated objections made us to check the answer, we found the same result. Here, brick is a rectangle obviously, and Pluto's surface is a part of a circle. It looks like a straight line, because PLUTO IS A PLANET, after all.<br /><br />If you see a "constructional" problem, please let us know. Otherwise, this looks OK.<br /><br />The answer is still classified as "not confirmed", since it has not been solved by someone else.<br /><br />-8foxesAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-30071543395102833352011-02-03T12:50:11.952-08:002011-02-03T12:50:11.952-08:00Are you sure something is not missing? Brick weigh...Are you sure something is not missing? Brick weight/height ratio maybeAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-67325347102547834652011-02-02T00:42:22.398-08:002011-02-02T00:42:22.398-08:00Oh no, no. This is not a balancing problem. Not ...Oh no, no. This is not a balancing problem. Not a Static problem in Physics. Assume that the brick is already fixed in the given configuration. Suppose, for example, the point it touches to the Plotu's ground holds the whole experiment firmly. <br /><br />This is a pure-geometry problem. If there is no computational errors, I don't see a problem.<br /><br />-Peace8foxeshttps://www.blogger.com/profile/09567328431908997738noreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-7062455515770897882011-02-02T00:28:31.287-08:002011-02-02T00:28:31.287-08:00Is this problem even solvable with the given infor...Is this problem even solvable with the given information? Consider the base length from the leftmost weight to the rightmost weight. I believe this length varies which could change the answer to the question depending on what it was. And if the rectangle were balancing, I think the weights vary too which would change the answer. Am I just missing something here?sixnoreply@blogger.com