tag:blogger.com,1999:blog-5730391.post144822316966369427..comments2024-05-22T08:40:05.849-07:00Comments on Blaine's Puzzle Blog: NPR Sunday Puzzle (Oct 7, 2012): Hexagon Diagonals - Count the TrianglesBlainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.comBlogger87125tag:blogger.com,1999:blog-5730391.post-58471238089222093442012-10-14T02:25:01.358-07:002012-10-14T02:25:01.358-07:00I've submitted an answer. I believe I figured...I've submitted an answer. I believe I figured out the property, but I had to admit that my 6th word was unknown.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-19796605990478913422012-10-14T01:59:03.052-07:002012-10-14T01:59:03.052-07:00It is up now and I solved it as I read the questio...It is up now and I solved it as I read the question. I am going back to bed though and will leave you all waiting until I get up in the morning for a clue, so please remain calm until then.skydiveboyhttps://www.blogger.com/profile/17174073226290431753noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-66006317471362871132012-10-14T01:48:24.317-07:002012-10-14T01:48:24.317-07:00This comment has been removed by the author.skydiveboyhttps://www.blogger.com/profile/17174073226290431753noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-73333535691420054772012-10-14T00:53:11.091-07:002012-10-14T00:53:11.091-07:00After perhaps the earliest update of the NPR Weeke...After perhaps the <b><i>earliest</i></b> update of the NPR Weekend Sunday Puzzle page last week, this week we get another <b><i>very late</i></b> update. The NPR website had <b><i>still</i></b> not yet been updated as I was typing this.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-15668945698398839002012-10-11T17:33:28.821-07:002012-10-11T17:33:28.821-07:00I think it will be many more than that. This puzz...I think it will be many more than that. This puzzle really wasn't tricky; it was completely solvable by brute force, and counting to 82 isn't all that hard. You just needed a lack of anything better to do with your life this week.janhttps://www.blogger.com/profile/05927176621372532733noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-29249080915346143032012-10-11T17:11:15.859-07:002012-10-11T17:11:15.859-07:00I hava a P.S. too!
AbqGuerrilla, don't pay her...I hava a P.S. too!<br />AbqGuerrilla, don't pay her too much mind, she's just a camp follower who sometimes gets her panties in a bunch over long posts. Maybe she just needs a long post to set her right. skydiveboyhttps://www.blogger.com/profile/17174073226290431753noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-34226765423806082792012-10-11T16:29:05.412-07:002012-10-11T16:29:05.412-07:00AbqGkuerrilla:
And you didn't notice Unknown n...AbqGkuerrilla:<br />And you didn't notice Unknown neglected to use a question mark at the end of his sentence? Yeah, like that one.<br /><br />I don't know about the hook-up; my harem is upset with me for using their names as points on the hexagon puzzle and a few were left over and not recognized. Anyway it was a disaster of confusion and put a hex on my attempt to solve it that way.skydiveboyhttps://www.blogger.com/profile/17174073226290431753noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-30914240716857437802012-10-11T16:26:23.280-07:002012-10-11T16:26:23.280-07:00My only comment on this puzzle is a question: how...My only comment on this puzzle is a question: how many people submitted the correct answer? I say fewer than 50.Ruthhttps://www.blogger.com/profile/08414999868454070433noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-53420477061788383852012-10-11T16:09:55.738-07:002012-10-11T16:09:55.738-07:00I just realized...
since BF is removed, the drawi...I just realized...<br /><br />since BF is removed, the drawing instructions in those cases should from start at one end of the disconnected line and end at the other end.<br /><br />So replace ACFBD with BDACF, and ACFBE becomes BEACF, and lastly ADFBE becomes BEADF.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-41009018985609893542012-10-11T15:59:55.868-07:002012-10-11T15:59:55.868-07:00I too wish to compliment your video.
I do have on...I too wish to compliment your video.<br /><br />I do have one suggestion. It's about how you <b><i>name</i></b> your 5-vertices star figures.<br /><br />Instead of naming them ABCDF, ABCEF, ABDEF<br />and ABCDE, ACDEF, BCDEF; I would name them with the letters arranged in the order in which connecting lines would draw the figures.<br /><br />Thus: ACFBD, ACFBE, ADFBE<br />and: ACEBD, ADFCE, BDFCE<br /><br />I've tried to arrange my suggested names to exactly correspond to your listing. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-75779855043094148522012-10-11T15:26:47.598-07:002012-10-11T15:26:47.598-07:00P.S. Can someone please sell Unknown an apostroph...P.S. Can someone please sell Unknown an apostrophe for his last posting. I'm overly possessive and I used all mine.AbqGuerrillahttps://www.blogger.com/profile/08439641600563812511noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-33807671949486265462012-10-11T15:21:37.899-07:002012-10-11T15:21:37.899-07:00Unknown: We ain't cloggin', bruddah. We ...Unknown: We ain't cloggin', bruddah. We be bloggin! <br /><br />Hey, SDB: Whaddya say? We could go Dutch this Saturday night. I know a place where checkout time is not until noon, so we could listen to the Puzzle Man over continental b'fast. I'll even take the foldout couch. AbqGuerrillahttps://www.blogger.com/profile/08439641600563812511noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-72516622754634484472012-10-11T14:37:48.807-07:002012-10-11T14:37:48.807-07:00Ok, this certainly isn't the classiest or most...Ok, this certainly isn't the classiest or most professional html file, but it displays the above mentioned diagram and then gives my list in a much better format.<br /><br /><a href="http://users.az.com/~jwaters/Puzzles/2012-10-07.htm" rel="nofollow">http://users.az.com/~jwaters/Puzzles/2012-10-07.htm</a><br /><br />Enjoy, everybody.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-74914345415650496442012-10-11T14:31:45.339-07:002012-10-11T14:31:45.339-07:00I got the same list. Using the alpha labels, and ...I got the same list. Using the alpha labels, and keeping the triangle names alphabetically sorted helped avoid miscounting, I found.janhttps://www.blogger.com/profile/05927176621372532733noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-40879922957555399382012-10-11T13:11:05.733-07:002012-10-11T13:11:05.733-07:00In all 10 possiblities, only the first option coul...In all 10 possiblities, only the first option could've been begun by either of the three involved vertices. <br />So we may throw out possibility #4 (just a repeat of #1), as well as all first options for #5 and #s 7-10.<br />Every possibility with only a single "move 2" or a single "move 4" needs only to deduct 1 from the possible starting vertices.<br />Every possibility with two "move 2"'s or a "move 2" and a "move 4" or two "move 4"'s needs to deduct 2 from their possible starting vertices.<br />The first of possibility #6's options has a count of exactly 1. <br />The count of double-cross triangles in #6 can be explained as the top 3 little triangles of the would've-been Star of David within the hexagon but for the removed line.<br />I count triangles for the puzzle as stated, and also for the hexagon WITH the removed line. <br />I consider the hexagon oriented with single top and bottom vertices and the removed line joining the bottom's neighbors.<br />In the second column, I keep tallies of added triangles above and below the removed line.<br /><br />So: Puzzle as stated: <br /> 1. adds 5<br /> 2. adds 5<br /> 3. adds 5<br /> 5. adds 4 (a "move 2" AND a "move 4") <br /> 6. adds 1 & 4 (two "move 2"'s), then adds 3, so adds 8<br /> 7. adds 4 (two "move 2"'s), then adds (10-1 = 9), so adds 13 <br /> 8. adds 6 & 5 (the "move 4"), so adds 11<br /> 9. adds 4 & 5, then also adds 9, so adds 18<br />10. adds 4, 5 & 4, so adds 13<br /><br />So I come up with 82. <br />Without that removed line:<br /> 1. adds 6 (+ 1 lower)<br /> 2. adds 6 (+ 1 upper)<br /> 3. adds 6 (+ 1 upper)<br /> 5. adds 6 (+ 1 upper & + 1 lower)<br /> 6. adds 2 & 6, then adds 2*3, so adds 14 (1^ & 2^, then 2^+1v, so + 5 upper + 1 lower)<br /> 7. adds 6, then adds 2*6, so adds 18 (2^, then 2^+1v, so + 4 upper + 1 lower)<br /> 8. adds 6 & 6, so adds 12 (+ 1 lower)<br /> 9. adds 6 & 6, then also adds 12, so adds 24 (2^ & 1^, then 2^+1v, so + 5 upper + 1 lower)<br /> 10. adds 6, 6 & 6, so adds 18 (1^+1v, 1v & 2v, so + 1 upper + 4 lower)<br /><br /><br />Total here would be 110. (+ 18 upper + 10 lower)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-84357881273156833082012-10-11T12:58:56.299-07:002012-10-11T12:58:56.299-07:00Perhaps the best way to count is to first ignore t...Perhaps the best way to count is to first ignore the exception and count ALL, then count the triangles REMOVED as the exception applies.<br /><br />Observation: Any triangle with all 3 of its vertices interior would require all 6 vertices of the hexagon; but that would result in the 3 opposing lines all crossing at the center, thus...<br /><br /> Any triangle can be described by either (1) initial vertix, vertix traveled to, 3rd vertix traveled to, and 4th vertix traveled to (possibly crossing 1st line),<br /> or (2) vertix angle, defined by first 3 vertices above and a line crossing BOTH lines to and from the angle vertix.<br /><br />Without loss of generality, let's move clockwise.<br /><br />From ANY vertix:<br /><br /> 1. Move 1, move 1, then ONLY possible triangle result ==> move 4, no double-cross triangles<br /> 2. Move 1, move 2, then ONLY possible triangle result ==> move 3, no double-cross triangles<br /> 3. Move 1, move 3, then ONLY possible triangle result ==> move 2, no double-cross triangles<br /> 4. Move 1, move 4, then ONLY possible triangle result ==> move 1, no double-cross triangles<br /> 5. Move 2, move 1, then TWO possible triangle results ==> move 3 OR move 4, no double-cross triangles<br /> 6. Move 2, move 2, then TWO possible triangle results ==> move 2 OR move 3, one double-cross triangle for 2 of the 3 vertices plus another double-cross for the would've-been triangle.<br /> 7. Move 2, move 3, then TWO possible triangle results ==> move 1 OR move 2, two double-crosses for each counted vertix, but one vertix lacks 1 of these counted double-crosses.<br /> 8. Move 3, move 1, then THREE possible triangle results ==> move 2, move 3, or move 4, no double-cross triangles<br /> 9. Move 3, move 2, then THREE possible triangle results ==> move 1, move 2, or move 3, double-cross count same as #7 above.<br />10. Move 4, move 1, then FOUR possible triangle results ==> move 1, move 2, move 3 or move 4, no double-cross triangles<br /><br />(to be continued)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5730391.post-50207152526984311982012-10-11T12:46:19.456-07:002012-10-11T12:46:19.456-07:0082 Total
Subtotals of sets in order of type and si...82 Total<br />Subtotals of sets in order of type and size from large to small:<br />10 Equilateral, 1, 6, 3<br />18 Isosceles, 4, 10, 4<br />54 Right, 10, 8, 8, 20, 8<br /><br />My wrong hint:<br />Blaine used the word "cube". 4³ + 10 = 74. (off by 8)<br /><br />The set of triangles I failed to see were the ones using 2 segments of the short diagonals as a hypotenuse.hughhttps://www.blogger.com/profile/16914509834442545746noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-64893539759821433502012-10-11T12:24:09.320-07:002012-10-11T12:24:09.320-07:00My first observation was that all triangles used a...My first observation was that all triangles used at least one of the outside vertices. Picking 3, 4 or 5 vertices ends up enumerating all the cases and provides some logical grouping, I felt. But I also did what you did and went back and did an exhaustive listing.Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-66753789753022241312012-10-11T12:17:41.242-07:002012-10-11T12:17:41.242-07:00Indeed, I was again inspired to make a "count...Indeed, I was again inspired to make a <a href="https://vimeo.com/51197832" rel="nofollow">"count the triangles" video</a> for this.Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-18694493834160273682012-10-11T12:13:25.074-07:002012-10-11T12:13:25.074-07:00Good video. For demonstration purposes, there'...Good video. For demonstration purposes, there's any number of ways to sort the triangles; in the end, I chose to go with the (frenetic from a geometric perspective) ordering that results from an alphabetic sorting. But in practice, I counted the triangles by exhaustively listing all (14 choose 3) = 364 possible combinations of line segments AB, AC, … , EF, and then ruling out the degenerate cases. I also employed a few other sorting techniques as a check to make sure I hadn't missed any.PlannedChaoshttps://www.blogger.com/profile/18106394176789111856noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-81689514129398345292012-10-11T12:08:00.794-07:002012-10-11T12:08:00.794-07:00I agree on the 110 triangles in a fully filled hex...I agree on the 110 triangles in a fully filled hexagon (based on the following <a href="https://cs.uwaterloo.ca/journals/JIS/sommars/newtriangle.html" rel="nofollow">Waterloo Journal Article</a>. There seems to be a discrepancy of 9 triangles in your count, so either you double-counted when adding triangles, or undercounted when counting triangles using the removed diagonal. It will be interesting to find out.Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-5073594989085578162012-10-11T12:07:45.840-07:002012-10-11T12:07:45.840-07:00Clues explained: My comment: "Didn't mea...Clues explained: My comment: "Didn't mean to lead anyone astray" contained the metal whose atomic number is 82. (Similarly, my use of the word "precious" in an earlier post referred to my initial (incorrect) count of 79.)Lorenzohttps://www.blogger.com/profile/08116815111532628378noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-69924025357930838652012-10-11T12:01:45.176-07:002012-10-11T12:01:45.176-07:00Answer = 82 triangles:
A video explanation is now ...Answer = 82 triangles:<br />A <a href="https://vimeo.com/51197832" rel="nofollow">video explanation</a> is now online and no longer password protected.Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-42152662856752344512012-10-11T12:00:02.821-07:002012-10-11T12:00:02.821-07:00EIGHTY-TWO (82)
16 triangles of one
23 triangles ...EIGHTY-TWO (82)<br /><br />16 triangles of one<br />23 triangles of two shapes<br />10 triangles of three shapes<br />14 triangles of four shapes<br /> 6 triangles of five shapes<br /> 6 triangles of six shapes<br /> 2 triangles of seven shapes<br /> 4 triangles of eight shapes<br /> 1 triangle of 10 shapes <br />__<br />82 Total Triangles<br /><br />My Hint:<br />"At 2 this afternoon I read Blaine's post that he'd just posted above in this thread and it almost made me give up, but now I'm glad I didn't."<br /><br />At 2 = 82 phonetically.<br /><br />My initial mistake with this puzzle was being hasty with counting the single triangles. I ended up counting 4 nonexistent triangles. What I did was to quickly count all the single enclosures thinking all of them were triangles and failing to notice that four had 4 sides. I was very careful with all the more complex triangles and even noticed the large one made up of ten smaller shapes. Later I happened to glance at the diagram and I realized my error. I wonder how many others made this same mistake.<br />skydiveboyhttps://www.blogger.com/profile/17174073226290431753noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-1346892592943223382012-10-11T12:00:01.153-07:002012-10-11T12:00:01.153-07:00The answer is 91 triangles. Of these, 16 have all ...The answer is 91 triangles. Of these, 16 have all three vertices at a vertex of the hexagon ("three diagonal endpoints") , 50 have one vertex inside the hexagon and two vertices at vertices of the hexagon ("four diagonal endpoints"), and 25 have two vertices inside the the hexagon and one vertex at a vertex of the hexagon ("five diagonal endpoints"). See Journal of Integer Sequences, Vol. 1 (1998), Article 98.1.5, The Number of Triangles Formed by Intersecting Diagonals of a Regular Polygon, by Sommars and Sommars. This paper shows that if all the diagonals are included there are 110 triangles formed (20, 60 and 30 of the above types). Removing one short interior diagonal of the hexagon removes 19 triangles, and leaves 91.<br />EKWhttps://www.blogger.com/profile/07620788732511183218noreply@blogger.com