wu :: forums « wu :: forums - Nonagon diagonals » Welcome, Guest. Please Login or Register. Apr 24th, 2024, 12:32pm

RIDDLES SITE WRITE MATH! Home Help Search Members Login Register

wu :: forums    riddles    easy (Moderators: Icarus, Eigenray, SMQ, towr, Grimbal, william wu, ThudnBlunder)    Nonagon diagonals « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Nonagon diagonals  (Read 2640 times) NickH Senior Riddler     Gender: Posts: 341 Nonagon diagonals   « on: Sep 4th, 2004, 9:16am » Quote Modify In regular nonagon ABCDEFGHI, show that AB + AC = AE. IP Logged Nick's Mathematical Puzzles Barukh Uberpuzzler     Gender: Posts: 2276 Re: Nonagon diagonals   NonagonDiagonals.zip « Reply #1 on: Sep 5th, 2004, 4:00am » Quote Modify The attached figure includes everything needed for the proof. Note: the blue triangle was built equilateral. IP Logged Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: Nonagon diagonals   nonagon.gif « Reply #2 on: Sep 5th, 2004, 12:16pm » Quote Modify You, sir, are a genius; such a lovely solution!   However, you have inspired me to (perhaps) an even simpler solution...   As each interior angle is 140o, ABC=140o, BCA=20o (isosceles), ACD=140-20=120o, and so JCD=180-120=60o. We know that triangle JAF is at least isosceles (AJ=AF), and as triangle JCD is equilateral, so too is triangle JAF; hence AJ=AF=AE. As AJ=AC+CJ=AC+AB, we prove that AE=AC+AB. IP Logged mathschallenge.net / projecteuler.net NickH Senior Riddler     Gender: Posts: 341 Re: Nonagon diagonals   « Reply #3 on: Sep 5th, 2004, 4:56pm » Quote Modify Excellent proof, Sir Col!   Here's a follow-up.  In regular heptagon ABCDEFG, show that 1/AB = 1/AC + 1/AD. IP Logged Nick's Mathematical Puzzles TenaliRaman Uberpuzzler I am no special. I am only passionately curious.     Gender: Posts: 1001 Re: Nonagon diagonals   hepta.gif « Reply #4 on: Sep 6th, 2004, 9:02am » Quote Modify ::Apply Ptolemy's Theorem on the quad ACDE or on any other suitable quad:: IP Logged Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: Nonagon diagonals   « Reply #5 on: Sep 7th, 2004, 4:56pm » Quote Modify The best I could prove was that AD = (AC2–AB2)/AB.   TenaliRaman, that is an inspired approach! I would have never have thought of using AE. In fact, it took me a while to figure that AD=AE!     Using the theorem we get AC.DE+CD.AE=AD.CE, but as AB=CD=DE, AD=AE, and AC=CE, it gives AC.AB+AB.AD=AD.AC. Therefore AB(AC+AD)=AD.AC, so 1/AB=1/AC+1/AD.     I must say that I had fun trying to prove Ptolemy's theorem too! IP Logged mathschallenge.net / projecteuler.net Barukh Uberpuzzler     Gender: Posts: 2276 Re: Nonagon diagonals   HeptagonDiagonals.JPG « Reply #6 on: Sep 8th, 2004, 11:26pm » Quote Modify Here’s another solution of heptagon diagonals that does not use any additional theorems.   Construct CX || AB. Clearly, CX is another big diagonal, therefore ABCX is a rhombus. Also, construct CY = CX = AB.     We have: ACD = 4[pi]/7; triangle YCX is isosceles, YCX = [pi]/7, so AYX = 4[pi]/7. Therefore, YX || CD, and triangles AYX, ACD are similar.   But then AY/AX  = AC/AD => (AC-AB)/AB = AC/AD, and the result follows. IP Logged NickH Senior Riddler     Gender: Posts: 341 Re: Nonagon diagonals   « Reply #7 on: Sep 12th, 2004, 12:19pm » Quote Modify Very cool, Barukh!   Note that the nonagon puzzle can be solved using Ptolemy by considering cyclic quadrilateral ABDG, but I much prefer Sir Col's argument by symmetry. IP Logged Nick's Mathematical Puzzles Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board