CTK Exchange Front Page Movie shortcuts Personal info Awards Reciprocal links Terms of use Privacy Policy Cut The Knot! MSET99 Talk Games & Puzzles Arithmetic/Algebra Geometry Probability Eye Opener Analog Gadgets Inventor's Paradox Did you know?... Proofs Math as Language Things Impossible My Logo Math Poll Other Math sit's Guest book News sit's Recommend this site         CTK Exchange

 Subject: "A challenging 'problemuzzle' involving a peculiar pentagon." Previous Topic | Next Topic
 Conferences The CTK Exchange College math Topic #648 Printer-friendly copy Email this topic to a friend Reading Topic #648
sarathian
Member since Sep-14-07
Sep-15-07, 09:36 AM (EST)    "A challenging 'problemuzzle' involving a peculiar pentagon." Let ABCDE be a convex but irregular pentagon such that each of the five peripheral triangles ABC, BCD, CDE, DEA and EAB has equal area , say x.The problem is to find the area of the pentagon in terms of x.P.S. - The condition of convexity is not required to be given as an additional condition; in fact the condition of each of the five peripheral triangles being equal in area, in itself, necessarily implies that the pentagon has to be convex.Science is the queen of human knowledge. Mathematics is the essence of science. Logic is the basis of mathematics. Geometry is the vehicle of intuition.

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

Subject     Author     Message Date     ID A challenging 'problemuzzle' involving a peculiar pentagon. sarathian Sep-15-07 TOP RE: A challenging 'problemuzzle' involving a peculiar ... Bui Quang Tuan Oct-17-07 1 RE: A challenging 'problemuzzle' involving a peculiar ... Bui Quang Tuan Nov-25-07 2 RE: A challenging 'problemuzzle' involving a peculiar ... alexb Nov-25-07 3 RE: A challenging 'problemuzzle' involving a peculiar ... alexb Nov-26-07 4 RE: A challenging 'problemuzzle' involving a peculiar ... Bui Quang Tuan Nov-26-07 5 RE: A challenging 'problemuzzle' involving a peculiar ... Bui Quang Tuan Nov-26-07 6 RE: A challenging 'problemuzzle' involving a peculiar ... alexb Nov-26-07 7 RE: A challenging 'problemuzzle' involving a peculiar ... Bui Quang Tuan Nov-29-07 8

 Conferences | Forums | Topics | Previous Topic | Next Topic
Bui Quang Tuan
Member since Jun-23-07
Oct-17-07, 11:13 AM (EST)    1. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #0

 Dear Radheyshyam Poddar and All,Could you and any other please give us some references about this interesting problem!Because I still don't know the name of this pentagon so for short please let me name it as Poddar pentagon.CONSTRUCTION OF PODDAR PENTAGON:- Construct any three collinear points X, Y, Z such that XY/YZ = (1 SquareRoot(5))/2 = Golden Ratio = GR. XY and YZ are directed segments. (or Y is on segment XZ)- A = reflection of Y in X.- Choose any point D not on the line XYZ- C = reflection of D in Z, C' = CY /\ AD, D' = DY /\ AC- B = reflection of D in midpoint of YD'- E = reflection of C in midpoint of YC'Some interesting results can be proved from this construction:1. ABCDE is Poddar pentagon.2. Area(ABCDE) = x*(5 SquareRoot(5))/2 = x*(GR 2)3. CY/YC' = DY/YD' = GR. Similar results with other similar segments in the pentagon.4. Area(DYC') = Area(CYD') = Area of other 3 respective triangles at A, B, E5. X is midpoint of BE and what is true with A, X, Y, Z can be true with similar points at other vertices of the polygon.6. Five diagonals of Poddar pentagon bound also one Poddar pentagon.Best regards,Bui Quang Tuan

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top Bui Quang Tuan
Member since Jun-23-07
Nov-25-07, 03:21 PM (EST)    2. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #1

 Dear All My Friends,Today I have found this interesting pentagon in problem 4.9 page 77, V. V. Prasolov, Zadachi po planimetrii, Part I, Moscow, 1st edition, 1986. I translate the problem and resolution here for your reference:Problem 4.9:Each diagonal of convex pentagon ABCDE cuts from it one triangle with area 1. Please calculate area of the pentagon ABCDE.Resolution (pp 85, 86):Because S(ABE) = S(ABC), points E and C are equidistant from AB and EC//AB. Similarly it can be shown that another diagonals are parallel with respective sides. Suppose P is intersection of BD and EC. If S(BPC) = x then S(ABCDE) = S(ABE) + S(EPB) + S(EDC) + S(BPC) = 3 + x. (S(EPB) = S(ABE) = 1 because ABPE is parallelogram). Now calculate x:S(BPC) / S(DPC) = BP / DP = S(EPB) / S(EPD) i.e. x / (1 - x) = 1/x. From this:x = (Sqrt(5) - 1)/2 and S(ABCDE) = (Sqrt(5) + 5)/2.Please note that in this book, S(X) denote as area of X.Best regards,Bui Quang Tuan

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top alexb
Charter Member
2141 posts
Nov-25-07, 07:55 PM (EST)    3. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #2

 Many thanks. Very simple and elegant.

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top alexb
Charter Member
2141 posts
Nov-26-07, 09:18 AM (EST)    4. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #3

 Are there irregular pentagons that satisfy the conditions of the problem?

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top Bui Quang Tuan
Member since Jun-23-07
Nov-26-07, 12:30 PM (EST)    5. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #4

 Dear Alex,Yes, they are many. I suggested one:- Construct any three collinear points X, Y, Z such that XY/YZ = (1 + Sqrt(5))/2 = Golden Ratio. XY and YZ are directed segments. (or Y is on segment XZ)- A = reflection of Y in X.- Choose any point D not on the line XYZ- C = reflection of D in Z, C' = CY /\ AD, D' = DY /\ AC- B = reflection of D in midpoint of YD'- E = reflection of C in midpoint of YC'ABCDE is one such pentagon.Best regards,Bui Quang Tuan

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top Bui Quang Tuan
Member since Jun-23-07
Nov-26-07, 01:19 PM (EST)    6. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #4

 Dear Alex,For more convenience and understanding we can construct this pentagon as following:On the median AZ of any triangle ADC construct one point Y such that:AY / YZ = (1+Sqrt(5))= 2*GoldenRatio(We can construct Y by use golden ratio construction)D' = intersection of DY, ACC' = intersection of CY, ADB = reflection of D in midpoint of YD'E = reflection of C in midpoint of YC'ABCDE is expected pentagonBest regards,Bui Quang Tuan

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top alexb
Charter Member
2141 posts
Nov-26-07, 10:03 PM (EST)    7. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #6

 Yes, thank you, Bui Quang Tuan.Sorry, I actually forgot about the problem and your original response.There is another simple construction:Let A, B, C be given.Draw Aa||BC and Cc||AB, and let K = Aa ^ Cc.Extend AK to D so that AD/AK = Phi andextend CK to E so that CE/CK = Phi.

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top Bui Quang Tuan
Member since Jun-23-07
Nov-29-07, 11:33 PM (EST)    8. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #2

 Dear Alex,I have found another solution for this problem, of course, not as nice as Prasolov's solution but it is interesting because it uses one wellknown fact:In a trapezoid: intersections of two sides, two diagonals and midpoints of two bottom are collinear and they divide the collinear segment by harmonic cross ratio.Suppose in the pentagon ABCDE:X = midpoint of BEY = intersection of BD, CEZ = midpoint of CDT = intersection of BC, DEIn trapezoid BCDE: T, X, Y, Z are collinear and XY/YZ = TX/TZ (*)ABYE is paralellogram so A, X, Y are collinear and AX = XYADTC is paralellogram so A, Z, T are collinear and AZ = ZTWe calculate (*)XY/YZ = TX/TZ = (TZ + ZY + YX)/TZ = 1 + (ZY + YX)/AZ =1 + (YZ + XY)/(2*XY + YZ)Denote XY/YZ = t then YX = t*YZ and:t = 1 + (YZ + t*YZ)/(2*t*YZ+YZ)t = 1 + (1+t)/(2*t+1)t*(2*t + 1) = 3*t+22*t^2 -2*t -2 = 0t^2 - t - 1 = 0From this t = Phi and we can get my construction of the pentagon.XY/YZ = Phi then S(BEY)/S(BYC) = EY/YC = PhiS(ABCDE) = S(ABE) + S(BEY) + S(CDE) + S(BYC) = 3 + 1/Phi = (5 + Sqrt(5))/2Best regards,Bui Quang Tuan

Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

 Conferences | Forums | Topics | Previous Topic | Next Topic
 Select another forum or conference Lobby The CTK Exchange (Conference)   |--Early math (Public)   |--Middle school (Public)   |--High school (Public)   |--College math (Public)   |--This and that (Public)   |--Guest book (Protected)   |--Thoughts and Suggestions (Public) Educational Press (Conference)   |--No Child Left Behind (Public)   |--Math Wars (Public)   |--Mathematics and general education (Public) You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.  