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Bui Quang Tuan
Member since Jun2307

Oct1707, 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, E 5. 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


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Bui Quang Tuan
Member since Jun2307

Nov2507, 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


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Bui Quang Tuan
Member since Jun2307

Nov2607, 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 

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alexb
Charter Member
2141 posts 
Nov2607, 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 AaBC and CcAB, and let K = Aa ^ Cc. Extend AK to D so that AD/AK = Phi and extend CK to E so that CE/CK = Phi. 

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Bui Quang Tuan
Member since Jun2307

Nov2907, 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 BE Y = intersection of BD, CE Z = midpoint of CD T = intersection of BC, DE In trapezoid BCDE: T, X, Y, Z are collinear and XY/YZ = TX/TZ (*) ABYE is paralellogram so A, X, Y are collinear and AX = XY ADTC is paralellogram so A, Z, T are collinear and AZ = ZT We 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+2 2*t^2 2*t 2 = 0 t^2  t  1 = 0 From this t = Phi and we can get my construction of the pentagon. XY/YZ = Phi then S(BEY)/S(BYC) = EY/YC = Phi S(ABCDE) = S(ABE) + S(BEY) + S(CDE) + S(BYC) = 3 + 1/Phi = (5 + Sqrt(5))/2Best regards, Bui Quang Tuan


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