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CTK Exchange
Bui Quang Tuan
Member since Jun-23-07
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Oct-17-07, 11:13 AM (EST) |
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1. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #0
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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 Jun-23-07
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Nov-25-07, 03:21 PM (EST) |
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2. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #1
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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 Jun-23-07
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Nov-26-07, 12:30 PM (EST) |
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5. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #4
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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 |
Nov-26-07, 10:03 PM (EST) |
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7. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #6
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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 and
extend CK to E so that CE/CK = Phi. |
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Bui Quang Tuan
Member since Jun-23-07
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Nov-29-07, 11:33 PM (EST) |
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8. "RE: A challenging 'problemuzzle' involving a peculiar ..."
In response to message #2
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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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