See complete Problem 63
Regular Heptagon, side and diagonals. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 63
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See complete Problem 63
Regular Heptagon, side and diagonals. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Let AB = CD = DE = s, AC = CE = c, AD = AE = d.
ReplyDeleteApplying Ptolemy's Theorem to cyclic quadrilateral ACDE, we obtain cs + ds = cd.
Then s(c + d) = cd, and so 1/s = (c + d)/cd = 1/d + 1/c.
Therefore, in regular heptagon ABCDEFG, 1/s = 1/c + 1/d
Problem 63 solution. This solution was submitted by Dan Suttin from San Antonio, TX
ReplyDeleteThanks Dan.
Apply inversion on A with any inversion radius (A -> A' ; B -> B' ...etc)
ReplyDeleteThen we can rewrite the statement by proving A'B'=A'C'+A'D'.
The inversion image is easy : AB,AC,AD remains unchanged while the circumcircle becomes a line B'C'D' not passing through A.
By the conformal properties, ∠BAC=∠B'A'C'=∠C'A'D'=∠A'B'C'=180/7
Construct a point E' on A'B' such that A'E' = A'C', so we simply need to prove E'B'=A'D'.
This is actually quite obvious.
A'C'=B'C'
∠E'B'C'=∠C'A'D'=180/7
∠B'E'C'=720/7=∠B'D'A'
So ΔB'C'E'=ΔA'C'D', and hence E'B'=A'D
Q.E.D.
Extend segments AB and DC to intersect outside of the circle in I (see graph).
ReplyDeleteABCD cyclic quadrilateral implies <IBC = <ADC = <BAD = <ICB = 2π/7, so ΔIAD and ΔIBC are similar and IB/BC = IA/AD (1).
Consider ΔCAE and ΔIAD : they are equal (isoceles with same base length d and same base angles 2π/7). So IA = CA = c and from (1) IB = s*c/d.
Also, IA = AB+IB = s+IB so IB = c-s.
So the following holds :
c-s = s*c/d
c = s+s*c/d
Dividing both terms by s*c : 1/s = 1/c + 1/d QED
Hello, the graph supporting my proof can be found there:
ReplyDeletehttps://drive.google.com/file/d/1MwCoNGh443qVvTIBlmLH4dbLLyRIYGMq/view
Greg
https://www.youtube.com/watch?v=u4m2Ewj1Vak
ReplyDelete