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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## Monday, May 19, 2008

###
Elearn Geometry Problem 63

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 63

Regular Heptagon, side and diagonals. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

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.