 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Pentagon Ratio (Posted on 2013-03-30) A piece of paper has the precise shape of a regular pentagon. The paper is folded such that a vertex of the pentagon coincides with the midpoint of that side of the pentagon which is farthest from the said vertex.

Determine the ratio of the length of the crease to the length of a side of the pentagon.

 See The Solution Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution. Comment 2 of 2 | Call the pentagon ABCDE and let its sides be length 1 unit.
Construct a line parallel to AE through C.  Extend sides AB and DE.
Call the intersection of AB and the parallel line point F.
Call the intersection of DE and the parallel line point G.
Call the intersection of AB and DE point H.
Construct the line thought C perpendicular to AE.
Call the intersection of this line and AE point I.
I is the midpoint of AE by symmetry.

If we crease the pentagon as instructed we create the perpendicular bisector of CI.
Call the points of intersection of this line with AB, CI and DE points X, Y and Z respectively.
We seek length XZ.

Since triangles BCF and DCG are isosceles, FC=1 and CG=1 so FG=2
Triangles FHG and AHE are similar with ratio of similitude 2, so I is the midpoint of CH.
Since Y is the midpoint of CI, YH= 3/2 * IH.
Triangles AHE and XHZ are similar with ratio of similitude 3/2, so XZ = 3/2 * AE

The ratio is 3/2.

 Posted by Jer on 2013-04-01 13:06:07 Please log in:
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