Problem:
A regular pentagon with area 1+5β is printed on paper and cut out. All five vertices are folded to the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
Answer Choices:
A. 4β5β
B. 5ββ1
C. 8β35β
D. 21+5ββ
E. 32+5ββ
Solution:
The folds are the perpendicular bisectors of the segments connecting the vertices of the pentagon to its center. Let C be the center of the pentagon, let G,A, and F be three consecutive vertices in order, let D be the intersection of the fold lines for vertices A and F, let B be the intersection of the fold lines for vertices A and G, and let E be the other intersection of fold lines on the fold line for F. Draw quadrilaterals ABCD and ABEF, as shown.
Let S be the area of the small pentagon. Because the two pentagons are similar, the ratio of their areas is the square of the ratio of their side lengths. Therefore
1+5βSβ=(AFBDβ)2
By symmetry, ABCD must be a rhombus, and AC bisects β BAD (and β GAF ). Then
β BAD=β BCD=5360ββ=72β so β BAC=272ββ=36β. Also, β GAF=53β 180ββ=108β, so β CAF=2108ββ=54β. Lastly, β BAF=β BAC+β CAF=36β+54β=90β.β
Therefore quadrilateral ABEF is a rectangle and BE=AF. It is known (and proved below) that the ratio of a diagonal of a regular pentagon to its side is 21+5ββ ( Ο, the golden ratio). Therefore
AFBDβ=BEBDβ=Ο1β=1+5β2β,
and
1+5βSβ=(AFBDβ)2=(1+5β2β)2.
The requested area is S=1+5β22β=(B)5ββ1β.
To prove that the diagonal to side ratio for a regular pentagon equals 21+5ββ, let regular pentagon PQRST have sides of length 1 and diagonals of length d, and let U be the point of intersection of the line segments PR and QTβ.
Because each diagonal is parallel to a side, parallelogram TURS is formed with sides of length 1, leaving segments PU and QUβ to have length dβ1. Because PQRT is a trapezoid, β³RTU and β³PQU are similar, and 1dβ=dβ11β. This implies that d2βdβ1=0, and the quadratic formula gives the positive solution d=21+5ββ=Ο.
OR
Inscribe the regular pentagon in a circle of radius r. Let A be the center of the circle, let C be a vertex of the pentagon, and let B be the midpoint of the minor arc from C to one of the vertices adjacent to C. Let D be the intersection of the bisector of β ABC with AC. Let E be the midpoint of AB, and let F be the intersection of AB with the pentagon. Note that AF is the center-to-side distance of the pentagon. See the figure.
The linear dimensions are reduced by the factor f=2β AFrβ, and the area is reduced by the factor f2. The scale factor f can be evaluated using similar triangles. Triangle β³ABC is isosceles with angles 36β,72β,72β and sides AB=AC=r. Segment BD bisects β ABC creating isosceles triangles β³ABD with two 36β angles and β³BDC with two 72β angles. Choose units so that 1=BC=BD=AD. Notice that β³BDCβΌβ³ABC and is smaller by a factor of r. Because AD+DC=AC, it follows that 1+r1β=r. Because r>1, this simplifies to r=21+5ββ. Now
