Problem:
A regular pentagon with area 1+5​ 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. 4−5​
B. 5​−1
C. 8−35​
D. 21+5​​
E. 32+5​​
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+5​S​=(AFBD​)2
By symmetry, ABCD must be a rhombus, and AC bisects ∠BAD (and ∠GAF ). Then
∠BAD=∠BCD=5360∘​=72∘ so ∠BAC=272∘​=36∘. Also, ∠GAF=53⋅180∘​=108∘, so ∠CAF=2108∘​=54∘. Lastly, ∠BAF=∠BAC+∠CAF=36∘+54∘=90∘.​
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+5​​ ( φ, the golden ratio). Therefore
AFBD​=BEBD​=φ1​=1+5​2​,
and
1+5​S​=(AFBD​)2=(1+5​2​)2.
The requested area is S=1+5​22​=(B)5​−1​.
To prove that the diagonal to side ratio for a regular pentagon equals 21+5​​, 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 QT​.
Because each diagonal is parallel to a side, parallelogram TURS is formed with sides of length 1, leaving segments PU and QU​ to have length d−1. Because PQRT is a trapezoid, △RTU and △PQU are similar, and 1d​=d−11​. This implies that d2−d−1=0, and the quadratic formula gives the positive solution d=21+5​​=φ.
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 ∠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=2⋅AFr​, and the area is reduced by the factor f2. The scale factor f can be evaluated using similar triangles. Triangle △ABC is isosceles with angles 36∘,72∘,72∘ and sides AB=AC=r. Segment BD bisects ∠ABC creating isosceles triangles △ABD with two 36∘ angles and △BDC with two 72∘ angles. Choose units so that 1=BC=BD=AD. Notice that △BDC∼△ABC and is smaller by a factor of r. Because AD+DC=AC, it follows that 1+r1​=r. Because r>1, this simplifies to r=21+5​​. Now
