AMC 10B/12B 2023 (Problem 25)A regular pentagon with area 1+51+\sqrt51+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?(A) 4−5\text{(A)}\;4-\sqrt5(A)4−5(B) 5−1\text{(B)}\;\sqrt5-1(B)5−1(C) 8−35\text{(C)}\;8-3\sqrt5(C)8−35(D) 5+12\text{(D)}\;\frac{\sqrt5+1}{2}(D)25+1(E) 2+53\text{(E)}\;\frac{2+\sqrt5}{3}(E)32+5Related TopicsToolkit 77 — sin, cos, tan, cot — Special AnglesToolkit 86 — Regular PolygonsHints (7)Hint 1OOO is the center of both ABCDEABCDEABCDE and A′B′C′D′E′A'B'C'D'E'A′B′C′D′E′.Hint 2OMOMOM corresponds to ONONON.Hint 3[A′B′C′D′E′][ABCDE]=(OMON)2\frac{[A'B'C'D'E']}{[ABCDE]}=\left(\frac{OM}{ON}\right)^2[ABCDE][A′B′C′D′E′]=(ONOM)2Hint 4OMON=12(2OMON)=12(OAON)\frac{OM}{ON}=\frac12\left(\frac{2OM}{ON}\right)=\frac12\left(\frac{OA}{ON}\right)ONOM=21(ON2OM)=21(ONOA)Hint 5∠EAB=180(5−2)5=108∘⟹∠OAL=54∘\angle EAB=\frac{180(5-2)}{5}=108^\circ\quad\Longrightarrow\quad\angle OAL=54^\circ∠EAB=5180(5−2)=108∘⟹∠OAL=54∘OAON=OAOL=1cos36∘\frac{OA}{ON}=\frac{OA}{OL}=\frac{1}{\cos36^\circ}ONOA=OLOA=cos36∘1Hint 6By Hints 3, 4, and 5 and Toolkit 77 — sin, cos, tan, cot — Special Angles,[A′B′C′D′E′][ABCDE]=(OMON)2=14(OAON)2=14⋅1cos236∘\frac{[A'B'C'D'E']}{[ABCDE]}=\left(\frac{OM}{ON}\right)^2=\frac14\left(\frac{OA}{ON}\right)^2=\frac14\cdot\frac{1}{\cos^2 36^\circ}[ABCDE][A′B′C′D′E′]=(ONOM)2=41(ONOA)2=41⋅cos236∘1=14(41+5)2=14⋅166+25=23+5=2(3−5)9−5=3−52=\frac14\left(\frac{4}{1+\sqrt5}\right)^2=\frac14\cdot\frac{16}{6+2\sqrt5}=\frac{2}{3+\sqrt5}=\frac{2(3-\sqrt5)}{9-5}=\frac{3-\sqrt5}{2}=41(1+54)2=41⋅6+2516=3+52=9−52(3−5)=23−5Hint 7[A′B′C′D′E′]=[ABCDE](3−52)=(1+5)(3−52)[A'B'C'D'E']=[ABCDE]\left(\frac{3-\sqrt5}{2}\right)=(1+\sqrt5)\left(\frac{3-\sqrt5}{2}\right)[A′B′C′D′E′]=[ABCDE](23−5)=(1+5)(23−5)=3−5+35−52=25−22=5−1=\frac{3-\sqrt5+3\sqrt5-5}{2}=\frac{2\sqrt5-2}{2}=\sqrt5-1=23−5+35−5=225−2=5−1Final Answer(B) 5−1\sqrt5-15−1
