MathDB
Pentagon Ratios

Source: 2015 AMC 10B #22

February 26, 2015
ratioAsymptotetrigonometryUSAMTSgeometryparallelogramangle bisector

Problem Statement

In the figure shown below, ABCDEABCDE is a regular pentagon and AG=1AG=1. What is FG+JH+CDFG+JH+CD? [asy] import cse5;pathpen=black;pointpen=black; size(2inch); pair A=dir(90), B=dir(18), C=dir(306), D=dir(234), E=dir(162); D(MP("A",A,A)--MP("B",B,B)--MP("C",C,C)--MP("D",D,D)--MP("E",E,E)--cycle,linewidth(1.5)); D(A--C--E--B--D--cycle); pair F=IP(A--D,B--E), G=IP(B--E,C--A), H=IP(C--A,B--D), I=IP(D--B,E--C), J=IP(C--E,D--A); D(MP("F",F,dir(126))--MP("I",I,dir(270))--MP("G",G,dir(54))--MP("J",J,dir(198))--MP("H",H,dir(342))--cycle); [/asy] <spanclass=latexbold>(A)</span>3<spanclass=latexbold>(B)</span>1245<spanclass=latexbold>(C)</span>5+253<spanclass=latexbold>(D)</span>1+5<spanclass=latexbold>(E)</span>11+11510<span class='latex-bold'>(A) </span> 3 \qquad<span class='latex-bold'>(B) </span> 12-4\sqrt5 \qquad<span class='latex-bold'>(C) </span> \dfrac{5+2\sqrt5}{3} \qquad<span class='latex-bold'>(D) </span> 1+\sqrt5 \qquad<span class='latex-bold'>(E) </span> \dfrac{11+11\sqrt5}{10}