2015 AMC 10B Problems/Problem 22

Problem

In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$ . What is $FG + JH + CD$ ? $[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("$A$",A,N); label("$B$",B,E); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E1,W); label("$F$",F,NW); label("$G$",G,NE); label("$H$",H,E); label("$I$",I,S); label("$J$",J,W); [/asy]$

Solution

Triangle $AFG$ is isosceles, so $AG=AF=1$ . Using the symmetry of pentagon $FGHIJ$ , notice that $\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG$ . Therefore, $JH=AF=1$ .

Since $\triangle AJH \sim \triangle AFG$ , $\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1$ . From this, we get $FG=\frac{\sqrt{5} -1}{2}$ .

Since $\triangle DIJ \cong \triangle AFG$ , $DJ=DI=AF=1$ . Since $\triangle AFG \sim ADC$ , $\frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}$ . Thus, $CD = (\sqrt{5} - 1) + (\frac{\sqrt{5} - 1}{2})^2 = \frac{\sqrt{5} +1}{2}$

Therefore, $FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2015 AMC 10B Problems/Problem 22

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