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$",J,W); [/asy]$ $\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10}$

Solution 1

Triangle $AFG$ is isosceles, so $AG=AF=1$ . $FJ = FG$ since $\triangle FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$ , notice that $\triangle JHG \cong \triangle AFG$ . Therefore, $JH=AF=1$ .

Since $\triangle AJH \sim \triangle AFG$ , $\frac{JH}{AF+FJ}=\frac{FG}{FA}$ . $\frac{1}{1+FG} = \frac{FG}1$ $1 = FG^2 + FG$ $FG^2+FG-1 = 0$ $FG = \frac{-1 \pm \sqrt{5} }{2}$

So, $FG=\frac{-1 + \sqrt{5}}{2}$ since $FG$ must be greater than 0.

Notice that $CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}$ .

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

Solution 2 (Trigonometry)

Note that since $ABCDE$ is a regular pentagon, all of its interior angles are $108^\circ$ . We can say that pentagon $FGHIJ$ is also regular by symmetry. So, all of the interior angles of $FGHIJ$ are $108^\circ$ . Now, we can angle chase and use trigonometry to get that $FG=2\sin18^\circ$ , $JH=2\sin18^\circ*(2\sin18^\circ+1)$ , and $DC=2\sin18^\circ*(2\sin18^\circ+2)$ . Adding these together, we get that $FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)$ . Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to $8\sin18^\circ*(1+\sin18^\circ)$ , but we can find that this is closest to $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$ .

Solution 3

When you first see this problem you can't help but see similar triangles. But this shape is filled with $36 - 72 - 72$ triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of $FG$ so we can apply similar triangles easily. To simplify the process lets write $FG$ as $x$ .

First what is $JH$ in terms of $x$ , also remember $AJ = 1+x$ : $\frac{JH}{1+x}=\frac{x}{1}$ $JH$ = ${x}^2+x$

Next, find $DC$ in terms of $x$ , also remember $AD = 2+x$ : $\frac{DC}{2+x}=\frac{x}{1}$ $DC$ = ${x}^2+2x$

So adding all the $FG + JH + CD$ we get $2{x}^2+4x$ . Now we have to find out what x is. For this, we break out a bit of trig. Let's look at $\triangle AFG$ By the law of sines: $\frac{x}{sin(36)}=\frac{1}{sin(72)}$ $x=\frac{sin(36)}{sin(72)}$

Now by the double angle identities in trig. $sin(72) = 2sin(36)cos(36)$ substituting in $x=\frac{1}{2cos(36)}$

A good thing to memorize for AMC and AIME is the exact values for all the nice sines and cosines. You would then know that: $cos(36)$ = $\frac{1 + \sqrt{5}}{4}$

so now we know: $x = \frac{2}{1+\sqrt{5}} = \frac{-1+\sqrt{5}}{2}$

Substituting back into $2{x}^2+4x$ we get $FG+JH+CD=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

Solution 4 (Answer choices)

Notice that $A$ is trisected, meaning that $AG=BH=EJ=JH=1$ . Since $JH=1$ , and the other lines we are supposed to solve for do not look like they can contribute to the whole number value, it is likely that $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$ is our answer.

Note: Only do this if low on time since there could potentially be a weird figure affecting the $1$ .

Solution 5

Because basically everything in pentagon stuff is in ratio of big Phi or small Phi, we use logic to find that the answer is big Phi $+ 1 +$ small Phi = $(\sqrt5+1)/2 + 1 + (\sqrt5-1)/2$ = $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

-Lcz

Solution 6

Notice that $\angle AFG=\angle AFB$ and $\angle FAG=\angle ABF$ , so we have $\bigtriangleup AFG=\bigtriangleup BAF$ . Thus $\frac{AF}{FG}=\frac{FG+GB}{AF}$ $\frac{1}{FG}=\frac{FG+1}{1}$ Solving the equation gets $FG=\frac{\sqrt{5}-1}{2}$ .

Since $\bigtriangleup AFG=\bigtriangleup AJH$ $\frac{AF}{FG}=\frac{AF+FJ}{JH}$ $\frac{AF}{FG}=\frac{AF+FG}{JH}$

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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