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Difference between revisions of "2015 AMC 10B Problems/Problem 22"

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==Solution 2==
 
==Solution 2==
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Note that since <math>ABCDE</math> is a regular pentagon, all of its interior angles are <math>108\circ</math>. We can say that pentagon <math>FGHIJ</math> is also regular by symmetry. So, all of the interior angles of <math>FGHIJ</math> are <math>108\circ</math>.
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Note that since <math>ABCDE</math> is a regular pentagon, all of its interior angles are <math>108^\circ</math>. We can say that pentagon <math>FGHIJ</math> is also regular by symmetry. So, all of the interior angles of <math>FGHIJ</math> are <math>108^\circ</math>. Now, we can angle chase and use trigonometry to get that <math>FG=2sin18^\circ</math>, <math>JH=2sin18^\circ*(2sin18^\circ+1)</math>, and <math>DC=2sin18^\circ*(2sin18^\circ+2)</math>. Adding these together, we get that <math>FG+JH+CD=2sin18^\circ*(4+4sin18^\circ)=8sin18^\circ*(1+sin18^\circ)</math>. Because calculators were permitted in the 2015 AMC 10B, we can use a calculator to find out which of the options is equal to <math>8sin18^\circ*(1+sin18^\circ)</math>. We can find that the option that matches is $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\$.
  
 
==See Also==
 
==See Also==

Revision as of 14:04, 29 December 2017

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

However, $FG=\frac{-1 + \sqrt{5}}{2}$ 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

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 $FG=2sin18^\circ$, $JH=2sin18^\circ*(2sin18^\circ+1)$, and $DC=2sin18^\circ*(2sin18^\circ+2)$. Adding these together, we get that $FG+JH+CD=2sin18^\circ*(4+4sin18^\circ)=8sin18^\circ*(1+sin18^\circ)$. Because calculators were permitted in the 2015 AMC 10B, we can use a calculator to find out which of the options is equal to $8sin18^\circ*(1+sin18^\circ)$. We can find that the option that matches is $\boxed{\mathbf{(D)}\ 1+\sqrt{5}$.

See Also

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
All AMC 10 Problems and Solutions

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