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

(Created page with "Solution Triangle <math>AFG</math> is isosceles so <math>AG</math>=<math>AF</math>=<math>1</math>. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\angl...")
 
(Formatting, heavy phrasing edit, added "problem" and "see also" sections, removed "solution by...")
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Solution
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==Problem==
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In the figure shown below, <math>ABCDE</math> is a regular pentagon and <math>AG=1</math>. What is <math>FG + JH + CD</math>?
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<asy>
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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));
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//(0,0) is a convenient point
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//E1 to prevent conflict with direction E(ast)
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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];
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draw(A--B--C--D--E1--A);
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draw(A--D--B--E1--C--A);
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draw(F--I--G--J--H--F);
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label("$A$",A,N);
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label("$B$",B,E);
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label("$C$",C,SE);
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label("$D$",D,SW);
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label("$E$",E1,W);
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label("$F$",F,NW);
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label("$G$",G,NE);
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label("$H$",H,E);
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label("$I$",I,S);
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label("$J$",J,W);
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</asy>
  
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Triangle <math>AFG</math> is isosceles so <math>AG</math>=<math>AF</math>=<math>1</math>. Using the symmetry of pentagon  <math>FGHIJ</math>, notice that <math>\angle{FAG}</math> is congruent to <math>\angle{FBI}</math>, so triangles <math>JHG</math> and <math>DIJ</math> are congruent to our original triangle <math>AFG</math>. Therefore, <math>JH</math> = <math>1</math>. Now, we still need to find the length of <math>FG</math> and <math>DC</math>. Also, we know that <math>FJ</math> = <math>FG</math> since pentagon <math>FGHIJ</math> is regular. Let's call the length of <math>FG</math> and <math>FJ</math> <math>x</math>.  Now we can solve for <math>x</math>. Triangles <math>AFG</math> and <math>AJH</math> are similar. So,
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==Solution==
  
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<cmath> \frac{1}{x+1} = \frac{x}{1}</cmath>
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Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG</math>. Therefore, <math>JH=AF=1</math>.
  
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From this, we get <math>x = \frac{\sqrt{5} -1}{2}</math>.
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Since <math>\triangle AJH \sim \triangle AFG</math>, <math>\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1</math>. From this, we get <math>FG=\frac{\sqrt{5} -1}{2}</math>.
  
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Now, we just have to find the length of <math>DC</math> which we'll call <math>y</math>. We already know that <math>DJ</math> = <math>AF</math> = <math>1</math>. Triangles <math>AFG</math> and <math>ADC</math> are similar so we have,
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Since <math>\triangle DIJ \cong \triangle AFG</math>, <math>DJ=DI=AF=1</math>. Since <math>\triangle AFG \sim ADC</math>, <math> \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac1{CD}</math>. Solving for <math>CD</math>, we get  <math>CD = \frac{\sqrt{5} +1}{2}</math>
  
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<cmath> \frac{1}{2+x} = \frac{x}{y}</cmath>
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Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
  
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Solving for <math>y</math> we get  <math> y = \frac{\sqrt{5} +1}{2}</math>
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==See Also==
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{{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}
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Adding up <math>FG</math> , <math>JH</math>, and <math>CD</math> gives us <math>1 + x + y</math> which is <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
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{{MAA Notice}}
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Solution by arebei2
 

Revision as of 22:48, 16 March 2015

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 $\triangle AFG \sim ADC$, $\frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac1{CD}$. Solving for $CD$, we get $CD = \frac{\sqrt{5} +1}{2}$

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

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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