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

(Solution)
Line 27: Line 27:
 
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>.
 
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>.
  
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}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}</math>. Solving for <math>CD</math>, we get  <math>CD = \frac{\sqrt{5} +1}{2}</math>
+
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}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}</math>. Thus, <math>CD = (\sqrt(5) - 1) + (\frac{\sqrt(5) - 1}{2})^2 = \frac{\sqrt{5} +1}{2}</math>
  
 
Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
 
Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>

Revision as of 15:39, 30 August 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 $\triangle AFG \sim ADC$, $\frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}$. Thus, $CD = (\sqrt(5) - 1) + (\frac{\sqrt(5) - 1}{2})^2 = \frac{\sqrt{5} +1}{2}$

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

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS