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

m (Solution 1)
(Solution 6)
 
(12 intermediate revisions by 3 users not shown)
Line 64: Line 64:
  
 
Substituting back into <math>2{x}^2+4x</math> we get <math>FG+JH+CD=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
 
Substituting back into <math>2{x}^2+4x</math> we get <math>FG+JH+CD=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
 +
 +
==Solution 4 (Answer choices)==
 +
 +
Notice that <math>A</math> is trisected, meaning that <cmath>AG=BH=EJ=JH=1</cmath>.
 +
Since <math>JH=1</math>, 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 <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> is our answer.
 +
 +
Note: Only do this if low on time since there could potentially be a weird figure affecting the <math>1</math>.
 +
 +
==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 <math>+ 1 +</math> small Phi = <math>(\sqrt5+1)/2 + 1 + (\sqrt5-1)/2</math> = <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
 +
 +
-Lcz
 +
 +
== Solution 6 ==
 +
 +
Notice that <math>\angle AFG=\angle AFB</math> and <math>\angle FAG=\angle ABF</math>, so we have <math>\bigtriangleup AFG=\bigtriangleup BAF</math>. Thus
 +
<cmath>\frac{AF}{FG}=\frac{FB}{JH}</cmath>
 +
<cmath>\frac{AF}{FG}=\frac{FG+GB}{AF}</cmath>
 +
<cmath>\frac{1}{FG}=\frac{FG+1}{1}</cmath>
 +
Solving the equation gets <math>FG=\frac{\sqrt{5}-1}{2}</math>.
 +
 +
Since <math>\bigtriangleup AFG=\bigtriangleup AJH</math>
 +
<cmath>\frac{AF}{FG}=\frac{AJ}{JH}</cmath>
 +
<cmath>\frac{AF}{FG}=\frac{AF+FJ}{JH}</cmath>
 +
<cmath>\frac{AF}{FG}=\frac{AF+FG}{JH}</cmath>
 +
<cmath>\frac{1}{\frac{\sqrt{5}-1}{2}}=\frac{1+\frac{\sqrt{5}-1}{2}}{JH}</cmath>
 +
Solving the equation gets <math>JH=1</math>.
 +
 +
Since <math>\bigtriangleup AFG=\bigtriangleup ADC</math>
 +
<cmath>\frac{AF}{FG}=\frac{AD}{DC}</cmath>
 +
<cmath>\frac{AF}{FG}=\frac{AD+FJ+JD}{DC}</cmath>
 +
<cmath>\frac{AF}{FG}=\frac{2AF+FG}{DC}</cmath>
 +
<cmath>\frac{1}{\frac{\sqrt{5}-1}{2}}=\frac{2+\frac{\sqrt{5}-1}{2}}{DC}</cmath>
 +
Solving the equation gets <math>DC=\frac{\sqrt{5}+1}{2}</math>
 +
 +
Finally adding them up gets <math>FG+JH+DC=\frac{\sqrt{5}-1}{2}+1+\frac{\sqrt{5}+1}{2}=1+\sqrt{5}</math> <math>\boxed{\mathrm{(D)}}</math>
 +
 +
Note: this solution might be a bit complicated but it definitely works when none of the cleverer symmetries in Solution 1 is noticed.
 +
 +
~ Nafer
  
 
==See Also==
 
==See Also==

Latest revision as of 12:05, 9 August 2019

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 \[\frac{AF}{FG}=\frac{FB}{JH}\] \[\frac{AF}{FG}=\frac{FG+GB}{AF}\] \[\frac{1}{FG}=\frac{FG+1}{1}\] Solving the equation gets $FG=\frac{\sqrt{5}-1}{2}$.

Since $\bigtriangleup AFG=\bigtriangleup AJH$ \[\frac{AF}{FG}=\frac{AJ}{JH}\] \[\frac{AF}{FG}=\frac{AF+FJ}{JH}\] \[\frac{AF}{FG}=\frac{AF+FG}{JH}\] \[\frac{1}{\frac{\sqrt{5}-1}{2}}=\frac{1+\frac{\sqrt{5}-1}{2}}{JH}\] Solving the equation gets $JH=1$.

Since $\bigtriangleup AFG=\bigtriangleup ADC$ \[\frac{AF}{FG}=\frac{AD}{DC}\] \[\frac{AF}{FG}=\frac{AD+FJ+JD}{DC}\] \[\frac{AF}{FG}=\frac{2AF+FG}{DC}\] \[\frac{1}{\frac{\sqrt{5}-1}{2}}=\frac{2+\frac{\sqrt{5}-1}{2}}{DC}\] Solving the equation gets $DC=\frac{\sqrt{5}+1}{2}$

Finally adding them up gets $FG+JH+DC=\frac{\sqrt{5}-1}{2}+1+\frac{\sqrt{5}+1}{2}=1+\sqrt{5}$ $\boxed{\mathrm{(D)}}$

Note: this solution might be a bit complicated but it definitely works when none of the cleverer symmetries in Solution 1 is noticed.

~ Nafer

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