Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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<math>\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} </math> | <math>\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} </math> | ||
− | ==Solution 1== | + | ==Solution 1 (no trig, just memorize golden ratio)== |
+ | |||
+ | Notice that <math>JH=BH=BG=AG=1</math>. Since a <math>36-72-72</math> triangle has the congruent sides equal to <math>\frac{\sqrt{5}+1}{2}</math> times the short base side, we have <math>FG=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2}</math>. Now notice that <math>CD=AB=AH</math>, and that <math>\bigtriangleup AJH</math> is <math>36-72-72</math>. So, <math>CD=\frac{\sqrt{5}+1}{2}</math> and adding gives <math>\boxed{1+\sqrt{5}}</math>, or <math>\boxed{\textbf{(D)}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <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> | ||
Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. <math>FJ = FG</math> since <math>\triangle FGJ</math> is also isosceles. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle JHG \cong \triangle AFG</math>. Therefore, <math>JH=AF=1</math>. | Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. <math>FJ = FG</math> since <math>\triangle FGJ</math> is also isosceles. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle JHG \cong \triangle AFG</math>. Therefore, <math>JH=AF=1</math>. | ||
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<cmath>FG = \frac{-1 \pm \sqrt{5} }{2}</cmath> | <cmath>FG = \frac{-1 \pm \sqrt{5} }{2}</cmath> | ||
− | + | So, <math>FG=\frac{-1 + \sqrt{5}}{2}</math> since <math>FG</math> must be greater than 0. | |
Notice that <math>CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}</math>. | Notice that <math>CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}</math>. | ||
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Therefore, <math>FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> | Therefore, <math>FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> | ||
− | ==Solution | + | ==Solution 3 (Trigonometry)== |
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=2\sin18^\circ</math>, <math>JH=2\sin18^\circ*(2\sin18^\circ+1)</math>, and <math>DC=2\sin18^\circ*(2\sin18^\circ+2)</math>. Adding these together, we get that <math>FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)</math>. 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 <math>8\sin18^\circ*(1+\sin18^\circ)</math>, but we can find that this is closest to <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>. | 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=2\sin18^\circ</math>, <math>JH=2\sin18^\circ*(2\sin18^\circ+1)</math>, and <math>DC=2\sin18^\circ*(2\sin18^\circ+2)</math>. Adding these together, we get that <math>FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)</math>. 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 <math>8\sin18^\circ*(1+\sin18^\circ)</math>, but we can find that this is closest to <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>. | ||
− | ==Solution | + | ==Solution 4== |
− | When you first see this problem you can't help but see similar triangles. But this shape is filled with | + | When you first see this problem you can't help but see similar triangles. But this shape is filled with <math>36 - 72 - 72</math> triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of <math>FG</math> so we can apply similar triangles easily. To simplify the process lets write <math>FG</math> as <math>x</math>. |
First what is <math>JH</math> in terms of <math>x</math>, also remember <math>AJ = 1+x</math>: <cmath>\frac{JH}{1+x}=\frac{x}{1}</cmath><math>JH</math> = <math>{x}^2+x</math> | First what is <math>JH</math> in terms of <math>x</math>, also remember <math>AJ = 1+x</math>: <cmath>\frac{JH}{1+x}=\frac{x}{1}</cmath><math>JH</math> = <math>{x}^2+x</math> | ||
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Next, find <math>DC</math> in terms of <math>x</math>, also remember <math>AD = 2+x</math>: <cmath>\frac{DC}{2+x}=\frac{x}{1}</cmath><math>DC</math> = <math>{x}^2+2x</math> | Next, find <math>DC</math> in terms of <math>x</math>, also remember <math>AD = 2+x</math>: <cmath>\frac{DC}{2+x}=\frac{x}{1}</cmath><math>DC</math> = <math>{x}^2+2x</math> | ||
− | So adding all the <math>FG + JH + CD</math> we get <math>2{x}^2+4x</math>. Now we have to find out what x is. For this, we break out a bit of trig. Let's look at <math>\triangle AFG</math> By the law of sines: | + | So adding all the <math>FG + JH + CD</math> we get <math>2{x}^2+4x</math>. Now we have to find out what x is. For this, we break out a bit of trig. Let's look at <math>\triangle AFG</math>. By the law of sines: |
− | <cmath>\frac{x}{sin(36)}=\frac{1}{sin(72)}</cmath> | + | <cmath>\frac{x}{\sin(36)}=\frac{1}{\sin(72)}</cmath> |
− | <cmath>x=\frac{sin(36)}{sin(72)}</cmath> | + | <cmath>x=\frac{\sin(36)}{\sin(72)}</cmath> |
− | Now by the double angle identities in trig. <math>sin(72) = | + | Now by the double angle identities in trig. <math>\sin(72) = 2\sin(36)\cos(36)</math> |
− | substituting in <cmath>x=\frac{1}{ | + | substituting in <cmath>x=\frac{1}{2\cos(36)}</cmath> |
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: | 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: | ||
− | <math>cos(36)</math>= <cmath>\frac{1 + \sqrt{5}}{4}</cmath> | + | <math>\cos(36)</math>= <cmath>\frac{1 + \sqrt{5}}{4}</cmath> |
so now we know: | so now we know: | ||
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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 5 (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 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 16:54, 29 September 2020
Contents
Problem
In the figure shown below, is a regular pentagon and . What is ?
Solution 1 (no trig, just memorize golden ratio)
Notice that . Since a triangle has the congruent sides equal to times the short base side, we have . Now notice that , and that is . So, and adding gives , or .
Solution 2
Triangle is isosceles, so . since is also isosceles. Using the symmetry of pentagon , notice that . Therefore, .
Since , .
So, since must be greater than 0.
Notice that .
Therefore,
Solution 3 (Trigonometry)
Note that since is a regular pentagon, all of its interior angles are . We can say that pentagon is also regular by symmetry. So, all of the interior angles of are . Now, we can angle chase and use trigonometry to get that , , and . Adding these together, we get that . 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 , but we can find that this is closest to .
Solution 4
When you first see this problem you can't help but see similar triangles. But this shape is filled with triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of so we can apply similar triangles easily. To simplify the process lets write as .
First what is in terms of , also remember : =
Next, find in terms of , also remember : =
So adding all the we get . Now we have to find out what x is. For this, we break out a bit of trig. Let's look at . By the law of sines:
Now by the double angle identities in trig. substituting in
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: =
so now we know:
Substituting back into we get
Solution 5 (Answer choices)
Notice that is trisected, meaning that . Since , 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 is our answer.
Note: Only do this if low on time since there could potentially be a weird figure affecting the .
Solution 6
Notice that and , so we have . Thus Solving the equation gets .
Since Solving the equation gets .
Since Solving the equation gets
Finally adding them up gets
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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.