Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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<asy> | <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)); | 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)); | ||
− | |||
− | |||
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]; | 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--B--C--D--E1--A); | ||
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==Solution 1== | ==Solution 1== | ||
+ | |||
+ | <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>. | ||
Since <math>\triangle AJH \sim \triangle AFG</math>, | Since <math>\triangle AJH \sim \triangle AFG</math>, | ||
− | <cmath>\frac{JH}{AF+FJ}=\frac{FG}{FA}</cmath> | + | <cmath>\frac{JH}{AF+FJ}=\frac{FG}{FA}</cmath> |
<cmath>\frac{1}{1+FG} = \frac{FG}1</cmath> | <cmath>\frac{1}{1+FG} = \frac{FG}1</cmath> | ||
<cmath>1 = FG^2 + FG</cmath> | <cmath>1 = FG^2 + FG</cmath> | ||
Line 33: | Line 51: | ||
<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>. | ||
Line 39: | Line 57: | ||
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> | ||
− | + | ||
− | Note | + | Note by Fasolinka: Alternatively, extend <math>FI</math> and call its intersection with <math>DC</math> <math>K</math>. It is not hard to see that quadrilaterals <math>FGCK</math> and <math>JHKD</math> are parallelograms, so <math>DC=DK+KC=JH+FG=1+\frac{-1+\sqrt{5}}{2}</math>, and the same result is achieved. |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 3== | ==Solution 3== | ||
− | 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>< | + | 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><cmath>JH = {x}^2+x</cmath> |
− | 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>< | + | 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><cmath>DC = {x}^2+2x</cmath> |
− | 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: | ||
− | < | + | <cmath>\cos(36)= \frac{1 + \sqrt{5}}{4}</cmath> |
so now we know: | so now we know: | ||
− | <cmath>x = \frac{2}{1+\sqrt{5}}</cmath> | + | <cmath>x = \frac{2}{1+\sqrt{5}} = \frac{-1+\sqrt{5}}{2}</cmath> |
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 == | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Note that: | ||
+ | <cmath>\frac{FG}{DC}=\frac{AG}{AC}</cmath> | ||
+ | <cmath>\frac{JH}{DC}=\frac{AH}{AC}</cmath> | ||
+ | Summing the equations, we have: | ||
+ | <cmath>\frac{FG + JH + CD}{CD} - 1 = \frac{2 + GH}{AC}</cmath> | ||
+ | We know that <cmath>AC = 2 + GH</cmath> | ||
+ | So we have | ||
+ | <cmath>\frac{FG+JH+CD}{CD} - 1 = 1 \Rightarrow FG + JH + CD = 2 \cdot CD</cmath> | ||
+ | All that remains is to find <math>CD.</math> By the law of cosines, we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | CD^2 &= 2 - 2 \cos 108\\ | ||
+ | &= 2 - \frac{1 - \sqrt{5}}{2} \\ | ||
+ | &= \frac{3 + \sqrt{5}}{2} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Thus, | ||
+ | <cmath>\begin{align*} | ||
+ | 2 \cdot CD &= 2 \cdot \sqrt{\frac{3 + \sqrt{5}}{2}} \\ | ||
+ | &= 2 \cdot \frac{1 + \sqrt{5}}{2} \\ | ||
+ | &= \boxed{\mathbf{(D)}\ 1 + \sqrt{5}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | I'll leave you to convince yourself about certain facts that were deduced in order to obtain these equations. | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=TpHZVbBGmVQ (Beauty of Math) | ||
+ | |||
+ | https://www.youtube.com/watch?v=HfoQjjO2OFk | ||
==See Also== | ==See Also== |
Latest revision as of 15:41, 22 October 2021
Contents
Problem
In the figure shown below, is a regular pentagon and . What is ?
Solution 1
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,
Note by Fasolinka: Alternatively, extend and call its intersection with . It is not hard to see that quadrilaterals and are parallelograms, so , and the same result is achieved.
Solution 2
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 3
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 4
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
Solution 5
Note that: Summing the equations, we have: We know that So we have All that remains is to find By the law of cosines, we have
Thus,
I'll leave you to convince yourself about certain facts that were deduced in order to obtain these equations.
~mathboy282
Video Solution
https://www.youtube.com/watch?v=TpHZVbBGmVQ (Beauty of Math)
https://www.youtube.com/watch?v=HfoQjjO2OFk
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.