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> | ||
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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 3== | + | 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 (Trignometry)== | ||
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>. | 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: | ||
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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 4 | + | == Solution 4 == |
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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 | 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{AF}{FG}=\frac{FG+GB}{AF}</cmath> | ||
<cmath>\frac{1}{FG}=\frac{FG+1}{1}</cmath> | <cmath>\frac{1}{FG}=\frac{FG+1}{1}</cmath> | ||
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Since <math>\bigtriangleup AFG=\bigtriangleup AJH</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+FJ}{JH}</cmath> | ||
<cmath>\frac{AF}{FG}=\frac{AF+FG}{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}= \boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</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 | ||
+ | |||
+ | ==Solution 6(Trignometry)== | ||
+ | |||
+ | Let's denote the length FG <math>x</math>. From similarity, GH, HI, JI, and JF are also x. Now, because <math>\triangle AFG \sim \triangle AJH \sim \triangle ACD</math> by SAS similarity, we can write the lengths of FG + JH + CD as <math>2x^2+4x</math>. We obtain this result by directly writing out the similarity statements and then multiplying. | ||
+ | |||
+ | So all we have to do is find x. Because a regular pentagon has angles of 108 degrees, with an easy angle chase we can find <math>\angle AGF = 72</math>. Now drop the altitude from A, and call the point T. Since <math>\angle GAT = 18</math> degrees, we can easily use sine18 deg to find TG = <math>\frac{\sqrt{5}-1}{4}</math> and therefore x = <math>\frac{\sqrt{5}-1}{4}</math>. Plugging this into <math>2x^2+4x</math>, we obtain the result <math>\boxed{\mathbf{(D)}\ 1 + \sqrt{5}}</math>. | ||
+ | |||
+ | ~MathCosine | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=TpHZVbBGmVQ (Beauty of Math) | ||
==See Also== | ==See Also== |
Latest revision as of 18:52, 18 August 2024
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 (Trignometry)
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
Solution 6(Trignometry)
Let's denote the length FG . From similarity, GH, HI, JI, and JF are also x. Now, because
by SAS similarity, we can write the lengths of FG + JH + CD as
. We obtain this result by directly writing out the similarity statements and then multiplying.
So all we have to do is find x. Because a regular pentagon has angles of 108 degrees, with an easy angle chase we can find . Now drop the altitude from A, and call the point T. Since
degrees, we can easily use sine18 deg to find TG =
and therefore x =
. Plugging this into
, we obtain the result
.
~MathCosine
Video Solution
https://www.youtube.com/watch?v=TpHZVbBGmVQ (Beauty of Math)
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.