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

Latest revision as of 18:47, 28 October 2024

Problem

In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$ . What is $FG + JH + CD$ ? $[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("$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

$[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("$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 $AFG$ 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, $FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

Note by Fasolinka: Alternatively, extend $FI$ and call its intersection with $DC$ $K$ . It is not hard to see that quadrilaterals $FGCK$ and $JHKD$ are parallelograms, so $DC=DK+KC=JH+FG=1+\frac{-1+\sqrt{5}}{2}$ , and the same result is achieved.

Solution 1 alternative (No trig, only similar triangles :) )

Further insight: I didn't use the obvious similar triangles AFG and AJH, and then AJH and ADC as that would lead to a cubic for x.

~mathboy282

Solution 2

Notice that $JH=BH=BG=AG=1$ . Since a $36-72-72$ triangle has the congruent sides equal to $\frac{\sqrt{5}+1}{2}$ times the short base side, we have $FG=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2}$ . Now notice that $CD=AB=AH$ , and that $\bigtriangleup AJH$ is $36-72-72$ . So, $CD=\frac{\sqrt{5}+1}{2}$ and adding gives $\boxed{1+\sqrt{5}}$ , or $\boxed{\textbf{(D)}}$ .

Solution 3 (Trignometry)

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. $\sin(72) = 2\sin(36)\cos(36)$ substituting in $x=\frac{1}{2\cos(36)}$

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

Notice that $\angle AFG=\angle AFB$ and $\angle FAG=\angle ABF$ , so we have $\bigtriangleup AFG \sim \bigtriangleup BAF$ . Thus $\frac{AF}{FG}=\frac{FB}{AF}$ $\frac{AF}{FG}=\frac{FG+GB}{AF}$ $\frac{1}{FG}=\frac{FG+1}{1}$ Solving the equation gets $FG=\frac{\sqrt{5}-1}{2}$ .

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

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

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

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: $\frac{FG}{DC}=\frac{AG}{AC}$ $\frac{JH}{DC}=\frac{AH}{AC}$ Summing the equations, we have: $\frac{FG + JH + CD}{CD} - 1 = \frac{2 + GH}{AC}$ We know that $AC = 2 + GH$ So we have $\frac{FG+JH+CD}{CD} - 1 = 1 \Rightarrow FG + JH + CD = 2 \cdot CD$ All that remains is to find $CD.$ By the law of cosines, we have $\begin{align*} CD^2 &= 2 - 2 \cos 108\\ &= 2 - \frac{1 - \sqrt{5}}{2} \\ &= \frac{3 + \sqrt{5}}{2} \end{align*}$

Thus, $\begin{align*} 2 \cdot CD &= 2 \cdot \sqrt{\frac{3 + \sqrt{5}}{2}} \\ &= 2 \cdot \frac{1 + \sqrt{5}}{2} \\ &= \boxed{\mathbf{(D)}\ 1 + \sqrt{5}} \end{align*}$

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 $x$ . From similarity, GH, HI, JI, and JF are also x. Now, because $\triangle AFG \sim \triangle AJH \sim \triangle ACD$ by SAS similarity, we can write the lengths of FG + JH + CD as $2x^2+4x$ . 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 $\angle AGF = 72$ . Now drop the altitude from A, and call the point T. Since $\angle GAT = 18$ degrees, we can easily use sine18 deg to find TG = $\frac{\sqrt{5}-1}{4}$ and therefore x = $\frac{\sqrt{5}-1}{4}$ . Plugging this into $2x^2+4x$ , we obtain the result $\boxed{\mathbf{(D)}\ 1 + \sqrt{5}}$ .

~MathCosine

Solution 7 (Knowledge of $\varphi$ )

In a pentagon, FG:JH=JH:CD=CD:AB=1: $\varphi$ . Also, JH=AG. FG= $\frac{1}{\varphi}=\varphi-1$ , so FG+JH+CD= $\varphi-1+1+\varphi=2\varphi=\boxed{(\textbf{D})~1+\sqrt{5}}$

Video Solution

https://www.youtube.com/watch?v=TpHZVbBGmVQ (Beauty of Math)

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

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

Latest revision as of 18:47, 28 October 2024

Contents

Problem

Solution 1

Solution 1 alternative (No trig, only similar triangles :) )

Solution 2

Solution 3 (Trignometry)

Solution 4

Solution 5

Solution 6(Trignometry)

Solution 7 (Knowledge of $\varphi$ )

Video Solution

See Also

@@ Line 3: / Line 3: @@
 <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);
@@ Line 23: / Line 21: @@
 ==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>.
 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>1 = FG^2 + FG</cmath>
@@ Line 33: / Line 51: @@
 <cmath>FG = \frac{-1 \pm \sqrt{5} }{2}</cmath>
-However, <math>FG=\frac{-1 + \sqrt{5}}{2}</math> since <math>FG</math> must be greater than 0.
+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>.
 Therefore, <math>FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>
+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 1 alternative (No trig, only similar triangles :) )==
+[[File:201510BSolmb282.png]]
+Further insight: I didn't use the obvious similar triangles AFG and AJH, and then AJH and ADC as that would lead to a cubic for x.
+~mathboy282
 ==Solution 2==
-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 permitted in the 2015 AMC 10B, we can use a calculator to find out which of the options is equal to <math>8\sin18^\circ*(1+\sin18^\circ)</math>. We can find that the option that matches is <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>.
+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>.
+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><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:
+<cmath>\frac{x}{\sin(36)}=\frac{1}{\sin(72)}</cmath>
+<cmath>x=\frac{\sin(36)}{\sin(72)}</cmath>
+Now by the double angle identities in trig. <math>\sin(72) = 2\sin(36)\cos(36)</math>
+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:
+<cmath>\cos(36)= \frac{1 + \sqrt{5}}{4}</cmath>
+so now we know:
+<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>
+== Solution 4 ==
+Notice that <math>\angle AFG=\angle AFB</math> and <math>\angle FAG=\angle ABF</math>, so we have <math>\bigtriangleup AFG \sim \bigtriangleup BAF</math>. Thus
+<cmath>\frac{AF}{FG}=\frac{FB}{AF}</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 \sim \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 \sim \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*}
+\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
+==Solution 7 (Knowledge of <math>\varphi</math>)==
+In a pentagon, FG:JH=JH:CD=CD:AB=1:<math>\varphi</math>. Also, JH=AG. FG=<math>\frac{1}{\varphi}=\varphi-1</math>, so FG+JH+CD=<math>\varphi-1+1+\varphi=2\varphi=\boxed{(\textbf{D})~1+\sqrt{5}}</math>
+== Video Solution ==
+https://www.youtube.com/watch?v=TpHZVbBGmVQ (Beauty of Math)
 ==See Also==

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

Latest revision as of 18:47, 28 October 2024

Contents

Problem

Solution 1

Solution 1 alternative (No trig, only similar triangles :) )

Solution 2

Solution 3 (Trignometry)

Solution 4

Solution 5

Solution 6(Trignometry)

Solution 7 (Knowledge of )

Video Solution

See Also

Solution 7 (Knowledge of $\varphi$ )