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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 24"
(there was an error in my solve.)
 
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==Problem==
 
==Problem==
−
Convex quadrilateral <math>ABCD</math> has <math>AB = 18, \angle{A} = 60 \textdegree</math>, and <math>\overline{AB} \parallel \overline{CD}</math>. In some order, the lengths of the four sides form an arithmetic progression, and side <math>\overline{AB}</math> is a side of maximum length. The length of another side is <math>a</math>. What is the sum of all possible values of <math>a</math>?
+
Convex quadrilateral <math>ABCD</math> has <math>AB = 18, \angle{A} = 60^\circ,</math> and <math>\overline{AB} \parallel \overline{CD}.</math> In some order, the lengths of the four sides form an arithmetic progression, and side <math>\overline{AB}</math> is a side of maximum length. The length of another side is <math>a.</math> What is the sum of all possible values of <math>a</math>?
  
 
<math>\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84</math>
 
<math>\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84</math>
  
−
==Solution==
+
==Solution 1==
 +
 
 
Let <math>E</math> be a point on <math>\overline{AB}</math> such that <math>BCDE</math> is a parallelogram. Suppose that <math>BC=ED=b, CD=BE=c,</math> and <math>DA=d,</math> so <math>AE=18-c,</math> as shown below.
 
Let <math>E</math> be a point on <math>\overline{AB}</math> such that <math>BCDE</math> is a parallelogram. Suppose that <math>BC=ED=b, CD=BE=c,</math> and <math>DA=d,</math> so <math>AE=18-c,</math> as shown below.
−
 
+
<asy>
−
<b>DIAGRAM WILL BE READY VERY VERY SOON ...</b>
+
/* Made by MRENTHUSIASM */
−
 
+
size(250);
 +
pair A, B, C, D, E;
 +
A = (0,0);
 +
B = (18,0);
 +
D = A+9*dir(60);
 +
C = D+(7,0);
 +
E = D+(B-C);
 +
dot("$A$",A,1.5*SW,linewidth(4));
 +
dot("$B$",B,1.5*SE,linewidth(4));
 +
dot("$C$",C,1.5*NE,linewidth(4));
 +
dot("$D$",D,1.5*NW,linewidth(4));
 +
dot("$E$",E,1.5*S,linewidth(4));
 +
draw(A--B--C--D--cycle);
 +
draw(D--E,dashed);
 +
label("$60^\circ$",A,2.5*dir(30),fontsize(10));
 +
label("$18-c$",midpoint(A--E),1.5*S,red);
 +
label("$c$",midpoint(E--B),2.25*S,red);
 +
label("$b$",midpoint(B--C),scale(1.5)*rotate(90)*dir(midpoint(B--C)--B),red);
 +
label("$b$",midpoint(D--E),scale(1.5)*rotate(90)*dir(midpoint(E--D)--E),red);
 +
label("$c$",midpoint(C--D),1.5*N,red);
 +
label("$d$",midpoint(D--A),scale(1.5)*rotate(90)*dir(midpoint(D--A)--D),red);
 +
</asy>
 
We apply the Law of Cosines to <math>\triangle ADE:</math>
 
We apply the Law of Cosines to <math>\triangle ADE:</math>
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 31: Line 53:
 
Note that <math>(\bigstar)</math> becomes <math>3k(4k-18)=(36-3k)(-k),</math> from which <math>k=2.</math> So, this case generates <math>(b,c,d)=(14,12,16).</math><p>
 
Note that <math>(\bigstar)</math> becomes <math>3k(4k-18)=(36-3k)(-k),</math> from which <math>k=2.</math> So, this case generates <math>(b,c,d)=(14,12,16).</math><p>
 
   <li><math>(b,c,d)=(18-3k,18-k,18-2k)</math></li><p>
 
   <li><math>(b,c,d)=(18-3k,18-k,18-2k)</math></li><p>
−
Note that <math>(\bigstar)</math> becomes <p>
+
Note that <math>(\bigstar)</math> becomes <math>k(3k-18)=(36-5k)(-k),</math> from which <math>k=9.</math> So, this case generates no valid solutions <math>(b,c,d).</math><p>
 
   <li><math>(b,c,d)=(18-3k,18-2k,18-k)</math></li><p>
 
   <li><math>(b,c,d)=(18-3k,18-2k,18-k)</math></li><p>
−
Note that <math>(\bigstar)</math> becomes <p>
+
Note that <math>(\bigstar)</math> becomes <math>2k(3k-18)=(36-4k)(-2k),</math> from which <math>k=18.</math> So, this case generates no valid solutions <math>(b,c,d).</math><p>
 
</ol>
 
</ol>
 +
Together, the sum of all possible values of <math>a</math> is <math>18+(13+3+8)+(14+12+16)=\boxed{\textbf{(E) } 84}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 2==
 +
 +
Let <math>b, c</math>, and <math>d</math> denote the sides <math>BC, CD</math>, and <math>AD</math> respectively. 
 +
<asy>
 +
size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C);
 +
pen pdot=linewidth(3)+fontsize(12);
 +
dot("$A$",A,SW,pdot); dot("$B$",B,SE,pdot); dot("$C$",C,NE,pdot); dot("$D$",D,NW,pdot); 
 +
draw(A--B--C--D--cycle); 
 +
label("$60^\circ$",A,5*dir(30),fontsize(10));
 +
label("$\theta$", B, 5*dir(155),fontsize(10));
 +
pen plabel=red+fontsize(12);
 +
label("$18$",midpoint(A--B),1.5*S,plabel);
 +
label("$b$", midpoint(B--C), scale(1.5)*rotate(90)* dir((B+C)/2--B), plabel);
 +
label("$c$", (C+D)/2,1.5*N, plabel);
 +
label("$d$",(D+A)/2, scale(1.5)*rotate(90)*dir((D+A)/2--D), plabel);
 +
</asy>
 +
Since <math>AB\parallel CD</math>, we get
 +
<cmath> \tfrac{\sqrt 3}{2}\ d = b\sin\theta \quad \textrm{and}\quad \tfrac 12 d + c + b\cos\theta = 18. </cmath>
 +
Using <math>b^2\sin^2\theta + b^2\cos^2\theta = b^2</math>, we eliminate <math>\theta</math> from above to get
 +
<math>(36-2c-d)^2+3d^2=4b^2</math>, which rearranges to <math>(36-2c-d)^2-d^2=4(b^2-d^2)</math>, and, upon factoring, yields
 +
<cmath>\begin{align}
 +
    (18-c)(18-c-d)=(b+d)(b-d).
 +
\end{align}</cmath>
 +
We divide into two cases, depending on whether <math>c</math> is the smallest side.
 +
 +
If <math>c</math> is not the smallest side then <math>18-c=\pm (b-d)</math>. If <math>c=18</math>, we get a rhombus of side <math>18</math>, so  one possible value is <math>a=18</math>. Otherwise, we can cancel the common factor from <math>(1)</math>. After rearranging we get<cmath>18-c=-b \quad \textrm{or}\quad  18-c=b+2d.</cmath>
 +
The first condition is false because <math>-b< 0 <18-c</math>; the second condition is false because <math>b+2d > |b-d| = 18-c</math>.
 +
 +
If <math>c</math> is the smallest side, then <math>18-c = \pm 3(b-d)</math>. Assuming <math>c<18</math> we can cancel common factors in <math>(1)</math> to get<cmath>8b=13d \quad \textrm{or}\quad 8b=7d.</cmath>  The first condition yields the solution <math>(c,d,b)=(3,8,13)</math> and the second condition yields the solution <math>(c,b,d)=(12,14,16)</math>.
 +
 +
Together, the sum of all possible values of <math>a</math> is <math>18+(3+8+13)+(12+14+16)=\boxed{\textbf{(E) } 84}.</math>
 +
 +
== Solution 3 ==
 +
Denote <math>x = AD</math>, <math>\theta = \angle B</math>.
 +
Hence, <math>BC = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sin \theta}</math>, <math>DC = 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta</math>.
 +
 +
<math>\textbf{Case 1}</math>: <math>DC = AD = BC = AB</math>.
 +
 +
This is a rhombus. So each side has length <math>18</math>.
 +
 +
For the following cases, we consider four sides that have distinct lengths.
 +
To make their lengths an arithmetic sequence, we must have <math>\theta \neq 120^\circ</math>.
 +
 +
Therefore, in the subsequent analysis, we exclude the solution <math>\theta = 120^\circ</math>.
 +
 +
<math>\textbf{Case 2}</math>: <math>DC < AD < BC < AB</math>.
 +
 +
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations:
 +
<cmath>
 +
\[
 +
AB - BC = BC - AD = AD - DC .
 +
\]
 +
</cmath>
 +
 +
Hence,
 +
<cmath>
 +
\begin{eqnarray*}
 +
&
 +
18 - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta}
 +
= \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - x
 +
= x - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) .
 +
&
 +
\end{eqnarray*}
 +
</cmath>
 +
 +
By solving this system of equations, we get <math>\left( \cos \theta , \sin \theta , x\right) = \left( \frac{11}{13} , \frac{4 \sqrt{3}}{13} , 8 \right)</math>.
 +
 +
Thus, in this case, <math>DC = 3</math>, <math>AD = 8</math>, <math>BC = 13</math>.
 +
 +
<math>\textbf{Case 3}</math>: <math>DC < BC < AD < AB</math>.
 +
 +
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations:
 +
<cmath>
 +
\[
 +
AB - AD = AD - BC = BC - DC .
 +
\]
 +
</cmath>
 +
 +
Hence,
 +
<cmath>
 +
\begin{eqnarray*}
 +
&
 +
18 - x
 +
= x - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta}
 +
= \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta}  - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) .
 +
&
 +
\end{eqnarray*}
 +
</cmath>
 +
 +
By solving this system of equations, we get <math>\left( \cos \theta , \sin \theta , x\right) = \left( - \frac{1}{7} , \frac{4 \sqrt{3}}{7} , 16 \right)</math>.
 +
 +
Thus, in this case, <math>DC = 12</math>, <math>AD = 16</math>, <math>BC = 14</math>.
 +
 +
<math>\textbf{Case 4}</math>: <math>BC < CD < AD < AB</math>.
 +
 +
By doing the similar analysis, we can show there is no solution in this case.
 +
 +
<math>\textbf{Case 5}</math>: <math>BC < AD < CD < AB</math>.
 +
 +
By doing the similar analysis, we can show there is no solution in this case.
 +
 +
<math>\textbf{Case 6}</math>: <math>AD < CD < BC < AB</math>.
 +
 +
By doing the similar analysis, we can show there is no solution in this case.
 +
 +
<math>\textbf{Case 7}</math>: <math>AD < BC < CD < AB</math>.
 +
 +
By doing the similar analysis, we can show there is no solution in this case.
 +
 +
Therefore, the sum of all possible values of <math>a</math> is
 +
<cmath>
 +
\begin{align*}
 +
18 + \left( 3 + 8 + 13 \right) + \left( 12 + 14 + 16 \right)
 +
& = 84 .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(E) } 84}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
−
<b>WILL COMPLETE VERY SOON. A MILLION THANKS FOR NOT EDITING THIS PAGE.</b>
 
  
−
~MRENTHUSIASM
+
==Video Solution==
 +
https://youtu.be/CMpnQ6I8AXc
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
 +
==Video Solution and Exploration by hurdler==
 +
 
 +
[https://www.youtube.com/watch?v=IA5fWLdvVQg& Video exploration and motivated solution]
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:22, 19 September 2025

Problem

Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$?

$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$

Solution 1

Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*SE,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*NW,linewidth(4)); dot("$E$",E,1.5*S,linewidth(4)); draw(A--B--C--D--cycle); draw(D--E,dashed); label("$60^\circ$",A,2.5*dir(30),fontsize(10)); label("$18-c$",midpoint(A--E),1.5*S,red); label("$c$",midpoint(E--B),2.25*S,red); label("$b$",midpoint(B--C),scale(1.5)*rotate(90)*dir(midpoint(B--C)--B),red); label("$b$",midpoint(D--E),scale(1.5)*rotate(90)*dir(midpoint(E--D)--E),red); label("$c$",midpoint(C--D),1.5*N,red); label("$d$",midpoint(D--A),scale(1.5)*rotate(90)*dir(midpoint(D--A)--D),red); [/asy] We apply the Law of Cosines to $\triangle ADE:$ \begin{align*} AD^2 + AE^2 - 2\cdot AD\cdot AE\cdot\cos 60^\circ &= DE^2 \\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\ (18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar) \end{align*} Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order. It is clear that $0\leq k<6.$

If $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid.

If $k\neq0,$ then we have six cases:

  1. $(b,c,d)=(18-k,18-2k,18-3k)$

    2. Note that $(\bigstar)$ becomes $2k(5k-18)=(36-4k)(2k),$ from which $k=6.$ So, this case generates no valid solutions $(b,c,d).$

  2. $(b,c,d)=(18-k,18-3k,18-2k)$

    3. Note that $(\bigstar)$ becomes $3k(5k-18)=(36-3k)k,$ from which $k=5.$ So, this case generates $(b,c,d)=(13,3,8).$

  3. $(b,c,d)=(18-2k,18-k,18-3k)$

    4. Note that $(\bigstar)$ becomes $k(4k-18)=(36-5k)k,$ from which $k=6.$ So, this case generates no valid solutions $(b,c,d).$

  4. $(b,c,d)=(18-2k,18-3k,18-k)$

    5. Note that $(\bigstar)$ becomes $3k(4k-18)=(36-3k)(-k),$ from which $k=2.$ So, this case generates $(b,c,d)=(14,12,16).$

  5. $(b,c,d)=(18-3k,18-k,18-2k)$

    6. Note that $(\bigstar)$ becomes $k(3k-18)=(36-5k)(-k),$ from which $k=9.$ So, this case generates no valid solutions $(b,c,d).$

  6. $(b,c,d)=(18-3k,18-2k,18-k)$

    7. Note that $(\bigstar)$ becomes $2k(3k-18)=(36-4k)(-2k),$ from which $k=18.$ So, this case generates no valid solutions $(b,c,d).$

Together, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=\boxed{\textbf{(E) } 84}.$

~MRENTHUSIASM

Solution 2

Let $b, c$, and $d$ denote the sides $BC, CD$, and $AD$ respectively. [asy] size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); pen pdot=linewidth(3)+fontsize(12); dot("$A$",A,SW,pdot); dot("$B$",B,SE,pdot); dot("$C$",C,NE,pdot); dot("$D$",D,NW,pdot); draw(A--B--C--D--cycle); label("$60^\circ$",A,5*dir(30),fontsize(10)); label("$\theta$", B, 5*dir(155),fontsize(10)); pen plabel=red+fontsize(12); label("$18$",midpoint(A--B),1.5*S,plabel); label("$b$", midpoint(B--C), scale(1.5)*rotate(90)* dir((B+C)/2--B), plabel); label("$c$", (C+D)/2,1.5*N, plabel); label("$d$",(D+A)/2, scale(1.5)*rotate(90)*dir((D+A)/2--D), plabel); [/asy] Since $AB\parallel CD$, we get \[\tfrac{\sqrt 3}{2}\ d = b\sin\theta \quad \textrm{and}\quad \tfrac 12 d + c + b\cos\theta = 18.\] Using $b^2\sin^2\theta + b^2\cos^2\theta = b^2$, we eliminate $\theta$ from above to get $(36-2c-d)^2+3d^2=4b^2$, which rearranges to $(36-2c-d)^2-d^2=4(b^2-d^2)$, and, upon factoring, yields \begin{align} (18-c)(18-c-d)=(b+d)(b-d). \end{align} We divide into two cases, depending on whether $c$ is the smallest side.

If $c$ is not the smallest side then $18-c=\pm (b-d)$. If $c=18$, we get a rhombus of side $18$, so one possible value is $a=18$. Otherwise, we can cancel the common factor from $(1)$. After rearranging we get\[18-c=-b \quad \textrm{or}\quad 18-c=b+2d.\] The first condition is false because $-b< 0 <18-c$; the second condition is false because $b+2d > |b-d| = 18-c$.

If $c$ is the smallest side, then $18-c = \pm 3(b-d)$. Assuming $c<18$ we can cancel common factors in $(1)$ to get\[8b=13d \quad \textrm{or}\quad 8b=7d.\] The first condition yields the solution $(c,d,b)=(3,8,13)$ and the second condition yields the solution $(c,b,d)=(12,14,16)$.

Together, the sum of all possible values of $a$ is $18+(3+8+13)+(12+14+16)=\boxed{\textbf{(E) } 84}.$

Solution 3

Denote $x = AD$, $\theta = \angle B$. Hence, $BC = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sin \theta}$, $DC = 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta$.

$\textbf{Case 1}$: $DC = AD = BC = AB$.

This is a rhombus. So each side has length $18$.

For the following cases, we consider four sides that have distinct lengths. To make their lengths an arithmetic sequence, we must have $\theta \neq 120^\circ$.

Therefore, in the subsequent analysis, we exclude the solution $\theta = 120^\circ$.

$\textbf{Case 2}$: $DC < AD < BC < AB$.

Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - BC = BC - AD = AD - DC . \]

Hence, \begin{eqnarray*} & 18 - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - x = x - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}

By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( \frac{11}{13} , \frac{4 \sqrt{3}}{13} , 8 \right)$.

Thus, in this case, $DC = 3$, $AD = 8$, $BC = 13$.

$\textbf{Case 3}$: $DC < BC < AD < AB$.

Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - AD = AD - BC = BC - DC . \]

Hence, \begin{eqnarray*} & 18 - x = x - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}

By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( - \frac{1}{7} , \frac{4 \sqrt{3}}{7} , 16 \right)$.

Thus, in this case, $DC = 12$, $AD = 16$, $BC = 14$.

$\textbf{Case 4}$: $BC < CD < AD < AB$.

By doing the similar analysis, we can show there is no solution in this case.

$\textbf{Case 5}$: $BC < AD < CD < AB$.

By doing the similar analysis, we can show there is no solution in this case.

$\textbf{Case 6}$: $AD < CD < BC < AB$.

By doing the similar analysis, we can show there is no solution in this case.

$\textbf{Case 7}$: $AD < BC < CD < AB$.

By doing the similar analysis, we can show there is no solution in this case.

Therefore, the sum of all possible values of $a$ is \begin{align*} 18 + \left( 3 + 8 + 13 \right) + \left( 12 + 14 + 16 \right) & = 84 . \end{align*}

Therefore, the answer is $\boxed{\textbf{(E) } 84}$.

~Steven Chen (www.professorchenedu.com)


Video Solution

https://youtu.be/CMpnQ6I8AXc

~MathProblemSolvingSkills.com

Video Solution and Exploration by hurdler

Video exploration and motivated solution

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_24&oldid=260889"