2021 Fall AMC 12A Problems/Problem 24

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),1.5*NE,red); label("$b$",midpoint(D--E),1.5*NE,red); label("$c$",midpoint(C--D),1.5*N,red); label("$d$",midpoint(D--A),1.5*NW,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:

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

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

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

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

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

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

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

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

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

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

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

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

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 and Exploration by hurdler

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2021 Fall AMC 12A (Problems • Answer Key • Resources)
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2021 Fall AMC 12A Problems/Problem 24

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Video Solution and Exploration by hurdler

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