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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 24"

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~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
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== Solution 4 (Quick Cheese) ==
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Drawing out the trapezoid, we first assume <math>BC=18-k</math>, <math>CD=18-3k</math>,<math>AD=18-2k</math>. Then, draw the altitude of <math>D</math> down to <math>AB</math>, call the foot of the altitude <math>E</math>. Similarly, draw the altitude of <math>C</math> down to <math>AB</math>, call this foot <math>F</math>.
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Then we have <math>AE=9-k</math> as <math>ADE</math> is a <math>30-60-90</math> triangle, and <math>EF=CD=18-3k</math>. Since <math>AB=18, BF=18-EF-AE=4k-9</math>. We know the heights <math>DE</math> and <math>CF</math> are equal, so <math>CF=DE=(9-k)\sqrt{3}</math>. By Pythagorean theorem on triangle <math>BCF</math>, we have <math>(4k-9)^2+((9-k)\sqrt{3})^2=(18-k)^2</math>. Solving for <math>k</math> gives <math>k=\frac{5 \pm \sqrt{17}}{2}</math>. Notice that the sum of all of the other sides given some <math>k</math> is <math>54-6k</math>. Plugging in the two possible values of <math>k</math> and summing gives <math>108-3(5+\sqrt{17}+5-\sqrt{17})=108-3(10)=78</math>.
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We're obviously missing some side length possibilities, but we know the answer exceeds <math>78</math>. Thus, the only possible answer is <math>\boxed{\textbf{(E) }84}</math>
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- Shadowleafy
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:18, 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)

Solution 4 (Quick Cheese)

Drawing out the trapezoid, we first assume $BC=18-k$, $CD=18-3k$,$AD=18-2k$. Then, draw the altitude of $D$ down to $AB$, call the foot of the altitude $E$. Similarly, draw the altitude of $C$ down to $AB$, call this foot $F$.

Then we have $AE=9-k$ as $ADE$ is a $30-60-90$ triangle, and $EF=CD=18-3k$. Since $AB=18, BF=18-EF-AE=4k-9$. We know the heights $DE$ and $CF$ are equal, so $CF=DE=(9-k)\sqrt{3}$. By Pythagorean theorem on triangle $BCF$, we have $(4k-9)^2+((9-k)\sqrt{3})^2=(18-k)^2$. Solving for $k$ gives $k=\frac{5 \pm \sqrt{17}}{2}$. Notice that the sum of all of the other sides given some $k$ is $54-6k$. Plugging in the two possible values of $k$ and summing gives $108-3(5+\sqrt{17}+5-\sqrt{17})=108-3(10)=78$.

We're obviously missing some side length possibilities, but we know the answer exceeds $78$. Thus, the only possible answer is $\boxed{\textbf{(E) }84}$

- Shadowleafy

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

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