Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (→‎Video Solution)
 
(8 intermediate revisions by 3 users not shown)
Line 61: Line 61:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 2 ==
+
==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>.
 
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>.
 
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>.
Line 67: Line 99:
 
<math>\textbf{Case 1}</math>: <math>DC = AD = BC = AB</math>.
 
<math>\textbf{Case 1}</math>: <math>DC = AD = BC = AB</math>.
  
−
This is a rhombus. So each side has length 18.
+
This is a rhombus. So each side has length <math>18</math>.
  
 
For the following cases, we consider four sides that have distinct lengths.
 
For the following cases, we consider four sides that have distinct lengths.
Line 146: Line 178:
 
</cmath>
 
</cmath>
  
−
Therefore, the answer is <math>\boxed{\textbf{(E) }84}</math>.
+
Therefore, the answer is <math>\boxed{\textbf{(E) } 84}</math>.
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
 +
 +
==Video Solution==
 +
https://youtu.be/CMpnQ6I8AXc
 +
 +
~MathProblemSolvingSkills.com
  
 
==Video Solution and Exploration by hurdler==
 
==Video Solution and Exploration by hurdler==

Latest revision as of 00:18, 29 August 2022

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_24&oldid=177644"