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"
Line 16: Line 16:
 
(18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar)
 
(18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Let <math>k</math> be the common difference of the arithmetic progression of the side-lengths. It follows that <math>b,c,</math> and <math>d</math> are <math>18-k, 18-2k,</math> and <math>18-3k,</math> in some order. Note that <math>0\leq k<6.</math>
+
Let <math>k</math> be the common difference of the arithmetic progression of the side-lengths. It follows that <math>b,c,</math> and <math>d</math> are <math>18-k, 18-2k,</math> and <math>18-3k,</math> in some order. It is clear that <math>0\leq k<6.</math>
  
 
If <math>k=0,</math> then <math>ABCD</math> is a rhombus with side-length <math>18,</math> which is valid.
 
If <math>k=0,</math> then <math>ABCD</math> is a rhombus with side-length <math>18,</math> which is valid.
Line 23: Line 23:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li><math>(b,c,d)=(18-k,18-2k,18-3k)</math></li><p>
 
   <li><math>(b,c,d)=(18-k,18-2k,18-3k)</math></li><p>
 +
Note that <math>(\bigstar)</math> becomes <math>2k(5k-18)=(36-4k)(2k),</math> from which <math>k=6.</math> So, this case generates no valid solutions <math>(b,c,d).</math><p>
 
   <li><math>(b,c,d)=(18-k,18-3k,18-2k)</math></li><p>
 
   <li><math>(b,c,d)=(18-k,18-3k,18-2k)</math></li><p>
 +
Note that <math>(\bigstar)</math> becomes <math>3k(5k-18)=(36-3k)k,</math> from which <math>k=5.</math> So, this case generates <math>(b,c,d)=(13,3,8).</math><p>
 
   <li><math>(b,c,d)=(18-2k,18-k,18-3k)</math></li><p>
 
   <li><math>(b,c,d)=(18-2k,18-k,18-3k)</math></li><p>
 +
Note that <math>(\bigstar)</math> becomes <math>k(4k-18)=(36-5k)k,</math> from which <math>k=6.</math> So, this case generates no valid solutions <math>(b,c,d).</math><p>
 
   <li><math>(b,c,d)=(18-2k,18-3k,18-k)</math></li><p>
 
   <li><math>(b,c,d)=(18-2k,18-3k,18-k)</math></li><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>
 
   <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>
 
</ol>
 
</ol>
  

Revision as of 23:57, 23 November 2021

Problem

Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60 \textdegree$, 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

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.

DIAGRAM WILL BE READY VERY VERY SOON ...

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

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

    7. Note that $(\bigstar)$ becomes

WILL COMPLETE VERY SOON. A MILLION THANKS FOR NOT EDITING THIS PAGE.

~MRENTHUSIASM

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=165853"