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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 24"
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==Solution==
 
==Solution==
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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>
+
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>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.
+
<b>DIAGRAM WILL BE READY VERY VERY SOON ...</b>
 +
 
 +
We apply the Law of Cosines to <math>\triangle ADE:</math>
 +
<cmath>\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*}</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>
  
 
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.

Revision as of 23:43, 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. Note 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. $(b,c,d)=(18-k,18-3k,18-2k)$

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

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

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

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

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

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