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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 24"
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==Solution==
 
==Solution==
−
Let <math>E</math> be a point on <math>\overline{AB}</math> such that <math>BCDE</math> is a parallelogram.
+
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>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.
 +
 
 +
If <math>k=0,</math> then <math>ABCD</math> is a rhombus with side-length <math>18,</math> which is valid.
 +
 
 +
If <math>k\neq0,</math> then we have six cases:
 +
<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,18-3k,18-2k)</math></li><p>
 +
  <li><math>(b,c,d)=(18-2k,18-k,18-3k)</math></li><p>
 +
  <li><math>(b,c,d)=(18-2k,18-3k,18-k)</math></li><p>
 +
  <li><math>(b,c,d)=(18-3k,18-k,18-2k)</math></li><p>
 +
  <li><math>(b,c,d)=(18-3k,18-2k,18-k)</math></li><p>
 +
</ol>
  
 
<b>WILL COMPLETE VERY SOON. A MILLION THANKS FOR NOT EDITING THIS PAGE.</b>
 
<b>WILL COMPLETE VERY SOON. A MILLION THANKS FOR NOT EDITING THIS PAGE.</b>

Revision as of 23:34, 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 $AE=18-c.$

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.

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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