Difference between revisions of "2021 Fall AMC 12A Problems/Problem 24"
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==Solution== | ==Solution== | ||
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| + | ===Solution 1=== | ||
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>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. | ||
<asy> | <asy> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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| + | ===Remark=== | ||
| + | Note that 4 of the 6 above cases can be eliminated by the triangle inequality (after removing parallelogram BCDE). ~hurdler | ||
| + | |||
| + | ===Video Solution and exploration by hurdler=== | ||
| + | |||
| + | [https://www.youtube.com/watch?v=IA5fWLdvVQg& Video exploration and motivated solution] | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2021 Fall|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:38, 24 November 2021
Contents
Problem
Convex quadrilateral 
has 
and 
In some order, the lengths of the four sides form an arithmetic progression, and side 
is a side of maximum length. The length of another side is 
What is the sum of all possible values of 
?
Solution
Solution 1
Let 
be a point on such that 
is a parallelogram. Suppose that 
and 
so 
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),1.5*NE,red); label("$b$",midpoint(D--E),1.5*NE,red); label("$c$",midpoint(C--D),1.5*N,red); label("$d$",midpoint(D--A),1.5*NW,red); [/asy]](http://latex.artofproblemsolving.com/7/d/b/7dba6eb322b654fdc911f3b9e09f649f10beaae3.png)
We apply the Law of Cosines to 
Let 
be the common difference of the arithmetic progression of the side-lengths. It follows that 
and 
are 
and 
in some order. It is clear that 
If 
then is a rhombus with side-length 
which is valid.
If 
then we have six cases:
Note that 
becomes 
from which 
So, this case generates no valid solutions 
Note that becomes 
from which 
So, this case generates 
Note that becomes 
from which So, this case generates no valid solutions
Note that becomes 
from which 
So, this case generates 
Note that becomes 
from which 
So, this case generates no valid solutions
Note that becomes 
from which 
So, this case generates no valid solutions
Together, the sum of all possible values of is 
~MRENTHUSIASM
Remark
Note that 4 of the 6 above cases can be eliminated by the triangle inequality (after removing parallelogram BCDE). ~hurdler
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. 
