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Difference between revisions of "2009 AMC 10B Problems/Problem 13"

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\text{(E) } \overline{EA}
 
\text{(E) } \overline{EA}
 
</math>
 
</math>
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[[Category: Introductory Geometry Problems]]
  
 
== Solution ==
 
== Solution ==
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And finally, the polygon rotates around <math>C</math> until point <math>D</math> hits the <math>x</math> axis at <math>(2014,0)</math>.
 
And finally, the polygon rotates around <math>C</math> until point <math>D</math> hits the <math>x</math> axis at <math>(2014,0)</math>.
 
At this point the side <math>\boxed{\overline{CD}}</math> touches the point <math>(2009,0)</math>. So the answer is <math>\boxed{C}</math>
 
At this point the side <math>\boxed{\overline{CD}}</math> touches the point <math>(2009,0)</math>. So the answer is <math>\boxed{C}</math>
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==Solution 2: Mod Arithmetic==
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The perimeter is <math>23</math> and <math>2009\equiv8(</math>mod <math>23)</math>, so it will end up on side <math>AB</math> + a total of 8 more units. <math>4<8</math>, but <math>4+6=10>8</math>, so it ends on side <math>CD</math> for an answer of  <math>\boxed{C}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2009|ab=B|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 17:39, 9 February 2023

Problem

As shown below, convex pentagon $ABCDE$ has sides $AB=3$, $BC=4$, $CD=6$, $DE=3$, and $EA=7$. The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$-axis. The pentagon is then rolled clockwise to the right along the $x$-axis. Which side will touch the point $x=2009$ on the $x$-axis?

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,0), Ep=7*dir(105), B=3*dir(0); pair D=Ep+B; pair C=intersectionpoints(Circle(D,6),Circle(B,4))[1]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(B--C--D--Ep--A); draw((6,6)..(8,4)..(8,3),EndArrow(3)); xaxis("$x$",-8,14,EndArrow(3)); label("$E$",Ep,NW); label("$D$",D,NE); label("$C$",C,E); label("$B$",B,SE); label("$(0,0)=A$",A,SW); label("$3$",midpoint(A--B),N); label("$4$",midpoint(B--C),NW); label("$6$",midpoint(C--D),NE); label("$3$",midpoint(D--Ep),S); label("$7$",midpoint(Ep--A),W); [/asy]

$\text{(A) } \overline{AB} \qquad \text{(B) } \overline{BC} \qquad \text{(C) } \overline{CD} \qquad \text{(D) } \overline{DE} \qquad \text{(E) } \overline{EA}$

Solution

The perimeter of the polygon is $3+4+6+3+7 = 23$. Hence as we roll the polygon to the right, every $23$ units the side $\overline{AB}$ will be the bottom side.

We have $2009 = 23 \times 87 + 8$. Thus at some point in time we will get the situation when $A=(2001,0)$ and $\overline{AB}$ is the bottom side. Obviously, at this moment $B=(2004,0)$.

After that, the polygon rotates around $B$ until point $C$ hits the $x$ axis at $(2008,0)$.

And finally, the polygon rotates around $C$ until point $D$ hits the $x$ axis at $(2014,0)$. At this point the side $\boxed{\overline{CD}}$ touches the point $(2009,0)$. So the answer is $\boxed{C}$

Solution 2: Mod Arithmetic

The perimeter is $23$ and $2009\equiv8($mod $23)$, so it will end up on side $AB$ + a total of 8 more units. $4<8$, but $4+6=10>8$, so it ends on side $CD$ for an answer of $\boxed{C}$.

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 10 Problems and Solutions

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

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