2009 AMC 10B Problems/Problem 13
Problem
As shown below, convex pentagon 
has sides 
, 
, 
, 
, and 
. The pentagon is originally positioned in the plane with vertex 
at the origin and vertex 
on the positive 
-axis. The pentagon is then rolled clockwise to the right along the -axis. Which side will touch the point 
on the -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]](http://latex.artofproblemsolving.com/f/d/6/fd6ac5c8a85c125cb27930bfe3c7aa4b9cd98e12.png)
Solution
The perimeter of the polygon is 
. Hence as we roll the polygon to the right, every 
units the side 
will be the bottom side.
We have 
. Thus at some point in time we will get the situation when 
and is the bottom side. Obviously, at this moment 
.
After that, the polygon rotates around until point 
hits the axis at 
.
And finally, the polygon rotates around until point 
hits the axis at 
.
At this point the side 
touches the point 
. So the answer is 
Solution 2: Mod Arithmetic
The perimeter is and 
mod 
, so it will end up on side 
+ a total of 8 more units. 
, but 
, so it ends on side 
for an answer of .
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 | ||
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