2010 AIME I Problems/Problem 13
Problem
Rectangle and a semicircle with diameter
are coplanar and have nonoverlapping interiors. Let
denote the region enclosed by the semicircle and the rectangle. Line
meets the semicircle, segment
, and segment
at distinct points
,
, and
, respectively. Line
divides region
into two regions with areas in the ratio
. Suppose that
,
, and
. Then
can be represented as
, where
and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
![[asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((126,0)--(63,109.1192),zzttqq); draw((63,109.1192)--(0,0),zzttqq); draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21)); draw((0,-154.31785)--(252,-154.31785)); draw((252,-154.31785)--(252,0)); draw((0,0)--(84,0)); draw((84,0)--(252,0)); draw((63,109.1192)--(63,0)); draw((84,0)--(84,-154.31785)); draw(arc((126,0),126,0,180)); /* points and labels */ dot((0,0)); label("$A$",(-16.43287,-9.3374),NE/2); dot((252,0)); label("$B$",(255.242,5.00321),NE/2); dot((0,-154.31785)); label("$D$",(3.48464,-149.55669),NE/2); dot((252,-154.31785)); label("$C$",(255.242,-149.55669),NE/2); dot((126,0)); label("$O$",(129.36332,5.00321),NE/2); dot((63,109.1192)); label("$N$",(44.91307,108.57427),NE/2); label("$126$",(28.18236,40.85473),NE/2); dot((84,0)); label("$U$",(87.13819,5.00321),NE/2); dot((113.69848,-154.31785)); label("$T$",(116.61611,-149.55669),NE/2); dot((63,0)); label("$N'$",(66.42398,5.00321),NE/2); label("$84$",(41.72627,-12.5242),NE/2); label("$168$",(167.60494,-12.5242),NE/2); dot((84,-154.31785)); label("$T'$",(87.13819,-149.55669),NE/2); dot((252,0)); label("$I$",(255.242,5.00321),NE/2); clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle); [/asy]](http://latex.artofproblemsolving.com/b/2/a/b2a9609890d33310178bc86dcdc5884b38279e7a.png)
Solution 1
The center of the semicircle is also the midpoint of . Let this point be O. Let
be the length of
.
Rescale everything by 42, so . Then
so
.
Since is a radius of the semicircle,
. Thus
is an equilateral triangle.
Let ,
, and
be the areas of triangle
, sector
, and trapezoid
respectively.
To find we have to find the length of
. Project
and
onto
to get points
and
. Notice that
and
are similar. Thus:
.
Then . So:
Let be the area of the side of line
containing regions
. Then
Obviously, the is greater than the area on the other side of line
. This other area is equal to the total area minus
. Thus:
.
Now just solve for .
Don't forget to un-rescale at the end to get .
Finally, the answer is .
Solution 2
Let be the center of the semicircle. It follows that
, so triangle
is equilateral.
Let be the foot of the altitude from
, such that
and
.
Finally, denote , and
. Extend
to point
so that
is on
and
is perpendicular to
. It then follows that
. Since
and
are similar,
Given that line divides
into a ratio of
, we can also say that
where the first term is the area of trapezoid , the second and third terms denote the areas of
a full circle, and the area of
, respectively, and the fourth term on the right side of the equation is equal to
. Cancelling out the
on both sides, we obtain
By adding and collecting like terms,
.
Since ,
, so the answer is
Solution 3
Note that the total area of is
and thus one of the regions has area
As in the above solutions we discover that , thus sector
of the semicircle has
of the semicircle's area.
Similarly, dropping the perpendicular we observe that
, which is
of the total rectangle.
Denoting the region to the left of as
and to the right as
, it becomes clear that if
then the regions will have the desired ratio.
Using the 30-60-90 triangle, the slope of , is
, and thus
.
is most easily found by
:
Equating,
Solving,
See Also
- <url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.
2010 AIME I (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.