Difference between revisions of "2010 AIME I Problems/Problem 13"
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<math>DA = 63 \sqrt{6} \longrightarrow \boxed {069}</math> | <math>DA = 63 \sqrt{6} \longrightarrow \boxed {069}</math> | ||
− | == Solution 4 (Coordinates) == | + | === Solution 4 (Coordinates) === |
Like above solutions, note that <math>ANO</math> is equilateral with side length <math>126,</math> where <math>O</math> is the midpoint of <math>AB.</math> Then, if we let <math>DA=a</math> and set origin at <math>D=(0,0),</math> we get <math>N=(63,a+63\sqrt{3}), U=(84,a).</math> Line <math>NU</math> is then <math>y-a=\sqrt{27}(x-84),</math> so it intersects <math>CA,</math> the <math>x</math>-axis, at <math>x=(a/\sqrt{27}+84),</math> giving us point <math>T.</math> Now the area of region <math>R</math> is <math>252a+\pi(126)^2 / 2,</math> so one third of that is <math>84a+\pi(126)^2 / 6.</math> | Like above solutions, note that <math>ANO</math> is equilateral with side length <math>126,</math> where <math>O</math> is the midpoint of <math>AB.</math> Then, if we let <math>DA=a</math> and set origin at <math>D=(0,0),</math> we get <math>N=(63,a+63\sqrt{3}), U=(84,a).</math> Line <math>NU</math> is then <math>y-a=\sqrt{27}(x-84),</math> so it intersects <math>CA,</math> the <math>x</math>-axis, at <math>x=(a/\sqrt{27}+84),</math> giving us point <math>T.</math> Now the area of region <math>R</math> is <math>252a+\pi(126)^2 / 2,</math> so one third of that is <math>84a+\pi(126)^2 / 6.</math> | ||
Revision as of 19:34, 8 February 2020
Contents
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,
Solution 4 (Coordinates)
Like above solutions, note that is equilateral with side length
where
is the midpoint of
Then, if we let
and set origin at
we get
Line
is then
so it intersects
the
-axis, at
giving us point
Now the area of region
is
so one third of that is
The area of the smaller piece of is
Setting this equal to
and canceling the
yields
so
and the anser is
~ rzlng
See Also
- <url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.
2010 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 14 | |
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