Difference between revisions of "Mock AIME 1 2010 Problems/Problem 3"
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== Solution == | == Solution == | ||
− | <math>\boxed{284}</math>. | + | |
+ | Let point <math>O</math> be on the perpendicular bisector of <math>\overline{AB}</math> such that <math>\overline{AO} \perp \overline{OB}</math>. We know that <math>AO=OB</math> (because <math>O</math> is on the perpendicular bisector of <math>\overline{AB}</math>), so we can draw circle <math>O_A</math>. Do the same on the other side of <math>AB</math> with point <math>Q</math> and draw circle <math>Q_A</math>. The central angle of <math>\widehat{AB}</math> in both circles is <math>90^{\circ}</math>. Thus, by the properties of [[inscribed angle|inscribed angles]], we know that the [[locus]] of all points <math>P</math> such that <math>\measuredangle APB = \tfrac{90^{\circ}}2=45^{\circ}</math> is the border of the area enclosed by the two circles. Going out of this area would force <math>\measuredangle APB<45^{\circ}</math>, so the inequality is not satified there. Conversely, going inside the area would force <math>\measuredangle APB > 45^{\circ}</math>, so the inequality is satisfied. Thus, we look for the area enclosed by the two circles in the diagram below. | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = origin; | ||
+ | point B = A + (20*sqrt(2),0); | ||
+ | |||
+ | point O = (A+B)/2+(0,10*sqrt(2)); | ||
+ | point Q = O-(0,20*sqrt(2)); | ||
+ | |||
+ | // Segment AB and Circles | ||
+ | draw(A--B); | ||
+ | draw(circle(O,length(segment(O,A)))); | ||
+ | draw(circle(Q,length(segment(Q,A)))); | ||
+ | |||
+ | // Radii | ||
+ | draw(O--A); | ||
+ | draw(O--B); | ||
+ | draw(Q--A); | ||
+ | draw(Q--B); | ||
+ | |||
+ | // Labelling Points | ||
+ | dot(A); | ||
+ | label("A",A,W); | ||
+ | dot(B); | ||
+ | label("B",B,E); | ||
+ | dot(O); | ||
+ | label("O",O,N); | ||
+ | dot(Q); | ||
+ | label("Q",Q,S); | ||
+ | |||
+ | // Right angle marks | ||
+ | markscalefactor=0.35; | ||
+ | draw(rightanglemark(A,O,B)); | ||
+ | draw(rightanglemark(A,Q,B)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | We know that <math>\triangle AOB</math> and <math>\triangle AQB</math> are isosceles right triangles, so their leg lengths are <math>\tfrac{20\sqrt2}{\sqrt2}=20</math>. Thus, the radius of the circles is <math>20</math>, and so the area of the enclosed region is <math>2 \cdot 20^2 \cdot \pi = 800\pi</math> minus the region enclosed by both circles. To find the area of this doubly enclosed region, we consider the square in the diagram below. | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point O = origin; | ||
+ | point A = dir(-135); | ||
+ | point B = dir(-45); | ||
+ | point C = dir(45); | ||
+ | point D = dir(135); | ||
+ | |||
+ | // Square ABCD and Circle | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(circle(O,1)); | ||
+ | |||
+ | // Diagonals | ||
+ | draw(A--C); | ||
+ | draw(B--D); | ||
+ | |||
+ | // Labelling Points | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("B",B,SE); | ||
+ | dot(C); | ||
+ | label("C",C,NE); | ||
+ | dot(D); | ||
+ | label("D",D,NW); | ||
+ | dot(O); | ||
+ | label("O",O,N); | ||
+ | |||
+ | // Right angle mark | ||
+ | markscalefactor=0.015; | ||
+ | draw(rightanglemark(A,O,B)); | ||
+ | |||
+ | // Length Labels | ||
+ | label("$20\sqrt2$",midpoint(C--D),N); | ||
+ | label("$20$",midpoint(O--B),SW); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | The entire square has an area of <math>(20\sqrt2)^2=800</math>, and the circle has area <math>\pi \cdot 20^2 = 400\pi</math>. Thus, each of the sections of the circle not inside the square has an area of <math>\tfrac{400\pi-800}4=100\pi-200</math>. Thus, looking back at the first diagram, we see that the doubly enclosed area (which we would count twice if we just added the areas of the two circles) is composed of two of these regions. Thus, it has area <math>2\cdot(100\pi-200)=200\pi-400</math>. Subtracting this from the area of the two circles, we see that the area of the enclosed region is <math>800\pi-(200\pi-400)=600\pi+400</math>. The greatest integer less than this amount is <math>600\cdot3.14+400=1884+400=2284</math>, so our answer is <math>\boxed{284}</math>. | ||
== See Also == | == See Also == |
Revision as of 08:32, 2 August 2024
Problem
Let be a line segment of length
, and let
be the set of all points
such that
. Find the last three digits of the largest integer less than the area of
.
Solution
Let point be on the perpendicular bisector of
such that
. We know that
(because
is on the perpendicular bisector of
), so we can draw circle
. Do the same on the other side of
with point
and draw circle
. The central angle of
in both circles is
. Thus, by the properties of inscribed angles, we know that the locus of all points
such that
is the border of the area enclosed by the two circles. Going out of this area would force
, so the inequality is not satified there. Conversely, going inside the area would force
, so the inequality is satisfied. Thus, we look for the area enclosed by the two circles in the diagram below.
We know that and
are isosceles right triangles, so their leg lengths are
. Thus, the radius of the circles is
, and so the area of the enclosed region is
minus the region enclosed by both circles. To find the area of this doubly enclosed region, we consider the square in the diagram below.
The entire square has an area of , and the circle has area
. Thus, each of the sections of the circle not inside the square has an area of
. Thus, looking back at the first diagram, we see that the doubly enclosed area (which we would count twice if we just added the areas of the two circles) is composed of two of these regions. Thus, it has area
. Subtracting this from the area of the two circles, we see that the area of the enclosed region is
. The greatest integer less than this amount is
, so our answer is
.