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Difference between revisions of "Mock AIME 1 2010 Problems/Problem 3"

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== See Also ==
 
== See Also ==
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*[[Mock AIME 1 2010 Problems]]
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*[[Mock AIME 1 2010 Problems/Problem 2|Followed by Problem 2]]
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[[Category:Intermediate Geometry Problems]]
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*[[Mock AIME 1 2010 Problems/Problem 4|Followed by Problem 4]]
 

Latest revision as of 08:45, 2 August 2024

Problem

Let $AB$ be a line segment of length $20 \sqrt{2}$, and let $S$ be the set of all points $P$ such that $m \angle APB \geq 45^{\circ}$. Find the last three digits of the largest integer less than the area of $S$.

Solution

Let point $O$ be on the perpendicular bisector of $\overline{AB}$ such that $\overline{AO} \perp \overline{OB}$. We know that $AO=OB$ (because $O$ is on the perpendicular bisector of $\overline{AB}$), so we can draw circle $O_A$. Do the same on the other side of $AB$ with point $Q$ and draw circle $Q_A$. The central angle of $\widehat{AB}$ in both circles is $90^{\circ}$. Thus, by the properties of inscribed angles, we know that the locus of all points $P$ such that $\measuredangle APB = \tfrac{90^{\circ}}2=45^{\circ}$ is the border of the area enclosed by the two circles. Going out of this area would force $\measuredangle APB<45^{\circ}$, so the inequality is not satified there. Conversely, going inside the area would force $\measuredangle APB > 45^{\circ}$, 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 $\triangle AOB$ and $\triangle AQB$ are isosceles right triangles, so their leg lengths are $\tfrac{20\sqrt2}{\sqrt2}=20$. Thus, the radius of the circles is $20$, and so the area of the enclosed region is $2 \cdot 20^2 \cdot \pi = 800\pi$ 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 $(20\sqrt2)^2=800$, and the circle has area $\pi \cdot 20^2 = 400\pi$. Thus, each of the sections of the circle not inside the square has an area of $\tfrac{400\pi-800}4=100\pi-200$. 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 $2\cdot(100\pi-200)=200\pi-400$. Subtracting this from the area of the two circles, we see that the area of the enclosed region is $800\pi-(200\pi-400)=600\pi+400$. The greatest integer less than this amount is $600\cdot3.14+400=1884+400=2284$, so our answer is $\boxed{284}$.

See Also

Mock AIME 1 2010 (Problems, Source)
Preceded by
Problem 2 		Followed by
Problem 4
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