Question 1199312
<br>
let O(0,0), A(6,0), B(6,6), c(0,6) be the vertices of a square OABC, and Let M be the midpoint of OB.<br>
Here is a sketch:
{{{drawing(400,400,-1,7,-1,7
,line(0,0,6,0),line(6,0,6,6),line(6,6,0,6),line(0,6,0,0),line(0,0,6,6)
,locate(0,-.2,"O(0,0)"),locate(6,-.2,"A(6,0)"),locate(6,6.5,"B(6,6)"),locate(0,6.5,"C(0,6)")
,circle(3,3,.1),locate(3.2,3.2,"M(3,3)")
)}}}
a) Find the probability that a point chosen at random from the square is farther from O than from M.<br>
To do this, determine where the points are that are the same distance from O as they are from M.  This is simple; they lie on the perpendicular bisector of segment OM.<br>
{{{drawing(400,400,-1,7,-1,7
,line(0,0,6,0),line(6,0,6,6),line(6,6,0,6),line(0,6,0,0),line(0,0,6,6)
,locate(0,-.2,"O(0,0)"),locate(6,-.2,"A(6,0)"),locate(6,6.5,"B(6,6)"),locate(0,6.5,"C(0,6)")
,circle(3,3,.1),locate(3.2,3.2,"M(3,3)")
,line(3,0,0,3)
)}}}
Note that this result can be obtained algebraically by solving the equation that says the distance from P(x,y) to O(0,0) is the same as the distance from P(x,y) to M(3,3):<br>
{{{sqrt(x^2+y^2)=sqrt((x-3)^2+(y-3)^2)}}}<br>
(I won't do that in my response; however, it would be a good exercise for the student to solve this equation and find that indeed the set of points equidistant from O and M are the points on the line x+y=3.)<br>
It should be easy to see from the sketch that the points in the square that are farther from O than from M constitute exactly 7/8 of the square.  So<br>
ANSWER to part a): 7/8<br>
b) Find the probability that a point chosen at random from the square is more than twice as far from O as from M.<br>
To solve this part, we could find the points that are EXACTLY twice as far from O as from M by solving the equation<br>
{{{sqrt(x^2+y^2)=2*(sqrt((x-3)^2+(y-3)^2))}}}<br>
(Again I won't do that here in my response; and again it would be a good exercise for the student to do so.  You should find that the circle is centered at (4,4) with radius 2*sqrt(2).)<br>
The set of points exactly twice as far from O as from M is a circle.  Two of those points are (2,2) and (6,6); and by symmetry the segment joining those two points is a diameter of the circle.<br>
Here is a sketch:
{{{drawing(400,400,-1,7,-1,7
,line(0,0,6,0),line(6,0,6,6),line(6,6,0,6),line(0,6,0,0),line(0,0,6,6)
,locate(0,-.2,"O(0,0)"),locate(6,-.2,"A(6,0)"),locate(6,6.5,"B(6,6)"),locate(0,6.5,"C(0,6)")
,circle(3,3,.1),locate(3.2,3.2,"M(3,3)")
,circle(2,2,.1),locate(2.2,2.2,"(2,2)"),circle(4,4,.1),locate(4.2,4.2,"(4,4)")
,circle(4,4,sqrt(8))
,circle(6,2,.1),locate(6,2.2,"(6,2)"),circle(2,6,.1),locate(2,6,"(2,6)")
)}}}
To answer this part of the problem, we need to find what fraction of the square is inside the circle.<br>
The region of the square inside the circle consists of a semicircle plus an isosceles right triangle determined by the segment from (6,2) to (2,6).  The radius of the circle is {{{2*sqrt(2))}}} and the lengths of the legs of the triangle are 4, so the area of the semicircle is {{{(1/2)(pi)((2*sqrt(2))^2)=4pi}}} and the area of the triangle is {{{(1/2)(4)(4)=8}}}.<br>
The area of the whole square is 36, so the probability that a random point in the square is more than twice as far from O as it is from M is<br>
ANSWER to part b): {{{(4pi+8)/36=(pi+2)/9}}}<br>