<li>What is the probability that two points on a circle of radius r are more than r units apart?</li>
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<p>i’ve asked my friend and 2 math tutors at the library and nothing</p>
<p>I’m just going to guess and say 50%…</p>
<p>is it 50?..</p>
<p>I think it should be more than 50% since if you move the line from straight and take to another part it should hit the circle at a closer place, hard to explain but ya I think more than 50%.</p>
<p>2/3
Law of Cosines</p>
<p>You’ll find that their distance apart is less than that whenever Point 2 is within π/3 radians of Point 1 (From -π/3 to π/3 if you set point 1 at 0 radians). This makes 2/3 of the circle too far.</p>
<p>c^2 = a^2+b^2 - 2abcosC
c^2 = 2r^2 - 2r^2cosC
1 - c^2/2r^2 = cosC</p>
<p>C = π/3 – This means that when c=r, C=π/3, therefore for any C>π/3, c>r</p>
<p>So then any Point 2 that is separated from Point 1 by an angle of less than or equal to π/3 (in either direction) it will be less than or equal to a distance r from Point 1. It is too far then from π/3 to 5π/3, a total angle of 4π/3 which is 2/3 ( (4π/3)/(2π)=2/3 ) of the total circle.</p>
<p>It would defnitely be greater than 50%. In fact, it would probably be around 70ish or so %.</p>
<p>i don’t know the answer</p>
<p>please draw on paint or scan paper!</p>
<p>thank you</p>
<p>Sure 1 sec, I’ll edit the scanned paper in.
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<p>it is soo small</p>
<p>I edited it again.</p>
<p>Wow! That’s very impressive kstrike!</p>
<p>That’s exactly what I was thinking, but I was too lazy to actually do the problem :P</p>
<p>Great Job!</p>
<p>im attempting to understand but i think it might be too complicated for me</p>
<p>In case you don’t want to use the law of cosines: </p>
<p>Consider any arbitrary point on a circle. Draw another point on the circle a distance r away from it. Connect the two points with a chord, then draw two segments from the center of the circle to each of these two points. </p>
<p>Notice that every segment you’ve drawn so far has length r, since that’s the radius of the circle. Therefore, your picture shows a triangle with all sides equal to r - in other words, it’s equilateral. </p>
<p>Now take your original point, draw another point on a circle a distance r away in the opposite direction from your first, and repeat the process. You should now have two equilateral triangles. Each equilateral triangle has angle measure 60, for a combined angle of 120, which is 1/3 of the total circle. Any point on the circle which does not lie on the arcs bounded by these two triangles is more than r away from your original point. </p>
<p>Therefore, your answer’s 1 - 1/3, or 2/3.</p>
<p>Good job javademon. I think that was much more elegant than my solution.</p>
<p>Ok i understood everything until your last sentence. So what you mean bounded by those two triangles? and why only use two triangles? why not 3 or 1</p>
<p>what book is this problem from?</p>
<p>no book</p>
<p>it part of an application to a local university summer program which i probably won’t be applying to now :(</p>