document.write( "Question 397657: Two points are chosen at random on a line segment whose length is a > 0. Find the probability that the 3 line segments thus formed can be the sides of a triangle. \n" ); document.write( "

Algebra.Com's Answer #281762 by robertb(5830) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Let two of the segments have lengths x, y. Then the third segment must have length a - x - y. Use the triangle inequality on 3 different instances:\r
\n" ); document.write( "\n" ); document.write( "(i)x + y > a - x - y <==> x + y > a/2.\r
\n" ); document.write( "\n" ); document.write( "(ii) x +a - x - y > y <==> a/2 > y\r
\n" ); document.write( "\n" ); document.write( "(iii) Similarly , from y +a -x - y > x we get a/2 > x.\r
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\n" ); document.write( "\n" ); document.write( "The initial conditions are x >0, y >0, and a - x -y >0, or a > x+y. In the Cartesian plane, this feasibility region is an (open) isosceles triangle with vertices at (0,0), (0, a), and (a,0). It has area $\"a%5E2%2F2\"$ .\r
\n" ); document.write( "\n" ); document.write( "Incidentally, the region defined by the instances (i), (ii), and (iii) above is also an open isosceles triangle with vertices (a/2, 0), (a/2, a/2), and (0, a/2). This triangle has area $\"%281%2F2%29%2A%28a%2F2%29%2A%28a%2F2%29+=+a%5E2%2F8\"$ .\r
\n" ); document.write( "\n" ); document.write( "Therefore the probability that the 3 line segments can be the sides of a triangle is $\"%28a%5E2%2F8%29%2F%28a%5E2%2F2%29+=+2%2F8+=+1%2F4\"$ . \n" ); document.write( "

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