MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_001C_01C25109.4AFC4F00" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2600.0000 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2600.0000 X-Virus-Scanned: by amavisd-milter (http://amavis.org/) X-Virus-Scanned: by amavisd-milter (http://amavis.org/) This is a multi-part message in MIME format. ------=_NextPart_000_001C_01C25109.4AFC4F00 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable MATHEMATICS PROBLEM 125 =20 Let A and B be fixed points in three dimensional space. = Find and describe the locus of all points P that are twice as far from = A as from B. (That is, |AP| =3D 2|BP|, where |RS| denotes the distance between the = points R and S.) =20 =20 =20 Each midshipman submitting a correct solution with a correct explanation = to Problem 125 by 1700 Monday 30 September 2002 will win a cookie. = Submit solutions to Prof. Wardlaw at mathprob@usna.edu (please no = attachments!) or via his mailbox in Chauvenet 301. No correct solutions to Mathematics Problem 124 were = submitted. My solution to Mathematics Problem 124 is posted on the = board and appears below. MATHEMATICS PROBLEM 124 =20 Three numbers, a, b, and c are each chosen randomly and = independently from the unit interval [0, 1] using the uniform = distribution. What is the probability that a triangle can be formed = with sides of length a, b, and c ? =20 =20 Solution. We can consider a to lie in the interval [0, 1] = on the x-axis, b to lie in the interval [0, 1] on the y-axis, = and c to lie in the interval [0, 1] on the z-axis. Thus choosing = a, b, and c is equivalent to randomly choosing a point in the unit = cube =20 B =3D [0, 1]=B4 [0, 1]=B4 [0, 1] =3D {(x, y, z) =CE R3 : 0 < x < 1, 0 < = y < 1, 0 < z < 1} =20 in the first octant of R3. A triangle can be formed with such sides of = length a, b, and c if and only if 0 < a < 1, 0 < b < 1, 0 < c < = 1, a < b + c, b < a + c, and c < a + b. Now it can be seen that the = point P =3D (a, b, c) =CE B satisfies these inequalities if it does = not lie in any of the three tetrahedra =20 T1 =3D {(x, y, z) =CE B : 0 < x - y - z}, T2 =3D {(x, y, z) =CE B : = 0 < - x + y - z}, =20 and T3 =3D {(x, y, z) =CE B : 0 < -x - y + z}. =20 ( T1 has vertices (0, 0, 0), (1, 0, 0), (1, 1, 0), and (1, 0, 1), = T2 has vertices (0, 0, 0), (0, 1, 0), (1, 1, 0), and (0, 1, 1), and = T3 has vertices (0, 0, 0), (0, 0, 1), (1, 0, 1), and (0, 1, 1).) = Each of these tetrahedra has volume 1/6, and the volume of the = intersection of any two of the tetrahedra is 0. Now we see that a triangle can be formed from the numbers a, = b, c chosen from the interval [0, 1] if and only if the point P =3D = (a, b, c) lies in the region S =3D B - T1 - T2 - T3 obtained by = deleting the three tetrahedra from the cube B. Thus the probability = that a triangle can be formed with sides of lengths a, b, and c = chosen from the interval [0, 1] is the volume of S, Vol(S) =3D Vol(B) - Vol(T1) - Vol(T2) - Vol(T3) =3D 1 - 1/6 - 1/6 - 1/6 = =3D 1/2 . =20 ------=_NextPart_000_001C_01C25109.4AFC4F00 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

MATHEMATICS =20 PROBLEM  = 125

 

         =20 Let  A  and  B =20 be fixed points in three dimensional space.  Find and describe the locus of = all=20 points  P  that are twice as far = from  A =20 as from =20 B.

(That is, =20 |AP| =3D 2|BP|, where =20 |RS|  denotes the = distance=20 between the points  R  and  = S.)

 

  

 

Each midshipman submitting a correct solution = with a=20 correct explanation to Problem 125 by 1700 Monday 30 September 2002 will = win a=20 cookie.  Submit solutions = to Prof.=20 Wardlaw at mathprob@usna.edu (please no attachments!) or via his mailbox = in=20 Chauvenet 301.

         =20 No correct solutions to Mathematics Problem 124 were = submitted.  My solution to Mathematics = Problem 124=20 is posted on the board and appears = below.

 

 

MATHEMATICS =20 PROBLEM =20 124

 

Three numbers, =20 a,  b, and  c =20 are each chosen randomly and independently from the unit = interval  [0, 1]  using the uniform = distribution.  What is the probability that a = triangle=20 can be formed with sides of =20 length  a, =20 b,  and  c = ?

 

 

         =20 Solution.   We can = consider  a =20 to lie in the interval  = [0, 1]  on the    x-axis,  b =20 to lie in the interval  = [0, 1]  on the  y-axis,  and   c =20 to lie in the interval  = [0, 1]  on the  z-axis.  Thus choosing  a, =20 b, and  c =20 is equivalent to randomly choosing a point in the unit=20 cube

 

B = =3D [0, = 1]=B4 [0, 1]=B4 [0, 1] =3D=20 {(x, y, z) =CE R3=20 : 0 < x < 1, 0 < y < 1, 0=20 < z < 1}

 

in the first octant of  R3.  A triangle can be formed with = such sides=20 of  length  a, =20 b,  and  c =20 if and only if  0 < a < 1,  0 < b < 1,  0 < c < 1,  a < b + c,  b < a + c,  and  c=20 < a + b.  Now it = can be seen=20 that the point  P =3D (a, b, c) = =CE B  satisfies these inequalities = if it does=20 not lie in any of the three tetrahedra

 

T1 =3D {(x, y, z) = =CE B :  0 < x -  y - z},  T2 =3D {(x, y, z) = =CE B :  0 < - x + y -=20 z},

 

and

T3 =3D {(x, y, z) = =CE B :  0 < -x - y +=20 z}.

 

(  T1  has vertices  (0, 0, 0), (1, 0, 0), (1, 1, 0), = and =20 (1, 0, 1),  T2  has vertices  (0, 0, 0), (0, 1, 0), (1, 1, 0), and (0, 1, 1), and  T3  has vertices  (0, 0, 0),      (0, 0, = 1), (1,=20 0, 1), and  (0, 1, = 1).)  Each of these tetrahedra has = volume 1/6,=20 and the volume of the intersection of any two of the tetrahedra is=20 0.

         =20 Now we see that a triangle can be formed from the numbers  a, b, c chosen from the = interval =20 [0, 1]  if and = only if=20 the point  P =3D (a, b, c)  lies in the region  S =3D B=20 - T1 - = T2=20 - T3  obtained by deleting the = three=20 tetrahedra from the cube  = B. =20 Thus the probability that a triangle can be formed with = sides=20 of  lengths  a, =20 b,  and  c =20 chosen from the interval =20 [0, 1]  is =20 the volume of  S,

Vol(S) =3D = Vol(B) =96 Vol(T1) =96 Vol(T2) =96 Vol(T3) =3D 1 - 1/6 - = 1/6 - 1/6 =3D=20 1/2 .

 

------=_NextPart_000_001C_01C25109.4AFC4F00--