MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0000_01C1D98D.6AC38860" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 5.50.4522.1200 X-MimeOLE: Produced By Microsoft MimeOLE V5.50.4522.1200 X-AntiVirus: scanned for viruses by AMaViS 0.2.1 (http://amavis.org/) This is a multi-part message in MIME format. ------=_NextPart_000_0000_01C1D98D.6AC38860 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable MATHEMATICS PROBLEM 124 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 Each midshipman submitting a correct solution with a correct = explanation to Problem 124 by 1700 on Tuesday 30 April 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 123 were submitted. My = solution to Mathematics Problem 123 is posted on the board and is on the = reverse side of this sheet. MATHEMATICS PROBLEM 123 Two long (length at least a + b + c units) horizontal halls meet at = right angles. The east-west hall is a units wide, and the north-south = hall is b units wide. Both have vertical walls and a horizontal = ceiling c units high. What is the length of the longest rigid rod (of = negligible diameter) which can be carried along the east-west hall, = around the corner, and along the north-south hall? Solution First we find the length LH of the longest rigid rod that can go = through the halls in a horizontal position. To find this length, let = L(() be the length of the horizontal line segment through the corner of = the halls which touches the opposite walls of the two halls and makes = an angle of ( with the east- west wall. Then L(() =3D A + B, where = A is that portion of the line segment crossing the east-west hall of = width a, and B is that portion of the line segment crossing the = north-south hall of width b. Then a/A =3D sin((), since A is the = hypotenuse of the right triangle with angle ( and opposite side a, = and b/B =3D cos((), since B is the hypotenuse of the right triangle = with angle ( and adjacent side b. Hence,=20 L(() =3D A + B =3D a csc(() + b sec((), is minimized when L((() =3D -a csc(()cot(() + b sec(()tan(() =3D 0, or = when ( =3D ( is the angle such that a/b =3D (sec(()tan(())/(csc(()cot(()) =3D tan3((). Hence, the length LH of the longest rigid rod that can go through the = halls in a horizontal position is the minimum value of L((), and so LH =3D L(() =3D a csc(() + b sec(() =3D a(1+cot2(())1/2 + = b(1+tan2(())1/2 =3D a(1+ (b/a)2/3)1/2 + b(1+(a/b)2/3)1/2 =3D (a2+ a4/3b2/3 )1/2 + = (b2+ a2/3b4/3 )1/2 . Now if we lay this length LH on the floor of the hall and connect one = end of LH to the point on the ceiling directly above the other end, we = see that the hypotenuse of this right triangle with sides LH and c = has length =20 LMAX =3D ( LH 2 + c2 )1/2=20 =3D [a2+ a4/3b2/3 + 2(2 a2b2+ a8/3b4/3 + a4/3b8/3 )1/2 + b2+ a2/3b4/3 = + c2 ]1/2 , which is the length of the longest rigid rod that can be carried through = the hallways. ------=_NextPart_000_0000_01C1D98D.6AC38860 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable MATHEMATICS PROBLEM #117

MATHEMATICS  PROBLEM  124

 

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 ?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

        = ;  Each midshipman submitting a correct solution with a correct explanation to = Problem 124 by 1700 on Tuesday 30 April 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 123 were submitted.  My solution to Mathematics = Problem 123 is posted on the board and is on the reverse side of this = sheet.

MATHEMATICS  PROBLEM  = 123

 

Two long (length at least  a + b + c  units) horizontal halls meet at right angles.  The east-west hall is  a  units wide, and the north-south hall is  b  = units wide.  Both have vertical = walls and a horizontal ceiling  = c  units high.  What is the length of the longest rigid rod (of = negligible diameter) which can be carried along the east-west hall, around the = corner, and along the north-south hall?

 

Solution

 

First we find the length  LH  of the longest rigid rod that = can go through  the halls in a = horizontal position.  To find this = length, let  L(q)  be the length of the horizontal line segment through = the corner of the halls which touches the opposite walls of  the two halls and makes an = angle of  q  with the east- west wall.  Then  = L(q) = =3D A + B, where  A  is that portion of the line segment crossing the = east-west hall of width  a,  and  B  is = that portion of the line segment crossing the north-south hall of width  b.  Then  = a/A =3D sin(q), = since  A  is the hypotenuse of the right triangle with angle  q  and opposite side  a, and  b/B =3D = cos(q), = since  B  is the hypotenuse of the right triangle with angle  q  and adjacent side  b.  Hence, =

 

L(q) = =3D  A + B =3D a = csc(q) + = b sec(q),=

 

is = minimized when  = L=A2(q) =3D  -a csc(q)cot(q) + b sec(q)tan(q) =3D 0, or = when  q  =3D  a  = is the angle such that

 

a/b =3D  = (sec(a)tan(<= font size=3D3 face=3DSymbol>a))/(csc(a)cot(<= font size=3D3 face=3DSymbol>a)) = =3D tan3(a).=

 

Hence, the length  LH  of the longest rigid rod that can go through  the halls in a horizontal position is the minimum = value of  L(a), and = so

 

LH  =3D L(a) =3D a = csc(a) + = b sec(a) = =3D  = a(1+cot2(a))1/2 + b(1+tan2(a))1/2

 

=3D  a(1+ = (b/a)2/3)1/2 + b(1+(a/b)2/3)1/2  =3D  = (a2+ a4/3b2/3 )1/2 + (b2+ a2/3b4/3 = )1/2 .

 

Now if we lay this length  LH  on the floor of the hall and connect one end of  LH  to the point on the ceiling directly above the other = end, we see that the hypotenuse of this right triangle with sides  LH  and  = c  has length 

 

LMAX  =3D  ( LH 2  +  = c2 )1/2

 

=3D = [a2+ a4/3b2/3 + 2(2 a2b2+ = a8/3b4/3 + a4/3b8/3 )1/2  + b2+ a2/3b4/3  +  c2  ]1/2 ,

 

 

which is the length of the longest = rigid rod that can be carried through the hallways.

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