Quadrilaterals and Inscribed Circle

Date: 05/06/99 at 07:57:30
From: Jayant
Subject: Permutations and Combinations

Ten sticks of lengths 1,2,3,....,10 are given. Four are selected to 
form the sides of a quadrilateral. Find the number of quadrilaterals 
so formed. Also find in how many of these quadrilaterals a circle can 
be inscribed.

Date: 05/07/99 at 11:19:20
From: Doctor Anthony
Subject: Re: Permutations and Combinations

The condition for a quadrilateral ABCD to have an inscribed circle is 
that

    AB + CD = AD + BC

You can prove this easily by considering that the tangent to the 
circle from point A along AB is equal to the length of the tangent 
along AD, similarly the tangents from B are equal along BA and BC, and 
so on with tangents from C along CB and CD and tangents from D along 
DC and DA. Adding up these equal lengths leads to the result quoted 
above.

So the four sticks must satisfy the condition that the sum of one pair 
is equal to the sum of the other pair

  So    1+10 = 2+9 = 3+8 = 4+7 = 5+6

This gives:                               C(5,2) = 10 quadrilaterals

Similarly 2+10 = 3+9 = 4+8 = 5+7  giving  C(4,2) = 6 quadrilaterals

    "  3+10 = 4+9 = 5+8 = 6+7       "     C(4,2) = 6       "

    "  4+10 = 5+9 = 6+8             "     C(3,2) = 3       "

    "  5+10 = 6+9 = 7+8             "     C(3,2) = 3       "

    "  6+10 = 7+9                   "     C(2,2) = 1       "

    "  7+10 = 8+9                   "     C(2,2) = 1       "
     
    "  1+9 = 2+8 = 3+7 = 4+6        "     C(4,2) = 6       "

    "  1+8 = 2+7 = 3+6 = 4+5        "     C(4,2) = 6       "

    "  1+7 = 2+6 = 3+5              "     C(3,2) = 3       "

    "  1+6 = 2+5 = 3+4              "     C(3,2) = 3       "

    "  1+5 = 2+4                    "     C(2,2) = 1       "      

    "  1+4 = 2+3                    "     C(2,2) = 1       "
                                      --------------------------
                                           Total = 50  quadrilaterals

Since the order must stay as opposite pairs, we could keep one pair 
fixed and swap over the other pair giving double this answer, but if 
we can view the quadrilateral from either side, we would divide by 2 
again, so we stick at 50 quadrilaterals which can have an inscribed 
circle.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/