Quadrilaterals and Inscribed CircleDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/