Another Isosceles TriangleDate: 10/27/1999 at 02:49:07 From: Imaad Dalal Subject: Geometry A triangle has sides of length 29, 29, and 40 cm. Find another isosceles triangle with the same perimeter and area that also has sides of integral length. I tried creating a pair of simultaneous equations but that did not work. I then tried to work with the formula for area of the triangle, for instance, double the height and halve the base so that the area remains the same. That changes the perimeter. Help! Date: 10/27/1999 at 05:19:55 From: Doctor Pete Subject: Re: Geometry Hi, This is a fairly tricky problem. Given an isosceles triangle with sides a and base 2b, the perimeter is P = 2a + 2b and the area is A = b Sqrt[a^2 - b^2] (you can check these). Then for a = 29, b = 20, corresponding to the given triangle, we have the system of equations a + b = 49, b Sqrt[a^2 - b^2] = 420. (Remember that b is half the base.) Substitution gives b Sqrt[(49-b)^2 - b^2] = b Sqrt[49(49-2b)] = 7b Sqrt[49-2b] = 420. Dividing by 7 and squaring both sides gives b^2 (49 - 2b) = 60^2 or 2b^3 - 49b^2 + 3600 = 0. Now here comes the tricky part: we already know that b = 20 is a solution to this equation, so this cubic polynomial has a factor of (b-20). Dividing this factor out gives a quadratic, 2b^2 - 9b - 180 = 0. From here, it is straightforward to factor it completely: We have (2b + 15)(b - 12) = 0 from which it follows that b = 12 is another solution. The resulting isosceles triangle has sides a = 49 - 12 = 37, 2b = 24. Finally, some questions for you: Are there non-isosceles triangles that also satisfy the conditions of the problem? How would you go about finding such a triangle or triangles? Also, does the value b = -15/2 have any geometric significance? And, can you find other pairs of integers for which there is only one other isosceles triangle with integer sides and the same area and perimeter? - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/