The perimeter of an isosceles triangle ABC with CA = CB, is 420. Find the area of this triangle if D is a point on the base of the triangle such that CD and CA are integers, and AD and DB are primes.

AB + BC + CA = 420

CA = CB = 150.

AB = (420 - (150 + 150)) = 120.

/ ACB = ArcCos( 150² + 150² - 120² )/ ( 2*150*150 ) ~= 0.823033692134976 radians.

AD = 17 (a prime).

CD = SQRT( 17² + 150² - 2*17*150*cos( 0.823033692134976 radians )) = 139.

DB = (120 - 17) = 103 (a prime).

Corrections made to identification of the line segments.

The brute force was taking the length of the line segment AB and iterating through each of the possible prime pairs that equal 120 to find where CD was an integer. (As the triangle is isoceles, the mirror image of the triangle where AD = 103 and DB = 17 will also be a valid solution.

 AB = AD  + DB    CD
120 =  7  + 113  ~145.33
    = 11  + 109  ~142.75
    = 13  + 107  ~141.48
    = 17  + 103   139   <---
    = 19  + 101  ~137.79
    = 23  +  97  ~135.41
    = 31  +  89  ~130.91
    = 37  +  83  ~127.75
    = 41  +  79  ~127.77
    = 47  +  73  ~122.97
    = 53  +  67  ~120.40
    = 59  +  61  ~118.09

The lengths of 120 and 150 were also found by brute force in the same manner.

Edited on November 24, 2007, 11:33 pm

Posted by Dej Mar on 2007-11-21 20:23:28