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Let a=CA,b=CD,p=AD, and q=DB. Draw a circle with center at C and a radius of a. Extend segment CD in both directions to form diameter ECDF. Now ED=a+b and DF=a-b -->
(a+b)(a-b) = pq
Since p and q are prime, there are only two possibilities.
Case I: a+b=p and a-b=q (if p>=q, otherwise switch p and q) --> 2a=p+q which is impossible since the sum of the lengths of two sides of a triangle is greater than the length of the third side.
Case II: a+b=pq and a-b=1 --> a=(pq+1)/2 and b=(pq-1)/2
Letting P represent the perimeter yields:
P=2a+p+q=(p+1)(q+1)=420
But the only possible pairs of integers with this product where p and q are prime are (3,140) and (14,30). Now q=2 and p=139 --> a=139.5 which is impossible. So q=13 and p=29 --> a=189 and b=188, forcing an area of 1764sqrt(5). |
