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.

let CA=CB=a, AB=c, AD=p1, DB=p2 with p1,p2 prime

(1) c=p1+p2

(2) a=(420-c)/2

CD^2=a^2+p1^2-2*a*p1*Cos(CAB)=a^2+p2^2-2*a*p2*Cos(CAB)

p1^2-2*a*p1*Cos(CAB)=p2^2-2*a*p2*Cos(CAB)

Cos(CAB)=(p1^2-p2^2)/(2*a*(p1-p2))=(p1+p2)/(2a)=c/(2a)

(3) CD^2=a^2+p1^2-p1*c

from herons formula

(4) area=(210-a)*Sqrt[ 210*(210-c)]

now to find the solution I simply programmed mathematical to try every possible combination of primes (p1,p2) whose sum is less than 420 and report back any fit the following criteria

i.   a is an integer

ii. x is an integer

iii.  area is non-zero (otherwise triangle doesn't exist)

the only result that was returned was the following

a=189  p1=29 p2=13 (interchangable) c=42 CD=188 area=1764*sqrt(5)

thus the desired answer is 1764*sqrt(5)

an interesting note is that there are several combinations of p1,p2 that work if a zero area triangle is allowed

Posted by Daniel on 2007-11-21 12:49:14