Bunuel wrote:
If the length of each of the three sides of a triangle is a positive integer, is the triangle right-angled ?
We know that the side are integer;
Assume sides of the triangle are a,b,c;
Let's use statement 1:
The perimeter of the triangle is a prime number.p=a+b+c; As p is prime; p has to be an odd number; Bcs all primes are odd except 2 and p cannot be 2;(All side are positive integers)
p(o) = a + b + c;
a(o) ,b(o) ,c(o) for p to be odd;
a(e),b(e) and c(o) for p to be odd
To be a right angle triangles it should follow the rule : \(a^2+b^2=c^2\)
As\( a^2 = o, b^2=o c^2=o\) ; \(a^2+b^2=c^2\) is not possible o+o has to be even;
//y As\( a^2 = e, b^2=e and c^2=o\); \(a^2+b^2=c^2\) is not possible because e+e can't be o;
Hence all the triangles with perimeter as prime has to be Not right angled triangles;
We can un ambiguously solve the question with Statement A; (2) The area of a circle circumscribing the triangle is \(9\pi^2\)We know that a/SinA =b/SinB =c/SinC= 2R;
a = 2R*SinA;
As we know that R= 3; [Since The area of a circle circumscribing the triangle is \(9\pi^2\)]
If sides of the triangle are integer then the area has to be integer as
[ABC]=abc/4R=ab/sinC=\(\sqrt{s(s−a)(s−b)(s−c)} \)
We can prove Ar{ABC] is integer
As the abc/4R is rational and \(\sqrt{s(s−a)(s−b)(s−c)}\) has to be Rational; We also know s(s−a)(s−b)(s−c) is integral;
Hence Ar[ABC] is Integer.
Ar[ABC] = abc/4R; So one of the side should be of kR form ; As R is prime(3);
Let say a = kR; k is Integer;
a/SinA = 2R ; kR = 2RSinA; So k/2 = SinA;
As SinA belongs to [0,1] k has to be {1,2};Let k=1; Then \(R^2 = b^2+c^2 - 2bcCosA(Cos A=\sqrt{3}/2);\) So it is not possible; (Cosine Theorem)
K has to be 2;
If K is 2 SinA=1; So A = 90';Hence if sides of triangle are integer and Circumradius is prime then the triangle has to be a right angled triangle;
IMO D