Suppose that the side lengths of a triangle are three consecutive integers and one of the angles is twice another. The number of such triangles is/are

Step-by-step solution

Let B = 2 A and B D be the bisector of angle B , then CD = ab a + c & AD = bc a + c Now, Δ ABC and Δ BDC are similar, so BC AC = CD BC ⇒ a 2 = ab a + c b ⇒ b 2 = a a + c ......(i) Since, b > a ⇒ Either b = a + 1 or b = a + 2 , if b = a + 1 , then [From Eq.(i)] a + 1 2 = a + c a ⇒ c = 2 + 1 a c is integer ⇒ a = 1 , b = 2 , c = 3 but then, no triangle will form. If b = a + 2 , then obviously c = a + 1 , and then, [from Eq. (i)] a + 2 2 = a 2 a + 1 ⇒ a 2 − 3 a − 4 = 0 or a = 4 ∴ a = 4 , b = 6 , c = 5 is the only possi