Wait – that only works if equilateral. But problem is known to be true for any triangle with ∠A=60°? Let's check with a non-equilateral example: b=7, c=8, ∠A=60° ⇒ a²=49+64-56=57 ⇒ a≈7.55, s≈11.275, s-a≈3.725, BC/2≈3.775 – close but not equal. Hmm – maybe I misremembered. Actually known result: In any triangle, EF = (b+c-a)/2, and BC/2 = a/2. Equality requires b+c=2a, which is not generally true. So problem likely assumes something else – perhaps triangle is right-angled? Let's recall actual 1993 RMO problem: I think it was "In triangle ABC, ∠A=60°, incircle touches BC at D. Prove that EF = BC/2." But that's false generally. Searching memory: The actual problem: "In triangle ABC, ∠A=60°, incircle touches BC at D. Prove that EF = AD/2" or something else.
Wait correct: For triangle ABC, transversal line through E (on AB), F (on AC), and D (on BC), Menelaus says: rmo 1993 solutions
Using the inclusion-exclusion principle, we find the number of solutions to be 42. Wait – that only works if equilateral
Factor pairs of 24: (1,24), (2,12), (3,8), (4,6). Hmm – maybe I misremembered
isosceles triangles that do not share a side with the polygon. Total isosceles triangles = . Answer: 640 scalene triangles. Problem 5: Divisibility Proof Problem: Show that is divisible by Solution: Parity: Both 199319 to the 93rd power 139913 to the 99th power are odd, so their difference is even (divisible by 2). Modulo 81: Use the expansion , the expression is divisible by Being divisible by both 2 and 81, it is divisible by Problem 7: Age Logic Puzzle
For $n$ to be a positive integer, the discriminant must be a perfect square.
