The ultimate guide for tackling divisibility conditions

[@JamshidOlmosiy-o5h]

good solution

[@Generalist18]

I have a challenge.In a triangle ABC with AB < AC, D is a point on segment AC such that BD = CD. A line parallel to BD meets segment BC at E and line AB at F. Point G is the intersection of AE and BD. Show that ∠BCG = ∠BCF. I Love your videos please they are so helpful , underrated and concise you do t even know how much you helped me .I please request solve this problem couldn’t crack it(it’s easy but I am a beginner)

[@calimath6701]

We are glad to hear that you find our videos helpful. It means a lot to us! Your problem is quite nice (hint: Consider the point F' on AC such that BC is parallel to FF'. What can you say about FF'E and BCG?). Maybe, we'll post the solution in a short :)

[@Generalist18]

@@calimath6701 thx thx thx!oh it’s actually from the pan African math Olympiad problem 1 2023,it can be solved using trig ceva but i don’t like trig solutions as they are nasty and don’t have that geometric elegance.I will consider your hint and attempt.

[@gobleturky6192]

Why is the last step enough? Since there are infinitely many primes, couldn’t each one divide finitely many terms of the sequence but it doesn’t become 1 past a point? For example if the jth prime divides the first j terms

[@calimath6701]

In our first step, we figured out that a_i divides k. Hence, there are only finitely many primes dividing our sequence members.