Triangle is an isosceles triangle with base , the included angle of is degree . A point is marked off on side such that . What is the angle of ?
Moderator edit: This is now an approved challenge question.
Hmmm... I can't see clearly what the continuation is without getting into heavy stuff (probably power series or something: I'm not sure), but it perhaps doesn't matter: the problem's title says geometry and, at least over here, this means the problem must be solved using exclusively tools from geometry, without any trigonometry, calculus, etc.
I wonder what the OP originally meant.
I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines .
Below is the trigonometric method which is similar to the contest 's .
I shouldn't give the solution now but if you really want the pure-geometric solution , i think it is okay to give the hints :
The solution has three additional lines but i am going to give you only one of the lines because this is exactly the key to this problem while the others are not very important .
A figure is added in the attachment to make the proof easier to read.
On the given triangle build a regular -gon such that is the center of it
(which is possible due to the fact that , ).
Now, we are not assuming . We'll come back to it later.
Continue AB until it gets to another vertex, call it , of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line as shown in the attached figure. Next, draw as shown, and call the intersection of with . All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle it follows that , and in triangle we have . Therefore , but , so . Next notice that (again, by considering the circumscribing circle).
We have thus proven that the conditions and are obtained simultaneously, from which the result follows.