consisting of a prime number 
and two positive integers 
and 
such that 
and 
are both powers of 
be integers. Suppose that for each 
there exists an integer 
such that 
is divisible by 
for some integer 
.
,
Y by chessgocube
Let 
be an isosceles triangle with 
and circumcircle 
. The point 
lies on the shorter arc of over the chord 
and is different from 
and 
. Let 
denote the intersection of 
and 
. Prove that the line through and is a tangent of the circumcircle of the triangle 
.
(Karl Czakler)
be an isosceles triangle with 
and circumcircle 
. The point 
lies on the shorter arc of over the chord 
and is different from 
and 
. Let 
denote the intersection of 
and 
. Prove that the line through and is a tangent of the circumcircle of the triangle 
.(Karl Czakler)
This post has been edited 2 times. Last edited by parmenides51, May 31, 2021, 10:58 PM
gives us 
.
,
.
is a cyclic quadrilateral, so 
is isosceles then
~
so we are done by Alternate Segment Theorem. 
flips 
,
on the unit circle with:
,
.
as:![\[ e = \frac{ab(c+d) - cd(a+b)}{ab-cd} = \frac{1+d - da - \frac{d}{a}}{1-d} = \frac{a + ad - da^2 - d}{a(1-d)} \]](http://latex.artofproblemsolving.com/6/8/2/682a79dc5ebc85dd21b075f050137d4c90fa7208.png)
![\[ \vert c-d \vert \cdot \vert c - e \vert = \vert c - b \vert ^2 \]](http://latex.artofproblemsolving.com/c/c/7/cc7ac5581c483da865307b0ee576ba7a07aa7ce1.png)
Indeed:![\[ \vert (1-d)(1-e) \vert^2 = \left\vert (1-d)\frac{a - ad - a - ad + da^2 + d}{a(1-d)} \right\vert^2 = \left\vert \frac{d(a-1)^2}{a} \right\vert^2 = \frac{d(a-1)^2}{a}\cdot \overline{\left(\frac{d(a-1)^2}{a}\right)} = \frac{d(a-1)^2 \cdot \frac{1}{d}(1-\frac{1}{a})^2}{a \cdot \frac{1}{a}} = \left(\left(1-\frac{1}{a} \right)(1-a)\right)^2\]](http://latex.artofproblemsolving.com/2/d/5/2d5cf482dd6bb934a791eb90bd8a354601cf5347.png)
![\[ \vert c-b \vert ^4 = \left(\left(1-\frac{1}{a}\right)(1-a)\right)^2\]](http://latex.artofproblemsolving.com/3/5/7/35709d6c2b304939a8bb2a124991ca9e0811216a.png)
,
.
which is equal to its conjugate
