[/quote]
,
.![\[
(2^x - 1)(5^x - 1) = y^n
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have no positive integer solution 
.
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x^2 + y) \geq f(x) + y.
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such that:
be a positive integer, and let 
denote the integers modulo 
satisfying 
for all 
.
such that 
divides 
for any integers 
and there exists a polynomial 
with integer coefficients, such that 
for all 
.
is call 
if it satisfies the following properties:
and 
for all indices 
.
.
for which: Every 
.
such that 
then find minimal value of 
and incircle 
,
at 
,
is an arbitrary point on 
meets 
at 
,
meets 
meets 
.
are colinear. Let 
and 
is tangent to 
Y by Adventure10, Mango247
In acute-angled triangle 
, 
is an altitude and 
is a median. Points 
and 
on lines 
and 
respectively are such that 
and 
. Prove that the midpoint of 
is equidistant from 
and 
.
Proposed by Danylo Khilko
, 
is an altitude and 
is a median. Points 
and 
on lines 
and 
respectively are such that 
and 
. Prove that the midpoint of 
is equidistant from 
and 
.Proposed by Danylo Khilko
This post has been edited 2 times. Last edited by Mindstormer, Jul 22, 2018, 8:49 PM
Reason: typo
Reason: typo
and 
- midpoints of 
a
so 
- midpoint of common hypotenuse of 
so 
and 
.
- midpoint of 
and so 
.
is the unit circle, then we know that 
So, the midpoint of 
For computing 
we know 
(unit circle chord) and 
(because 
Using symmetry, 
Then the midpoint of 
,
We only need to show that the foot from 
And hence we're done!
