with circumcenter 
and orthocenter 
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
be the point on the circumcircle of triangle 
is perpendicular to 
be the point on line 
such that 
.
and 
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
.
and 
are perpendicular.
and 
are perpendicular.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
sastifying 
are there such that 
and 
?
, where 
is a real number. Prove that for all positive 
then prove
and 
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
let 
and 
be the circles passing through 
and 
respectively such that 
and 
are perpendicular.
weights, each weighing 
and another 
weights with 
each. Write a algorithm that determines the ways in which a scale can be balanced with a weight 
on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers 
.
Y by centslordm, Adventure10, Mango247
In a parallelogram 
, let 
be the point on the 
side such that 
and let 
be the point of side 
such that 
. If the distance from point 
to the line 
is 
, calculate the distance from point to the line .
, let 
be the point on the 
side such that 
and let 
be the point of side 
such that 
. If the distance from point 
to the line 
is 
, calculate the distance from point to the line .This post has been edited 1 time. Last edited by parmenides51, Sep 10, 2018, 6:39 PM
Reason: level edit
Reason: level edit
![$[ADC]=[ABCD]/2= [ABC]= [BCD] = [ADB]$](http://latex.artofproblemsolving.com/8/e/9/8e9ac0a26c2a82db2c7da639525dd0029cd42de5.png)
![$BM= 1/3 BC\Rightarrow [ABM ] = 1/3 [ABC] = 1/3 * 1/2 [ABCD] = 1/6 [ABCD]$](http://latex.artofproblemsolving.com/7/c/6/7c646a62f082e4c24b69a61f5789332a32833e55.png)
(1)![$\Rightarrow [AMC] = 2 [ABM] = 2/3 [ABC] = 1/3 [ABCD]$](http://latex.artofproblemsolving.com/b/5/9/b59b415435b755a27518ef8ea2822ef7b5e80997.png)
![$[ADN] = 1/6 [ABCD]$](http://latex.artofproblemsolving.com/6/f/5/6f5f06b5b5539c5255d233b17dac8c2849320717.png)
and ![$[ANC] = 1/3 [ABCD]$](http://latex.artofproblemsolving.com/6/8/a/68a1f8e60a80708b062205b60da0f63879c8998b.png)
are similar so ![$[MNC] / [CNB] = (2DN/ 3DN)^2= 4/9 $](http://latex.artofproblemsolving.com/b/a/a/baae9490c17a94b4e5ccf887cb905dd9a26ce62e.png)
![$\Rightarrow [MCN] = 4/9 [BCD] = 4/9 * 1/2 [ABCD] = 2/9 [ABCD]$](http://latex.artofproblemsolving.com/4/9/c/49c1c5f3947eb6219193610b82fbfe1c9fa0467b.png)
![$\Rightarrow [AMN]= [ABCD] - [ADN] - [ABM] - [MCN] = [ABCD] (1-1/6 - 1/6 - 2/9 ) = [ABCD] (18-3-3-4)/18= 4/9 [ABCD]$](http://latex.artofproblemsolving.com/b/3/e/b3e56c8d53048d251bfb3aa4891b7975f9f2986b.png)
![$[AMN] = 4/9 [ABCD] = 4/9 * 6 [ABM]= 8/3 [ABM]$](http://latex.artofproblemsolving.com/c/1/6/c161403620bb5d72c8d6016cb1a10092cb4f426d.png)
due to (1)![$[AMN] / [ABM] = 8/3 = 3/ x\Rightarrow x=9/8$](http://latex.artofproblemsolving.com/3/c/c/3cc9af793c2b0fef9e70e5a8bb1eeb4bbf58ec11.png)
