# Tough question - Coordinate geometry.

• Aug 5th 2009, 03:30 AM
cyanide911
Tough question - Coordinate geometry.
Quote:

Six points are taken at a circle centered at the origin, such that the mean if it's abcissas is 4/3 and that of it's oordinates is 2/3. Line joining the orthocenter of a triangle formed by any 3 points and centroid of a triangle formed by the remaining 3 pass through a fixed point h,k. Find h+k
Can someone solve this?

Thanks.
• Aug 5th 2009, 02:50 PM
aidan
Quote:

Originally Posted by cyanide911
Six points are taken at a circle centered at the origin, such that the mean if it's abcissas is 4/3 and that of it's oordinates is 2/3. Line joining the orthocenter of a triangle formed by any 3 points and centroid of a triangle formed by the remaining 3 pass through a fixed point h,k. Find h+k

Can someone solve this?
Thanks.

Select any three points, calculate orthocenter[1] and centroid[1]; those two points define line[1].

Use the other three points and calculate orthocenter[2] and centroid[2]; those two points define line[2].

The intersection of those two lines will give you the (h,k) coordinates.

The real difficulty of your problem is the generation of six coordinate points on a circle that average to (4/3,2/3).

If you supply the coordinates for the six points, the remainder is easy.
• Aug 6th 2009, 06:52 AM
cyanide911
Yeah, that's what makes it tough.
• Aug 13th 2009, 09:51 AM
aidan

Attached are three images.
From image 1, IF you have only 1 point, then that
point (4/3, 2/3) occurs on the circle with
R = sqrt( (4/3)^2 + (2/3)^2 ).

The line, from the orgin to the point, is defined
by y = x/2

---
For two points:
(see image 2)

(1) to be equidistant from the origin or center of the circle
&
(2) the sum of the x_coordinates to average 4/3 = (x1+x2)/2
&
(3) the sum of the y-coordinates to average 2/3 = (y1+y2)/2

They must be on a line perpendicular to y = x/2
that passes through the point (4/3, 2/3) and be equidistant (distance w) from the point (4/3,2/3).

That line is defined by y = -2x + 10/3

With those two points (x1,y1) and (x2,y2) the Radius can be uniquely determined.

---
for six points:
(see image 3)

(1) to be equidistant from the origin or center of the circle
&
(2) the sum of the x_coordinates to average 4/3 = (x1+x2+x3+x4+x5+x6)/6
&
(3) the sum of the y-coordinates to average 2/3 = (y1+y2+y3+y4+y5+y6)/6

The points (3 through 6),
must be on a line parallel to y = -2x + 10/3

That parallel offset is dimension "v".

(Note the possibility of a difficult situation if dimension v, causes (x4,y4) and (x5,y5) to exceed the length of radius determined above.)

With that, you can find the necessary 6 coordinate points to construct an answer to the first post.

Hope that helps.
• Aug 30th 2009, 08:38 PM
aidan
http://www.mathhelpforum.com/math-he...lers-line.html

If it weren't for the above reference I would not have associated this
problem with the 9-point circle.

Quote:

a circle centered at the origin (=circumcenter),
such that the mean if it's coordinates is (4/3, 2/3).
A line joining the orthocenter of a triangle formed by any 3 points and centroid of a triangle formed by the remaining 3 pass through a fixed point h,k.
Find h+k
The distance from the circumcenter to the centroid of a triangle is twice the distance from the centroid to the 9 point circle point.

$\displaystyle \dfrac{3}{2} \times (4/3,2/3) = (2,1) = (h,k)$

so h+k = 3