Question

show that centroid G divides the join of its orthocentre H and circumcentre S in ratio 2:1

Answer 1

I am thinking that you can possibly model this problem on the coordinate plane with a triangle with one vertex on the origin and one vertex on the x-axis (ie. A(0,0), B(2a,0), C(2b,2c)). From there, find the centroid, orthocenter, and circumcenter and use distance formula to confirm.

I have not checked this myself yet, so I will work on that. It would be a fairly long and tedious process to work through.. :P

Answer 2

did u get it?

Answer 3

The relation between these two lengths that I found is not directly obvious to me...
\[ HG = \sqrt{(2c - \frac{2a+2b}{3}))^{2} + ((2a-2b) - \frac{2c}{3})^{2}} \]
\[ GS = \sqrt{(\frac{2a+2b}{3} - a)^{2} + (\frac{2c}{3} - (\frac{-ab+b^{2}}{c} + c^{2}))^{2}} \]

I'm doubting the possibility that these two numbers would really satisfy a 2:1 ratio, so I'm first going to check over the work leading to this.

Answer 4

isnt there any simpler method!!

Answer 5

I have not learned any other methods to approaching a problem like this (though maybe some do exist).

I found my mistake (I used the wrong x-value for point H) and was able to work it into the correct form HG = 2GS. I will show my work (although I have to type it all up now).

Answer 6

k

Answer 7

show pls

Answer 8

I start with a triangle with vertices A(0,0), B(2a, 0), and C(2b, 2c)
I found the three midpoints for AB, BC, and AC: M1(a,0) (for AB), M2(a+b,c) (for BC), and M3(b,c) (for AC).

I found the equations for lines CM1, AM2, and BM3 by finding the slopes between the points and then using point-slope form with the point that was on the x-axis.
CM1: y - 0 = ((2c) / (2b-a)) (x - a)
AM2: y - 0 = ((c) / (a+b)) (x - 0)
BM3: y - 0 = ((c) / (b-2a)) (x - 2a)

By setting CM1 = AM2, I was able to find the x and y values for the centroid to be...
G( (2a + 2b) / 3 , 2c/3 )

For the orthocenter, I found the slopes of AB, BC, and AC. Since we need the perpendicular lines that intersect the opposite vertex from the side (the altitude), I found the opposite reciprocals of these values and once again used point-slope, this time using the opposite vertex as the point.
Perpend. to AB, on C: x = 2b
Perpend. to BC, on A: y - 0 = ( (a-b) / (c) ) (x - 0)
Perpend. to AC, on B: y - 0 = (-b/c) (x - 2a)

Since we have an x-value and we are finding the intersection, I just plugged it in for the second equation and found the orthocenter to be...
H( 2b , (2ab - b^2) / c )

Lastly, for the circumcenter, I used those perpendicular slopes again with the point-slope form, except I used the midpoints instead (because the circumcenter is on the perpendicular bisectors of each line).
Perpend. to AB, on M1: x = a
Perpend. to BC, on M2: y - c = ((a-b) / c) (x - (a+b)) <=> y = ((a-b) / c) (x - (a+b)) + c)
Perpend. to AC, on M3: y - c = (-b/c) (x - b) <=> y = (-b/c)(x - b) + c

We once again have an x-value; so to find where the lines intersect, we can plug that in to the second equation to find the circumcenter to be...
S( a , (-ab + b^2) / c + c )

... continuing...

Answer 9

We know that the centroid, the orthocenter, and the circumcenter are all collinear. We can use the distance formula to show that the distance between the points H and G is proportional to the distance between G and S.

\[ d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \]
\[ HG = \sqrt{(2b - \frac{2a+2b}{3})^{2} + (\frac{2ab - 2b^{2}}{c} - \frac{2c}{3})^{2}} \]
\[ GS = \sqrt{(a - \frac{2a+2b}{3})^{2} + ((\frac{-ab + b^{2}}{c} + c) - \frac{2c}{3})^{2}} \]

At this point, we can simplify by factoring out a 2^2 from each term under the square root in HG and moving it outside, and taking the opposite of both terms under the square root in GS (by opposite, I mean something like: (a - b)^2 = (-a + b)^2 (-1)^2 = (-a + b)^2 ).

Answer 10

Summary:
* Define some arbitrary vertices of a triangle.
* Find the midpoints.
* Find the slopes for the sides of the triangle and the slopes from one vertex of the triangle to the opposite midpoint.
* Find the equations for (A) the lines for the centroid, (B) the lines for the orthocenter, and (C) the lines for the circumcenter.
* Find the intersection of two equations in each to define the points of the centroid, orthocenter, and circumcenter.
* Use the distance formula to find the distance between the orthocenter and centroid, and the circumcenter and centroid.
* Use algebra to show that the former distance is twice the latter distance.

Answer 11

i salute to ur sincere and powerful working

Answer 12

thank u

Answer 13

this working is too big i think there is an easier mthod

Answer 14

wats ur opinion?

Answer 15

the work done by AccessDenied seems all correct. not sure if there is a simpler way - let me think over it - I'll come back to it.

Answer 16

k can u post it here if u gtter a better method with lesser steps?

Answer 17

sure

Answer 18

thx

Answer 19

u r vry gud in geometry part?

Answer 20

we'll soon find out... :)

Answer 21

I basically used the same method as AccessDenied:

place the triangle ABC at the origin such that:
A is at (0,0)
B is at (\(x_b,y_b\))
C is at (2b,0)
and lets call the lengths:
AB=2c
BC=2a
AC=2b

Then you can show that the coordinates of G, H and S are:
G is at (\(\frac{x_b+y_b}{3},\frac{y_b}{3}\))
H is at (\(x_b,-\frac{x_b(x_b-2b)}{y_b}\))
S is at (\(b,\frac{1}{2}(y_b+\frac{x_b(x_b-2b)}{y_b})\))

Then, using Pythagorus, you can show that:\[GH^2=(x_b-\frac{x_b+2b}{3})^2+(-\frac{x_b(x_b-2b)}{y_b}-\frac{y_b}{3})^2\]\[\qquad=\frac{4(x_b-b)^2}{9}+\frac{(3x_b-6bx_b+y_b^2)^2}{9y_b^2}\]\[GS^2=(b-\frac{x_b+2b}{3})^2+(\frac{1}{2}(y_b+\frac{x_b(x_b-2b)}{y_b})-\frac{y_b}{3})^2\]\[\qquad=\frac{(x_b-b)^2}{9}+\frac{(3x_b-6bx_b+y_b^2)^2}{36y_b^2}\]\[\qquad=\frac{1}{4}(\frac{4(x_b-b)^2}{9}+\frac{(3x_b-6bx_b+y_b^2)^2}{9y_b^2})\]\[\qquad=\frac{1}{4}GH^2\]therefore:\[\frac{GH^2}{GS^2}=4\]therefore:\[\frac{GH}{GS}=2\]