In triangle ABC, AB=3, AC=5, BC=4. The medians AD, BE, and CF of triangle ABC intersect at the centroid G. Let the projections of G onto BC, AC, and AB be P, Q, and R, respectively. Find GP+GQ+GR.\

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Guest Mar 13, 2017

#4**+10 **

Note, hectictar that the centroid is given by [ (x1 + x2 + x3)/ 3, (y1 + y2 + y3)/3 ]

Let B= (0,0) A = (0,3) and C = (4,0)....so

The centroid is [ (0 + 0 + 4)/3 , ( 0 + 3 + 0)/3] = [ 4/3, 1]

And as you pointed out, triangles ABG, BCG, and CAG, are all equal

So...area of ABG = (1/2)BH = (1/2)(AB)(GR) = (1/2)(3)(4/3) = 2 units^2

And the area of BCG = (1/2)(BC)(GP) = (1/2)(4)(1) = 2 units^2

And, by default, the area of the remaining triangle must be 2 units^2

So.....GQ kinda' "falls out" as

2 = (1/2)(AC)(GQ) = (1/2)(5)(GQ) = (5/2)GQ

So (5/2)(GQ) = 2

GQ = (2/5)(2) = 4/5

So......GR = 4/3 , GP = 1 and GQ = 4/5

So....your heights were correct.....

And their sum is = 47/15......just as you found...!!!

CPhill
Mar 13, 2017

#1**+11 **

Since 3^{2} + 4^{2} = 5^{2}, ABC is a right triangle with 3 as the height and 4 as the base.

So the area of ABC is (1/2)(4)(3) = 6 un.^{2}

The lines AG, BG, and GC split the big triangle, ABC, into 3 equal pieces since they go from the vertices to the centroid.

The areas of ABG, BCG, and CAG, are all equal. They are all 6/3 = 2 un^{2}.

GR, GQ, and GP are all the heights of the 3 triangles. We know the bases and the areas, so just solve for the heights.

Triangle ABG:

2 = (1/2)(3)(GR)

__4/3 = GR__

Triangle BCG:

2 = (1/2)(4)(GP)

__1 = GP__

Triangle CAG:

2 = (1/2)(5)(GQ)

__4/5 = GQ__

4/3 + 1 + 4/5 = 47/15 = 31 and 2/15

hectictar
Mar 13, 2017

#4**+10 **

Best Answer

Note, hectictar that the centroid is given by [ (x1 + x2 + x3)/ 3, (y1 + y2 + y3)/3 ]

Let B= (0,0) A = (0,3) and C = (4,0)....so

The centroid is [ (0 + 0 + 4)/3 , ( 0 + 3 + 0)/3] = [ 4/3, 1]

And as you pointed out, triangles ABG, BCG, and CAG, are all equal

So...area of ABG = (1/2)BH = (1/2)(AB)(GR) = (1/2)(3)(4/3) = 2 units^2

And the area of BCG = (1/2)(BC)(GP) = (1/2)(4)(1) = 2 units^2

And, by default, the area of the remaining triangle must be 2 units^2

So.....GQ kinda' "falls out" as

2 = (1/2)(AC)(GQ) = (1/2)(5)(GQ) = (5/2)GQ

So (5/2)(GQ) = 2

GQ = (2/5)(2) = 4/5

So......GR = 4/3 , GP = 1 and GQ = 4/5

So....your heights were correct.....

And their sum is = 47/15......just as you found...!!!

CPhill
Mar 13, 2017