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# 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,

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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.\

https://latex.artofproblemsolving.com/f/e/2/fe244b1629a2e0e2234971692900b954dd71e952.png

Guest Mar 13, 2017

#4
+79867
+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
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#1
+5565
+11

Since 32 + 42 = 52, 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 un2.

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
#2
+79867
+5

Very nice, hectictar...!!!

CPhill  Mar 13, 2017
#3
+5565
+6

Aww you spent all that time typing and didn't post it?

Also I wasn't really really sure that they were the heights but it just kind of looked like they were.

Is there a way to tell for sure?

hectictar  Mar 13, 2017
#4
+79867
+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
#5
+5565
+11

Ohh... thank you

hectictar  Mar 13, 2017

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