In triangle ABC, AB = 3, AC = 5, and 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.

10 answers

the answer is 47/15

8 isn't correct.

no it is not correct

it is not 16

You are correct. I calculated using 2/3 of the projection of the median, which is wrong.

I'll think on it some more. You do the same...

wow stop cheating on acumbulos

To find GP + GQ + GR, we need to first find the lengths of GP, GQ, and GR.

The centroid of a triangle divides each median into two segments, with the ratio of the lengths of the smaller segment to the larger segment being 1:2.

In triangle ABC, the lengths of the medians are AD = (2/3) * AG, BE = (2/3) * BG, and CF = (2/3) * CG.

Now, let's find the length of AG:
Since G is the centroid, we can find AG by dividing the length of the median into two segments in the ratio 2:1. In triangle ABC, AD is a median, and let's assume that AG is divided into two segments as AG = 2x and GD = x. Let's use the fact that AD is also a median, so it divides BC into two segments in the ratio 2:1.

BC = 4, so BD = (2/3) * BC = (2/3) * 4 = 8/3.
Similarly, CD = (1/3) * BC = (1/3) * 4 = 4/3.

Using the Pythagorean theorem on triangle BGD:
(BD)^2 + (GD)^2 = (BG)^2
(8/3)^2 + x^2 = (2x)^2
(64/9) + x^2 = 4x^2
(9/4)(64/9) + (9/4)x^2 = (9/4)(4x^2)
(256/16) + (9/4)x^2 = (36/4)x^2
16 + (9/4)x^2 = 9x^2
(9/4)x^2 - 9x^2 = -16
(9 - 36/4)x^2 = -16
(9 - 9)x^2/4 = -16
(-x^2/4) = -16
x^2/4 = 16
x^2 = 16 * 4
x^2 = 64
x = √64
x = 8

So, GD = x = 8.
AG = 2x = 2 * 8 = 16.

Now, we can find the lengths of GP, GQ, and GR:

GP = (1/3) * GD = (1/3) * 8 = 8/3.
GQ = (1/3) * GE
GE = CD = 4/3
GQ = (1/3) * (4/3) = 4/9.
GR = GP = (1/3) * GD = (1/3) * 8 = 8/3.

Finally, GP + GQ + GR = (8/3) + (4/9) + (8/3) = 32/9 + 4/9 + 24/9 = 60/9 = 20/3.

So, GP + GQ + GR = 20/3.

since the centroid is 2/3 of the way from each vertex to the opposite side, the projections on the sides are 2/3 the length of the sides.

Their sum is thus 2/3 of the perimeter of ABC = 8