Triangle A B C has centroid G . Triangles A B G , B C G , and C A G have centroids G 1 , G 2 , G 3 respectively.
The value of [ A B C ] [ G 1 G 2 G 3 ] can be represented by q p , for positive coprime integers p , q .
Find p + q .
Details and Assumptions:
[ A B C D ] denotes the area of A B C D .
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Fantastic solution including simple logic
we can take M as the midpoint of BG,then MG1G2 is similar to MAC.this gives G1G2 as AC/3 similarly for the others this gives its area as 1/9 of total area.so 10.
I don't think that it's THAT obvious that △ D E F is 1/4 △ A B C
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Its the simple application of midpoint theorem
Well, △ D E F ≅ △ A E F ≅ △ D B F ≅ △ D E C soo...
We know that a centroid divides a median in ratio 2:1. Hence,
Let AD = x, BE = y, CF = z.
G D = 3 1 x , G E = 3 1 y , G F = 3 1 z
G G 1 = 9 2 x , G G 2 = 9 2 y , G G 3 = 9 2 z
Now, [ G 1 G 2 G 3 ] = [ G G 1 G 2 ] + [ G G 1 G 3 ] + [ G G 2 G 3 ]
[ G G 1 G 2 ] = 2 1 ∗ 9 2 x ∗ 9 2 y ∗ s i n θ 1
[ G G 1 G 3 ] = 2 1 ∗ 9 2 y ∗ 9 2 z ∗ s i n θ 2
[ G G 2 G 3 ] = 2 1 ∗ 9 2 z ∗ 9 2 x ∗ s i n θ 3
if θ 1 , θ 2 , θ 3 are angles G 1 G G 2 , G 1 G G 3 , G 2 G G 3 respectively.
Similarly, [ A B C ] = 2 1 ∗ 3 2 ∗ 3 2 ( x y s i n θ 1 + y z s i n θ 2 + z x s i n θ 3 )
As a result, [ G 1 G 2 G 3 ] : [ A B C ] = 2 1 ∗ 3 2 ∗ 3 2 ( x y s i n θ 1 + y z s i n θ 2 + z x s i n θ 3 ) 2 1 ∗ 9 2 ∗ 9 2 ( x y s i n θ 1 + y z s i n θ 2 + z x s i n θ 3 )
= 1 : 9
So, p + q = 1 0
It does not reduce generalization, assume ABC is a equilateral triangle. So: I K G 1 G 3 = 3 2 B C G 1 G 3 = 3 2 × 2 1 = 3 1 [ A B C ] [ G 1 G 2 G 3 ] = 9 1 ⇒ p + q = 1 0
Although the problem could be approached in your way, when facing constraint of time, it no doubt defeats the motto of the problem!
First, let's extend A G , B G and C G to meet △ A B C at E , D and F respectively. Since each median of the larger triangle bisects the side of one of the smaller triangles, we can say that ( A , G 2 ) , ( C , G 1 ) and ( B , G 3 ) are collinear. Also, note that G 1 G 3 ∣ ∣ B C by midpoint theorem.
Therefore, ∠ B G E = ∠ H G G 3 and ∠ B E G = ∠ G 3 H G ⇒ △ G E B ≈ △ G H G 3 . Now, we know that 2 B G = G D and 3 2 G D = G G 3 . So, 3 B G = G G 3 . Which implies -
s 1 = 3 s 2 . . . . . . . . . . . . . . . . . . ( 1 )
(Where s 1 is the semiperimeter of △ G E B and s 2 is the semiperimeter of △ G H G 3 ).
By the law of cotangents r 1 = cot 2 ∠ G B E 3 s − 3 G H and r 2 = cot 2 ∠ G G 3 H s − G H .
(Where r 1 is the inradius of △ G E B and r 2 is the inradius of △ G H G 3 ). Which gives us -
r 1 = 3 r 2 . . . . . . . . . . . . ( 2 ) .
Using the formula [ A B C ] = inradius ⋅ semiperimeter, we can say from ( 1 ) and ( 2 ) that -
9 [ G H G 3 ] = [ G E B ] . Observe that G is the centroid of △ G 1 G 2 G 3 .It can also be proven that the medians and centriod of a triangle divide it into six triangles of equal area (but not necessarily congruent ones).
So we can conclude that, [ A B C ] [ G 1 G 2 G 3 ] = 9 1 ⇒ p + q = 1 0
Note - because △ B E G ≈ △ G H G 3 , △ G E C ≈ △ G H G 1 and B E = C E , we also proved that G is the centroid of G 1 G 2 G 3 .
Let M be the midpoint of BG,we get M A G 1 M = 3 1 and C M G 2 M = 3 1 so we get A C G 2 G 1 = 3 1 by the same approach each side of t r i a n g l e G 1 G 2 G 3 is one-third of its corresponding side in ABC so the area is one-ninth so p+q=10
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Let D , E , F be the midpoints of B C , C A , A B respectively.
Note that obviously [ A B C ] [ D E F ] = 4 1
Since G 1 is the centroid of △ A B G , then G G 1 = 2 G 1 F
Similarly, G G 2 = 2 G 2 D G G 3 = 2 G 3 E
Thus, △ G 1 G 2 G 3 maps to △ F D E by a homothety of ratio 3 2 .
Therefore, [ D E F ] [ G 1 G 2 G 3 ] = ( 3 2 ) 2 = 9 4
Finally, [ A B C ] [ G 1 G 2 G 3 ] = [ D E F ] [ G 1 G 2 G 3 ] ⋅ [ A B C ] [ D E F ] = 9 4 ⋅ 4 1 = 9 1 and our answer is 1 + 9 = 1 0