Coordinated Trouble #3

Geometry Level 4

G i s t h e c e n t r o i d o f t h e t r i a n g l e h a v i n g v e r t i c e s A ( 3 , 12 ) , B ( 6 , 6 ) , C ( 9 , 18 ) . F i n d t h e v a l u e o f P A 2 + P B 2 + P C 2 − G A 2 − G B 2 − G C 2 , w h e r e P i s ( 5 , 10 ) . G\quad is\quad the\quad centroid\quad of\quad the\\ triangle\quad having\quad vertices\quad \\ A(3,12),\quad B(6,6),\quad C(9,18).\\ Find\quad the\quad value\quad of\\ { { PA }^{ 2 } }{ +PB }^{ 2 }{ +PC }^{ 2 }{ -GA }^{ 2 }{ -GB }^{ 2 }{ -GC }^{ 2 },\\ where\quad P\quad is\quad (5,10).


The answer is 15.

Soummo Mukherjee by Soummo Mukherjee , 25, India

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1 solution

G ( 3 + 6 + 9 3 , 12 + 6 + 18 3 ) = G ( 6 , 12 ) . Change of axis does not alter the relative distances between the points, so the result in this problem also is not altered. Let G’ be the new origin. S o A ′ ( − 3 , 0 ) , B ′ ( 0 , − 6 ) , C ′ ( 3 , 6 ) , G ′ ( 0 , 0 ) , P ′ ( − 1 , − 2 ) . R e q u i r e d v a l u e = P ′ A ′ 2 + P ′ B ′ 2 + P ′ C ′ 2 − ( G ′ A ′ 2 + G ′ B ′ 2 + G ′ C ′ 2 ) R e q u i r e d v a l u e = 2 2 + 2 2 + 1 2 + 4 2 + 4 2 + 8 2 − ( 3 2 + 0 + 0 + 6 2 + 3 2 + 6 2 ) R e q u i r e d v a l u e = 8 + 17 + 80 − ( 9 + 36 + 45 ) = 15. G \Big(\dfrac{3+6+9} 3, \dfrac{12+6+18} 3 \Big)=G(6,12).\\ \text{Change of axis does not alter the relative distances between the points, so the result in this problem also is not altered.} \\ \text{Let G' be the new origin.} ~\\ So~~~A' (-3,0),~~~~~ B' (0,-6),~~~~~ C' (3,6),~~~~~ G' (0,0),~~~~~ P' (-1,-2).\\ Required ~value=P'A'^2+P'B'^2+P'C'^2 - (G'A'^2+G'B'^2+G'C'^2) \\ Required ~value=2^2+2^2 ~~~+~1^2+4^2~~~+~4^2+8^2 ~~~~-~~~~(3^2+0~~~~+~~0+6^2~~~~+~~3^2+6^2)\\ Required ~value=8 + 17 + 80 - (9 + 36 + 45)=\Large~~~\color{#D61F06}{15}.

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