See complete Problem 227 at:
gogeometry.com/problem/p227_triangle_centroid_perpendicular.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Tuesday, January 20, 2009
Elearn Geometry Problem 227: Triangle, Centroid, Exterior line
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Let DF be px+qy+k=0 and let A:(0,0),B:(a,b) & C:(c,0). Therefore,d=k/(p^2+q^2)^(1/2),e=(ap+bq+k)/(p^2+q^2)^(1/2)and f=(cp+k)/(p^2+q^2)^(1/2)
ReplyDeleteRHS = ((a+c)p+bq+3k)/(p^2+q^2)^(1/2) ----(1)
Now G is [(a+c)/3,b/3]Hence, LHS= 3g =3[(a+c)p/3+bq/3+k]/(p^2+q^2)^(1/2)= ((a+c)p+bq+3k)/(p^2+q^2)^(1/2)=RHS by (1). Hence,3g = d + e + f
Ajit: ajitathle@gmail.com
let be M , point BG meet AC
ReplyDeletedraw A'MC' // DF , meet g on G' and e on B'
tr MGG' and MBB' are similar( ang MGG' = ang MBB')
=> (g - (d+f)/2 )/1 = ( e - (d+f)/2)/3 , (1=GM, 3=BM)
2g-d-f = (2e-d-f)/3
6g-3d-3f = 2e-d-f
3g = d+f+e
See the drawing : Drawing
ReplyDelete- Draw a segment D’F’ parallel to DF passing through G
- Define D’A=d’, E’B=e’ and CF’=f’
- e’=d’+f’ (pb 226) => e’-d’-f’=0
- DFF’D’ rectangle
- g=d+d’=f+f’ => d=g-d’ and f=g-f’
- e=e’+g
- e+f+d=e’+g+g-f’+g-d’
- Therefore e+f+d=3g