## Wednesday, January 21, 2009

### Elearn Geometry Problem 229: Triangle, Centroid, Transversal

See complete Problem 229 at:
gogeometry.com/problem/p229_triangle_centroid_transversal_perpendicular.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

#### 3 comments:

1. Let A be (0,0), B:(a.b) and C:(c,0) Hence, G:((a+c)/3,b/3). Let DF be px+qy+k=0. By convention, distances above this line are considered positive and those below negative. Thus, e = (ap+bq+k)/(p^2+q^2)^(1/2), d=-k/(p^2+q^2)^(1/2), f = -cp/(p^2+q^2)^(1/2) and g= -((a+c)p/3+bq/3 +k)/(p^2+q^2)^(1/2). RHS = -k/(p^2+q^2)^(1/2)-(cp+k)/(p^2+q^2)^(1/2)-(ap+bq+k)/(p^2+q^2)^(1/2)
or RHS =-((a+c)p+bq+3k)/(p^2+q^2)^(1/2)--(1)
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) --(2)
By (1) & (2) LHS=RHS or 3g = d + f - e
Ajit: ajitathle@gmail.com

2. like P227
draw A'MC'// DF ( M middle of AC)

tr MGG' and MBB" are similar

((d+f)/2-g)/1 = ((d+f)/2+e)/3 give the result

3. See the drawing

- Define a segment D2F2//DF passing through G with D2 in AD and F2 in FC
- Define d1=DD2 and d2=D2A with d1+d2=d
- Define f1=FF2 and f2=F2C with f1+f2=f
- Define L intersection of D2F2 and BE
- From problem 226, d2+f2=e2+e
- By construction, d1=g=e2=f1
- d+f=d1+d2+f1+f2=g+e2+e+g=3g+e
- Therefore d + f – e = 3g