## Thursday, March 12, 2009

### Problem 269. Right Triangle, Altitude, Perpendicular, Projection

Proposed Problem

See complete Problem 269 at:
gogeometry.com/problem/p269_right_triangle_altitude_orthogonal_projection.htm

Level: High School, SAT Prep, College geometry

#### 1 comment:

1. From similar triangles it's easy to prove that: a^2=BH*c, b^2=AH*c & h^2=BH*AH. Likewise h^2=CE*a=b*CF. Hence c*CE*CF = c*h^2/a*h^2/b. Triangle ABC=1/2*c*h =1/2*a*b or c*h=a*b
Therefore,c*CE*CF = c*h^2/a*h^2/b =ch^4/ab =ch^4/ch = h^3 ---------------------------(1)
Similarly, BH^2 =a*BE & AH^2=b*AF So ab*AF*BE=(AH*BH)^2 = h^4 since h^2=BH*AH. Thus, AF*BE =h^4/ab =h^4/ch =h^3/c or c*AF*BE = h^3 --(2)
All right triangles within the outer triangles are similar. Thus, AF/AH=b/c while AH/b=b/c
So we have:AF = b * b/c * b/c = b^3/c^2 -----(3)
Likewise, BE = a^3/c^2 --------------------(3)
Using (3) we say that (AF^2)^(1/3)+(BE^2)^(1/3)=
(b^6/c^4)^(1/3)+(a^6/c^4)^(1/3)= (a^2+b^2)/c^(4/3)=c^2/(c^(4/3)= c^(2/3) ------------- (4)
We've HM/HE = b/c and HE/h = a/c So HM=HE*b/c= hab/c^2. Similarly, HN =hba/c^2 or HM=HN ---(5)
Finally, 2(BM*BN)^(1/2) = 2*BE*AF/(AH*BH)^(1/2) since BE^2=BM*BH and AF^2=AN*AH.
So LHS = 2[(ab)^3/c^4]/h =2(ch)^3/hc^4 =2h^2/c
Now MN =2HM=2hab/c^2 as shown above.
Or MN=2h(ch)/c^2=2h^2/c =LHS=2(BM*BN)^(1/2)--(6)
Ajit
ajitathle@gmail.com