See complete Problem 183 at:

www.gogeometry.com/problem/p183_right_triangle_hypotenuse.htm

Right Triangle, Hypotenuse Trisection Points, Squares of the Distances. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Tuesday, September 23, 2008

### Elearn Geometry Problem 183

Labels:
distance,
hypotenuse,
Pythagoras,
right triangle,
square,
Stewart's Theorem,
trisection

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by Cosine Law

ReplyDeleted²+(b/3)²+2d(b/3)Cos[BDC]=AB²

d²+(2b/3)²-2d(2b/3)Cos[BDC]=BC²

so

3d²+2/3b²=2AB²+BC²----(1)

and similarly

3e²+2/3b²=2BC²+AB²----(2)

(1)+(2)

3(d²+e²)+4/3b²=3(AB²+BC²)=3b²

d²+e²=5/9b²

a geometry solution

ReplyDeleteDraw DF, EG perpendicular to BC

mark x = BF = FG = GC ( thales theor )

mark y = EG => FD = 2y ( middle line )

tr BDF => d'2 = x'2 + 4y'2 (1)

tr BEG => e'2 = y'2 + 4x'2 (2)

from (1) and (2)

d'2 + e'2 = 5x'2 + 5y'2 = 5( x'2 + y'2 ) (3)

tr EGC => x'2 + y'2 = 1/9 b'2 (4)

from (3) and (4)

d'2 + e'2 = 5/9 b'2

P.S. d'2 mean d*d ( power 2 )

Hi,I'm INDSHAMAT from Srilanka.My solution has been denoted as follows

ReplyDeleteAB2={2d2+2(b2/9)}-e2 (1)(By Apollonius theorem)

BC2={2e2+2(b2/9)}-d2 (2)(By Apollonius theorem)

AB2+BC2=b2 (3)(By pythogorous theorem)

So {2d2+2(b2/9)}-e2+{2e2+2(b2/9)}-d2=b2

Therefore d2+e2=5(b2/9)is the Answer

Let AB=c, BC=a, b/3=f.--->

ReplyDelete0) a^2+c^2=b^2

1) c^2+e^2=2*(d^2+f^2), and

2) a^2+d^2=2*(e^2+f^2) --->

(1)+(2):

a^2+c^2+d^2+e^2=2*(d^2+e^2+2*f^2)

b^2+d^2+e^2=2*d^2+2*e^2+4*f^2

b^2=d^2+e^2+4*(b/3)^2

--->

d^2+e^2=b^2*(1-4/9)

Choose a coordinate system so that B(0,0), A(a,0) and C(0,c).

ReplyDeleteThen we find that D has coordinates (2a/3, c/3) and that E has coordinates (a/3, 2c/3).

So by the formula for the distance between two points: d^2 = (a/3)^2 + (2c/3)^2 (1)

e^2 = (2a/3)^2 + (c/3)^2 (2)

From (1) + (2) we get: d^2 + e^2 = 5a^2/9 + 5c^2/9 (3) and by Pythagoras' theorem a^2 + c^2 = b^2 (4).

From (3) and (4) follows that d^2 + e^2 = (5/9)b^2.

Let M be the midpoint of AC (DE).

ReplyDeleteIn triangle BDE:

BD^2 + BE^2 = 2(BM^2 + ME^2)

(i.e.)d^2 + e^2 = 2(BM^2 + b^2/36).....(i)

In triangle ABC:

b^2 = AC^2 = AB^2 + BC^2 = 2(BM^2 + MC^2)

(i.e.) b^2 = 2(BM^2 + b^2/4)..... (ii)

From (i) and (ii);

d^2 + e^2 - b^2 = 2[(b^2/36) - (b^2/4)]

d^2 + e^2 = b^2 + (b^2/18) - (b^2/2)= 5b^2/9

Post Script:

ReplyDeleteNotice that BM = AM = MC = b/2 (since triangle ABC is right angled at B)

So BD^2 + BE^2 = 2(BM^2 + ME^2)implies

d^2 + e^2 = 2[(b^2/4) + (b^2/36)]

=(b^2/2) + (b^2/18) = 5b^2/9