Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 864.

## Friday, March 29, 2013

### Problem 864: Parallelogram, Diagonal, Congruence, Similarity, Metric Relations

Labels:
congruence,
metric relations,
parallelogram,
similarity,
triangle

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if i worng, please do not sanction my post

ReplyDeletebut if my solution is true:

BF=Y=GE

FG=Z

DE=X-3

3/X=y/(z+y)

y/(z+y)=(x-3)/3

3/x=(x-3)/3

9=x^2-3x

x^2-3x-9=0

x1,2=(3+-sqrt(9-4*1*(-9)))/2=

=1.5+sqrt(45)/2

Constrct a line PQ through F parrallel to AB,CD such that P lies on BC and Q lies on AD.

ReplyDeleteFP:FQ = 3:x

Since triangle(BFP) = triangle(EGD), FP = GD, so CG = FQ.

By triangle(BCG) ~ triangle(EDG),

GD/CG = DE/BC

=> FP/FQ = DE/BC

=> 3/x = (x - 3)/3

=> x = 3*(1 + sqrt5)/2

From ΔFBC~ΔFEA

ReplyDelete3/x = BF/FE := a/b

From ΔGBC~ΔGED

3/(x-3) = BG/GE = b/a

⇒ x(x-3) = 9

⇒ x^2 - 3x - 9 = 0

⇒ x = 3(1 + √5)/2

We use Barycentric Coordinates.

ReplyDeleteDefinition: Capital letters will denote vectors. P = x A + y B + z C will be written as P = (x, y, z). By definition, x + y + z = 1

Let A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1).

ABCD is a parallelogram, hence D + B = A + C, which results in D = (1, -1, 1)

Let F = q B + (1 – q) G. As FB = GE, we have G = (1 – q) F + q E

Let F = (p, 0, 1 – p). This leads to G = (p, -q, 1 – p) / (1 – q)

Consider the fact that G is the intersection of CD and BF. If (x, y, z) is on a specified straight line, then 0 = ux + vy + wz for some constant u, v, w. After solving the systems of equations, it’s clear that line BF is x (1 – p) – z p = 0 and CD is x + y = 0. To find G, we also need x + y + z = 1. With this, it’s clear that G = (p / (1 – p), -p / (1 – p), 1).

Using the two expressions of G, we get that p = q. Using these facts leads to finding E = (p – 1 + 1/(1-p), -1/(1-p), -p + 2).

But we also know that E is the intersection of AD [z + y = 0] and BF [x(1-p) – zp = 0]. So E = (1, -(1 – p)/p, (1 – p)/p). Combining the two expressions for E, we get p = 0.5 (3 + sqrt(5)) or 0.5(3 – sqrt(5)). This gives two possible coordinates for E: 0.5 (2, sqrt(5) – 1, 1 – sqrt(5)) or 0.5 (2, -sqrt(5) – 1, 1 + sqrt(5)).

D is between A and E. Thus, must be possible to express D as mA + nE, such that m + n = 1 for positive m and n. From the formula of the division of an interval, AD : DE = n : m.

Straight away, it is clear that E = 0.5 (2, -sqrt(5) – 1, 1 + sqrt(5)) only, as the alternative would produce negative m or n. Simple arithmetic tells us that (sqrt(5) – 1) A / (sqrt(5) + 1) + 2 E / (1 + sqrt(5)) = D.

Hence, AD : DE = 2 : sqrt(5) – 1, which means that AD:AE = 2 : 1 + sqrt(5)

It’s given that AD = 3

So quite obviously, AE = 1.5 (1 + sqrt(5)).