Triangle ABC is right-angled at C, a = |BC|, b = |CA|, with a < b.
Squares ABDE, BCFG, and CAHI are erected externally to triangle ABC.
Lines HE and DG intersect at P, lines DG and FI intersect at Q, and lines FI and HE intersect at R.

If triangle PQR is right-angled, determine the value of b/a.

One I put this on the coordinate plane I found 3 ways of solving it:

Put C at the origin.  WLOG let a=1 and then b=x is the ratio sought.
Put B at (0,1) and A at (x,0)
The big square D = (1,x+1), E = (x+1,x)
The little square F=(-1,0), G=(-1,1)
The last square H=(x,-x), I=(0,-x)

The point Q could be probably be found in terms of x but isnt needed.  It is enough to know Q is in line with GD and also with FI.

Method 1: Slopes
The slope of GD = (x+1 - 1)/(1 - -1) = x/2
The slope of FI =  (-x - 0)/(0 - -1) = -x
To be perpendicular x/2 * -x = -1
x^2/2 = 1
x^2 = 2
so x=sqrt(2)

Method 2: Vectors
GD = 2i + xj
FI = 1i - xj
These vectors are orthogonal if the dot product is 0
GD.FI = (2*1) + (x*-x) = 2 - x^2
2 - x^2 = 0
x^2 = 2
x = sqrt(2)

Method 3: Trigonometry
Call the angle GD forms with the x axis by alpha
Call the angle FI forms with the x axis by beta
angle PQR = alpha + beta = 90
x/2 = tan(alpha)
x = tan(beta) = tan(90 - alpha) = 1/tan(alpha)
2tan(alpha) = 1/tan(alpha)
2[tan(alpha)]^2 = 1
tan(alpha) = sqrt(1/2) = sqrt(2)/2

but tan(alpha) = x/2
so x/2 = sqrt(2)/2
x = sqrt(2)

Posted by Jer on 2010-01-22 15:05:10