Crazy Geometry (1)

With ref. to the diagram below:

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/May11/Crazygeom1.jpg


Note that QS, QR, QX and RT are perp. bisectors of BP, AP, AB, and PC resp.

Let AB = AC = x, then from the given we have BC = x√2

Then ST = x/√2 since SP = BP/2 and PT = PC/2

Considering sine law in △ABP and △ACP:

In △ABP: AP/sin 45 = √2/sin θ

In △ACP: AP/sin 45 = x/sin (45 + θ)

So x/sin (45 + θ) = √2/sin θ

x = [√2 sin (45 + θ)]/sin θ

= (sin θ + cos θ)/sin θ

= 1 + 1/tan θ

tan θ = 1/(x - 1)

Thus, sin2 θ = 1/[(x - 1)2 + 1] and cos2 θ = (x - 1)2/[(x - 1)2 + 1]

Now, ∠BQP = 2∠BAP = 2θ (∠at centre = 2 x ∠ at sircum)

Similarly: ∠PRC = 2∠CAP = 180 - 2θ

With △QSB cong. to △QSP and △RTP cong. to △RTC:

∠PQS = θ and ∠PRT = 90 - θ

Then ∠RPT = θ

Consider △QSP: QP = SP/sin θ = 1/(√2 sin θ)

Consider △RPT: PR = PT/cos θ = (ST - SP)/cos θ = (x/√2 - 1/√2)/cos θ = (x - 1)/(√2 cos θ)

Also ∠QPR = 180 - ∠QPS - ∠RPT = 90

Hence by Pyth. thm, QP2 + PR2 = QR2 = 4

1/(2 sin2 θ) + (x - 1)2/(2 cos2 θ) = 4

1/sin2 θ + (x - 1)2/cos2 θ = 8

(x - 1)2 + 1 + (x - 1)2 + 1 = 8

(x - 1)2 = 3

x = √3 + 1

Therefore we have:

BC = (√3 + 1)√2

PC = (√3 + 1)√2 - √2 = √6

My method is using co-ordinate geometry.
Let A be (0,0). Side of triangle = 2a, so B(2a, 0) and C is (0, 2a).
Co-ordinates of R is (r, a) and Q is (a, q) since R and Q are on the perpendicular bisector of AB and AC respectively.
Since PB = sqrt 2, so the perpendicular distance of P to AB = 1, so the co-ordinates of P is (2a - 1, 1)
From here we set up 3 equations:
AQ = PQ = radius of circumcircle APB. That is
(a - 2a + 1)^2 + (q - 1)^2 = a^2 + q^2
Simplifying we get q = 1 - a ......................(1)
AR = PR = radius of circumcircle APC. That is
(r - 2a + 1)^2 + (a - 1)^2 = a^2 + r^2
Simplifying we get r = a - 1 .................(2)
Since distance between Q and R is 2, so
(a - q)^2 + (r - a)^2 = 2^2 ..............(3)
Sub. (1) and (2) into (3), we get
(2a - 1)^2 + (a - 1 - a)^2 = 4
(2a - 1)^2 = 4 - 1 = 3
2a - 1 = sqrt 3
2a = side of triangle = sqrt 3 + 1
so BC = sqrt 2(sqrt 3 + 1) = sqrt 6 + sqrt 2.
so PC = BC - BP = sqrt 6 + sqrt 2 - sqrt 2 = sqrt 6.


2011-05-21 04:15:19 補充：
Correction : Line 4 should be " AC and AB respectively."