數學大難題

4. Let M be the mid-point of QS.
Then AM is perpendicular to QS. (property of sqaure)
AM = 1/2 x sqrt(2 x AB)^2
　　= AB/sqrt(2)

In triangle AMS,
AM/MS = tan 60
[AB/sqrt(2) ]/ MS = sqrt(3)
AB = sqrt(6) x MS
MS = AB/sqrt(6)

Hence,
PS = sqrt(MS^2 + AM^2)
　　= sqrt (2 MS ^2)
　　= sqrt (AB^2/3)
　　= AB / sqrt(3)
So PQ = PS = AB / sqrt(3) (definition of square)
So AB:PQ = 1: 1/sqrt(3) = sqrt(3) : 1

5a. In △ABE and △CDF,
AB = CD (opp. sides of //gram)
angle BAE = angle DCF (alt. angles, AB//CD)
AE = CF (given)
so △ABE ~= △CDF (SAS)

b. Let BD and AC intersects at G.
BG=GD (diag. of //gram)
AG=GC (diag. of //gram)
AE + EG = GF + FC
Since AE = CF (given)\
So EG = GF
So BEDF is a parallelogram (diag. bisects each other)

6a.One question: Is BCS a straight line? If so,

In △PQR and △SCR,
PR = SR (given)
angle PQR = angle SCR (alt. angles, PQ//BS)
angle QPR = angle CSR (alt. angles, PQ//BS)
So △PQR ~= △SCR.
So QR=CR=RC (corr. sides, ~=△s)

b. Something wrong in the question!
AR>QR but QR=RC!
so AR =/= RC!

Did you type it wrongly?

2007-07-18 18:11:04 補充：
PS = sqrt(MS^2 AM^2)sorry, it should be:PS = sqrt(MS^2 PM^2) "A" should be "P"

2007-07-19 09:14:16 補充：
6b) AQ:QC = AP:PB = 1:1 (intercept theorem)AQ:(QR RC) = 1:1 Since QR=CR (corr. sides, ~=△s)AQ:2RC = 1:1AQ=2RCSo, AR = AQ QR = 3RCHence, AR:RC = 3:1