Licang property oa office system

Then CN=3-t, AM=4-2t,

∠∠BCA =∠MAQ = 45，

∴QN=CN=3-t，

∴PQ= 1+t，

∴S△AMQ= 12AM？ PQ = 12(4-2t)( 1+t)=-T2+t+2。

(2) From the meaning of the question, CN=NQ=3-t, QP= 1+t, AM=4-2t.

∴s△bcq= 12×3(3-t),s△aqm= 12(4-2t)( 1+t)，

∫s△bcq:s△AQM = 3:2, namely 3 (3-t): (4-2t) (1+t) = 3: 2,

Solution: t= 1,

That is, when t= 1, s △ bcq: s △ aqm = 3: 2.

(3) existence.

Let t seconds pass, NB=t, OM=2t,

Then CN=3-t, AM=4-2t,

∴∠BCA=∠MAQ=45，

① If ∠ AQM = 90, PQ is the height of isosceles Rt△MQA on the base MA,

Pq is the center line of the bottom MA,

∴PQ=AP= 12MA，

∴ 1+t= 12(4-2t)，

Solution: t = 12.

② If ∠ QMA = 90, then QM and QP coincide,

∴QM=QP=MA，

∴ 1+t=4-2t

∴t= 1.