Answer

British Broadcasting Corporation

C A B C A B C A B C a b c

So * * * has nine situations: (AAA)(AAB)(AAC)

(American Bar Association) (ABB) (American Broadcasting Corporation)

ACB anti-corruption Committee

(1) When three people go to high school, the probability of (AAA) is 1/9.

(2) Only two people choose the same thing (AAB)(AAC)(ABA)(ABB)(ACA)ACC).

The probability is 6/9 and the simplification is 2/3.

The second question:

(1) It is proved that ∫△∴mb=mc is an equilateral triangle ∠MBC=∠MCB.

BC ∴∠MBC=∠AMB

* am = Deutsche Mark ∴△AMB≌△DMC ∴AB=DC.

So the trapezoid ABCD is an isosceles trapezoid.

(2) Solution: ∫△MBC is an equilateral triangle ∠ MPQ = 60 unchanged.

∴∠bmp+∠bpm= 120∠cpq+∠BPM = 120 ∴∠bmp=∠cpq

∴△ BMP ∴△ CPQ ∴ BM/PC = BP/CQ, which means 4/x=(4-x)/(4-y).

Deformation can get y= 1/4 x? - x + 4

(3)① When y takes the minimum value, according to y= 1/4 x? The minimum value of -x+4 is (4ac-b? )/4a=3

At this time ∴ QC=4-3= 1.

∫△BMP∽△cpq ∴bm/pc=bp/cq, that is, 4/x=(4-x)/ 1, and x=2.

At this time, ∴, PC=BP MP is the midline on BC, and MP is also the angular bisector of ∠BMC according to the three-line unity theorem.

∴∠ PMC = 30 and ∴ ≈mpq = 60∴∠mqp =∠pqc = 90.

∴△PQC is a right triangle

② When BP= 1, AM is parallel and equal to BP, so the quadrilateral ABPM is a parallelogram.

Because ∠ mpq = 60 remains unchanged, there is only one parallelogram that meets the requirements.