Moving point problem in junior middle school mathematics

Math moving point problem in junior high school: In Rt△ABC, ∠ A = 90, AB=4, AC=3, M is the moving point on AB (not coincident with A and B), and MN∨BC, △AMN and △PMN are symmetrical about MN. Let am = X. (1) When p falls on BC, find x; (2) Let the overlapping area of △PMN and quadrilateral MNCB be y, find the functional relationship between y and x ... and find out what the maximum value of y is when the value of x is?

(1) analysis: ∫Rt△ABC, ∠ A = 90, AB=4, AC=3.

Let the fixed point M move from A to AB, let MN∑BC cross AC to N when crossing M, ⊿AMN and ⊿PMN are symmetric about Mn, AM=x, and P falls right on BC.

Connect AP, cross MN to O.

∴AP⊥MN,MA=MP,NA=NP,AP⊥BC

∴tanb=ac/ab=3/4； sinB = AC/BC = 3/5； cosB=AB/BC=4/5

tanC = AB/AC = 4/3； sinC = AB/BC = 4/5； cosC=AC/BC=3/5

AP=AbsinB=4*3/5= 12/5

∴x=am=oa/sinb= 1/2ap/sinb=6/5*5/3=2

When p falls on BC, X=2.

(2) Analysis: Let the overlapping area of △PMN and quadrilateral MNCB be y.

According to the meaning of the question, when M goes from A to P and just falls on BC, the overlapping area of △PMN and quadrilateral MNCB is equal to the area of △AMN.

AN=AMtanB=3/4x

s(⊿anm)= 1/2*x*3/4x=3/8x^2

When M continues to move forward, point P will move out of m', p' in ⊿ABC,vk diagram.

Let P'M', BC and T intersect, p' n', BC and V intersect.

At this time, the overlapping area of △PMN and quadrilateral MNCB is equal to the area of △ △AM' n '-δP' TV.

O 'a = am 'sinb = 3/5x = > o 'p' = 3/5 times.

∴pp'=o'p'-o'p=3/5x-( 12/5-3/5x)=6/5x- 12/5

⊿P'TV∽⊿P'M'N'

∴TV/M'N'=PP'/P'O'

M'N'=AM'/sinB=5/4x

∴tv=( PP '/p ' o ')* m ' n ' =(6/5x- 12/5)/(3/5x)*(5/4x)= 5/2x-5

∴s(△p'tv)= 1/2*tv*pp'= 1/2(5/2x-5)*(6/5x- 12/5)=3/2x^2-6x+6