D is the bisector of BC, so it is easy to get the coordinates of point D as D ([x2+2 (x3-x2)/3], [y2+2 (y3-y2)/3).
M is the midpoint of AD, and the set point M=M(m, n).
M = [x 1+x2+2 (x3-x2)/3]/2，n = [y 1+y2+2 (y3-y2)/3]/2。
Vector AB=(x2-x 1, y2-y 1)), vector AC=(x3-x 1, y3-y 1),
The coordinates of set points p and q are P (S 1, T 1) and Q (S2, T2).
Then there are vectors AP = (S 1-X 1, T 1-Y 1) and AQ=(s2-x 1, t2-y 1).
The vector PM=(m-s 1, n-t 1) and the vector QM=(m-s2, n-t2).
It is known that there are vector AP=x* vector AB, vector AQ=x* vector AC.
∴ have s1-x1= x * (x2-x1), t1-y/= x * (y2-y1)? ( 1),(2)
s2-x 1=y*(x3-x 1)，t2-y 1=y*(y3-y 1)？ (3),(4)
The three-point * * line of p, m and q, ∴ (T2-t1)/(S2-s1) = (n-t1)/(m-s1)? (5)
(1)~(5), five equations, * * with s 1, S2; t 1，T2； X and y, where m and n are known.
Eliminate s2, t 1, t2, and solve the expression of x and y containing s 1.
The value of 2x+y is a unary function containing only the variable s 1, and its extreme value is easy to find.
It's too late, sorry, I'll continue counting tomorrow; You can also calculate it yourself when you are free.)
(Through drawing, I found that when AP = 1/3 * AB and AQ = 2/3 * AC, 2x+y takes the minimum value of 4/3.
At this time, x= 1/3, y=2/3, ∴2x+y=2/3+2/3=4/3).
(2)θ∈(π/4，π/2)，sinθ& gt； 0，cosθ& gt； 0，x/y = sinθ/cosθ& gt； 0
Sinθ/x=cosθ/y, => sin? θ/x？ =cos？ θ/y？
( 1-cos？ θ)/x？ =cos？ θ/y？
1/cos？ θ=(x？ +y？ )/y？
=> Because? θ=y？ /(x？ +y？ )
Sin? θ= 1-cos？ θ=x？ /(x？ +y？ )
Because? θ/x？ +sin? θ/y？ =y？ /[x？ (x？ +y？ )]+x？ /[y？ (x？ +y？ )]= 10/[3(x？ +y？ )]
= & gty？ /x？ +x？ /y？ = 10/3
Let t=x/y, then there is 1/t? +t？ = 10/3
Easy to solve t=√3, 1/√3? (negative root abandonment)
And θ∈(π/4, π/2), ∴t=x/y=tanθ∈( 1,+∞).
∴t= 1/√3<； 1 abandon, that is, t=x/y=√3.
The value of x/y is ∴ 3.