Given that ab = 20, points m and N are the golden section points of the line segment near AB respectively, then Mn =?

Given that ab = 20, points m and N are the golden section points of the line segment near AB respectively, then Mn =?

MN=(2*0.618-1)*20=4.72

It is known that a and B are two points on the line Mn, Mn = 4, Ma = 1, MB > 1. Take a as the center, rotate clockwise to point m, and take B as the center, rotate counterclockwise to point n, so that two points m and N re form a point C to form a triangle ABC. Let AB = X
1 to find the value range of X
2 if △ ABC is a right triangle, find the value of X
3. Explore when △ ABC gets the maximum area and what is the maximum area

1. AB = x, a = 1, the value of B changes between 0-3, AB should be (0,3), that is, X changes between 0-3, excluding 0 and 3; 2, a may be a right angle side, or a hypotenuse, may Pythagorean theorem equation, let AB = y, then BN = 3-y can get y value

(1 / 2) it is known that a and B are two points on the line Mn, Mn = 4, Ma = 1, MB > 1
(1 / 2) it is known that a and B are two points on the line Mn, Mn = 4, Ma = 1, MB > 1. Take point a as the center, rotate point m clockwise, and take point B as the center, rotate point n counterclockwise, so that, m, n

1. Because the length of the three sides of the triangle is 1,3-x, X
So x + 1 > 3-x gets x > 1
3-x + 1 > x gets X

A. B is on the line Mn, Mn = 4, Ma = 1, MB > 1
A. B on the line Mn, Mn = 4, Ma = 1, MB > 1. Take a as the center, rotate m clockwise, take B as the center, rotate n counterclockwise, so that the coincidence point of M and N is C, get △ ABC, let AB = X. (1) find the value range of X. (2) if △ ABC is RT △ find the value of X. (3) explore the maximum area of △ ABC

1. Because the length of the three sides of the triangle is 1,3-x, X
So x + 1 > 3-x gets x > 1
3-x + 1 > x gets X