The square problem, a square puzzle

1.Yes (the point is the center of the square).

Proof: Let the side length of the square be a and the length of ap be b(0<=b<=a).

Put the square into a planar Cartesian coordinate system, so that the coordinates of point A are (0,0),b(a,0),c(a,a),d(0,a).

p(b,0),q(a,b),e(a-b,a),f(0,a-b)

From the above, the equation for the PE line is (a-2x)*(y-a 2)=a*(x-a 2).

Therefore, the straight line passes through the fixed point (a 2, a 2), that is, the center of the square.

Use the formula a*a+b*b>=(a+b)*(a+b) 2

Since a is certain, s>=a*a 2

The equal sign is taken if and only if b=a2, i.e., when point p (and all other points) is the midpoint, there is a minimum area a*a2

And because s=2b*b-2ab+a*a is a subtraction function in [0,a 2] and an increase function in [a 2,a].

So when b=0 or a, there is a maximum.

That is, when p, q, e, and f are respectively located at the vertices of the four sides of the square, their areas are the largest, which is a*a

1.Not. When they are located at the midpoint of the four sides of the square, the area is the smallest, which is 1 2s regular ABCD.

When P, Q, E, and F are located at the vertices of the four sides of the square, they have the largest area, which is S-positive ABCD.

According to the characteristics of the square, it can be seen that the diagonals are equal, and the diagonals are bisected and perpendicular to each other.

So ao=bo, angle aob=90 degrees, angle oab=45 degrees.

1 Place 5 points arbitrarily in a square with a side length of 1 to prove that two points can be found at a distance of no more than (1 2).

The side length of the square ABCD is 1, and the midpoints of AB, BC, CD, and Da are P, Q, R, and S., respectively

Even PR, QS, two line segments are handed over to O

Obviously o is the center of the square abcd, and the quadrilaterals apos, bqop, croq, dsor are all squares with sides of 1 2.

Since the five points M1, M2, M3, M4, and M5 are all within the large square ABCD, at least two of them are within the same small square.

Since the distance between any two points in a small square, including the inner boundary, is the largest distance between the two opposite vertices, and the side length of the small square is 1 2, and the length of its diagonal is (1 2).

Therefore, placing 5 points arbitrarily in a square with a side length of 1 proves that two points must be found at a distance of no more than (1 2).

Question: The side length of the square ABCD is 2, E and F are the midpoint of the AB and BC line segments, respectively, and the following is 0 points.

Connecting de, df, and ce, where ce crosses df and g, find the area of the quadrilateral ebfg. Answer, with a similar triangle.

Do the symmetry point Q2 with respect to AC, connect Bq2 to AC at a point, and this point is the P point that is sought.

The value is root number 5