Sx=x squared +(8-x) squared, that is, Sx=2x squared-16x+64 Since P is a moving point, the sum of areas is not a fixed value. Its function is like a parabola with an upward opening, with an inherent minimum. Derivation, its derivative function is zero, and its extreme point is also the maximum point. When x=4, there is a minimum value, resulting in S(4)=32.
(2) There are two triangles whose areas are always equal, that is, S triangle APK=S triangle DFK.
(Don't draw, just prove. )
It is proved that if AP=x continues, BP=8-x, PK=a and DK = X-A..
Get: sapk = ax/2，sadk = x (x-a)/2，sdfk = (x-a) (8-x)。
Obviously, SAPK and SADK, SADK and SDFK are not exactly the same.
Suppose SAPK is constant equal to SDFK, ax/2 is constant equal to (x-a)(8-x), which is simplified to the square of X -8x+8a=0.
According to similar triangles property, PK/BF=AP/AB, that is, there is always a/(8-x)=x/8, that is, there is always: the square of x -8x+8a=0.
So the hypothesis holds. So there are two triangles whose areas are always equal.
(3) The path length is 8π.
Because the point O passes through the midpoint of each side, the tangent is symmetrical about the midpoint of the square ABCD and symmetrical about the horizontal axis and the vertical axis. Therefore, its trajectory must be inscribed circle or inscribed square. Note that the inscribed circle here also includes the case that its arc is concave inward. First, try to draw an inscribed square. If the intersection point is not on its side, its trajectory is an inscribed arc. Then, verify whether the vertex is in a or the center of the square, that is, verify whether the distance to the fixed point is equal to a fixed length. This problem has been verified as four circular arcs, that is, four semi-circular arcs are tangent. Just do p and move to one side, and finally multiply by 4. Of course, you can also see O's trajectory by drawing n PQ.
The verification method can adopt hypothesis, and then prove whether the hypothesis is true by sine and cosine theorem of triangle.
(4) The path length of point O is Mn/2 = 3. When P moves to the midpoint of AB, OM+ON has a minimum value, and the minimum value can be found in Rt triangle as 2 times 5= 10.
Playing by hand is really tiring. If you think there is something wrong, please point it out and we will discuss it and seek to adopt it ~