Question 1: Find a point p on the straight line L to minimize the value of PA+PB.
A summary of the shortest path problem in junior high school mathematics
Exercise: Connect AB, and the intersection with straight line L is point P. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. The minimum value of PA+PB is AB.
Question 2: ("General horse drinking problem") Find a point P on the straight line L to minimize the value of PA+PB.
A summary of the shortest path problem in junior high school mathematics
Practice: Make point B' the symmetrical point of point B about straight line L, and the intersection point connecting AB' and L is point P. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. The minimum value of PA+PB is AB'.
Question 3: Find points M and N on the straight lines l 1 and l2 respectively to minimize the perimeter of △PMN.
A summary of the shortest path problem in junior high school mathematics
Practice: Make points P' and P'' symmetrical to two straight lines respectively, connect P'P'', and the intersection with the two straight lines is point M, n. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. The minimum value of PM+MN+PN is the length of line segment P'P'.
Question 4: Find points M and N on straight lines l 1 and l2 respectively to minimize the perimeter of quadrilateral PQMN.
A summary of the shortest path problem in junior high school mathematics
Exercise: Do some Q's respectively? The symmetrical points Q' and P' of the straight line l 1 and l2 connect Q'P', and the intersection with the two straight lines is point M, n. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. The minimum value of the perimeter of the quadrilateral PQMN is the length of the line segment Q'P'+PQ.
Question 5: ("bridge site selection problem") straight line m∑n, find points m and n on m and n respectively, so that MN⊥m,
The value of AM+MN+BN is the smallest.
A summary of the shortest path problem in junior high school mathematics
Exercise: Move point A downward by the length unit of MN to get a', connect a'b, cross with n at point N, and cross with n at point M to make nm ⊥ m. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. The minimum value of AM+MN+BN is A'B+MN.
Question 6: Find two points M and N on the straight line L (M is on the left) to minimize the values of MN = a and AM+MN+NB.
A summary of the shortest path problem in junior high school mathematics
Practice: translate point A to the right by one length unit to get a', make a symmetrical point A' about the straight line L, and connect A''B with the straight line L at point N,
Translate point n to the left by one unit m.
A summary of the shortest path problem in junior high school mathematics
Principle: The shortest line segment between two points. The minimum value of AM+MN+NB is a "b+Mn.
Question 7: Find point A on l 1 and point B on l2 to minimize the value of PA+AB.
A summary of the shortest path problem in junior high school mathematics
Practice: Make point P' the symmetrical point of point P about l 1, and make point B ⊥ L2 pass through l 1 at point A.
A summary of the shortest path problem in junior high school mathematics
Principle: The distance from a point to a straight line is the shortest. The minimum value of PA+AB is the length of line segment p' b.
Question 8: A is the fixed point on l 1, and B is the fixed point on l2. Find the m point on l2 and the n point on l 1.
Minimize the value of AM+MN+NB.
A summary of the shortest path problem in junior high school mathematics
Practice: Make point A' the symmetry point of l2 and point B' the symmetry point of l 1. Connect A'B' to intersection l2 and point M and intersection l 1 and point N. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. The minimum value of AM+MN+NB is the length of line segment A'B'.
Question 9: Find a point P on the straight line L to minimize the value of | PA-PB |.
A summary of the shortest path problem in junior high school mathematics
Exercise: the intersection of the vertical line connecting AB and AB and the straight line L is point P. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The distance from a point on the perpendicular bisector to both ends of the line segment is equal. | Pa-Pb | = 0。
Question 10: find a point p on the straight line l to maximize the value of | PA-PB |.
A summary of the shortest path problem in junior high school mathematics
Practice: Make a straight line AB, and the intersection with the straight line L is point P. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The difference between any two sides of a triangle is smaller than the third side. | PA-PB | ≤ AB, and | pa-Pb | = the maximum value of ab.
Question 1 1: find a point p on the straight line l to maximize the value of | PA-PB |.
A summary of the shortest path problem in junior high school mathematics
Practice: Make the straight line AB' from point B to line L symmetrical to point B, and the intersection with line L is point P. 。
A summary of the shortest path problem in junior high school mathematics
Principle: The difference between any two sides of a triangle is smaller than the third side. | PA-PB |≤ AB', the maximum value | PA-Pb | = AB'.
Question 12: ("fermat point") Every internal angle in △ABC is less than 120. Find a point p in △ABC.
Minimize the value of PA+PB+PC.
A summary of the shortest path problem in junior high school mathematics
Practice: Seek fermat point, that is, satisfy ∠ APB = ∠ BPC = ∠ APC = 120.
Take AB and AC as sides, make equilateral △ABD and △ACE outward, and connect CD and BE at point P, which is the demand.
A summary of the shortest path problem in junior high school mathematics
Principle: The line segment between two points is the shortest. Pa+Pb+PC = minimum value of CD.
Second, "fermat point"-the point with the smallest sum of three points.
Fermat point tectonic method;
(1) given the connection of three points to form a triangle (△ABC), and each internal angle of this triangle is less than120;
(2) As shown in the following figure: Give three points, A, B and C,
A summary of the shortest path problem in junior high school mathematics
Make a regular triangle with AC as the side to get point D, and make a regular triangle with BC as the side to get point E,
Connecting BD and AE at point O, we assert that point O is "fermat point".
Fermat point's proof method:
Prove △ AEC △ DBC.
△AEC rotates 60 clockwise around point C to get △DBC, so △ AEC △ DBC.
So ∠OBC = ∠OEC, so O, B, E and C are four * * * cycles.
Expanding knowledge: four-point * * * circle judgment method
If the angle between two points on the same side of a line segment and the two endpoints of the line segment is equal, then these two points and the two endpoints of the line segment are * * * circles.
A summary of the shortest path problem in junior high school mathematics
So ∠ BOE = ∠ BCE = 60, ∠ COE = ∠ CBE = 60,
So ∠ BOC = ∠ BOE+∠ Coe = 120, and likewise ∠AOC = ∠AOB = 120.
So ∠ BOC = ∠ AOC = ∠ AOB = 120.
A summary of the shortest path problem in junior high school mathematics
Take point O as a point on AE, and use △AEC to rotate 60 clockwise around point C to get point O2.
So ∠ o CO2 = 60, OC = O2C, OA = O2D,
So △OCO2 is an equilateral triangle, so there is OO2 = OC.
So BD = OA+OB+OC.