Fermat has never received a special mathematics education in his life, and mathematics research is just a hobby. However, in France in the17th century, no mathematician can match it: he was one of the inventors of analytic geometry; The contribution to the birth of calculus is second only to Newton, Leibniz, the main founder of probability theory, and the person who inherited the world of number theory in17th century. In addition, Fermat also made important contributions to physics. Fermat, the master of mathematics, was the greatest mathematician in France in the17th century. In particular, his Fermat's Last Theorem has puzzled the wise people in the world for 358 years.
There are three villages, A, B and C, and a water supply station should be built in the middle to deliver water to the three villages. Now, it is necessary to determine the location of the water supply station to minimize the total length of the required pipeline? This problem is abstracted by mathematical model as follows:
Determine a point P in △ ABC to minimize the sum of the distances from P to three vertices PA+PB+PC.
The solution is as follows: take AB and AC as sides, make a regular triangle ABD, and ACE connects CD and BE at a point, then this point is the required point.
Proof: It is discussed in the following three situations:
(1) When ∠ BAC < 120, as shown in the following figure. Connect PA, PB and PC, in △ABE and △ACD, AB = ADAE = AC ∠ BAE = ∠ BAC+60 ∠ DAC = ∠ BAC+60 = ∠ BAE △ ABE congruence △ACD.
∴ ∠ABE=∠ADC So that A, D, B and P are four * * * cycles.
∴∠APB= 120，∠APD=∠ABD=60
Similarly: ∠ APC = ∠ BPC = 120.
Make a circle with p as the center and PA as the radius. PD is at point f, connecting AF,
It is proved that ∠ APD = 60 by rotating △ABP clockwise by 60 with A as the axis.
△ APF is a regular triangle. ∴ It is not difficult to find that when∠ BAC =120 ABP coincides with △ADF.
∴BP=DF PA+PB+PC=PF+DF+PC=CD
In addition, take any point G different from P in △ABC and connect GA, GB, GC and GD with B as the axis.
Rotate △ ABG 60 counterclockwise, and record the rotation of point G to point M. 。
Then △ABG coincides with △BDM, and m is either on the line DG or outside DG.
g b+ GA = GM+MD≥GDGA+g b+ GC≥GD+GC & gt； DC .
So CD is the shortest line segment.
(2) When ∠ BAC = 120, it can be seen from the above practice that the point to be found is point A. 。
(3) when ∠ BAC >; At 120, if the method in (1) is followed again, the point p will be outside of △ABC, so that PA+PB+PC will become larger again. Therefore, in this case, point A is the point that meets the meaning of the question.
The above is a simple fermat point question. By extrapolating this problem to four points, it can be verified that the intersection of diagonal lines of quadrilateral is the point to be found.
fermat point
Open classification: science, mathematics
Fermat point definition
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In a polygon, the point with the smallest sum of distances to each vertex is called the fermat point of the polygon.
In a plane triangle:
(1). Three triangles with internal angles less than 120, with AB, BC and CA as sides, make regular triangles ABC 1, ACB 1, BCA 1 on the outside of the triangle, and then connect AA 1.
(2) If the internal angle of the triangle is greater than or equal to 120 degrees, then the vertex of this obtuse angle is the demand.
(3) When △ABC is an equilateral triangle, the outer center coincides with fermat point.
(1) In an equilateral triangle, BP=PC=PA, and BP, PC and PA are the heights of the three sides of the triangle and the bisector of the triangle respectively. Is the center of inscribed circle and circumscribed circle. △BPC?△CPA?△PBA .
(2) When BC=BA but CA≠AB, BP is the bisector of the height and the median line on the triangle CA and the angle on the triangle.
certificate
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(1) The opposite opening angle of fermat point is 120 degrees.
△CC 1B and △AA 1B, BC = ba 1, Ba = bc 1, ∠ CBC 1 = ∠ b+60 degrees = ∠ ABA/kloc.
△CC 1B and △AA 1B are congruent triangles, and ∠PCB=∠PA 1B is obtained.
In the same way, ∠CBP=∠CA 1P can be obtained.
From ∠PA 1B+∠CA 1P=60 degrees, ∠PCB+∠CBP=60 degrees, so ∠CPB= 120 degrees.
Similarly, ∠APB= 120 degrees, ∠APC= 120 degrees.
(2)PA+PB+PC=AA 1
Rotate △BPC 60 degrees around point B to coincide with △BDA 1 and connect PD, then △PDB is an equilateral triangle, so ∠BPD=60 degrees.
And ∠BPA= 120 degrees, so a, p and d are on the same straight line.
And ∠CPB=∠A 1DB= 120 degrees, ∠PDB=60 degrees, ∠PDA 1= 180 degrees, so a, p, d, a.
(3)PA+PB+PC is the shortest.
Take any point M (not coincident with point P) in △ABC, connect AM, BM and CM, rotate △BMC 60 degrees around point B to coincide with △BGA 1, connect AM, GM, A 1G (same as above), and then AA 1
Plane quadrilateral fermat point
Fermat point's proof in a quadrilateral is easier to learn than in a triangle.
(1) In the convex quadrilateral ABCD, fermat point is the intersection point p of two diagonal lines AC and BD.
(2) In the concave quadrilateral ABCD, fermat point is the concave vertex D(P).