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 AC as the edge and make a regular triangle ABD ACE to connect with the outside. If we meet a point, this point is what we are looking for.
Proof: as shown in the figure below. 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 △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.
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.
(1) For any triangle △ABC, if there is a point E in or on the triangle, and EA+EB+EC has the minimum value, then E is the fermat point when it reaches the minimum value.
Fermat point's calculation
(2) If the interior angle of a triangle is greater than or equal to 120, the vertex of this interior angle is the fermat point; If all three internal angles are less than 120, then three points with an opening angle of 120 inside the triangle are the fermat point of the triangle.
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It has been proved that only fermat point can,
If you look on the coordinate axis, you already know the points in a triangle whose three sides are 120.
This is naturally easy to determine.
When all three internal angles of the triangle ABC are less than 120, make regular triangles ABC 1, ACB 1, BCA 1 with AB, BC and CA as sides respectively, and then connect AA 1, BB 1, and.