The lengths of three sides of △ ABC are 3, 4 and 5 respectively. Point P is the point on its inscribed circle. Find the maximum and minimum of the sum of the areas of three circles with diameters PA, Pb and PC

The lengths of three sides of △ ABC are 3, 4 and 5 respectively. Point P is the point on its inscribed circle. Find the maximum and minimum of the sum of the areas of three circles with diameters PA, Pb and PC

Δ ABC is a right triangle because its side lengths are 3, 4 and 5 respectively. The radius of the inscribed circle can be obtained as 1. The rectangular coordinate system is established by taking two right angles as X and Y axes respectively. Assuming that the longer right angles coincide with X axis, let s = PA & sup2; + Pb & sup2; + PC & sup2; then s = x ^ 2 + y ^ 2 + (4-x) ^ 2 + y ^ 2 + x ^ 2 + (3-y) ^ 2 = 3 (x ^ 2 + y ^ 2)

It is known that the length of ABC triangle is 3,4,5, P is the point on its inscribed circle, PA, Pb, PC are the diameters, and the maximum and minimum area of three circles?

Establish coordinate system
Let a (3,0) B (0,4) C (0,0) P (x, y) be the inscribed circle radius R
Triangle ABC area s = 1 / 2Ab * AC = 1 / 2 (AB + AC + BC) r = 12, the solution is r = 1
The coordinates of the center of the inscribed circle (1,1)
P has (x-1) ^ 2 + (Y-1) ^ 2 = 1 on the inscribed circle
The sum of squares of the distances from point P to a, B, C is d = x ^ 2 + y ^ 2 + (x-3) ^ 2 + y ^ 2 + x ^ 2 + (y-4) ^ 2
=3(x-1)^2+3(y-1)^2-2y+19=22-2y
Obviously, 0 ≤ y ≤ 2 means 18 ≤ D ≤ 229 π / 2 ≤ π D / 4 ≤ 11 π / 2
That is to say, the maximum sum of the areas of the three circles with PA, Pb and PC as diameters is 11 π / 2;
The minimum value is 9 π / 2

As shown in the figure, let the distance between P and the two vertices a and B of equilateral triangle ABC be 2 and 3 respectively, and find the maximum value of PC
This graph is like this: the equilateral triangle ABC, with a point P on the outside of AB, connects PC. It is not simply Pythagorean theorem!

Rotate PA 60 ° counterclockwise around point a to get ad, then Da = PA, connect CD, DP, CP, as shown in the figure,
∵△ ABC is equilateral triangle ABC,
∴∠BAC=60°,AC=AB
∴∠DAC=∠BAP,
∴△DAC≌△PAB,
∴DC=PB,
And Pb = 3, PA = 2,
∴DC=3,
∵PC≤DP+DC,
∴PC≤5,
So the maximum PC can achieve is 5