ABCD is a unit square with movable point E on side CD.
P
_{1} is on the same side of Line AE as D and P
_{2} is on the other side such that AEP
_{1} and AEP
_{2} are equilateral triangles.
(A) Find the smallest possible distance P_{1}D.
(B) Find the distance CE that minimizes P_{2}B.
(A) Call measure of angle DAE = A.
AD = 1
AP1 = 1/cos(A)
Angle DAP1 = 60°  A
By the law of cosines:
(DP1)^2 = 1 + (1/cos(A))^2  (2/cos(A))*cos(60°A)
The derivative using the corresponding radian measure pi/6 for 30°:
2 sec(A) (tan(A) (sec(A)  sin(A + pi/6))  cos(A + pi/6))
which becomes zero at A = arctan(sqrt(3)/2) ~= 40.893394649131°
The square of the smallest distance evaluates to 1/4, so the smallest distance is 1/2.
But we want CE = 1  tan(A) = 1  sqrt(3)/2 ~= .13397459621556.
(B) Call measure of angle DAE = A.
Angle BAP2 = 30°  A
BA = 1
AP2 =1 /cos(A)
By the law of cosines:
(BP2)^2 = 1 + (1/cos(A))^2  (2/cos(A))*cos(30°A)
The derivative using the corresponding radian measure pi/6 for 30°:
2 sec(A) (tan(A) (sec(A)  cos(pi/6  A))  sin(pi/6  A))
which becomes zero at A = arctan(1/2) ~= 26.565051177078 °, making about a 3.4° angle between the side of the triangle and side AB.
The square of the smallest distance evaluates to 7/4  sqrt(3), so the smallest distance is sqrt(7/4  sqrt(3)) ~= .133974596215551.
But we want CE = 1  tan(A) = 1/2.
Between (A) and (B), the size of CE and the shortest distance from Pn have interchanged.

Posted by Charlie
on 20220114 11:07:23 