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 P1D.
(B) Find the distance CE that minimizes P2B.
(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.
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Posted by Charlie
on 2022-01-14 11:07:23 |
solution
