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Closest approaches (Posted on 2022-01-14) Difficulty: 3 of 5
ABCD is a unit square with movable point E on side CD.
P1 is on the same side of Line AE as D and P2 is on the other side such that AEP1 and AEP2 are equilateral triangles.

(A) Find the smallest possible distance P1D.

(B) Find the distance CE that minimizes P2B.

No Solution Yet Submitted by Jer    
Rating: 3.0000 (1 votes)

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Solution solution | Comment 1 of 2
(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 2022-01-14 11:07:23
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