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Missouri State University’s Advanced Problem # 104 Vincent Pantaloni (Orléans, France) 1

Problem : A circle one foot in radius is placed on a ramp that makes an angle of 30 degrees with the horizontal so that a point P on the circle touches the ramp 10 feet from the base of the ramp. As the circle rolls down the ramp, what is the maximum height that P attains ?

P 10

30°

Solution : Here’s a picture of what is happening. We know that point P will draw an arch of cycloid (dotted curve). It’s highest point will be reached when we have a horizontal tangent to that arch. We need to find that point P ′ and then compute the length BP ′ , that is BA + AP ′ . I don’t want to find an equation to that arch on a slope, I’d rather turn my head 30 degrees to the left and look at it this way :

P' Horizontal tangent to the arch

P A

30°

B

O

b = 4.44 After this 30° rotation our "horizontal tangent" has a slope of tan(−30) = − √13 . In this situation, Wikipedia reminds me that such a cycloid arch has parametric equation :  x = t − sin t , t ∈ [0, 2π] y = 1 − cos t

2

P′ b

1

b b

1

2

3

4A

b

5

6P

We are now searching the value of t for which we have a tangent vector whose coordinates are proportional to (1, − √13 ). The tangent vector has coordinates : 

x′ = 1 − cos t y ′ = sin t

, t ∈ [0, 2π]

These coordinates are proportional iff the cross products are equal, so we have to solve : 1 sin t = − √ (1 − cos t) 3 Hopefuly we’ll make it look better. . . All of the following equations are equivalent : √ (1) ⇐⇒ cos t − 1 − 3 sin t = 0 √ 1 3 1 ⇐⇒ cos t − sin t = 2 2 2  π 1 = ⇐⇒ cos t + 3 2 π π ⇐⇒ t + = ± [2π] 3 3

(1)

(2) (3) (4) (5) (6)

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4π The only solution with t ∈]0, 2π[ is − 2π 3 + 2π = 3 . By√substituting in the cycloid’s parametric equation, 3 3 we get the coordinates of P ′ on the graph : P ′ ( 4π 3 + 2 , 2 ) which is approximately (5.05, 1.5) and that seems quite consistent with our graph. Now in order to find AP ′ we have to search for the coordinates of point A. Now it’s quite easy, just a matter of straight lines : √ √ The line (AP ′ ) is perpendicular to our tangent, so it has a slope of 3. (AP ′ ) : y = 3x + b. We find the y intersect by using the fact that P ′ is on this line so : √ ! √ 3 3 √ 4π +b + yP ′ = 3xP ′ + b that is = 3 2 3 2

Solving for b gives b = −

√ 4π 3 . Now A has coordinates A(xA , 0), we look for xA : 3 yA =

√

3xA + b

b 4π xA = − √ = 3 3

gives

We can now compute AP ′ : v u √ !2  2 q u √ 3 3 2 2 ′ AP = (xP ′ − xA ) + (yP ′ − yA ) = t + = 3 2 2

2π Now we need AB so first we have AP = xP − xA = 2π − 4π 3 = 3 and then knowing that point P on the circle touches the ramp 10 feet from the base of the ramp we have OA = OP − AP = 10 − 2π 3 . Considering the right angle triangle OAB we have :   2π 1 π AB = OA sin(30) = 10 − × = 5− 3 2 3

And finally adding AB and AP ′ we find BP ′ which is what we were looking for. The maximum height that P attains is : π √ BP ′ = BA + AP ′ = 5 − + 3 3 √ π That is just 3 − 3 ≃ 0.68 ft above it’s starting point. Remarks. The result seems quite simple compared to the computations I had to do, so there might be a simpler geometrical solution. One might think that P ′ is the first position where the point P is "on top" of the circle. Actually it is not (if my computations are correct) because when this happens the length 2π AP is equal to π − π6 = 5π 6 which is not the same as what we found : AP = 3 . You can see that this first guess is wrong on the picture below (done using GeoGebra) : b b

b

b

b

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b

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