2

In curling, it is often necessary to hit and displace an opponent’s stone to win the end. Olivia would like to hit her opponent’s stone with her own stone. If she releases her stone at the hog line, it needs to travel another 87 feet before reaching her opponent’s stone. Olivia can be off by 11 inches in either direction and still hit her opponent’s stone. Assuming her stone doesn’t curl (change direction), within what angle, $\theta$, to the nearest tenth, must Olivia throw her stone to hit her opponent’s stone? enter image description here

With this question, do you need to convert the 11 inches to feet, and then solve from there? And since both triangles are the same, do you only solve for one? $11$ inches = $0.916667$ Feet

$\tan (a) = a / b$

$\tan (a) = 0.916667 / 87$

$a = \arctan (0.105364022989)$

$a = 0.6037^\circ$

gebruiker
  • 6,154
Madison
  • 395

1 Answers1

1

There is something not quite right about this question. If we assume that

  • you stand on the hog line
  • the line between you and the centre of the yellow stone is perpendicular to the hog line
  • the red stone doesn't curl,

then you cannot actually come to the situation in the picture. The red stone would have to pass throug the yellow one, which is not very realistic. Nevertheless, the picture still shows a right angled triangle, with certain side lengths, so we can say that in the picture $\theta =\arctan \left (\frac{0.916667}{87}\right )\approx0.6037$.

Now I would like to actually solve what the maximum value for $\theta$ can be, for the red stone to still hit the yellow one, under the same assumtions as listed above.

Now I will drop all units and introduce

  • $Y:=$ you (standing on the hog line)
  • $A:=$ centre of the yellow stone
  • $A':=$ centre of the red stone
  • $r:=$ radius of the stones we use
  • $d:=\vert YA\vert$
  • $m:= $ the line through $Y$ and $A$
  • $l:=$ a line through $Y$ with angle $\theta$ from $m$

  • $B:=$ point on $l$, such that $d(A,l)=d(A,B)$, implying $\angle YBA=90^\circ$. enter image description here We may obviously assume that $0\leq\theta < 90^\circ$. Note that $\sin (\theta)$ is stricly increasing on this interval.

    Now, as $A'$ moves along $l$, the two stones will collide, if and only if $\vert AB\vert\leq 2r$. This means that $$ \sin (\theta) =\frac{\vert AB\vert}d \leq\frac{2r}d\implies \theta \leq \arcsin\left( \frac{2r}d \right).$$ And of course, the complete answer will be $\left\vert\theta\right \vert\leq \arcsin\left(\frac {2r} d\right) $.

gebruiker
  • 6,154