An equilateral triangle is inscribed in the parabola x 2 = 8 y such that one of its vertices is at the vertex of this parabola.
The side length of the triangle can be written as q p ; where q and p are natural numbers and p is a prime. Find p + q .
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To get the side of the triangle we can also double the x coordinate of the point of intersection.
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Oh, right, because the side makes an angle of 60 degrees to the x-axis. Good point. :)
Just a small variation of @brian charlesworth 's good solution:
In the first quadrant, the squared length of the inscribed triangle's side can be expressed as
x 2 + ( 8 1 x ) 2
The squared length of the inscribed triangle's side crossing the y -axis can be expressed as
( 2 x ) 2
The triangle is equilateral, hence
x 2 + ( 8 1 x 2 ) 2 = ( 2 x ) 2 ⇒ x 2 ( ( 8 1 x ) 2 − 3 ) = 0 ⇒ x = 8 3
So the length of any side in the triangle is 2 x = 1 6 3 . This means p = 3 and q = 1 6 , and the solution to the problem is
p + q = 1 9
I prefer your approach; more concise and no trig. :)
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The equilateral triangle will be inverted and symmetric about the y -axis. The side in the first quadrant will thus make an angle of 6 0 degrees with the positive x -axis, and thus have the equation y = 3 x , (as it also has one end at the origin).
We then need to find where this side and the parabola intersect. This will occur when
3 x = 8 x 2 ⟹ x = 8 3 ,
where we have disregarded the trivial solution x = 0 .
Now for this value of x we have that the point of intersection is ( 8 3 , 2 4 ) , so the side length of the equilateral triangle is the distance between this point and the origin. This distance is
( 8 3 ) 2 + 2 4 2 = 1 6 3 .
Thus q = 1 6 , p = 3 and p + q = 1 9 .