Dynamic Geometry: P7

Geometry Level 4

The diagram shows a black parabola y = x 2 y=x^{2} and a blue equilateral triangle moving so that it's tangent to the parabola and to the line y = 0 y=0 . As the triangle is moving, its center (cyan) is following the path of a pink curve. At some moment, the sum of the coordinates of the triangle's center is equal to 2 + 2 3 2+2\sqrt{3} . At this moment the perimeter of the triangle can be expressed as a b a\sqrt{b} where a a and b b are positive integers, and b b is square-free. Find a + b a+b .


The answer is 9.

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1 solution

K T
Feb 2, 2021

Let ( x , y ) (x, y) be a point on the parabola. The centre of the triangle is given by ( x + 1 2 s , 2 3 y ) (x+\frac12 s, \frac23 y) where s s is the side of the triangle. We have y = x 2 y=x^2 and s = 2 3 3 y s=\frac23 \sqrt{3} y . The condition becomes x + 1 3 3 x 2 + 2 3 x 2 = 2 + 2 3 x+\frac13 \sqrt{3} x^2+ \frac23 x^2=2+2\sqrt{3} which is met when x = 3 x=\sqrt{3} . Then p = 3 s = 6 3 p=3s=6\sqrt{3} so submit the answer 6 + 3 = 9 6+3=\boxed{9}

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