Dynamic Geometry: P7

Geometry Level 4

The diagram shows a black parabola $y=x^{2}$ and a blue equilateral triangle moving so that it's tangent to the parabola and to the line $y=0$ . 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 $2+2\sqrt{3}$ . At this moment the perimeter of the triangle can be expressed as $a\sqrt{b}$ where $a$ and $b$ are positive integers, and $b$ is square-free. Find $a+b$ .

The answer is 9.

1 solution

K T
Feb 2, 2021

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

Dynamic Geometry: P7

The answer is 9.

1 solution

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