Line $y=c$ where $c$ is a constant intersects $y=x^2$ with part of it reflected at $y=1$ at $A, B, C, D$ where $AB=BC=CD$. What is $AB$?

Question

An M-shaped curve is created by graphing the parabola $y=x^2$ in the coordinate plane, and then reflecting the part of the parabola that is above the line $y=1$ across the line $y=1$. There is a horizontal line that intersects the M-shaped curve at four points $A, B, C$, and $D$ so that $AB=BC=CD$. As a fraction in simplest radical form, what is the distance $\overline{AB}$?

I begin by drawing a diagram:

The M-shaped curve is traced in red, and the four yellow dots represent A, B, C, D, from left to right, respectively.

I find the reflected parabola's equation to be $y=-x^2+2$ because it is a parabola opening downward shifted 2 up.

So I can say for $|x|\le1$, $y=x^2$; for $|x|\ge1$, $y=-x^2+2$. So it's a piecewise function.

I am stuck here. I assume I need to find solutions for $x$ from both parabola's equations, but I am not sure how. Help is appreciated!

Thanks!

Max0815

Answer 1

HINT I would consider finding a straight line $$r : y = k$$ with $0<k<1$, such that the intersection between $r_1$ and $$\gamma_1 : y=x^2$$ generates a chord of some length $L$ and the intersection between $r$ and $$\gamma_2:y =-x^2+2$$ generates a chord of length $3L$.

Intersecting $r$ and $\gamma_1$ yields $$ \begin{cases} y = k\\ y=x^2 \end{cases} $$ and thus the chord lenght is $$L = 2\sqrt{k}.$$ Similarly, by the intersection between $r$ and $\gamma_2$ you get $$3L = 2\sqrt{2-k}.$$ Equating the last two expressions and solving for $k$ will bring you close to your final result.

Answer 2

Let the line that intersects the M-shaped curve to give points $A, B, C$ and $D$ be $y=c$ where $c$ is a constant.

We can see that points $A$ and $D$ are part of the parabola $y=-x^2+2$.
We can also see points $B$ and $C$ are part of the parabola $y=x^2$.

We can set equations now: $$c=-x^2+2$$$$\implies2-c=x^2$$$$\implies x=\pm\sqrt{2-c}$$

Thus, we can see that point $A$(the least greatest) has $x$ coordinate of $-\sqrt{2-c}$ and $D$(the greatest) has $x$ coordinate of $\sqrt{2-c}$.

$$c=x^2$$$$\implies x=\pm\sqrt{c}$$

Thus, we can see that point $B$(second to last least greatest) has $x$ coordinate of $-\sqrt{c}$ and point $C$ has $x$ coordinate of $\sqrt{c}$.

So, generalizing, we have $$A=-\sqrt{2-c}$$$$B=-\sqrt{c}$$$$C=\sqrt{c}$$$$D=\sqrt{2-c}$$

Now, we can set equations and solve for $c$. In order to avoid getting an equation with unlimited solutions, we put the "$\sqrt{c}$"s on one side, and the "$\sqrt{2-c}$"s on the other.

We can see that $$D-A=3(C-B)$$

Thus, $$\implies\sqrt{2-c}--\sqrt{2-c}=3(\sqrt{c}--\sqrt{c})$$$$\implies2\sqrt{2-c}=6\sqrt{c}$$$$\implies\sqrt{2-c}=3\sqrt{c}$$$$\implies 2-c=9c$$$$\implies 10c=2$$$$\implies c=\frac{1}{5}$$

Since we are trying to find the distance between two points, we will find the distance between points $C$ and $B$ and that will be our answer:

$$C-B=\text{answer}$$$$\implies 2\sqrt{c}=\text{answer}$$$$\implies 2\sqrt{\frac{1}{5}}=\text{answer}$$$$\implies \frac{2}{\sqrt{5}}=\text{answer}$$$$\implies \boxed{\frac{2\sqrt{5}}{5}}$$