Beautiful Curve Stitching

Part of the solution is given in the problem since we are told that the shape that emerges from all those tangent is a parabola. The problem is that the different lines we are given are tangents to the parabola, but we don't know where is the point these line are tangent to.

The first step is to consider the full family of lines which go through the points $(t,0)$ and $(0,10-t)$ (where $t$ is any real number). These lines are in symmetry with the 45° in this system. The parabola also has one line of symmetry. Thanks to this we conclude $(\tfrac{5}{2},\tfrac{5}{2})$ is the vertex of the parabola.

Next because it's a parabola, the tangent line only touch the graph (and do not cross it). This will give us a second point on the parabola. Indeed for $t$ close to 0 but positive, the lines have negative slope while for $t$ close to 0 but positive they have positive slope. Furthermore, these line all intersect very close to the point $(0,10)$ . This means that $(0,10)$ is also on the parabola (see below for details). By symmetry $(10,0)$ is also on the parabola.

So we have the vertex at $(\tfrac{5}{2},\tfrac{5}{2})$ and another point at $(10,0)$ . We just need to rotate (so that the symmetry axis is vertical) and translate (so that the vertex is at $(0,0)$ ) . $\begin{array}{c} (\tfrac{5}{2},\tfrac{5}{2}) \\ (10,0) \end{array} \overset{\text{rotate by 45°}}\to \begin{array}{c} (0,\tfrac{5}{2} \sqrt{2}) \\ (5 \sqrt{2},5 \sqrt{2}) \end{array} \overset{\text{translate down by } \tfrac{5}{2} \sqrt{2} }{\to} \begin{array}{c} (0,0) \\ (5 \sqrt{2}, \tfrac{5}{2} \sqrt{2}) \end{array}$ So we have a point on the end-parabola: $x = 5 \sqrt{2}, y= \tfrac{5}{2} \sqrt{2}$ . Pluging in will show that the equation for the parabola is $y = \frac{\sqrt{2}}{20} x^2$ so that $a = 2$ and $b = 20$ . The answer is thus $\fbox{200}$ .

How do we know it's a parabola?

The idea is already given above: we have a family of lines between the points $(t,0)$ and $(0,10-t)$ . Say $L_t$ is this line, then the equation for $L_t$ is $y = -\frac{10-t}{t} x + (10-t)$ . Note that any two such lines have exactly one intersection point. Want to know which point on $L_t$ is on the parabola, then just look at the point $P_{t,\epsilon}$ which is at the intersection of $L_t$ and $L_{t+\epsilon}$ . Then as $\epsilon \to 0$ , $P_{t,\epsilon}$ will converge to a point $\bar{P}_t$ on $L_t$ and this point will be on the curve. You then get a parametrisation $t\mapsto \bar{P}_t$ and you can check that this parametrises a parabola. Caution: the parametrisation can be singular; here for example the equation for $L_t$ is already singular when $t=0$ .

As an example, for $t=10$ , $L_{10}$ is the line $y =0$ and $L_{10+ \epsilon}$ is $y = \frac{\epsilon}{10+\epsilon} x - \epsilon$ . The intersection of $L_{10}$ and $L_{10+ \epsilon}$ has a $y$ -coordinate which is obviously 0 and the $x$ -coordinate is obtained by solving $0 = \frac{\epsilon}{10+\epsilon} x - \epsilon$ . This gives $x=10+\epsilon$ . Hence $P_{10,\epsilon} = (10+\epsilon,0)$ . Letting $\epsilon \to 0$ shows that the point $(10,0)$ was on the curve.

From there to enveloppes... [an edit motivated by Julian Poon's comment]

In general, you have a familiy of curve $C_t$ given as the solution of an equation $f(t,x,y) =0$ (in this problem $C_t$ are the lines $L_t$ and $f(t,x,y) = -y -\frac{10-t}{t} x + (10-t)$ ). The intersection point of $C_t$ and $C_{t+\epsilon}$ will satisfy the equations $f(t,x,y) = 0 \text{ and } f(t+\epsilon,x,y) = 0$ or equivalently $f(t,x,y) = 0 \text{ and } f(t+\epsilon,x,y) - f(t,x,y)= 0$ Now the equation on the right will lose meaning if $\epsilon \to 0$ , which is a bugger because [since we have three variables] we need two independent equations to get a curve. So the trick is to rewrite it as $f(t,x,y) = 0 \text{ and } \frac{f(t+\epsilon,x,y) - f(t,x,y)}{\epsilon}= 0$ If $f(t,x,y)$ is differentiable in $t$ , then as $\epsilon \to 0$ you get two independent equations (hurray!): $f(t,x,y) = 0 \text{ and } \frac{ \mathrm{d}}{\mathrm{d}t} f(t,x,y) = 0$ See Mark Henning's answer for the implementation of this solution method.

We want to identify the envelope of the family of straight lines $(10-u)x + uy \;= \; u(10-u) \hspace{2cm} 0 < u < 10$ To do this we differentiate partially with respect to $u$ $y - x \; = \; 10 - 2u$ and eliminate $u$ from the two equations, obtaining $\begin{aligned} \frac{10-x+y}{2}x + \frac{10+x-y}{2}y & = \; \frac{(10+x-y)(10-x+y)}{4} \\ 5(x+y) - \tfrac12(x-y)^2 & = \; 25 - \tfrac14(x-y)^2 \\ 5(x+y) & = \; 25 + \tfrac14(x-y)^2 \end{aligned}$ which becomes $Y \; = \; \tfrac{5}{\sqrt{2}} + \tfrac{\sqrt{2}}{20}X^2$ where $X \; = \; \tfrac{x-y}{\sqrt{2}} \hspace{1cm} Y \; = \; \tfrac{x+y}{\sqrt{2}}$ In other words, we obtain the parabola $y \; = \; \tfrac{\sqrt{2}}{20}x^2$ after rotation about the origin through $45^\circ$ and a further translation. Thus $a=2$ , $b=20$ , so that $\tfrac{b^2}{a} = \boxed{200}$ .

It is enough to know the slope and point of tangency to the smooth curve of one line. You can use that along with the derivation of sqrt(a)/b x^2 to get the solution. I used the line connecting 0 and 10, giving me that the derivation of sqrt(a)/b x^2 at x=10/sqrt(2) has to be 1.

(1) Take line x+y=5 and it's perpendicular distance from point (0,10)=5/√2 as "y" value in parabola y=(√a/b)x^2

(2) Find distance of point (5/2,5/2) along the line x+y=5, where, the perpendicular from point (0,10) cuts x+y=5 as equal to 5√2. Take this as the "x" value in the parabola y=(√a/b)x^2.

(3) Using (5/√2)=(√a/b){(5√2)^2}, we get the answer for b^2/a=200.

Answer=200