# Puzzles

## 10 December

For all values of \(x\), the function \(f(x)=ax+b\) satisfies

$$8x-8-x^2\leqslant f(x)\leqslant x^2.$$

What is \(f(65)\)?

Edit: The left-hand quadratic originally said \(8-8x-x^2\). This was a typo and has now been corrected.

#### Show answer

#### Hide answer

If we plot the two quadratics, we see that they meet at \(x=2\) where they both have gradient 4.

The line \(f(x)\) must therefore pass through the point \((2,4)\) and have gradient 4, so its equation is \(f(x)=4x-4\).

This means that \(f(65)\) is **256**.

## 6 December

\(p(x)\) is a quadratic with real coefficients. For all real numbers \(x\),

$$x^2+4x+14\leq p(x)\leq 2x^2+8x+18$$

\(p(2)=34\). What is \(p(6)\)?

## Two tangents

Find a line which is tangent to the curve \(y=x^4-4x^3\) at 2 points.

#### Show answer

#### Hide answer

At \(x=a\), \(y=a^4-4a^2\) and \(\frac{dy}{dx}=4a^3-12a^2\). Therefore the equation of the tangent at \(x=a\) will be \(y=(4a^3-12a^2)x+8a^3-3a^4\).

Taking this away from \(y=x^4-4x^3\) gives \(y=x^4-4x^3-(4a^3-12a^2)x-8a^3+3a^4\). We can now look at where this curve is tangent to \(y=0\) and look for a value of \(a\) that makes it tangent at two points.

If this curve is tangent to the \(x\)-axis at \(x=b\), then it will have a repeated root at \(x=b\). We know it is tangent at \(x=a\), so dividing \(x^4-4x^3-(4a^3-12a^2)x-8a^3+3a^4\) by \((x-a)\) twice gives \(x^2+(2a-4)x+3a^2-8a\). We want this to have a repeated root, hence the discriminant, \((2a-4)^2-4(3a^2-8a)\), must be 0.

Solving this gives \(a=1\pm\sqrt3\).

Therefore the equation of the line is \(y=-8x-4\).

## Between quadratics

\(p(x)\) is a quadratic polynomial with real coefficients. For all real numbers \(x\),

$$x^2-2x+2\leq p(x)\leq 2x^2-4x+3$$

\(p(11)=181\). Find \(p(16)\).

#### Show answer

#### Hide answer

$$x^2-2x+2=(x-1)^2+1$$
$$2x^2-4x+3=2(x-1)^2+1$$

Therefore the minimum point of both of these quadratics is \((1,1)\). \(p(x)\) will only be between these if:

$$p(x)=a(x-1)^2+1\quad\text{where }1\leq a\leq 2$$

We know that \(p(11)=181\), so:

$$\begin{array}{rl}
181&=p(11)\\
&=a(11-1)^2+1\\
&=100a+1
\end{array}$$

Therefore \(a=1.8\). This means that:

$$\begin{array}{rl}
p(16)&=1.8(16-1)^2+1\\
&=1.8\times225+1\\
&=406
\end{array}$$

## Bézier curve

1) A set of points \(P_0\), ..., \(P_n\) are chosen (in the example \(n=4\)).

2) A set of points \(Q_0\), ..., \(Q_{n-1}\) are defined by \(Q_i=t P_{i+1}+(1-t) P_i\) (shown in green).

3) A set of points \(R_0\), ..., \(R_{n-2}\) are defined by \(R_i=t Q_{i+1}+(1-t) Q_i\) (shown in blue).

.

.

.

\(n\)) After repeating the process \(n\) times, there will be one point. The Bézier curve is the path traced by this point at \(t\) varies between 0 and 1.

What is the Cartesian equation of the curve formed when:

$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$

#### Show answer & extension

#### Hide answer & extension

If

$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$

then

$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$
$$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$

and so

$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$

which means that

$$x=t^2$$$$y=(1-t)^2$$

or

$$y=(1-\sqrt{x})^2$$

#### Extension

What should \(P_0\), \(P_1\) and \(P_2\) be to get a curve with Cartesian equation

$$y=(1-\sqrt{2x})^2$$

## Parabola

On a graph of \(y=x^2\), two lines are drawn at \(x=a\) and \(x=-b\) (for \(a,b>0\). The points where these lines intersect the parabola are connected.

What is the y-coordinate of the point where this line intersects the y-axis?

#### Show answer & extension

#### Hide answer & extension

The co-ordinates of the points where the lines intersect the parabola are \((a,a^2)\) and \((-b,b^2)\). Hence the gradient of the line between them is:

$$\frac{a^2-b^2}{a-(-b)}=\frac{(a+b)(a-b)}{a+b}=a-b$$

Therefore the y-coordinate is:

$$b^2 + b(a-b) = ba$$

Ferdinand Möbius, who discovered this property called the curve a *Multiplicationsmaschine* or 'multiplication machine' as it could be used to perform multiplication.

#### Extension

How could you use the graph of \(y=x^2\) to divide 100 by 7?

## Two lines

Let A and B be two straight lines such that the gradient of A is the y-intercept of B and the y-intercept of A is the gradient of B (the gradient and y-intercept of A are not the same). What are the co-ordinates of the point where the lines meet?

#### Show answer & extension

#### Hide answer & extension

Let A have the equation \(y = mx + c\). B will have the equation \(y = cx + m\).

Therefore, \(mx + c = cx + m\).

Which rearranges to \(x(m - c) = m - c.\)

So \(x = 1\).

Substituting back in, we find \(y=m+c\).

The co-ordinates of the point of intersection are \((1,m+c)\).

#### Extension

Let \(a\), \(b\) and \(c\) be three distinct numbers. What can you say about the points of intersection of the parabolas:

$$y = ax^2 + bx + c\mathrm{,}\\
y = bx^2 + cx + a\mathrm{,}\\
\mathrm{and\ }y = cx^2 + ax + b$$