Monthly Archives: September 2024

September 29, 2024 · 10:18 am

Intersecting Circles

Two circles of radius L and 2L intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}

cos\theta=\frac{5L^2}{4\sqrt{2}L^2}

cos\theta=\frac{5}{4\sqrt{2}}

\theta=0.487

cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}

cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}

cos\alpha=\frac{11}{8\sqrt{2}}

\alpha=0.236

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are 2\theta and 2\alpha (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

A=\frac{1}{2}r^2(\theta-sin\theta)

A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2

Area of yellow segment

A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035

Total Area =0.0733L^2+0.0035L^2=0.108L^2

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September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

  • Find the function to optimise (in terms of one variable).
  • Find the stationary points.
  • Find the nature of the stationary points.
  • Find the maximum or minimum.
(a) Volume of the cylinder V=\pi r^2h
42=2r+h
h=42-2r
\therefore V_C=\pi r^2(42-2r)
Volume of spherical decorations V_S=\frac{4}{3}\pi( r_s)^3 where r_s=\frac{r}{3}
V_S=\frac{4\pi r^3}{81}
Volume unused space V=\pi r^2(42-2r)-20(\frac{4\pi r^3}{81})
V=2\pi (21r^2-r^3-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{81r^3}{81}-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{121r^3}{81})

(b) V=2\pi (21r^2-\frac{121r^3}{81})
\frac{dV}{dr}=2\pi (42r-\frac{121r^2}{27})
\frac{dV}{dr}=0
0=42r-\frac{121r^2}{27}
0=r(42-\frac{121r}{27})
r=0 or r=\frac{1134}{121}=9.372

\frac{d^2V}{dr^2}=2\pi (42-\frac{242r}{27})
(\frac{d^2V}{dr^2})_{|r=9.372}=-42
\therefore r=9.372 is a maximum.

Dimensions of the vase, internal diameter=18.7cm internal height=23.3cm

(c) Maximum volume of empty space =2\pi (21r^2-\frac{121r^3}{81})=3863.08cm^3
Volume of one sphere =\frac{4}{3}\pi r^3=3448.03cm^3

There is enough unused space for one extra decoration, but it would depend on how they are packed.

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Filed under Differentiation, Optimisation, Year 12 Mathematical Methods

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September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let x be the number of cats and y be the number of dogs.
Then 0.9x+0.1y think they are cats.
But we also know 20% of the total think they are cats.
0.2(x+y)
Therefore, 0.9x+0.1y=0.2(x+y)
0.9x+0.1y=0.2x+0.2y
0.7x=0.1y
7x=y
Percentage of cats is \frac{x}{x+y}\times100
Substitute 7x for y
\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%
\therefore 12.5% of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

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September 19, 2024 · 9:43 am

Solving Cubic Functions

I have been thinking about cubics a bit lately because some of my students are solving and then sketching cubics. Plus I am reading An Imaginary Tale by Paul Nahin, which talks about solving cubics and complex numbers.

Cubics must have at least one real root. If one of the roots is a rational number, then we can use the Factor and Remainder Theorem.

For example,

Solve 2x^3-3x^2-3x+2=0

We know the root(s) must be a factor of 2\times3=6.
I always start with 1 or -1
2(1)^3-3(1)^2-3(1)+2=2-3-3+2=-2 \therefore x\neq=1
Try -1
2(-1)^3-3(-1)^2-3(-1)+2=-2-3+3+2=0 \therefore x=-1 and (x+1) is a factor.
Then we can do polynomial long division.

Now we know that 2x^3-3x^2-3x+2=(x+1)(2x^2-5x+2)
And we can factorise the quadratic (or using the quadratic equation formula)
2x^2-5x+2=2x^2-4x-x+2
=2x(x-2)-1(x-2)
=(2x-1)(x-2)
\therefore x=-1, \frac{1}{2}, 2

But what if it is not factorisable?

For example,

Solve 2x^3+5x^2-2x+4=0

How many roots does this equation have?

We could find the derivative and find out how many stationary points the function has.

f'(x)=6x^2+10x-2

This is a quadratic function. Find the discriminant to determine the number of roots.

\Delta=b^2-4ac=100-4(6)(-2)=148

As \Delta>0, there are two stationary points, which means we could have 1, 2 (one root is repeated) or three roots, depending on if the function crosses the x-axis between stationary points. So not much use.

We could try the discriminant of a cubic.

\Delta=18abcd-4b^3d+b^2c^2-4ac^2-27a^2d^2

\Delta=18(2)(5)(-2)(4)-4(5^3)(4)+(5^2)(-2)^2-4(2)(-2)^2-27(2)^2(4)^2=-5100

The discriminant is negative so there is one real root.

From my reading, we need to turn the cubic into a depressed cubic (cubics of the form x^3+px+q=0).

We can do this by using a change of variable.

Let x=t-\frac{5}{6}
2(x^3+\frac{5}{2}x^2-x+2)=0
\therefore x^3+\frac{5}{2}x^2-x+2=0
Substitute t-\frac{5}{6} into the cubic.
(t-\frac{5}{6})^3+\frac{5}{2}(t-\frac{5}{6})^2-(t-\frac{5}{6})+2
(t^3-3(\frac{5}{6})t^2+3(\frac{5}{6})^2t-(\frac{5}{6})^3+\frac{5}{2}(t^2-2(\frac{5}{6})t+(\frac{5}{6})^2)-t+\frac{5}{6}+2
t^3-\frac{5}{2}t^2+\frac{25}{12}t-\frac{125}{216}+\frac{5}{2}t^2-\frac{25}{6}t+\frac{125}{72}-t+\frac{5}{6}+2
t^3-\frac{37}{12}t+\frac{431}{108}

We can then use Cardano’s formula

x=\sqrt[3]{-\frac{q}{2}+\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}

p=-\frac{37}{12} and q=\frac{431}{108}
\frac{p}{3}=\frac{-37}{36}
\frac{q}{2}=\frac{431}{216}
(\frac{431}{216})^2+(\frac{-37}{36})^3=\frac{139}{48}
t=\sqrt[3]{\frac{-431}{216}+\sqrt{\frac{139}{48}}}+\sqrt[3]{\frac{-431}{216}-\sqrt{\frac{139}{48}}}
t=-2.21095...
\therefore x=-2.21095...-\frac{5}{6}=-3.044

We can see from the sketch below that there is only one solution and it is about -3.

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September 14, 2024 · 9:45 am

Sum and Product of the Roots of Polynomials

There is a relationship between the sum and product of the roots of a polynomial and the co-efficient of the polynomial.

Let’s start with a quadratic.

The general form for a quadratic (polynomial of degree 2) is

y=ax^2+bx+c

Use the quadratic equation formula to find the roots

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Hence the roots are

x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} and x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

Sum of the roots:

\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}

Product of the roots:

\frac{-b+\sqrt{b^2-4ac}}{2a}\times\frac{-b-\sqrt{b^2-4ac}}{2a}

\frac{b^2}{4a^2}-\frac{b^2-4ac}{4a^2}

\frac{4ac}{4a^2}

\frac{a}{c}

Worked Example
The equation 4x^2+bx+c=0 has two distinct roots. The product of the roots is \frac{3}{4} and the sum is 2. Find b and c.
\frac{-b}{a}=2
a=4
\frac{-b}{4}=2
b=-8

\frac{c}{a}=\frac{3}{4}
\frac{c}{4}=\frac{3}{3}
c=3
The equations is 4x^2-8x+3

Solve the equation to prove the roots do in fact sum to 2 and multiply to \frac{3}{4}
4x^2-8x+3
4x^2-6x-2x+3
2x(2x-3)-1(2x-3)
(2x-1)(2x-3)
x_1=\frac{1}{2} and x_2=\frac{3}{2}
\frac{1}{2}+\frac{3}{2}=2 and \frac{1}{2}\times\frac{3}{2}=\frac{3}{4}

Let’s move to a cubic function.

The general equation is f(x)=ax^3+bx^2+cx+d

Let’s say the roots of this cubic are \alpha, \beta, \gamma

Then ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)

=a(x^2-\beta x - \alpha x+\alpha\beta)(x-\gamma)

=a(x^2-(\alpha+\beta)x+\alpha\beta)(x-\gamma)

=a(x^3-\gamma x^2-x^2(\alpha+\beta)+\gamma(\alpha+\beta)x-\alpha\beta\gamma)

=a(x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma)

The sum of the roots

\alpha+\beta+\gamma=\frac{-b}{a}

The product of the roots

\alpha\beta\gamma=\frac{-d}{a}

Also, it can be handy to know

\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}

Worked example
f(x)=x^3-6x^2+4x+12, the roots are \alpha, \beta and \gamma
Find
(a) \alpha+\beta+\gamma
(b) \alpha^2+\beta^2+\gamma^2
(c) \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}

(a) \alpha+\beta+\gamma=\frac{-b}{a}
\alpha+\beta+\gamma=\frac{6}{1}
\alpha+\beta+\gamma=6

(b)\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2\alpha\beta-2\alpha\gamma-2\beta\gamma
=6^2-2(4)
=28

(c)\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta\gamma}{\alpha\beta\gamma}+\frac{\alpha\gamma}{\alpha\beta\gamma}+\frac{\alpha\beta}{\alpha\beta\gamma}
=\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}
=\frac{4}{-13}
=\frac{-1}{3}

We can extend the method we used for finding the sum and product of the roots of cubic to polynomials of greater degree.

If the four roots of a quartic are \alpha, \beta, \gamma and \delta, and the general equation is ax^4+bx^3+cx^2+dx+e, then

\alpha+\beta+\gamma+\delta=\frac{-b}{a}

\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}

\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\frac{-d}{a}

\alpha\beta\gamma\delta=\frac{e}{a}

Worked Example (just one more)
The roots of the cubic equation x^3-4x^2-3x-2 are \alpha, \beta and \gamma. Find the cubic equation whose roots are \alpha+\beta, \alpha+\gamma, and \beta+\gamma

\alpha+\beta+\gamma=4
\alpha\beta+\alpha\gamma+\beta\gamma=-3
\alpha\beta\gamma=2

\frac{-b}{a}=\alpha+\beta+\alpha+\gamma+\beta+\gamma
\frac{-b}{a}=2(\alpha+\beta+\gamma
\frac{-b}{a}=2(4)
\frac{-b}{a}=8

\frac{c}{a}=(\alpha+\beta)(\alpha+\gamma)+(\alpha+\beta)(\beta+\gamma)+(\alpha+\gamma)(\beta+\gamma)
=\alpha^2+\alpha\gamma+\beta\gamma+\gamma^2+\alpha\beta+\alpha\gamma+\beta^2+\beta\gamma+\alpha\beta+\alpha\gamma+\gamma\beta+\gamma^2
=\alpha^2+\beta^2+\gamma^2+3\alpha\gamma+3\beta\gamma+3\alpha\beta
=(\alpha+\beta+\gamma)^2-2\alpha\gamma-2\beta\gamma-2\alpha\beta+3\alpha\gamma+3\beta\gamma+3\alpha\beta
=(\alpha+\beta+\gamma)^2+\alpha\gamma+\alpha\beta+\beta\gamma
=4^2-3
\frac{c}{a}=13

\frac{-d}{a}=(\alpha+\beta)(\alpha+\gamma)(\beta+\gamma)
=(\alpha^2+\alpha\gamma+\beta\alpha+\beta\gamma)(\beta+\gamma)
=\alpha^2\beta+\alpha^2\gamma+\alpha\gamma\beta+\alpha\gamma^2+\beta^2\alpha+\beta\alpha\gamma+\beta^2\gamma+\beta\gamma^2
=2\alpha\beta\gamma+\alpha^2\beta+\beta^2\alpha+\alpha^2\gamma+\gamma^2\alpha+\beta^2\gamma+\gamma^2\beta
=2(2)+\alpha\beta(\alpha+\beta)+\alpha\gamma(\alpha+\gamma)+\beta\gamma(\beta+\gamma)
=-24+\alpha\beta(\alpha+\beta+\gamma)-\alpha\beta\gamma+\alpha\gamma(\alpha+\gamma+\beta)-\alpha\beta\gamma+\beta\gamma(\beta+\gamma+\alpha)-\beta\gamma\alpha
=4+(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-3(2)
=-2+(4)(-3)
\frac{-d}{a}=-14

If a=1 then b=-8, c=13 and d=14
The cubic is x^3-8x^2+13x+14

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September 7, 2024 · 9:31 am

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at O and P, we know there is a circle, which has the diagonal of the square as its diameter (see second image below).

\angle{RSP} is 45^{\circ} (Angle between the diagonal and side of a square)

PORS is a cyclic quadrilateral.

In cyclic quadrilaterals opposite angles are supplementary.

Hence, \angle{ROP}=180^{\circ}-45^{\circ}=135^{\circ}

As \angle{ROS}=90^{\circ}, \angle{SOP} must be 45^{\circ}

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