Preliminary mathematics marathon (1 Viewer)

AAEldar · Apr 13, 2010

hscishard said:
Yea..why did you repear so many times?

It is right however

My bad, it screwed up when I first tried to put it in, had url in there for some reason :S Fixed it now

Good to know it's right, was confused for a sec there hahaha.

Gussy Booo · Apr 13, 2010

hscishard said:
For Q2.
It's funny.
For the first part A and B could be 0
But for the Tan part, A can be 60 and B can be 30. Or 0

But they have to satisfy both equations.

Also, an angle of 0, is not an acute angle

hscishard · Apr 13, 2010

I know gay.

hscishard · Apr 14, 2010

Can someone do Q23 Chp 14B in Cambridge 3unit, Year 11.
Can't type this question up.
I don't know how to find the perimeter of the bigger hexagon.

Drongoski · Apr 14, 2010

hscishard - pl type out question at leat for benefit of readers

$\100dpi \small internal\ regular\ hexagon\ is\ made\ up\ of\ 6\ equilateral\ triangles\ of\ side\ =\ r\\ \\ By\ dropping\ perp\ bisector\ of\ a \ typical\ inner\ equilateral\ triangle\ from\ centre\ of\ circle\\ \\ as\ well\ as\ the\ associated\ outer\ equilateral\ triangle\ the\ length\ of\ inner\ perp\\ \\ =\ r\ cos30^{\circ}\ =\ \frac{\sqrt{3}r}{2}\ \ and\ \ for\ outer\ triangle\ =\ r.\ \ By\ similar\ triangles\ and\ proportionality,\ outer\ \ triangles\\ \\ have\ sides\ =\ \frac{2r}{\sqrt{3}}. The\ perimeters\ of\ inner\ and\ outer\ hexagons\ =\ 6\times r\\ \\ and\ \ 6\times \frac{2r}{\sqrt{3}}\ resp.\ \ The\ perimeter(i.e.\ circumference)\ of\ the\ inclosed\ circle\ =\ 2\pi r\\ \\ and\ \ \ \ 6r\ \ < \ \ 2\pi r\ \ \ <\ \frac{12r}{\sqrt{3}} \ \ \ \ \Rightarrow \ \ \ \ 3\ \ \ < \ \ \pi \ \ <\ \ 2\sqrt{3}\\ \\ QED$

hscishard · Apr 14, 2010

Drongoski said:
hscishard - pl type out question at leat for benefit of readers

$\100dpi \small internal\ regular\ hexagon\ is\ made\ up\ of\ 6\ equilateral\ triangles\ of\ side\ =\ r\\ \\ By\ dropping\ perp\ bisector\ of\ a \ typical\ inner\ equilateral\ triangle\ from\ centre\ of\ circle\\ \\ as\ well\ as\ the\ associated\ outer\ equilateral\ triangle\ the\ length\ of\ inner\ perp\\ \\ =\ r\ cos30^{\circ}\ =\ \frac{\sqrt{3}r}{2}\ \ and\ \ for\ outer\ triangle\ =\ r.\ \ By\ similar\ triangles\ and\ proportionality,\ outer\ \ triangles\\ \\ have\ sides\ =\ \frac{2r}{\sqrt{3}}. The\ perimeters\ of\ inner\ and\ outer\ hexagons\ =\ 6\times r\\ \\ and\ \ 6\times \frac{2r}{\sqrt{3}}\ resp.\ \ The\ perimeter(i.e.\ circumference)\ of\ the\ inclosed\ circle\ =\ 2\pi r\\ \\ and\ \ \ \ 6r\ \ < \ \ 2\pi r\ \ \ <\ \frac{12r}{\sqrt{3}} \ \ \ \ \Rightarrow \ \ \ \ 3\ \ \ < \ \ \pi \ \ <\ \ 2\sqrt{3}\\ \\ QED$

Thx man. Yea, I don't know how to post pics on BOS. Teach me that too

.

ohexploitable · Apr 14, 2010

You can use image tags or just type it up.

hscishard · Apr 14, 2010

Where are the image tags. I need to post a pic

ohexploitable · Apr 14, 2010

Insert the link between

.

Alternatively, just click "Go Advanced" and click on the insert image button.

hscishard · Apr 14, 2010

WHat if it's not on the internet...I have it as a file, but when I click upload i get the error page

ohexploitable · Apr 14, 2010

Upload it elsewhere.

I suggest Welcome to whirltools image hosting -

hscishard · Apr 14, 2010

Woohoo!

Drongoski · Apr 14, 2010

hscishard said:
Thx man. Yea, I don't know how to post pics on BOS. Teach me that too .

If I knew how to post figures, diagrams, graphs . . . I wouldn't have posted such a long-winded solution.

ohexploitable · Apr 16, 2010

I don't want this thread to die

New Question:

A piece of wire is 70 cm long. One section is bent into an equilateral triangle and the other into a rectangle that is three times as long as it is wide. Find the lengths of the two pieces of wire if the sum of the areas of the two shapes is a minimum.

hscishard · Apr 17, 2010

Sorry exploit, I got a trig question. How can you show in b i) without using the double angle forumale?

hscishard · Apr 17, 2010

ohexploitable said:
I don't want this thread to die

New Question:

A piece of wire is 70 cm long. One section is bent into an equilateral triangle and the other into a rectangle that is three times as long as it is wide. Find the lengths of the two pieces of wire if the sum of the areas of the two shapes is a minimum.

Wth..I got 34.54 and 35.46.

ohexploitable · Apr 17, 2010

$\begin{align*} \text{Let the sides of the triangle = }x \\\text{Let the shorter sides of the rectangle = }y \end{align*}$

$\begin{align*} \therefore 3x+8y&=70 \\y&=\frac{70-3x}{8} \end{align*}$

$\begin{align*} A_{tri}&=\frac{\sqrt{3}x^{2}}{4} \\A_{rect}&=\frac{3(70-3x)}{64} \\A&=\frac{\sqrt{3}x^{2}}{4}+\frac{3(70-3x)}{64} \end{align*}$

Then make it into a general quadratic expression and find the minimum value of the parabola.

hscishard · Apr 17, 2010

ohexploitable said:
$\begin{align*} \text{Let the sides of the triangle = }x \\\text{Let the shorter sides of the rectangle = }y \end{align*}$

$\begin{align*} \therefore 3x+8y&=70 \\y&=\frac{70-3x}{8} \end{align*}$

$\begin{align*} A_{tri}&=\frac{\sqrt{3}x^{2}}{4} \\A_{rect}&=\frac{3(70-3x)}{64} \\A&=\frac{\sqrt{3}x^{2}}{4}+\frac{3(70-3x)}{64} \end{align*}$

Then make it into a general quadratic expression and find the minimum value of the parabola.

I did it in a similar way. Except I let x = perimeter of triangle.
Therefore each side was x/3.

Not sure which is more complicated.
But I didn't get
$\frac{x^{2}\sqrt{3}}{4}$
From the way you did it the 4 should be a 2.

I somehow got this for that.
$\frac{x^{2}\sqrt{3}}{36}$

EDIT: Wait no I didn't I had 3(70-x)^2/64. ?????

ohexploitable · Apr 17, 2010

A = 1/2 ab sin C
A = x²/2 * √3/2
A = x²√3/4

hscishard · Apr 17, 2010

ohexploitable said:
A = 1/2 ab sin C
A = x²/2 * √3/2
A = x²√3/4

What the hell. Then where is my flaw?

$(\frac{x}{3})^{2} - (\frac{x}{6})^{2} = c^{2}$
$\frac{x^{2}}{12} = c^{2}$
$c = \frac{x}{2\sqrt{3}}$
$Area = (\frac{x}{2\sqrt{3}})(x/6)$
$= x^{2}/12\sqrt{3} = x^2\sqrt{3}/36$
Latex takes.. aggeeessssss

Preliminary mathematics marathon (1 Viewer)

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