Maths questions help please :(

**authenticity** · 21 Jun 2012, 12:47 AM

1. Ten equal circular discs can just be placed tightly within a frame in the form of an equilateral triangle. Nine of them touch the frame and the tenth is in the centre, with six other discs touch it. If the sides of the frame are each the length of R, express the radius of one disc in terms of R. Also show that the outside perimeter of the figure which remains when the frame is removed is $\frac{11}{12}(3-\sqrt{3})\pi R$

2. An open rectangular tank $\alpha$ units deep and $\beta$ units wide holds water and is tilted so that the base BC makes an angle θ with the horizontal. When BC is returned to the horizontal, show that the depth of the water is $\frac{\alpha ^{2}\cot \theta }{2b}$

Thanks in advance!

**Fus Ro Dah** · 21 Jun 2012, 2:07 AM

Very cool question! I really like the second one. Very original.

Will try tomorrow if I get time after school.

**SpiralFlex** · 21 Jun 2012, 2:57 AM

$1) Always draw a pretty diagram, I failed to draw a decent one on paper.....Also, let the mini-circle's radius be r and of course our equilateral triangle side length as mentioned is R.$

$Now, we need to manipulate these circles somehow, form a recognisable shape and see if we can find a relationship between r and R and we can only utilise the side lengths.$

$Whenever I see a circle and a tangent brushing it, there is almost always a right angle involved. We simply draw it through the centre, (circle geometry)$

$We now have this diagram,$

$Ahh, we have effectively made blocks of rectangular pieces, except for the two outer outliers which I labelled as x. So, the relationship between r and R is almost complete.$

$R=2r+2r+2r+x+x$

$=6r+2x....(1)$

$To find x, we need a bit of trigonometry, lets zoom in into a smaller sectional piece. VROOOM, I mean ZOOOM!!!$

$We can find x easily!$

$\tan 60 = \frac{x}{r}$

$x=r\sqrt{3}$

$From (1)$

$R=6r+2r\sqrt{3}....(1)$

$r=\frac{R}{6+2\sqrt{3}}...(2)$

$This is your new relationship, now we have to derive the equation given$

$When the frame is removed, your diagram looks something like this,$

$This is our so called perimeter, notice that I have not included the centre because it's excluded and feels a sense of not belonging kekeke.$

$Also, notice that 6 circles have got the same arc length. However three on the outside don't.$

$So our perimeter will consist of$

$P=3*(outside circles) + 6*(circle edges)$

$For the edges circle:$

$$This should be easy! Because it's a semi-circular arc, so that means, arc length =half the circumference of the circle. ie.$ \;\pi r.$

$For the circles on the corners. We can inscribe a equilateral triangle in it. So, we get 5/6 th of the circumference.$

$P=5\pi r +6\pi r$

$\therefore P=11\pi r$

$Still remember equation 2?$

$r=\frac{R}{6+2\sqrt{3}}...(2)$

$Multiply by 11 pie$

$11\pi r = \frac{11\pi R}{6+2\sqrt{3}}$

$P =\frac{11\pi R}{6+2\sqrt{3}}$

$P=\frac{11\pi R}{6+2\sqrt{3}}*\frac{6-2\sqrt{3}}{6-2\sqrt{3}}$

$=\frac{11 \pi R(6-2\sqrt{3})}{24}$

$\therefore P =\frac{11\pi}{12}(3-\sqrt{3})$

$\square$

$*Falls asleep* (Note I got Q2 first, Q1 took me longer.)$

$2) Let h be the depth of the water and W be the water content$

$W = \frac{1}{2}*\alpha * x....(1) \;$(Area of triangle)$$

$When the water is laid horizontally, the water content remains the same but we get a rectangular shape as BC returns to its horizontal position$

$W=h*\beta ....(2)\;$(Area of a rectangle)$$

$By equating (1) and (2)$

$\frac{1}{2}\alpha x=h \beta$

$h=\frac{\alpha x}{2\beta}....(3)$

$Clearly, we need to introduce the cotangent function, so to do this it is obvious we need to do some simple trigonometry.$

$Let O be the horizontal line through A cutting BC.$

$$In$\; \triangle AOB$

$\angle AOB =\theta$

$\cot \theta=\frac{x}{\alpha}$

$\therefore x=\frac{\cot \theta}{\alpha}$

$$Hence from$\; (3)$

$h=\frac{\alpha^2 \cot \theta}{2\beta}$