# Last maths question in the Oxford Physics Aptitude Test

• Nov 10th 2008, 09:33 AM
AAKhan07
Last maths question in the Oxford Physics Aptitude Test
http://i299.photobucket.com/albums/m...07/Q12jpeg.jpgHi everyone, I know it doesn't look it but AB = BC = CD. The triangle is an equilateral and you need to find the ratio of the area of the triangle to the area of the 2/3 ring sorrounding it.

On the Oxford Physics Aptitude Test, this was the last question in the maths section.

I got
$
\frac{9\sqrt{3}}{8\pi(3+2\sqrt{3})}
$
• Nov 10th 2008, 09:47 AM
Opalg
Is √2 a typo for √3? I get the answer as $\frac{9\sqrt{3}}{8\pi(3+2\sqrt{3})}$.
• Nov 10th 2008, 09:56 AM
AAKhan07
... yes! You're right, I meant to write $\sqrt{3}$, I sure hope I didn't make the same mistake in the exam... it's doubtful that I did that. I'll edit the first post to fix the typo.

Anyway, thanks for confirming, I just wanted to know whether I had it right, it was an 8 mark question!

Thanks,

KHAN
• Nov 21st 2008, 08:00 PM
David24
Hey mate,

just wondering, do you know whether your solution is correct? (no offence intended) I have a slightly different solution to yours

Cheers,

David
• Nov 21st 2008, 08:16 PM
whipflip15
I got the same answer also. I like the question. Why is there a circle inside the triangle?
• Nov 22nd 2008, 04:57 AM
David24
Hi guys,

As I was saying before I get a slightly different answer to everyone else, I have attached my working? If someone gets a chance to read it could you please let me know where I am going wrong,

Thanks,

David
• Nov 22nd 2008, 06:22 AM
whipflip15
• Nov 22nd 2008, 06:52 AM
Opalg
I can't read that attachment, but here's how I did the calculation. Let AB=BC=CD=2a. Then the triangle has base 2a and height √3a, and so

$\boxed{\text{Area of triangle }=\sqrt3a^2}$.

Let O denote the centre of the circle. Then $\frac a{OC} = \cos(\pi/6) = \frac{\sqrt3}2$, so that $OC = \frac2{\sqrt3}a$, and $OD = \Bigl(\frac2{\sqrt3}+2\Bigr)a$. Therefore the area of the C-shaped ring is $\tfrac23\pi(OD^2-OC^2) = \frac{8\pi a^2}9\bigl((1+\sqrt3)^2-1\bigr)$.

$\boxed{\text{Area of C-ring }=\tfrac89\pi a^2(3+2\sqrt3)}$.

That gives the ratio of the two areas as $\frac{9\sqrt{3}}{8\pi(3+2\sqrt{3})}$, as before.
• Nov 22nd 2008, 11:43 AM
David24
Quote:

Originally Posted by Opalg
I can't read that attachment, but here's how I did the calculation. Let AB=BC=CD=2a. Then the triangle has base 2a and height √3a, and so

$\boxed{\text{Area of triangle }=\sqrt3a^2}$.

Let O denote the centre of the circle. Then $\frac a{OC} = \cos(\pi/6) = \frac{\sqrt3}2$, so that $OC = \frac2{\sqrt3}a$, and $OD = \Bigl(\frac2{\sqrt3}+2\Bigr)a$. Therefore the area of the C-shaped ring is $\tfrac23\pi(OD^2-OC^2) = \frac{8\pi a^2}9\bigl((1+\sqrt3)^2-1\bigr)$.

$\boxed{\text{Area of C-ring }=\tfrac89\pi a^2(3+2\sqrt3)}$.

That gives the ratio of the two areas as $\frac{9\sqrt{3}}{8\pi(3+2\sqrt{3})}$, as before.

Opalg,

Thanks for your repsonse (I apologise for the attachment - will have to get a better scanner). I now see where my error lies,
I was doing a ratio comparison of the triangle to outer ring as oppossed to (2/3)*outer ring

My solution now agrees with yours,

Thanks for the detailed explanation,

Regards,

David
• Nov 22nd 2008, 11:57 AM
AAKhan07
My Solution
"I got the same answer also. I like the question. Why is there a circle inside the triangle?"

I think they put a circle there so that the radius of the circle can be used to relate the two areas, that's certainly the way I did it. My solution is probably not as elegant as yours or Opalg's but here it is:

Let radius of small circle = r
Imagine drawing lines from the three points where the side of the triangle is tangential to the circle, each to the centre of the circle. This forms three sectors in the triangle. Take one sector, half it along the line which goes from one corner of the triangle to the centre of the circle and you can find the radius of the larger circle (which is smaller than the largest circle)

$
cos60= \frac{r}{R}
$

Let R denote the radius of the inner side of the big ring.
R' will denote the radius of the outer side of the big ring.

$
\frac{1}{2} = \frac{r}{R}
$

$
R = 2r
$

Now to find the length of one side of the triangle, adding this to 2r will give R', the radius of the outer side of the ring. Let L be the length of the side of the triangle.

$
tan60 = \frac{0.5L}{r}
$

$
\sqrt{3} = \frac{0.5L}{r}
$

$
L=2r\sqrt{3}
$

So the radius of the outer-cirlce of the ring is $2r + 2r\sqrt{3} = 2r(1+\sqrt{3})$

The area of the ring must be area of outer-circle minus the area of the inner-circle multiplied by 2/3

$
(\pi(2r(1+\sqrt{3}))^2 - \pi(2r)^2) \times 2/3
$

$
(\pi(4r^2(4+2\sqrt{3}) - 4\pi r^2) \times 2/3
$

$
(16\pi r^2 +8\pi r^2 \sqrt{3} - 4\pi r^2) \times 2/3
$

$
(4\pi r^2 (3 + 2\sqrt{3})) \times 2/3
$

$
\frac{8\pi r^2 (3 + 2\sqrt{3})}{3}
$

Now to find the area of the triangle, split it into 3 sections again but this time the sections are formed by drawing lines from the vertices to the centre. The area of each of these triangles, A, can be found.

$
A = \frac{1}{2} (2r)^2 sin120
$

$
A = r^2 \sqrt{3}
$

Multiply this by 3 to get $3r^2\sqrt{3}$

The ratio we need is Area of triangle : Area of ring so $3r^2\sqrt{3}$ divided by $\frac{8\pi r^2 (3 + 2\sqrt{3})}{3}
$
which yields my original answer and the same answer as everyone else:

$\frac{9\sqrt{3}}{8\pi(3+2\sqrt{3})}
$

I'm just a 17-year old student so that's as elegant as my knowldge will allow. I just sincerely hope I got the same answer during the test. A good mark in the Oxford Physics Aptitude Test means I score an interview! Then, hopefully, a place at Oxford! Oxford is among the greatest Universities on Earth (and the oldest in the English-speaking world) so studying Physics there would be great!