I recently wrote the second round of the South African Mathematics Olympiad Round 2. Only the top 100 students in the country progress to the final round.

Could you be ranked amongst the best? PROVE IT!

All answers are integer values between 000 and 999.

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`*italics*`

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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TopNewestPROBLEM 20(hardest problem of the paper)There are 300 white boxes and

nred boxes in storage. Each box contains the same number of soccer balls. The total number of soccer balls in all of the boxes isn^2 + 290n - 2490. Determinen.Log in to reply

Problem 14

ABCD is a quadrilateral with parallel sides AB and CD. AB is longer than CD. If angle D is twice angle B, AD = 5 and CD = 3, then find the length of AB.

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Extend \(CD\) to meet \(AD\) at \(E\), by the parallels, \(\angle CED=\angle D-\angle ECD=\angle D-\angle B=\angle B=\angle ECD\). Hence \(ED=DC=3,AB=AE=AD+DE=5+3=8\)

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Problem 19

A and B ride at constant speeds in opposite directions around a circular track, starting at diametrically opposite points. If they start at the same time, meet one another for the first time after B has ridden 100 metres, and meet a second time 60 metres before A completes one lap, what is the circumference of the track in metres?

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Let \(A\) and \(B\) be on the opposite ends of the diameter which is horizontal ( though it does not matter but for convenience sake and without loss of originality we can assume it.) with \(A\) on the left while \(B\) on the right. Since they move in opposite directions, we can take \(A\) to be moving in clockwise direction and \(B\) in counter-clockwise direction. Let the radius of the circle be \(R\) and angular speeds of \(A\) and \(B\) be \(\omega_{1}\) and \(\omega_{2}\) respectively. If the meet for the first time after \(t\) seconds and for second time after \(t_{1}\) seconds from the start, we can write the following equations:

\( R\omega_{2} t = 100 \dots (1)\)

\( \omega_{1} t + \omega_{2} t = \pi \dots (2)\)

Using \((1)\) and \((2)\), we can write

\( \displaystyle \frac{\pi R \omega_{2}}{\omega_{1} + \omega_{2}} = 100 \Rightarrow \frac{\pi R}{100} = 1 + \frac{\omega_{1}}{\omega_{2}} \dots (5) \)

Similarly, for second encounter, we can write,

\( R\omega_{1} t_{1} = 2\pi R - 60 \dots (3) \)

\( R\omega_{2} t_{1} = \pi R + 60 \dots(4) \)

Using \((3) \text{and} (4)\), we can write,

\(\displaystyle \frac{\omega_{1}}{\omega_{2}} = \frac{2\pi R - 60}{\pi R + 60} \dots (6)\)

Substituting eqn. \((6)\) in eqn. \( (5)\), we obtain,

\(\displaystyle \frac{\pi R}{100} = 1 + \frac{2\pi R - 60}{\pi R + 60} \Rightarrow \pi R = 240\)

Hence, the circumference of the circular path is \(480\).

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A very unique approach!

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Problem 11

If a die is rolled once, the probability that a 4 shows is (1/6). If three dice are rolled the probability that a 4 shows up exactly once is

x / 72. What is the value ofx?Log in to reply

The probability of a 4 shows up is : \(\frac{1}{6}\)

The probability of a 4 doesn't show up is: \(\frac{5}{6}\)

Hence, the probability is

\(\displaystyle 3\big (\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6} \big )\)

We need to times three because there are three different case on which die should be showing 4. I'll try the other problems tomorrow, it's quite late already here....

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My exact reasoning! At the end of the day, x = 25.

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Thanks for sharing these problems! Why don't you make it as a problem set instead of sharing a note?

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Hi Christopher! I'll definitely share the other problems in a problem set. I've posted these questions here because I'm not sure if my answers are correct. I was hoping that I could compare my answers with your solutions so that I can achieve some peace of mind. Please help!

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I think that a set of notes would likely make more sense (esp if you don't know what the answer is). This way, you can present several problems, and it would be easy for others to find / discuss.

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Oh ok!

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I agree @Christopher Boo

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