2012 Year 9 &10 Mathematics Marathon (1 Viewer)

HSC2014

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One day you will be able to see into the future Fawun. Ain't even lying lol. With more experience, the better you can tell if you're heading down the wrong road when trying to solve a question. At this point, you rethink your approach and start over.
 

kazemagic

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i give up question 9, always end up with some polynomial shit. How the hell did u guys do it O_O
 

enoilgam

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^ and how do i acquire that experience
Tommorow I will post a simple but tedious surd question - you should try it.

I just realised that for 2012, this is the biggest maths marathon thread. Interesting.
 

HSC2014

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Tommorow I will post a simple but tedious surd question - you should try it.

I just realised that for 2012, this is the biggest maths marathon thread. Interesting.
I blame too much free time ^^
 

Sy123

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i give up question 9, always end up with some polynomial shit. How the hell did u guys do it O_O
Notice that they are asking for an integral answer. So one in the form of integers. But if we split our isosceles triangle in half, we get two right angled triangles. Each of these sides have an integer side (we can assume height is an integer since the area 12 is an integer, so height must be an integer too). Thus we due to the two triangles are right angled and all have integer sides then we can assume that they are part of a Pythagorean triad. If we test the first triad.

3, 4, 5.
5 is obviously the hypotenuse, so that means the other two sides are going to be 3 and 4.

But hang on, lets take the area of this specific triangle. its (1/2 * 3 *4)*2=12!
But does this triad satisfy our base is 1 more than the two other sides?
Lets try 4 being the bottom, we get 4+4=8 as the length of the base which is NOT 1 more than 5, so that is invalid
But if 3 is the bottom then we get the base as 3+3=6 Which IS 1 more than 5. So that means its valid.

I have probably said this in too many words, my verbosity is very weak lol
 

Shazer2

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I seriously haven't known how to do 1 question in this thread and I'm 2U year 11. Is this some sort of accelerated maths? :/
 

kazemagic

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Notice that they are asking for an integral answer. So one in the form of integers. But if we split our isosceles triangle in half, we get two right angled triangles. Each of these sides have an integer side (we can assume height is an integer since the area 12 is an integer, so height must be an integer too). Thus we due to the two triangles are right angled and all have integer sides then we can assume that they are part of a Pythagorean triad. If we test the first triad.

3, 4, 5.
5 is obviously the hypotenuse, so that means the other two sides are going to be 3 and 4.

But hang on, lets take the area of this specific triangle. its (1/2 * 3 *4)*2=12!
But does this triad satisfy our base is 1 more than the two other sides?
Lets try 4 being the bottom, we get 4+4=8 as the length of the base which is NOT 1 more than 5, so that is invalid
But if 3 is the bottom then we get the base as 3+3=6 Which IS 1 more than 5. So that means its valid.

I have probably said this in too many words, my verbosity is very weak lol
O_O so theres no other methods to do this unless trial and error?
 

HSC2014

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Notice that they are asking for an integral answer. So one in the form of integers. But if we split our isosceles triangle in half, we get two right angled triangles. Each of these sides have an integer side (we can assume height is an integer since the area 12 is an integer, so height must be an integer too). Thus we due to the two triangles are right angled and all have integer sides then we can assume that they are part of a Pythagorean triad. If we test the first triad.

3, 4, 5.
5 is obviously the hypotenuse, so that means the other two sides are going to be 3 and 4.

But hang on, lets take the area of this specific triangle. its (1/2 * 3 *4)*2=12!
But does this triad satisfy our base is 1 more than the two other sides?
Lets try 4 being the bottom, we get 4+4=8 as the length of the base which is NOT 1 more than 5, so that is invalid
But if 3 is the bottom then we get the base as 3+3=6 Which IS 1 more than 5. So that means its valid.

I have probably said this in too many words, my verbosity is very weak lol
I overlooked this bit :( So that's why going into equations and stuff didn't work so well. Integers integers integer.

Does a math question ever include redundant information which is not necessary to like... trick someone? Like for the triangle question they could have included 12m^2 as the area but make it so that it is not necessarily required to achieve the answer, and thus your thinking would be stuffed up due to incorporating that piece of info.
 

SpiralFlex

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I seriously haven't known how to do 1 question in this thread and I'm 2U year 11. Is this some sort of accelerated maths? :/
Nah mate, I reckon some questions are appropriate for Year 11. Just keep practicing.
 

enoilgam

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So far, the petal question and the last two I posted all came from the same year 9 5.3 test. However, that year 9 test is easily the hardest I have ever seen and the time limit was ridiculous (45 minutes).
 

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