2012 Year 9 &10 Mathematics Marathon (1 Viewer)

Fawun

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lol you still trying to do the petal question.

It's no where near as hard as you think, you are probably overthinking it.
I swear to god it's so hard... it's like legit impossible for me. Same thing applies to Carrots love heart question.
 

RealiseNothing

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I swear to god it's so hard... it's like legit impossible for me. Same thing applies to Carrots love heart question.
Both this petal question and carrot's love heart one can be done in about 3 lines.

Hope that makes you feel better haha.
 

enoilgam

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Both this petal question and carrot's love heart one can be done in about 3 lines.

Hope that makes you feel better haha.
It took me about 7 or 8 (when I looked at carrots it also took him a few lines, although his proof was way more efficient then mine).
 

RealiseNothing

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It took me about 7 or 8 (when I looked at carrots it also took him a few lines, although his proof was way more efficient then mine).
*Spoiler for solution*

It's just:

area of square = 4A + 8B

B = area of quadrant - A

therefore area of square = 4A + 8(area of quadrant - A)

Solve for A.

This is what I did anyway.
 

Fawun

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*Spoiler for solution*

It's just:

area of square = 4A + 8B

B = area of quadrant - A

therefore area of square = 4A + 8(area of quadrant - A)

Solve for A.

This is what I did anyway.
None of that registered in my brain

but then again it's 1AM and who the hell is thinking about maths at 1AM?
 

HSC2014

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Re: 2012 Year 9 &10 Mathematics Marathon

Good idea! Here's a fun one to add on top of TerenceM's:

"An equilateral triangle with side length 1 is placed inside a circle such that all the vertices of the triangle touch the circumference of the circle (also known as 'circumscribing'). What is the radius of the circle?"
Can someone tell me why my method is wrong.

The side of the triangle with 1/3 of the circumference of the outer circle forms a semi-circle with a radius of 0.5.
Now we can find the area of these semi-circle's using pi*r^2 = pi*0.5^2 = 0.25pi
Since there are three of them, and they are only semi-circles, the area of all 3 would be 0.75pi/2

Now we find the area of the equilateral triangle (bh/2) = 0.5*root0.75

So therefore the total area of the circle (pi*r^2)= (0.75pi + root0.75)/2
So thereforeee..... r = root{" /pi}

= which gives me something like plusminus 0.716. (answer is 0.577 to 3dp) :(

If it is too hard to follow my working out don't worry about it!
 

RealiseNothing

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Re: 2012 Year 9 &10 Mathematics Marathon

Can someone tell me why my method is wrong.

The side of the triangle with 1/3 of the circumference of the outer circle forms a semi-circle with a radius of 0.5.
Now we can find the area of these semi-circle's using pi*r^2 = pi*0.5^2 = 0.25pi
Since there are three of them, and they are only semi-circles, the area of all 3 would be 0.75pi/2

Now we find the area of the equilateral triangle (bh/2) = 0.5*root0.75

So therefore the total area of the circle (pi*r^2)= (0.75pi + root0.75)/2
So thereforeee..... r = root{" /pi}

= which gives me something like plusminus 0.716. (answer is 0.577 to 3dp) :(

If it is too hard to follow my working out don't worry about it!
I would re-think that.
 

HSC2014

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Re: 2012 Year 9 &10 Mathematics Marathon

I would re-think that.
Yeah that was the part I was flimsy about. It seemed (and still seems) perfectly logical to me and that there is no error there :( Since the vertices of triangle only make contact with 1 point on the circumference, and lines on a triangle are straight, it would form a semi circle).
 

RealiseNothing

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Re: 2012 Year 9 &10 Mathematics Marathon

Yeah that was the part I was flimsy about. It seemed (and still seems) perfectly logical to me and that there is no error there :( Since the vertices of triangle only make contact with 1 point on the circumference, and lines on a triangle are straight, it would form a semi circle).
It's more like a semi-ellipse, not a semi-circle lol. Which is why it doesn't work.
 

HSC2014

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Ah true true true. Farout I feel so stupid LOL. /on a mission to nail out all misconceptions and be exceptional at math
 
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