Mathematics marathon

[aronpm]

Just wondering, did you do the test.

I haven't had the opportunity to do these math-test-things in years

 

[qqwref]

I guess the official solution would be

Spoiler

to just write out a bunch of equations:
4w = x+y+2011+z
4x = w+y+1+z
4y = w+x+2012
4z = w+x+2012

From the last two we get y = z, and we can insert those equations into the first two to get:
4w = x+2011+(w+x+2012)/2
4x = w+1+(w+x+2012)/2

Simplifying a bit:
7w = 3x+6034
7x = 3w+2014

And solving for x:
7x = 3(3x+6034)/7+2014
49x = 9x+18102+14098
40x = 32200
x = 805

 

[hic0057]

Thanks qqwref for that

 

[hic0057]

Another one...

A 40x40 white square is divided into 1x1 squares by lines parallel to its sides. Some of these 1x1 squares are coloured red so that each of the 1x1 squares, regardless of whether it is coloured red or not, shares a side with at most one red square (not counting itself.) What is the largest possible number of red squares?

The answer is between 000 and 999.

Also can you please explain it so that a 15 years old could understand. Thanks

 

[Erdos]

This is a simple case of using induction to find the formula for an nxn white square. In other words, try the coloring with a 1x1 square, then 2x2, 3x3,... You should find a pretty simple pattern.

 

[JonWhite]

Another one...

A 40x40 white square is divided into 1x1 squares by lines parallel to its sides. Some of these 1x1 squares are coloured red so that each of the 1x1 squares, regardless of whether it is coloured red or not, shares a side with at most one red square (not counting itself.) What is the largest possible number of red squares?

The answer is between 000 and 999.

Also can you please explain it so that a 15 years old could understand. Thanks

do you know what year AIME this problem was taken from

Also, try coloring one red square and see what other squares MUST be colored white.

 

[Tim Major]

The 40x40 grid was on the yr 10/11 test I did. I got 400 though not in a mathematical way, I just drew a much smaller grid and found the percentage.
I might have had the adjacent one too, but I left about 15 blank near the start to do the 10 point ones, and then had my test taking off me for talking (albeit it was a totally unrelated topic)
Did you get the one about climbing the mountain, 2 people 12km apart? I was stupid, I drew 20~ dashes and moved 2 torn pieces along them. I got the right answer though

 

[asportking]

Ok, I'm not sure if this is where I should ask it, but does anyone know a good way to understand normal form of a linear equation? I've been taking pre-calculus online, and I just learnt that there are a few things I need to know to pass the pre-calculus test that aren't on the online course, like normal form of a linear equation. I looked through a pre-calculus textbook, but I just can't understand what it's talking about. If anyone knows a video or a website that explains it well (or maybe could explain it themselves? ), it would be much appreciated.

 

[cmhardw]

Ok, I'm not sure if this is where I should ask it, but does anyone know a good way to understand normal form of a linear equation? I've been taking pre-calculus online, and I just learnt that there are a few things I need to know to pass the pre-calculus test that aren't on the online course, like normal form of a linear equation. I looked through a pre-calculus textbook, but I just can't understand what it's talking about. If anyone knows a video or a website that explains it well (or maybe could explain it themselves? ), it would be much appreciated.

I have to be honest that I have never studied normal form of a linear equation before today. You learn something new every day I suppose.

Ok, so here is what I have gathered from the Wikipedia article as well as looking into it myself.

I don't want to give too much away, as the self discovery of math is beautiful for one, and also you will remember this better if you derive most of this yourself.

Here is what I did to study normal form:

Spoiler

Start with the equation:
\( Ax+By=C \)

We seek to find the point on this line that also lies on the normal of the line. Wikipedia defines the normal as the shortest distance between the given line and the origin. The shortest distance between a given point and a given line is the line through that point which is perpendicular to the given line.

The slope of the equation:
\( Ax+By=C \)

is \( \frac{-A}{B} \)

Therefore the normal will have slope: \( \frac{B}{A} \)

The normal must go through the origin, so the equation on which the normal lies is:
\( y=\frac{B}{A}x \)

We now seek to find the point on the given line, which also lies on the normal. This point is the solution to the following system of equations.

\( Ax+By=C \)
\( y=\frac{B}{A}x \)

I strongly suggest you solve this system yourself, but the solution comes out to be:
\( \left(\frac{AC}{A^2+B^2} , \frac{BC}{A^2+B^2}\right) \)

The length of the normal, \( p \), is the distance from the above point to the origin. Using the distance formula you should get:
\( p=\sqrt{\left(\frac{AC}{A^2+B^2}-0\right)^2+\left(\frac{BC}{A^2+B^2}-0\right)^2} \)

Again, I strongly suggest you work this out on your own, but this simplifies to:
\( p=\left|\frac{C}{\sqrt{A^2+B^2}}\right| \)

To find the angle \( \theta \) between the normal and the x-axis, we simply need to realize that the normal line has slope \( \frac{B}{A} \) and so:
\( \theta=\arctan\left(\frac{B}{A}\right) \)

Now, Normal Form is listed on Wikipedia as:
\( y\sin\theta+x\cos\theta-p=0 \)

Substitute in what we know:
\( y\sin\left(\arctan\left(\frac{B}{A}\right)\right)+x\cos\left(\arctan\left(\frac{B}{A}\right)\right)-\left|\frac{C}{A^2+B^2}\right|=0 \)
and simplify to get:

\( y\frac{B}{\sqrt{A^2+B^2}}+x\frac{A}{\sqrt{A^2+B^2}}-\left|\frac{C}{\sqrt{A^2+B^2}}\right|=0 \)

Notice that this looks very similar to the standard form equation, just divided by \( \sqrt{A^2+B^2} \). You will also have to pay attention to the sign of the C term as well, since p is defined as the distance (so positive) of the normal, and C might be negative.

I'll leave you to connect the final dots, but basically given a standard form equation:
\( Ax+By=C \) you should at the very least be able to find \( sin\theta \), \( cos\theta \), and \( p \)

 

[asportking]

I have to be honest that I have never studied normal form of a linear equation before today. You learn something new every day I suppose.

Ok, so here is what I have gathered from the Wikipedia article as well as looking into it myself.

I don't want to give too much away, as the self discovery of math is beautiful for one, and also you will remember this better if you derive most of this yourself.

Here is what I did to study normal form:

Spoiler

Start with the equation:
\( Ax+By=C \)

We seek to find the point on this line that also lies on the normal of the line. Wikipedia defines the normal as the shortest distance between the given line and the origin. The shortest distance between a given point and a given line is the line through that point which is perpendicular to the given line.

The slope of the equation:
\( Ax+By=C \)

is \( \frac{-A}{B} \)

Therefore the normal will have slope: \( \frac{B}{A} \)

The normal must go through the origin, so the equation on which the normal lies is:
\( y=\frac{B}{A}x \)

We now seek to find the point on the given line, which also lies on the normal. This point is the solution to the following system of equations.

\( Ax+By=C \)
\( y=\frac{B}{A}x \)

I strongly suggest you solve this system yourself, but the solution comes out to be:
\( \left(\frac{AC}{A^2+B^2} , \frac{BC}{A^2+B^2}\right) \)

The length of the normal, \( p \), is the distance from the above point to the origin. Using the distance formula you should get:
\( p=\sqrt{\left(\frac{AC}{A^2+B^2}-0\right)^2+\left(\frac{AC}{A^2+B^2}-0\right)^2} \)

Again, I strongly suggest you work this out on your own, but this simplifies to:
\( p=\left|\frac{C}{\sqrt{A^2+B^2}}\right| \)

To find the angle \( \theta \) between the normal and the x-axis, we simply need to realize that the normal line has slope \( \frac{B}{A} \) and so:
\( \theta=\arctan\left(\frac{B}{A}\right) \)

Now, Normal Form is listed on Wikipedia as:
\( y\sin\theta+x\cos\theta-p=0 \)

Substitute in what we know:
\( y\sin\left(\arctan\left(\frac{B}{A}\right)\right)+x\cos\left(\arctan\left(\frac{B}{A}\right)\right)-\left|\frac{C}{A^2+B^2}\right|=0 \)
and simplify to get:

\( y\frac{B}{\sqrt{A^2+B^2}}+x\frac{A}{\sqrt{A^2+B^2}}-\left|\frac{C}{\sqrt{A^2+B^2}}\right|=0 \)

Notice that this looks very similar to the standard form equation, just divided by \( \sqrt{A^2+B^2} \). You will also have to pay attention to the sign of the C term as well, since p is defined as the distance (so positive) of the normal, and C might be negative.

I'll leave you to connect the final dots, but basically given a standard form equation:
\( Ax+By=C \) you should at the very least be able to find \( sin\theta \), \( cos\theta \), and \( p \)

Thanks, that really helped! I've been really stressed out lately, since I've got so much new stuff to learn now, no where to learn it from, and only a week to do it. It's awfully scary thinking that I might have just wasted the whole summer studying this only to have to take it all over again at school. I got a hold of last years lesson plan for pre-calc, so it helps knowing exactly what I need to learn.
Although I wonder, what exactly is the point of knowing normal form? I imagine it will become clear once I get into more advanced stuff, but I can't help but questioning the usefulness of this...

 

[qqwref]

Although I wonder, what exactly is the point of knowing normal form? I imagine it will become clear once I get into more advanced stuff, but I can't help but questioning the usefulness of this...

It isn't really useful, and I'm not sure why you're learning it. It probably has a few very specific applications, but in general you'll never see it. There is sort of an unofficial rule in actual math where you use whatever requires the least work for the problem/application at hand (hence the whole idea of 'simplest form') so you will see the ax+by+c=0 or ax+by=c forms almost 100% of the time.

 

[asportking]

It isn't really useful, and I'm not sure why you're learning it. It probably has a few very specific applications, but in general you'll never see it. There is sort of an unofficial rule in actual math where you use whatever requires the least work for the problem/application at hand (hence the whole idea of 'simplest form') so you will see the ax+by+c=0 or ax+by=c forms almost 100% of the time.

Oh thank god. That formula did NOT look fun to use all the time.

 

[Erdos]

I've never seen this before, but it seems like a trivial result of Gram-Schmidt, i.e. producing an orthogonal vector given another vector. Yeah don't think you'll ever need to know that for high school or even college. However, the concept is pretty important in all linear algebra: producing linearly independent equations.

 

[hic0057]

do you know what year AIME this problem was taken from

Also, try coloring one red square and see what other squares MUST be colored white.

This question was in the 2011 AIME test that was taken about 2 weeks ago in the Intermediate (year 9 and 10) and Senior (year 11 and 12) division.

The 40x40 grid was on the yr 10/11 test I did. I got 400 though not in a mathematical way, I just drew a much smaller grid and found the percentage.
I might have had the adjacent one too, but I left about 15 blank near the start to do the 10 point ones, and then had my test taking off me for talking (albeit it was a totally unrelated topic)
Did you get the one about climbing the mountain, 2 people 12km apart? I was stupid, I drew 20~ dashes and moved 2 torn pieces along them. I got the right answer though

There wasn't a year 10/11 AIME test but a Year 9/10 test and a 11/12 test.

I found the mountain one fairly easy.

Two tourist are walking 12km apart along a flat track at a constant speed of 4km/h. When each tourist reaches the slope of a mountain, she begins to climb a constant speed of 3km/h.

What is the distance, in km, between the two tourist during the climb.

They way I worked it out was finding how long the second person would take to get to the bottom of the mountain after the person ahead. It was 3hrs than I multiply that to the person ahead speed per hour to get the distance of 9km.

 

[vcuber13]

Find an equation to solve the area of any regular polygon.
what i got:

Spoiler

\( \frac{b^2 n \cdot cot(\frac{180}{n})}{2} \)

where b is the length of the base, and n is the number of sides.

doesnt work, time to retry

 

[Erdos]

Area of n-gon.

 

[vcuber13]

Spoiler

\( \frac{b^2 n \cdot cot(\frac{180}{n})}{2} \)

where b is the length of the base, and n is the number of sides.

doesnt work, time to retry

ive been thinking why this wasnt working, i figured it out:

Spoiler

the way i worked it out i got the 180/n and then tanned it, but because its angles and 180 degree its pi radians and so it should be:
\( \frac{b^2 n \cdot cot(\frac{\pi}{n})}{4} \)
oh, and i also missed a *.5 so it should be over 4 not over 2

 

[Keroma12]

Anybody know how to prove det(AB) = det(A)*det(B) using only the permutation definition of the determinant function? Completely forgotten how to type in math here, otherwise I would type out the definition. http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

(I am aware that this is a terrible way to prove this and that there are much much nicer ways.)

 

[jblake17]

Did anyone do the CIMC 2013?