Sister's math problem X-post

[J_V]

My sister is trying to pass this basic math course with this really annoying teacher and the only way she is going to do it is if she gets the extra credit. The only hitch is you have to get all the problems totally correct or you get no credit. This is the only one that is simple for me. If you guys could help, it would make my day /images/graemlins/smile.gif. It's not really a math problem.

http://www.photodump.com/direct/jvspree/triangle.jpg


How many triangles are in this diagram?

And yes, it really is my "sister's" /images/graemlins/smile.gif.

Is there a mathematical way to solve this? Maybe one of you google experts could find the answer for me. I found nothing, but I'm no google expert.

[ihardlyknowher]

The correct answer is clearly - a sh*tload.

[UCF THAYER]

[ QUOTE ]

How many triangles are in this diagram?


[/ QUOTE ]

Alot

[jakethebake]

If it's simple for you, why do you need help? Your sister should do her own homework, and if she can't pass "basic math" w/o extra credit, whe's an imbecile anyway.

[mslif]

Wrong forum, you should have put this in the math, science forum.

[jakethebake]

[ QUOTE ]
Wrong forum, you should have put this in the math, science forum.

[/ QUOTE ]

it already is. he's cross-posted.

00t was his back-up plan in case David didn't come through. /images/graemlins/tongue.gif

[tbach24]

85 I think

[ihardlyknowher]

I think it is 3*(7+6+5+4+3+2+1)=84.

[jason_t]

I posted the solution in your thread in SMP before you posted this thread.

[jason_t]

[ QUOTE ]
85 I think

[/ QUOTE ]

Off by one.

[jakethebake]

[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

It's more.

[jason_t]

[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[ihardlyknowher]

[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

Yay! What do I win?

[jakethebake]

[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[captswifty]

[ QUOTE ]
This is the only one that is simple for me.

[/ QUOTE ]

Do you mean not simple?

This math teacher sucks. With the amount of triangles that overlap, I can't think of a way to solve this without taking a couple hours and counting them all.

[kyro]

[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

[ihardlyknowher]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[/ QUOTE ]

Are you sure? Remember a trapezoid is not a triangle.

[jakethebake]

[ QUOTE ]
[ QUOTE ]
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I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[/ QUOTE ]

Are you sure? Remember a trapezoid is not a triangle.

[/ QUOTE ]

Yes. I'm sure.

[J_V]

Blow me dildo. Nobody in 2+2 history annoys me more than you. I can only take solice in the fact that you spend all day on a poker forum and you still suck at it.

[BreakfastBurrito]

i'm getting only 84

[J_V]

Yes, I meant not simple.

[captswifty]

[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

Incorrect. You're counting triangles from one side to the other, but what about just counting the ones in the middle? There are a lot more than 84 there.

[jason_t]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[/ QUOTE ]

No. No. No. No.

[GuyOnTilt]

My first guess is 84, which is just the summation of x for 1-7 times 3. Times three beecause of the three sections of triangles.

PS. This place really needs to support LaTeX (for the 8th time).

GoT

[jason_t]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

Incorrect. You're counting triangles from one side to the other, but what about just counting the ones in the middle? There are a lot more than 84 there.

[/ QUOTE ]

I have no idea what you're talking about, but if you understood how the solution

(7 + 6 + 5 + 4 + 3 + 2 + 1) * 3

is counting the triangles you would see that it's counting the "ones in the middle."

[ihardlyknowher]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
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[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[/ QUOTE ]

Are you sure? Remember a trapezoid is not a triangle.

[/ QUOTE ]

Yes. I'm sure.

[/ QUOTE ]

Here's is my reasoning. In the top tier there are 7 small triangles (triangle that cross zero lines). 6 triangles that cross 1 line. 5 that cross 2 lines. .... and 1 that crosses 6 lines.

Each tier adds a new set of such triangles. Thus, the times 3 part.

Please enlighten us as to yours?

[Patrick del Poker Grande]

[ QUOTE ]
With the amount of triangles that overlap, I can't think of a way to solve this without taking a couple hours and counting them all.

[/ QUOTE ]
How long does it take for you to count to 84? Or how about to 28?

[DrPublo]

[ QUOTE ]

PS. This place really needs to support LaTeX

[/ QUOTE ]

That would be very awesome.

The Doc

[captswifty]

I think the solution is 3*(26!)

(26!) is 26+25+24+23 etc...

and that answer is 1.20987438×10^27

I got that by counting all of the trianges in the top section (26) and multiplying it by 3... I could be way off...

[J_V]

Much appreciated.

[GuyOnTilt]

[ QUOTE ]
I can't think of a way to solve this without taking a couple hours and counting them all.

[/ QUOTE ]

Hurray for Sigma. Lower-bound x=1, upper-bound 7, solve for x. Multiply by 3. Woot.

PS. LaTeX please.

GoT

[jakethebake]

[ QUOTE ]
Blow me dildo. Nobody in 2+2 history annoys me more than you.

[/ QUOTE ]

Good luck in your sister's big upcoming race.



http://www.tricitypartnership.org/special_olympics_files/image004.jpg

[TheWorstPlayer]

Way I see it, there are three potential 'layers' of triangles: ones using just the top section, ones with the top two sections, and ones with all three sections. There are the same number of triangles in each layer (hopefully this is obvious). So we count in just the top layer and we multiply by 3.

There are 7 1-part triangles, 6 2-part triangles, 5 3-part triangles, etc. So it is, indeed, (7+6+5+4+3+2+1)*3=84 triangles.

Simple enough.

[captswifty]

[ QUOTE ]
Here's is my reasoning. In the top tier there are 7 small triangles (triangle that cross zero lines). 6 triangles that cross 1 line. 5 that cross 2 lines. .... and 1 that crosses 6 lines.

[/ QUOTE ]

There are more than 7 in the top tier, you're not counting the ones in the middle... I think there are 26 in the top tier. If you take the factoral of that and mulitply it by 3, I think that's the answer...

[ihardlyknowher]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[/ QUOTE ]

Are you sure? Remember a trapezoid is not a triangle.

[/ QUOTE ]

Yes. I'm sure.

[/ QUOTE ]

Here's is my reasoning. In the top tier there are 7 small triangles (triangle that cross zero lines). 6 triangles that cross 1 line. 5 that cross 2 lines. .... and 1 that crosses 6 lines.

Each tier adds a new set of such triangles. Thus, the times 3 part.

Please enlighten us as to yours?

[/ QUOTE ]

Another way to look at it:

You need one horizontal line and 2 non-horizontal lines to make a triangle. There are 8 non-horizontal lines. Thus, 3*(8choose2)=84 triangles.

[Sponger15SB]

way to totally [censored] up the frames retard

[jason_t]

[ QUOTE ]
I think the solution is 3*(26!)

(26!) is 26+25+24+23 etc...

and that answer is 1.20987438×10^27

I got that by counting all of the trianges in the top section (26) and multiplying it by 3... I could be way off...

[/ QUOTE ]

Yes, absurdly off.

[Bulldog]

[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

I concur.

[jason_t]

[ QUOTE ]
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[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

It's more. A lot more. C'mon jason_t. Are you stoned?

[/ QUOTE ]

Are you sure? Remember a trapezoid is not a triangle.

[/ QUOTE ]

Yes. I'm sure.

[/ QUOTE ]

Here's is my reasoning. In the top tier there are 7 small triangles (triangle that cross zero lines). 6 triangles that cross 1 line. 5 that cross 2 lines. .... and 1 that crosses 6 lines.

Each tier adds a new set of such triangles. Thus, the times 3 part.

Please enlighten us as to yours?

[/ QUOTE ]

Another way to look at it:

You need one horizontal line and 2 non-horizontal lines to make a triangle. There are 8 non-horizontal lines. Thus, 3*(8choose2)=84 triangles.

[/ QUOTE ]

Nice.

[kyro]

Interesting. The t-shirt that the girl is wearing is from the highschool my sister graduated from.

Oh, and jake, awfully ballsy making these comments considering you got the problem wrong yourself.

[jakethebake]

[ QUOTE ]
Oh, and jake, awfully ballsy making these comments considering you got the problem wrong yourself.

[/ QUOTE ]

I don't know what you're talking about. I'm up to 263 triangles and counting. I'll get it right. I just haven't finished yet.

[GuyOnTilt]

Jake,

Go. The. Fu[/b]ck. Away.

GoT

[citanul]

figure out the total number of shapes of triangles, and then multiply each by the number of those that there are. it's not that hard. the bottom trapezoids aren't triangles or anything like that, that make the "how many squares are here" ones so hard.

[Sifmole]

I see it this way....

There is 1 way to group 7 "columns"
2 ways for 6
3 ways for 5
4 ways for 4
5 ways for 3
6 ways for 2
7 ways for 1
and finally 1 big triangle

so I get
(((1*7)+(2*6)+(3*5)+(4*4)+(5*3)+(6*2)+(7*1))*3)+1
((7+12+15+16+15+12+7)*3)+1
((84)*3)+1
252+1
253

The only thing your solution over looks is that there is more ways to make a 6 column ( or 5, or 4 etc ) than 1.

[jason_t]

[ QUOTE ]
[ QUOTE ]
Oh, and jake, awfully ballsy making these comments considering you got the problem wrong yourself.

[/ QUOTE ]

I don't know what you're talking about. I'm up to 263 triangles and counting. I'll get it right. I just haven't finished yet.

[/ QUOTE ]

Do you know what a triangle is? Have you successfully pointed out an error in the so-called repeatedly verified solutions?

[kyro]

[ QUOTE ]
figure out the total number of shapes of triangles, and then multiply each by the number of those that there are. it's not that hard. the bottom trapezoids aren't triangles or anything like that, that make the "how many rectangles are here" ones so hard.

[/ QUOTE ]

[J_V]

[ QUOTE ]
Go. The. [censored]. Away.

[/ QUOTE ]

Give him some of that green powder!

[FouTight]

[ QUOTE ]
[ QUOTE ]
Oh, and jake, awfully ballsy making these comments considering you got the problem wrong yourself.

[/ QUOTE ]

I don't know what you're talking about. I'm up to 263 triangles and counting. I'll get it right. I just haven't finished yet.

[/ QUOTE ]

A agree with jakethebake on this matter.

There are middle triangles that, for some reason, nobody is taking into consideration.

[jason_t]

Mod, remove this image so frames aren't broken. Thanks.

[jason_t]

[ QUOTE ]
I see it this way....

There is 1 way to group 7 "columns"
2 ways for 6
3 ways for 5
4 ways for 4
5 ways for 3
6 ways for 2
7 ways for 1
and finally 1 big triangle

so I get
(((1*7)+(2*6)+(3*5)+(4*4)+(5*3)+(6*2)+(7*1))*3)+1
((7+12+15+16+15+12+7)*3)+1
((84)*3)+1
252+1
253

The only thing your solution over looks is that there is more ways to make a 6 column ( or 5, or 4 etc ) than 1.

[/ QUOTE ]

What you mean by your logic is almost correct but the translation of the logic into mathematics is not.

[ihardlyknowher]

For OOT extra credit:

How many trapezoids are there?

[kyro]

[ QUOTE ]
Mod, remove this image so frames aren't broken. Thanks.

[/ QUOTE ]

No way. Spaulding High School is making its mark in OOT. Make me proud!

[captswifty]

[ QUOTE ]
I see it this way....

There is 1 way to group 7 "columns"
2 ways for 6
3 ways for 5
4 ways for 4
5 ways for 3
6 ways for 2
7 ways for 1
and finally 1 big triangle

so I get
(((1*7)+(2*6)+(3*5)+(4*4)+(5*3)+(6*2)+(7*1))*3)+1
((7+12+15+16+15+12+7)*3)+1
((84)*3)+1
252+1
253

The only thing your solution over looks is that there is more ways to make a 6 column ( or 5, or 4 etc ) than 1.

[/ QUOTE ]

I think this is the solution I was going for, except I had the most retarded method ever, since it's been years since I've tried to figure anything like this out.

[jason_t]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Oh, and jake, awfully ballsy making these comments considering you got the problem wrong yourself.

[/ QUOTE ]

I don't know what you're talking about. I'm up to 263 triangles and counting. I'll get it right. I just haven't finished yet.

[/ QUOTE ]

A agree with jakethebake on this matter.

There are middle triangles that, for some reason, nobody is taking into consideration.

[/ QUOTE ]

Take the image, trace out one of these middle triangles, post it and I'll show you that we already counted it.

[jason_t]

[ QUOTE ]
[ QUOTE ]
I see it this way....

There is 1 way to group 7 "columns"
2 ways for 6
3 ways for 5
4 ways for 4
5 ways for 3
6 ways for 2
7 ways for 1
and finally 1 big triangle

so I get
(((1*7)+(2*6)+(3*5)+(4*4)+(5*3)+(6*2)+(7*1))*3)+1
((7+12+15+16+15+12+7)*3)+1
((84)*3)+1
252+1
253

The only thing your solution over looks is that there is more ways to make a 6 column ( or 5, or 4 etc ) than 1.

[/ QUOTE ]

I think this is the solution I was going for, except I had the most retarded method ever, since it's been years since I've tried to figure anything like this out.

[/ QUOTE ]

You had the most retarded method ever because you're wrong.

[kyro]

On a side note.

I wonder how many of the people pushing the "more than 84 triangles" side picked "above average" in the intelligence poll.

[J_V]

You OTTers own the math and science forum.

[ChromePony]

[ QUOTE ]
I think the solution is 3*(26!)

(26!) is 26+25+24+23 etc...

and that answer is 1.20987438×10^27

I got that by counting all of the trianges in the top section (26) and multiplying it by 3... I could be way off...

[/ QUOTE ]

I hope you realize how totally absurd this answer is and that its all just a big joke.

By the way your reasoning was somewhat sound, counting all the triangles in the top level and multiplying by 3, unfortunately there are 28 in the top level, not 26! (no idea where the ! came from).

[captswifty]

[ QUOTE ]
[ QUOTE ]
I think the solution is 3*(26!)

(26!) is 26+25+24+23 etc...

and that answer is 1.20987438×10^27

I got that by counting all of the trianges in the top section (26) and multiplying it by 3... I could be way off...

[/ QUOTE ]

I hope you realize how totally absurd this answer is and that its all just a big joke.

By the way your reasoning was somewhat sound, counting all the triangles in the top level and multiplying by 3, unfortunately there are 28 in the top level, not 26! (no idea where the ! came from).

[/ QUOTE ]

I think it has something to do with the fact that I just woke up with a hangover...

[jakethebake]

[ QUOTE ]
You OTTers own the math and science forum.

[/ QUOTE ]

00ti0ts!!! Dammit! 00ti0ts!!!!

This is an otter!

http://library.thinkquest.org/6068/pictures/looking%20otter.jpg

[wacki]

[ QUOTE ]
Mod, remove this image so frames aren't broken. Thanks.

[/ QUOTE ]

Done.

[Justin A]

With out looking at the replies I'm saying 84 as my first guess.

[ChromePony]

[ QUOTE ]
--------------------------------------------------------------------------------

I see it this way....

There is 1 way to group 7 "columns"
2 ways for 6
3 ways for 5
4 ways for 4
5 ways for 3
6 ways for 2
7 ways for 1



[/ QUOTE ]

This is right.


[ QUOTE ]
and finally 1 big triangle

[/ QUOTE ]

This is the one way to group 7 columns, dont recount it.

[ QUOTE ]
(((1*7)+(2*6)+(3*5)+(4*4)+(5*3)+(6*2)+(7*1))*3)+1
((7+12+15+16+15+12+7)*3)+1
((84)*3)+1
252+1
253

[/ QUOTE ]

Where did all this multiplication come from, there are 6 ways to group 2 so there are 6*2=12 possibilities??? NO, There are 6 you just said so.

[jakethebake]

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Mod, remove this image so frames aren't broken. Thanks.

[/ QUOTE ]

Done, next time just hit the button though.

[/ QUOTE ]

I hit the button. How about moving it to the right forum?

[durron597]

[ QUOTE ]
[ QUOTE ]
I think it is 3*(7+6+5+4+3+2+1)=84.

[/ QUOTE ]

Correct.

[/ QUOTE ]

This is right. And the 3 comes from bigger triangles, not from trapezoids.

[Justin A]

[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

I get 56.

[ChromePony]

fair enough /images/graemlins/laugh.gif

[jakethebake]

[ QUOTE ]
[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

I get 56.

[/ QUOTE ]

It's more. A lot more. C'mon Justin A. Are you stoned?

[jason_t]

[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

Same method.

(7 + 6 + 5 + 4 + 3 + 2 + 1) * 2 = 56

[jason_t]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

I get 56.

[/ QUOTE ]

It's more. A lot more. C'mon Justin A. Are you stoned?

[/ QUOTE ]

jake, you're an argumentative guy, that is obvious. But you should stick to arguing over matters of opinion, not fact.

[ihardlyknowher]

[ QUOTE ]
[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

Same method.

(7 + 6 + 5 + 4 + 3 + 2 + 1) * 2 = 56

[/ QUOTE ]

I get more.

[jakethebake]

[ QUOTE ]
jake, you're an argumentative guy, that is obvious. But you should stick to arguing over matters of opinion, not fact.

[/ QUOTE ]

Pregnant man gives birth. That's a fact.

[jakethebake]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

Same method.

(7 + 6 + 5 + 4 + 3 + 2 + 1) * 2 = 56

[/ QUOTE ]

I get more.

[/ QUOTE ]

See! I told you!

[bosoxfan]

wouldn't it be 85 if you count the big triangle that is the perimeter of the diagram?

[jakethebake]

[ QUOTE ]
wouldn't it be 85 if you count the big triangle that is the perimeter of the diagram?

[/ QUOTE ]

it was already counted.

[Patrick del Poker Grande]

[ QUOTE ]
wouldn't it be 85 if you count the big triangle that is the perimeter of the diagram?

[/ QUOTE ]
Yes, but only if you count it twice.

[ChromePony]

[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?


--------------------------------------------------------------------------------



I get 56.

[/ QUOTE ]

Yup.

This is basically the same problemm except the top row cant make trapazoids. 28*2 = 56.

[bosoxfan]

ummm...oops.
this is why I stay out of threads like this

[ihardlyknowher]

[ QUOTE ]
the top row cant make trapazoids.

[/ QUOTE ]

This is right, but your answer is wrong.

[BOTW]

This might actually be more interesting than it appears on the surface.

Each vertical line added to a triangle gives us the next triangular number of triangles, ((n+1)*(n+2))/2. Third diagonal of Pascal's triangle.

Ok, maybe not.

[Sifmole]

[ QUOTE ]
[ QUOTE ]
--------------------------------------------------------------------------------

I see it this way....

There is 1 way to group 7 "columns"
2 ways for 6
3 ways for 5
4 ways for 4
5 ways for 3
6 ways for 2
7 ways for 1



[/ QUOTE ]

This is right.


[ QUOTE ]
and finally 1 big triangle

[/ QUOTE ]

This is the one way to group 7 columns, dont recount it.

[ QUOTE ]
(((1*7)+(2*6)+(3*5)+(4*4)+(5*3)+(6*2)+(7*1))*3)+1
((7+12+15+16+15+12+7)*3)+1
((84)*3)+1
252+1
253

[/ QUOTE ]

Where did all this multiplication come from, there are 6 ways to group 2 so there are 6*2=12 possibilities??? NO, There are 6 you just said so.

[/ QUOTE ]

Uh, yeah..... /images/graemlins/blush.gif whoops..... been a long day.

[Tom1975]

I think it's 84 also for trapazoids. You can't use the top row, but you can use row 2 by itself, 3 by itself, and 2&3 combined.

[ihardlyknowher]

[ QUOTE ]
I think it's 84 also for trapazoids. You can't use the top row, but you can use row 2 by itself, 3 by itself, and 2&3 combined.

[/ QUOTE ]

Ding, ding, ding, ding, ding!!! Gutter, tell her what's she's won.

Good job guys! Now try this one:

http://www.mathpuzzle.com/lotsoftriangles.gif

[lucas9000]

[ QUOTE ]
[ QUOTE ]
Blow me dildo. Nobody in 2+2 history annoys me more than you.

[/ QUOTE ]

Good luck in your sister's big upcoming race.



http://www.tricitypartnership.org/special_olympics_files/image004.jpg

[/ QUOTE ]

holy crap i laughed my ass off at that.

[ghostface]

Its 84. I just counted it in 30 seconds. Count number in top row. Multiply by 3. Done. Basic math is easy.

This doesnt qualify as basic or math. It's counting.

[Worrots]

[ QUOTE ]
(26!) is 26+25+24+23 etc...

[/ QUOTE ]

No. n! = n*(n-1)*(n-2)*...*1

It also has nothing to do with solving this problem.

[Tom1975]

While brute force counting will solve the problem, there are quicker ways to solve it mathematically. What if there had been 100 columns instead of 7?

[Wintermute]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Blow me dildo. Nobody in 2+2 history annoys me more than you.

[/ QUOTE ]

Good luck in your sister's big upcoming race.



http://www.tricitypartnership.org/special_olympics_files/image004.jpg

[/ QUOTE ]

holy crap i laughed my ass off at that.

[/ QUOTE ]

it gave me a boner.

[GFunk911]

It's obviously 84, but I'm really really interesting in seeing jakethebake or one of his minions graphically point out one of the hidden triangles. Will it be a trapezoid? Will it kill you in your sleep? Stay tuned!!

[Worrots]

[ QUOTE ]
While brute force counting will solve the problem, there are quicker ways to solve it mathematically. What if there had been 100 columns instead of 7?

[/ QUOTE ]

c = # columns
r = # rows

# triangles = (sum of x where x={1,2,...,c})*r

This would be easier to show in summation notation.

[ThaSaltCracka]

9.

[durron597]

[ QUOTE ]
Good job guys! Now try this one:

http://www.mathpuzzle.com/lotsoftriangles.gif

[/ QUOTE ]

Looks like 8292 to me. That's ballpark.

[ThaSaltCracka]

well, I was way off.

[NutzyClutz]

Zero triangles. Trick question.
If you notice all non-horizontals are made up of jagged line segments.

[ihardlyknowher]

[ QUOTE ]
Zero triangles. Trick question.
If you notice I am an idiot.

[/ QUOTE ]

FYP.

[CallMeIshmael]

[ QUOTE ]
Zero triangles. Trick question.
If you notice all non-horizontals are made up of jagged line segments.

[/ QUOTE ]

You must be the life of the party.

[mmbt0ne]

Am I the only person who read this as "Sister's meth problem?"

[jakethebake]

[ QUOTE ]
Am I the only person who read this as "Sister's meth problem?"

[/ QUOTE ]

Meth + Retards = /images/graemlins/tongue.gif

[The Truth]

Not only is jason_t correct, but im shocked and appaled that there is debate over this.

[jedi]

I'm still waiting for jake's "more" explanation.

[J_V]

My sister passed her math class with a 69.9%. I will not have to spend next semester studying with her and I owe it all to OTT.

Thanks a lot.