Diagonals of a square

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Mrnaa7

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1. a. If the diagonals of a square are drawn in, how many triangles of all sizes are formed?
b. Describe how Polya's four steps were used to solve part a.
 
Polya's "four steps" were
1) Understand the problem.
2) Devise a plan to solve it.
3) Carry out the plan.
4) Verify that your result is correct.

George Polya (1887-1965) was a Hungarian born mathematician who wrote a book titled "How to Solve It" that was specifically about methods of solving general problems.
 
Mrnaa7 said:
1. a. If the diagonals of a square are drawn in, how many triangles of all sizes are formed?
b. Describe how Polya's four steps were used to solve part a.
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There are many ways you might solve this problem; the details are entirely up to you, and we can't help with that other than to take whatever initial plan you propose, and guide you to a way that makes it work.

So you need to follow the steps! First, tell us your understanding of the problem, which might be expressed in terms of a picture and a description of what you need to accomplish. Then, suggest a possible plan (such as an orderly way to count), and we can work with that.

The important thing is that it isn't Polya's method if you aren't thinking for yourself. If you read his book, you'll see that it is largely about how teachers can guide your thinking by asking the right questions, which exactly matches our goal here.
 
I've not read Polya for a very long time, but one thing he emphasizes (if I remember correctly) is that forming a plan may involve experimentation, false starts, simpler problems of a similar kind. It's not that you sit down and say "Oh, here's the plan lying ready to use." The word "devise" is intentional.

In this case, I might start by thinking about two simpler but similar problems.
 
You really can't draw a square with all its diagonal and count the triangles you created. To be honest I don't believe that. Please try it and see who is correct.
 
I've done it. What are you talking about?

Are you just trying to hint at an ambiguity in the question? Or assuming something about how the counting has to be done?
 
A square has four vertices. No three of which are colinear.
Therefore those vertices form \(\dbinom{4}{3}=4\) different triangles. SEE HERE
Now add the two diagonals and we get a fifth point. But that means of the \(\dbinom{5}{3}=10\) there are two triples that are colliear.
 
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Dr.Peterson said:
I've done it. What are you talking about?

Are you just trying to hint at an ambiguity in the question? Or assuming something about how the counting has to be done?
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I believe Jomo has "mis-syntaxed" his response(#6). He should have used a "?!" instead of a "." as shown below.

​
Jomo said:
You really can't draw a square with all its diagonal and count the triangles you created?!​
​
To be honest I don't believe that. Please try it and see who is correct.​
Click to expand...
​
 
I thought about this a little differently than pka did. One diagonal divides the square into two triangle. If the diagonal goes from "bottom left to upper right", they are in the "upper-left" and "lower-right". The second diagonal divides the square into another two diagonals. If the diagonal goes from "bottom right to upper left", they are in the "upper-right" and "lower-left". That is a total of four triangles.

With both diagonals in place there are another four triangles, upper-right, lower-right, upper-left, and lower-left. That makes a total of eight triangles.
 
HallsofIvy said:
I thought about this a little differently than pka did. One diagonal divides the square into two triangle. If the diagonal goes from "bottom left to upper right", they are in the "upper-left" and "lower-right". The second diagonal divides the square into another two diagonals. If the diagonal goes from "bottom right to upper left", they are in the "upper-right" and "lower-left". That is a total of four triangles.

With both diagonals in place there are another four triangles, upper-right, lower-right, upper-left, and lower-left. That makes a total of eight triangles.
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This is what I was trying to hint at. I must admit pka's approach never remotely occurred to me. But then he really knows combinatorics.
 
My guess is that the problem is at a somewhat lower level than combinatorics; I've seen such problems (especially with the Polya reference) in Math for Elementary Teachers courses, where what is expected is an "orderly way of counting", such as by size. But we won't know the context, or the available methods, until @Mrnaa7 replies.
 
Like Jomo suggested I'd definitely draw it. And then I'd count on my fingers. I guess this shows that I'm not a true mathematician! I like creating graphics so I couldn't resist this...

diag.gif
 
Dr.Peterson said:
My guess is that the problem is at a somewhat lower level than combinatorics; I've seen such problems (especially with the Polya reference) in Math for Elementary Teachers courses, where what is expected is an "orderly way of counting", such as by size. But we won't know the context, or the available methods, until @Mrnaa7 replies.
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Thank you I figured out the math problem
 
JeffM said:
This is what I was trying to hint at. I must admit pka's approach never remotely occurred to me. But then he really knows combinatorics.
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Thank you very much, I figured it was 4 also, but since I am not good at math I was doubting my answer. Thank you
 
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