+0  
 
0
932
8
avatar+283 

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a  square array of dots like the one shown below?
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Two triangles are congruent if they have the same traced outline, possibly up to rotating and flipping. This is equivalent to having the same set of 3 side lengths.

 

I've already tried 3 and then 9 and they weren't the answer. 

 Jan 11, 2019
 #1
avatar+6251 
+4

I get 6

 

 Jan 11, 2019
 #2
avatar+283 
0

There are a lot more than that, I have a hint that says it involves square roots but that's honestly all I know, I might just give up if nobody here understands it either lol (Thanks for trying though :3)

 Jan 11, 2019
edited by ANotSmartPerson  Jan 11, 2019
 #3
avatar+118673 
+2

Hi ANotSoSmartPerson.  laugh

I've tried looking at it logically and I get the same answer as Rom.

You think there are many more and you have seen other answers from different people.

Can you give any example of a triangle that Rom has missed?

(noting that this array is just 4 dots by 4 dots)

Melody  Jan 11, 2019
 #4
avatar
+1

Meoldy: Here are a couple of answers to the same question(I think!):

 

https://web2.0calc.com/questions/please-help_50163
https://web2.0calc.com/questions/help-fast_79662

 Jan 12, 2019
 #5
avatar+6251 
+1

the 2nd link is a 5x5 grid, being odd it's quite different

Rom  Jan 12, 2019
 #6
avatar+118673 
+3

The first one is right though.

It is a 4 by 4 dot grid.

 

Ok I'll give it another go.

Here are the options. for the 2 sides that are the same.

 

across 0 up 1   (1 unit)                 1 possibility as seen on circle below

across 0 up 2  (2 units)                 1 possibility as seen on circle below

across 0 up 3   (3 units)                1 possibility as seen on circle below

 

across 1 up 1  (sqrt2  units)           1 possibility as seen on circle below

across 1 up 2    (sqrt5 units)           3 possibilities as seen on circle below

across 1 up 3    (sqrt10 units)         2 possibilities as seen on circle below

 

across 2 up 2     (sqrt8 units)          1 possibility as seen on circle below

across 2 up 3      (sqrt13 units)        2 possibilities as seen on circle below

 

So unless I made a careless oversight (which is quite possible),  there are  12 possibilities.   

 

 

Melody  Jan 12, 2019
edited by Melody  Jan 12, 2019
 #7
avatar+283 
+2

I've tried 3, 6, 9, and 12 and none of them worked xd 

 

I gave up on the problem and apparently the answer is 11, thanks anyways Melody and Rom.

Solution:

We can construct segments of length 1,sqrt2 ,sqrt5 ,sqrt10 ,2 ,2sqrt2 ,sqrt13 ,3 ,3sqrt2  (these can be obtained systematically by considering all lengths of segments from the top left dot to other dots). The following results can be obtained either by inspection or analysis of slope.

For each of 1, sqrt5, sqrt13, 3, there are zero isosceles triangles having it as its base length.
For sqrt10, there is one.
For each of 2sqrt2, 3sqrt2, there are two.
For each of 2, sqrt2, there are three.
This gives a total of 1 + 2 + 2 +3 + 3 = 11  non-congruent isosceles triangles.

 Jan 13, 2019
edited by ANotSmartPerson  Jan 13, 2019
 #8
avatar+118673 
+3

Hi aNotSoSmartPerson

I just did that.

But you are right I added one that was too big.

 

 

equal side lengththird sidenumber of triangles 
1sqrt21 
22sqrt21 
33sqrt21 
sqrt221 
sqrt5 sqrt2, 2 ,sqrt(10) 3 
sqrt102, 2sqrt22 
sqrt8 = 2sqrt24\(\color{red}{\not{1}}\)Here is my error - this one is too big.   
sqrt13sqrt2, 42 
TOTAL 11 

 

There are 11   - just like you said   laugh

Melody  Jan 13, 2019
edited by Melody  Jan 13, 2019
edited by Melody  Jan 13, 2019

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