We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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?

- - - -

- - - -

- - - -

- - - -

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.

ANotSmartPerson Jan 11, 2019

#2**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)

ANotSmartPerson Jan 11, 2019

#3**+2 **

Hi ANotSoSmartPerson.

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**+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**

Guest Jan 12, 2019

#6**+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

#7**+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.

ANotSmartPerson Jan 13, 2019

#8**+2 **

**Hi aNotSoSmartPerson**

**I just did that.**

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

equal side length | third side | number of triangles | |

1 | sqrt2 | 1 | |

2 | 2sqrt2 | 1 | |

3 | 3sqrt2 | 1 | |

sqrt2 | 2 | 1 | |

sqrt5 | sqrt2, 2 ,sqrt(10) | 3 | |

sqrt10 | 2, 2sqrt2 | 2 | |

sqrt8 = 2sqrt2 | 4 | \(\color{red}{\not{1}}\) | Here is my error - this one is too big. |

sqrt13 | sqrt2, 4 | 2 | |

TOTAL | 11 |

**There are 11 - just like you said **

Melody
Jan 13, 2019