plotting 7 points on a surface so that

Hi, sorry if I can't explain you the question properly. We need to plot 7 point on a surface so that: after picking any 3 point, the distance between at least two points should be exactly 1 unit

for e.g I thought hexagon and a point in the center. But it does not be true for every 3 points. I can't find solution more than 6 points

#Geometry

Easy Math Editor

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Comments

You can see this works in every case except, when we choose A,D and G. Now if we rotate the right side in this way $\overline{AG}=1$ then we solved the problem:

A Former Brilliant Member - 10 months, 3 weeks ago

Actually this looks logical but is it true for BFG

Emre TUZCU - 10 months, 3 weeks ago

Its totally correct answer

Sahar Bano - 10 months, 3 weeks ago

@Sahar Bano – Right fool me :)

Emre TUZCU - 10 months, 3 weeks ago

What does BFG mean? I found only Big F $***$ ing Gun

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – That was right. I thought wrong for a sec

Emre TUZCU - 10 months, 3 weeks ago

$\leq1$ or exactly one unit?

A Former Brilliant Member - 10 months, 3 weeks ago

Exactly one unit. Also I think there may not be a solution

Emre TUZCU - 10 months, 3 weeks ago

If exactly one this is impossible. If you draw an equilateral triangle and write the circles with radius=1, than there won't be engraving of the three lones in one point, so the number of points should be less than 4

A Former Brilliant Member - 10 months, 3 weeks ago

Maybe in higher dimensions

A Former Brilliant Member - 10 months, 3 weeks ago

Thanks for reply. Can you explain your proof withmore details. I didn't get it It may be impossible for 7 points but not for 5 or 6 points. If you draw a hexagon, it satisfies the conditions. I think it's impossible for >=7 points

Emre TUZCU - 10 months, 3 weeks ago

If the task says that the distance between all of the points should be 1 unit, then the biggest number of points is 3. Only the mentioned triangle can satisfy the conditions. But if the task says that we can choose two from the three points, then the hexagon won't work. If the name of the hexagon is ABCDEF, than the distance between A, C and E isn't one unit.

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – That's right but its true for square or pentagon

Emre TUZCU - 10 months, 3 weeks ago

@Emre Tuzcu – Yeah. Here is an interesting problem: Given 2006 points. In every case when we choose 3 points, then the area of the triangle with the selected points won't be bigger than 1 unit. Prove that we can cover these points with a triangle with 4 units area!

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – From XV. NNMV 9. class.

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – okay but actually this is totally different question

Emre TUZCU - 10 months, 3 weeks ago

@A Former Brilliant Member – anyway, thanks for you attention

Emre TUZCU - 10 months, 3 weeks ago

@Emre Tuzcu – You are wolcome!

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – *welcome

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – I have a figure with 6 points which satisfies the condition mentioned in the notes. I can share it if you want

Sahar Bano - 10 months, 3 weeks ago

@Sahar Bano – I doubt it

A Former Brilliant Member - 10 months, 3 weeks ago

@Sahar Bano – Share with us please

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member –

In the above diagram $ACDE$ is a square with side of one unit length. $\triangle ABC$ and $\triangle FED$ are equilateral.

Now in parallelogram $ABFE$ , $\overline{FB}$ is also of 1 unit length.

Sahar Bano - 10 months, 3 weeks ago

@Emre TUZCU?

A Former Brilliant Member - 10 months, 3 weeks ago

I tried to show it with dots on the keyboard but the blanks does not count here. If you think 2 equilateral triangle touching each other at the bottom (one's right bottom corner touches the others left bottom corner) This makes 5 point and 2 on the upside and 3 on the bottom, so draw last point to make 2 upside down triangle on the upside

Emre TUZCU - 10 months, 3 weeks ago

I don't understand what does it mean :( The maximum number of dots is 5.

A Former Brilliant Member - 10 months, 3 weeks ago

I edited it to the question

Emre TUZCU - 10 months, 3 weeks ago

Nice work! I will write a c++ program when I have time.

A Former Brilliant Member - 10 months, 3 weeks ago

I solved it with 7 points

A Former Brilliant Member - 10 months, 3 weeks ago

Maybe incorrect, but I'm testing it yet

A Former Brilliant Member - 10 months, 3 weeks ago

Thank you so much for your attention, now I can sleep in peace :) @Pàll Màrton and @Sahar Bano

Emre TUZCU - 10 months, 3 weeks ago

You posted it 3 days ago :)

A Former Brilliant Member - 10 months, 3 weeks ago

it was rough time :)

Emre TUZCU - 10 months, 3 weeks ago

@Emre Tuzcu – No! The rough time, when you must eat the food at school.

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – Ahahah Actually I eat at the college no problem every time I go to school. Why its bothering you

Emre TUZCU - 10 months, 3 weeks ago

@Emre Tuzcu – The school has no money. That's something it looks like:

A Former Brilliant Member - 10 months, 3 weeks ago

@A Former Brilliant Member – oh, dude. I get it now. We usually eat better than this

Emre TUZCU - 10 months, 3 weeks ago

@A Former Brilliant Member – also, enjoy your meal. If you can :)

Emre TUZCU - 10 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$