Constraint on a length of a triangle's side

tush shared this question 5 months ago
Answered

I have a triangle ABC.

I want the side BC to be equal to (AB + AC)/3.

The closest I get to build such triangle is create a 4th point D such that

D = Point( Circle (B , (Distance(A, B) + Distance(A, C)) / 3))
i.e., a segment of length (AB + AC)/3 beginning at point B. But then I wished I could attach point B to point D, but it doesn't work since you obviously can't tell B to be have 2 rules.


I couldn't find a way to tell GeoGebra to build such a triangle.

Any help?

Best Answer
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perhaps you can begin with B C and do A in a ellipse doing AB+AC=3BC

Files: foro.ggb

Comments (6)

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3

Ce n'est pas toujours possible d'obtenir un tel triangle


Utiliser 2 cercles s'ils sont sécants, c'est gagné

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1

That's very elegant. I will take a few moments and try to understand better what you've done here. Thanks a lot!

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If AC is fixed and you want to locate B we can work from mathmagic's ellipse.

When A meets the minor axis, AC = 3/2 BC = AB.

When A meets the major axis (on the left), AC = 2 BC and AB = BC.

When A is between these two positions, i.e. in the top left quadrant of the ellipse

let AC = (3/2 + x)BC = (3+2x)/2 BC where x is between 0 and 1/2.

Then AB = (3/2 - x)BC = (3-2x)/2 BC.

Solving for BC we get BC = (2/(3+2x))AC and BC = (2/(3-2x))AB.

Equating these and solving for AB we get AB = ((3-2x)/(3+2x))AC.

Now we know the distance of B from A and from C in terms of x

so we can construct two intersecting circles centred at A and at C to locate B.

In the scenario above, AC > AB. If we consider when A is in the top right quadrant then AC < AB.

By doing a similar analysis we find BC = (2/(3-2x))AC and AB = ((3+2x)/(3-2x))AC.

So for each x we get two locations for B.

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File attached.

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3

perhaps you can begin with B C and do A in a ellipse doing AB+AC=3BC

Files: foro.ggb
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3

or find all the points C with locusequation()

Files: foro.ggb
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