high school geometry: A=(4,-2), B=(1,9), C=(10,2); AC midpt at (7,0), centroid at (5,3), median...

[davee]

A = (4,-2)
B = (1,9)
C = (10,2)

Midpoint of AC = (7,0)
Centroid of the triangle = (5,3)
Median AC = y=-3/2+10.5

The distance from the centre of the hole with a radius of 1 centimetre is drilled with the centre at the centroid of the triangle to the edge AC needs to be at least 2 centimetres.

Assume that the centre of the hole is on the median from B to AC. How much closer towards the edge AC could the hole be drilled?

Could you write an answer and an explanation of the answer?

 

[The Highlander]

A = (4,-2) B = (1,9) C = (10,2)
Midpoint of AC = (7,0)
Centroid of the triangle = (5,3)
Median AC = y=-3/2+10.5
Median AC: y=-3/2x+10.5                                                              
                                                                                                                          The distance from the centre of the hole with a radius of 1 centimetre is drilled with the centre at the centroid of the triangle to the edge AC needs to be at least 2 centimetres.

Assume that the centre of the hole is on the median from B to AC. How much closer towards the edge AC could the hole be drilled?

Could you write an answer and an explanation of the answer?

We are here (and are happy) to provide help but not to just provide answers!
What have you, yourself, tried in order to solve the problem?
For example, have you tried drawing the triangle & its median on squared paper?
(You could then assume that the distance between the lines on the paper represents 1 cm; that might help you think about how to reach for a solution.)
Please come back and show us your attempt at finding the answer.
Further assistance will then be provided (if necessary. ?)

Edit:-
Hint: Further to my suggestion (above), if you name a point D (where the median meets AC) then Pythagoras' theorem would enable you to find the distance from the centroid to D. The rest should be fairly straightforward after that. ?

Please come back and show us what you have done with the help I've offered you.

 

[Dr.Peterson]

Median AC = y=-3/2+10.5

Presumably you mean that the equation of the median from B to the midpoint of A y = -3/2 x + `0.5.

The distance from the centre of the hole with a radius of 1 centimetre is drilled with the centre at the centroid of the triangle to the edge AC needs to be at least 2 centimetres.

This doesn't make sense grammatically. If this is a problem you were given, please show us the exact wording of the problem, ideally as an image including any picture that came with it.

 

[The Highlander]

@davee

As well as some kind of attempt by you at answering the question (or at least giving us your thoughts on it) we need to see the original question.

If you cannot type out the original text exactly a picture will do (even if it's not in English).

 

[The Highlander]

This looks like another one we'll never see again. ?‍
Got the answer elsewhere (or, less likely, by themself), was shown (in school?) or just lost interest.
Since s/he hasn't bothered to respond to our requests and the statutory four days has now passed, I will post a solution for the benefit of any future viewers (who couldn't solve this simple problem themselves) and for the sake of 'completeness' (so the thread has a 'resolution in it).

Following my initial advice (at Post #2), drawing a sketch on squared paper clarifies the problem and shows that Pythagoras' theorem may be used to arrive at a solution...

​

If any viewer still can't see how the answer is arrived at then here is how to proceed...

The lines on the paper may be taken to be 1 cm apart. Thus, the distance from the Centroid to D may be calculated as \(\displaystyle \rm{\sqrt{13}}\) ≈ 3.61 cm. (By Pythagoras, the distance from the centroid to D is: \(\displaystyle \sqrt{2^2+3^2}=\sqrt{4+9} =\sqrt{13}\)) See pic below.

​

So, in order to maintain a distance of at least 2 cm from D (on the "edge" of the triangle, AC), the centre of the hole may be moved up to 1.61 cm or 16.1 mm (rounded to 3 significant digits) closer to D.

NB: This analysis assumes that the centre of the hole may only be moved along "the median from B to AC" and not into anywhere else in the triangle, as that would provide much greater distances (along the line y = ⅔x - 2⅔?) still maintaining a 2 cm gap between the hole's centre and the side AC, however, I believe that is implied in the question and is, therefore, a fair assumption to make.