This is impossible to solve as it isn’t given that the space these points are located in is Euclidean.
Not really, you just have to solve for all possible geometries, including non-euclidian. Should be trivial.
I’m glad I wasn’t the only one teased into trying to solve this. If you plot them on a straight line it’s a fairly straightforward 10". But the problem doesn’t tell you that, so yes, I don’t think we have enough information to solve
I guess maybe there’s an image on the sheet he’s looking at. Otherwise, yeah, not possible to solve with just that info.
Anywhere from 3-1/3" to 10" depending on where point A is located in space relative to B and C…
Yeah, isn’t this unsolvable to a single answer based on the data provided?
Yeah, B can be located anywhere on a sphere centered on A with a radius of 1/2 the distance from A to C.
Or I suppose you could just call it a circle since three points define a plane.
Really bold of them to give a triangle problem without specifying any angles.
With the law of cosine, we have:
X^2 =25/(5-4cosα)
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I had this posted over my when I was at engineering college!
You guys learn inches in school?
Yup, and most electronics worldwide are still based on 0.1 inch increments.
I’m a designer and had to learn points and pico in art school. Those are inch based too (and totally useless in digital layouts). Metric is so easy that we all get it and use it when it makes sense. Largely with 3D printing and automotive fasteners (yeah even most of the “American” cars).
The real problem is the US and UK dominated manufacturing, and electronics in the early days and both used imperial units (though not always the same units). And because of that old used machine tools are all inch standard and how we learn and do hobbies. That and metric hardware is way more expensive in the states. I can get 1/3 of the fasteners in metric for the same money.