Hello. Is this thread meant as a puzzle? That is, do you already have an answer in mind?I am experimenting with 13 more ways to calculate/approximate circumference without using classic 2*π*r or π*d formulae.
Are we to assume this is an isosceles triangle, or just any inscribed triangle?How can we approximate the circumference using the triangle inscribed in it in such a way that no part of the triangle touches the center of this circle?
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No, it's not a puzzle. I don't have an answer to my question. I am seeking guidance to prove something. Thanks.Hello. Is this thread meant as a puzzle? That is, do you already have an answer in mind?
If so, then let us know that this thread is a challenge for forum members to solve.
If not, then please share your beginning thoughts so that members can know where to begin helping you. Thanks.
Hello Dr. Peterson,Are we to assume this is an isosceles triangle, or just any inscribed triangle?
If the intention is to use only the three measured side lengths, then you are asking about finding the circumradius of a triangle given its side lengths.
Since it is easy to find the circumradius from an angle and its opposite side, you could find an angle by the Law of Cosines, and then use the Law of Sines to find the circumradius. Since you're looking for the circumference, you'll use pi for the last step.
Is that what you have in mind?
Please post the problem AGAIN with CORRECT problem statement. It will be useful if you can include a photograph of the problem as it was presented to you.How can we approximate the circumference using the triangle inscribed in it in such a way that no part of the triangle touches the center of this circle?
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Actually, a formula can be made that will work whether or not the triangle is isosceles, and whether or not the center of the circle happens to be inside the triangle. I don't need to assume anything, and didn't.Hello Dr. Peterson,
Yes, you have rightly assumed the triangle to be an isosceles one.
I am sorry, I should have also specified that the triangle can not be placed over the center of the circle, in other words-the circle center can not be anywhere inside this triangle
Regards
I have told you how you can derive such a formula. Have you tried? (Use the law of cosines, then the law of sines, then the formula for circumference.)No, it's not a puzzle. I don't have an answer to my question. I am seeking guidance to prove something. Thanks.
Thanks, Dr. Peterson,Actually, a formula can be made that will work whether or not the triangle is isosceles, and whether or not the center of the circle happens to be inside the triangle. I don't need to assume anything, and didn't.
But I take it you are saying I rightly interpreted your vague question, that you want to find the circumference given only the lengths of the triangle's three sides?
I have told you how you can derive such a formula. Have you tried? (Use the law of cosines, then the law of sines, then the formula for circumference.)
And what is it that you want to prove?
I don't see how that helps at all. How is it related to the triangle in the question?Thanks, Dr. Peterson,
I tried to derive a formula but couldn't find radius as the triangle is not touching the circle center. Hence, couldn't find circumference. Meanwhile, I am working to establish ratio of AB to BC and AB to AC, where AB is the radius of the circle transformed in to a right-angled scalene triangle, as shown in the attached figure.
Very big help, Dr. Peterson.I don't see how that helps at all. How is it related to the triangle in the question?
Here are the theorems I suggested using:
Using these, you can use a, b, and c to determine angle A, then use that and a to find R. From that, get the circumference as you want. The result, with a little work, is an algebraic function of a, b, and c.
- https://www.mathsisfun.com/algebra/trig-cosine-law.html (Use "easier version for angles")
- https://artofproblemsolving.com/wiki/index.php/Law_of_Sines (use "extended" form shown first)
It applies to any triangle. Why not isosceles?I was wondering is the Cosine law applicable to my isosceles triangle too?
Why? It applies to any triangle, wherever it is. And my figure fits that (unnecessary) condition, doesn't it? I don't see what you are concerned about.Secondly, doesn't law of Sines violate the condition in my problem, i.e. no part of the triangle must touch the circle center and the center must be outside the triangle?