×

# shortest perimeter

Between the various triangles on same base and between the same parallel lines which one would have the shortest perimeter... isosceles, equilateral or other? how can we prove it?

Note by Rahul Vernwal
4 years, 3 months ago

Sort by:

Sreejato, I would disagree with you. The main issue lies in understanding what Rahul meant in his problem, which could be stated clearer.

My interpretation of the problem is as follows: We are given 2 points $$A$$ and $$B$$ and a line $$\ell$$ which is parallel to $$AB$$. Consider all triangles $$ABC$$ with $$C$$ on $$\ell$$. Which triangle has minimal perimeter? Staff · 4 years, 3 months ago

Sir then is the length $$AB$$ and the perpendicular distance from $$\ell$$ constant? · 4 years, 3 months ago

Comment deleted Apr 04, 2013

I don't really think this is correct. I believe you meant $$A=(0,0), B=(c,0), C=(x,h).$$ Perimeter$$=c+\sqrt{x^2+h^2}+\sqrt{(c-x)^2+h^2}$$. Ignore the $$c$$ since it is a constant and by AM-GM the minimum is obtained when $$\sqrt{x^2+h^2}=\sqrt{(c-x)^2+h^2} \Rightarrow x=c-x \Rightarrow x=c/2$$, which is an isosceles triangle. Anyway I prefer Gabriel's solution. · 4 years, 3 months ago

Yes. There is a much more direct approach to this problem, where the triangle inequality hides all the algebraic expressions that you use.

Hint: Reflect $$B$$ across line $$\ell$$. Staff · 4 years, 3 months ago

Using Calvin's intepretation, we have: Villages A, and B lie on a straight line. There is a river parallel to the line containing A and B. What is the fastest way to run from A to the river and back to B? The answer is simple: reflect B across the river to obtain B'. The distance from the river to B and distance to B', for any point on the river, is the same. So the best point (C) on the river is the intersection of AB' and the river, which makes triangle ABC iscoceles. · 4 years, 3 months ago

Between the various triangles on same base and between the same parallel lines which one would have the shortest perimeter... isosceles or other? how to prove it? think of the case when base length is less than the perpendicular distance between parallel lines? · 4 years, 3 months ago

Comment deleted Apr 04, 2013

and what if the triangle cannot be equilateral? i mean if it cannot be formed that way · 4 years, 3 months ago

Please clarify how it can be formed. · 4 years, 3 months ago

when perpendicular distance between two parallel lines is greater then the height of equilateral triangle formed by using base length as length of sides....... · 4 years, 3 months ago

According to your specified constraints, an isosceles triangle is the best choice for minimizing the perimeter. · 4 years, 3 months ago

what does it mean?? type the full question · 4 years, 3 months ago

@ rahul · 4 years, 3 months ago

Please pardon me if I don't understand this properly but the perpendicular distance between two parallel lines when perpendicular distance between the two parallel lines is the height of equilateral triangle formed by using base length as length of sides. · 4 years, 3 months ago

equilateral<isoceles<abstract<right · 4 years, 3 months ago

yep · 4 years, 3 months ago

I'm going to manually edit the votes such that your post doesn't appear at the top. Staff · 4 years, 3 months ago

Fantastic solution!!. I also got the same answer but after using triogonometry, but you got it so effortlessly · 4 years, 3 months ago

right sreejato · 4 years, 3 months ago

it depends on lengths of rest two sides · 4 years, 3 months ago