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?
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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 ℓ which is parallel to AB. Consider all triangles ABC with C on ℓ. Which triangle has minimal perimeter?
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.
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?
Let the length of the base be taken as x. If the perpendicular distance is x√3 / 2 , then, the perimeter of equilateral triangle is 3x. The perimeter of right angled triangle is 3.025x . In this case Equilateral triangle and isosceles triangle is the same. So between 3x and 3.025x, 3x is smaller and equilateral triangle has shorter perimeter.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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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 ℓ which is parallel to AB. Consider all triangles ABC with C on ℓ. Which triangle has minimal perimeter?
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Sir then is the length AB and the perpendicular distance from ℓ constant?
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.
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?
it depends on lengths of rest two sides
Let the length of the base be taken as x. If the perpendicular distance is x√3 / 2 , then, the perimeter of equilateral triangle is 3x. The perimeter of right angled triangle is 3.025x . In this case Equilateral triangle and isosceles triangle is the same. So between 3x and 3.025x, 3x is smaller and equilateral triangle has shorter perimeter.