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How many trees can you plant on Earth so that each tree is equidistant from any other tree
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How many trees can you plant on Earth so that each tree is equidistant from any other tree
13,872 Views | (29 votes, avg: 3.62)
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Three!!
taking a section of earth, we have a circle and can draw a equalateral triangle in that circle. The vertices would be equidistant from each other.
Assuming that the whole surface of the earth is land where you can plant trees and earth is a sphere.
Assuming we plant a tree at a distance of every 20 km which means each tree can be considered as a circle (whose centre is a tree) with radius 10km. The distance from one tree (centre of circle) to another tree will be 20km.
The surface area of a sphere is 4*pi*r^2
The Diameter of the earth is 40000/pi km.
Therefore the surface area of earth is approx: 16 *pi * 10 ^ 8 sq km
Now the trees are placed at a distance of 10km from each other. Each tree is like placing a circle on the surface of the earth. The number of trees that can be placed across the surface of the earth would be : Surface area of earth/ Surface area of circle of radius 10km.
Therefore number of trees that can be placed: 16*10^6 trees
Earth is not a ball so we cannot take a section of it and treat it as a circle. Therfore there are just to points that suit North and South Pole.
Anwser = 2 trees
2 - they have to be placed as far as possilbe from eachother, the north and south pole would be ideal, but i don’t think a tree would be able to be planted in ice. therefore you should plant them on the equator, one half way around the world.
I tend to agree with Imti’s way of thinking (1-st answer), but we have to take into account the fact that earth is 3-dimentional. Also I’d like to mention that the question itself (”How many …”) doesn’t asks for the biggest amount of trees but for any amount (semantics). Thus the right answers are: 0, 1, 2, 3 and 4. I think that there is no need in explaining first 3 answers (0, 1 and 2), the fourth (3) was explained by Imti and the fifth (4) is placing trees at the tips of eqidilateral pyramid ensphered in Earth.
Answer = 0,1,2,3,4
The only 3-dimensional shape that allows points to be equidistant from each other is an equilateral triangle. So part of the solution is to see how many equilateral trangles you can fit on a sphere.
I can imagine 4.
Imagine a ball put on a tripod, and with one extra leg balancing on the top of the ball. The bottom 3 legs will connected to the ball in the shape of an equilateral triangle.
Any two bottom legs plus the top leg will also make an equilateral triangle of the same size.
In the end there should be 4 equilateral triangles that are made out from the 4 legs.
The correct answer is four. Four trees on the vertices of a tetrahedron would be equidistant each from the other.
The question states the earth, which is a sphere. Projecting the earth to a circle changes both the problem and the answer.
Actually Earth is not perfectly spherical. It is more of an ellipsoid. The fact that the distance of earth’s surface from earth’s core is more near the equator than near the poles (which explains for the variation in the magnitude of acceleration due to gravity ‘g’
) proves this.
So, while placing the trees over the vertices of a tetrahedron would make them equidistant from each other, it is impossible to find a way to superimpose the tetrahedron over ellipsoid (the earth) in such a way that the vertices of the tetrahedron touches the surface of ellipsoid.
Going by these lines, I feel 0, 1, 2 and 3 are the only possible solutions.
The correct answer is 3.
The reason why the question states “earth” is to make the question 2 dimentional - like a sheet of paper. In any 2-D surface, only 3 items can be placed equidistant from each other.
Now if the question were 3-D (meaning you could plant trees in the sky above), then you can have 4 trees that can be planted equidistant from each other! Consider an equilateral triangle as your starting 3 trees, then simply grow one “above” (in the sky) that is equidistant from all the three trees - the final geometric shape being akin to a pyramid with equidistant sides.
- good luck!
Aiiyy…the correct answer is 4…
you simply grow 3 trees like an equilateral triangle…and then grow a mound in the middle and grow a fourth tree on top of the mound so that its equidistant from the other three trees
the correct answer is…2, 3 and 4
0 can no way be the answer.
1 can not be equidistant from any other tree.
2 trees on the equator on opposite sides would be equidistant from each other.
3 trees on the equator (which would form an equilateral triangle) would be equidistant from each other.
4 trees(which would form a tetrahedron) would be equidistant from each other.
more than 4 trees cannot be equidistant from each other.
4: reason being simple, earth can be cut open only into 2 equal halves if and only if, it is cut open at the equator .i.e., 0 deg lattitude, or along the prime meridian.i.e., 0 deg longitude.
lets assume we cut it along the equator and the prime meridians(0 deg longitude), we have in hand 4 pcs of the earth of similar shapes.
what we have is a solid block in the facade shape of a triangle. now place a tree at the vertices of each block.
now combine these blocks once again to form a complete earth(spheriod) ,this makes it 4 trees for the earth.
1st at the north pole
2nd at the south pole
3rd on the equator
4th on the equator, diametrically opposite to the third one
an addition or substraction of a tree will void the distance
2 - eg at each Pole
4 on the equator are not all equidistant.
and neither are 3 on the equator.
Maybe 3 could be done some other way? almost certainly.
Most of Earth’s surface is water. There is no guarantee that an equilateral triangle, tetrahedron, etc. could be rotated to touch land at each corner. For trees = 2 you can look up Antipodes in Wikipedia to see where to plant them. Guess you’d have to write your own program to find spots in Google Earth for larger numbers.
I think that the use of the word earth may have sent things in the wrong direction. First, we don’t need to be concerned with two trees because there cannot be an issue of equal distances there - there is only one distance - A to B. Second, in order to plant trees in soil - no hanging plants - we could plant three in an equilateral triangle. Thus, A to B = B to C = C to A. If there was a hill that existing or a mound that we created (someone else’s suggestion) then we could plant a tree at the top of it. If that point D was equidistant to points A, B, and C, the points would form a tetrahedron. The general shape of the earth - or any planet - is not really the issue of the question. The specific shape in a specific area is more relevant.
4 would be the max number of trees as answered. Plant the trees at the four corners of a tetrahedron whose centroid is the center of the earth and whose corners are on the surface of the earth. This result is irrespective of whether earth is a sphere or not: Say earth is not a sphere(compressed sphere), take a random equilateral triangle parallel to the equatorial circle(a side of the triangle is now larger than the distance of a vertex to the north pole).Now go on pulling the eq triangle triangle southwards so that the side of the triangle decreases and the distance of a vertex to the north pole increases till they become equal. You’ll finally get a perfect tetrahedron will all 4 corners on the surface of the earth.
Regards,
J1g54w
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