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Old Maths Puzzle
7 Answers
Put this puzzle to one side ages ago as I couldn't do. Just found it and still can't!
3 Trees A, B & C are at the corners of a triangle which form a garden. Each
side of the garden is a whole number of feet. One side is 100ft long. A
straight fence ( a whole number of feet long) runs from the Apple tree to a
post on the opposite side & divides the garden into a lawn & a play area.
The distance from the cherry tree (which is on the lawn side) to the post
equals that from the Apple tree to the Beech tree.
The shape of the lawn is precisely that of the whole garden.
How many feet is the Beech tree from the post?
Thanks
3 Trees A, B & C are at the corners of a triangle which form a garden. Each
side of the garden is a whole number of feet. One side is 100ft long. A
straight fence ( a whole number of feet long) runs from the Apple tree to a
post on the opposite side & divides the garden into a lawn & a play area.
The distance from the cherry tree (which is on the lawn side) to the post
equals that from the Apple tree to the Beech tree.
The shape of the lawn is precisely that of the whole garden.
How many feet is the Beech tree from the post?
Thanks
Answers
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The beech tree is 45 feet from the post. First, draw a diagram. The triangle ABC represents the garden, with A, B and C the trees. Call the post P and position it somewhere along the side BC. Then, the fence is the line AP. The lawn is the triangle PAC. If we call the length of side AB x feet, then call BC = (x+y) feet, since PC = x feet and we are trying to find y. Then call AC...
09:41 Sun 01st May 2016
The beech tree is 45 feet from the post.
First, draw a diagram. The triangle ABC represents the garden, with A, B and C the trees. Call the post P and position it somewhere along the side BC. Then, the fence is the line AP. The lawn is the triangle PAC.
If we call the length of side AB x feet, then call BC = (x+y) feet, since PC = x feet and we are trying to find y. Then call AC z feet and AP (the fence) f feet. Mark these lengths on the diagram.
We are told that triangles PAC and ABC are the same shape ie congruent. Therefore, the sides must all be in the same proportion.
The sides of the triangles are: for triangle PAC f, z and x, and for triangle ABC x, (x+y), and z. We are told that these are all an exact number of feet.
Angle ACB is common to both triangles, so we can deduce that the following rations must all be equal:
f:x = z:(x+y) = x:z
We are told that one of the sides of the garden is 100 feet long. This must be either x, (x+y), or z. We can try each of these possibilities in turn.
First, assume x = 100 feet.
Then, from the second two ratios z:(100+y) = 100:z. Alternatively,
z/(100+y) = 100/z
or z*z = 100*(100+y)
We can draw up a table of possible values for Y, solve for z, and find values of z which are whole numbers. It will be found that the only possibilities are
(y = 21, z = 110), (y = 44, z = 120), (y = 69, z = 130), (y = 96, z = 140) and (y = 125, z = 150).
The next solution would be y = 189, z = 170 but these could not form a triangle, as the sides of the garden would have to be 100, 289, and 170. and since (100 + 170) < 289, this can be excluded.
We also know that f:x = x:z, ie
f/x = x/z
or f = x*x/z
Since x is assumed to be 100, and z must be 110, 120, 130, 140 or 150 we can solve for f. We find that there is no solution where f is a whole number of feet. Therefore, x cannot be 100 feet.
Second, assume that side (x+y) = 100 feet. x is therefore (100-y) feet.
From the ratios z:(x+y) = x:z we get
z/100 = (100-y)/z
or z*z = 100*(100-y)
We again draw up a table of possible values for Y, solve for z, and find values of z which are whole numbers. This time the only possibilities are
(x = 91, y = 19, z = 90), (x = 64, y = 36, z = 80) (x = 49, y = 51, z = 70) (x = 36, y = 64, z = 60).
The next solution would be (x = 25, y = 75, z = 50), but these also cannot form a triangle since 25 + 50 = 75.
Going back to f:x = x:z, ie
f/x = x/z
or f = x*x/z
we can solve again for f with the values of x and z above. We find that there is no solution where f is a whole number of feet. Therefore, (x + y) cannot be 100 feet.
Thirdly assume z = 100 feet. Then,
z:(x+y) = x:z ie
100/(x+y) = x/100
or 100*100 = x*(x+y)
ie y = (100*100 - x*x) / x. The only solution which gives a possible triangle with a whole number of feet is x = 80, y = 45, which gives f = 64.
Therefore the two triangles have sides 125, 100, 80 (for the garden) and 100, 80, 64 (for the lawn), in perfect proportion 5:4, with the fence 64 feet long and the post 45 feet from B.
First, draw a diagram. The triangle ABC represents the garden, with A, B and C the trees. Call the post P and position it somewhere along the side BC. Then, the fence is the line AP. The lawn is the triangle PAC.
If we call the length of side AB x feet, then call BC = (x+y) feet, since PC = x feet and we are trying to find y. Then call AC z feet and AP (the fence) f feet. Mark these lengths on the diagram.
We are told that triangles PAC and ABC are the same shape ie congruent. Therefore, the sides must all be in the same proportion.
The sides of the triangles are: for triangle PAC f, z and x, and for triangle ABC x, (x+y), and z. We are told that these are all an exact number of feet.
Angle ACB is common to both triangles, so we can deduce that the following rations must all be equal:
f:x = z:(x+y) = x:z
We are told that one of the sides of the garden is 100 feet long. This must be either x, (x+y), or z. We can try each of these possibilities in turn.
First, assume x = 100 feet.
Then, from the second two ratios z:(100+y) = 100:z. Alternatively,
z/(100+y) = 100/z
or z*z = 100*(100+y)
We can draw up a table of possible values for Y, solve for z, and find values of z which are whole numbers. It will be found that the only possibilities are
(y = 21, z = 110), (y = 44, z = 120), (y = 69, z = 130), (y = 96, z = 140) and (y = 125, z = 150).
The next solution would be y = 189, z = 170 but these could not form a triangle, as the sides of the garden would have to be 100, 289, and 170. and since (100 + 170) < 289, this can be excluded.
We also know that f:x = x:z, ie
f/x = x/z
or f = x*x/z
Since x is assumed to be 100, and z must be 110, 120, 130, 140 or 150 we can solve for f. We find that there is no solution where f is a whole number of feet. Therefore, x cannot be 100 feet.
Second, assume that side (x+y) = 100 feet. x is therefore (100-y) feet.
From the ratios z:(x+y) = x:z we get
z/100 = (100-y)/z
or z*z = 100*(100-y)
We again draw up a table of possible values for Y, solve for z, and find values of z which are whole numbers. This time the only possibilities are
(x = 91, y = 19, z = 90), (x = 64, y = 36, z = 80) (x = 49, y = 51, z = 70) (x = 36, y = 64, z = 60).
The next solution would be (x = 25, y = 75, z = 50), but these also cannot form a triangle since 25 + 50 = 75.
Going back to f:x = x:z, ie
f/x = x/z
or f = x*x/z
we can solve again for f with the values of x and z above. We find that there is no solution where f is a whole number of feet. Therefore, (x + y) cannot be 100 feet.
Thirdly assume z = 100 feet. Then,
z:(x+y) = x:z ie
100/(x+y) = x/100
or 100*100 = x*(x+y)
ie y = (100*100 - x*x) / x. The only solution which gives a possible triangle with a whole number of feet is x = 80, y = 45, which gives f = 64.
Therefore the two triangles have sides 125, 100, 80 (for the garden) and 100, 80, 64 (for the lawn), in perfect proportion 5:4, with the fence 64 feet long and the post 45 feet from B.
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