A triangle with each of its sides a whole number of inches has an area of 7446 square inches. Moreover the side lengths form an arithmetic progression with the middle term even. What are the lengths of the sides?
Cinelli
Got a credit card? use our Credit Card & Finance Calculators
Thanks to eyeball08,Wondergirly,bofh,johnstevens77,Bhoddhisatva, for Donating to support the site
Triangle
-
UncleEbenezer
- The full Lemon
- Posts: 10783
- Joined: November 4th, 2016, 8:17 pm
- Has thanked: 1470 times
- Been thanked: 2993 times
Re: Triangle
Um, what am I missing?
7446 is a multiple of 6. So trivially we have a multiple of the lowest Pythogorean Triad of all.
7446 is a multiple of 6. So trivially we have a multiple of the lowest Pythogorean Triad of all.
-
Gengulphus
- Lemon Quarter
- Posts: 4255
- Joined: November 4th, 2016, 1:17 am
- Been thanked: 2628 times
Re: Triangle
UncleEbenezer wrote:Um, what am I missing?
7446 is a multiple of 6. So trivially we have a multiple of the lowest Pythogorean Triad of all.
You seem to be missing the fact that the Pythagorean triangles you describe have sides 3n, 4n, 5n for some whole number n and area half the product of the two smaller sides, i.e. 6n^2, and while 7446 is a multiple of 6, it is not 6 times the square of an integer.
Gengulphus
-
Gengulphus
- Lemon Quarter
- Posts: 4255
- Joined: November 4th, 2016, 1:17 am
- Been thanked: 2628 times
Re: Triangle
Spoiler...
First, the triangle is not a Pythagorean triangle (I rather suspect that the fact that the 3,4,5 Pythagorean triangle is a near match is a deliberate red herring by the puzzle-setter!). To prove that, note that three whole numbers in arithmetic progression with the middle term even must be 2n-m, 2n, 2n+m for whole numbers n and m. If they're also a Pythagorean triplet, then (2n-m)^2 + (2n)^2 = (2n+m)^2. Expanding the squares gives 4n^2 - 4nm + m^2 + 4n^2 = 4n^2 + 4nm + m^2, and cancelling equal terms in that reduces it to 4n^2 - 4nm = 4nm, or 4n(n-2m) = 0. So a Pythagorean triplet in arithmetic progression must either have n = 0, which isn't possible except for the totally degenerate triplet (0,0,0) because sides of triangles cannot be negative, or have n = 2m, which makes it a multiple of (2*2-1, 2*2, 2*2+1) = (3,4,5). Since (0,0,0) is a multiple of (3,4,5), it must therefore always be a multiple of (3,4,5) - but my previous post shows such triplets cannot be solutions to this problem.
The sides of the triangle must however still be of the form 2n-m, 2n, 2n+m to be in arithmetic progression, and the fact that they are sides of a triangle implies that 2n-m is non-negative, i.e. 2n >= m. Furthermore, the length 2n+m of the longest side must be at most the sum 4n-m of the lengths of the two shorter sides, so 2n+m <= 4n-m, which implies m <= n. Since the triangle has positive area, we know it is not degenerate, so that can actually be tightened up a bit, to m < n.
At this point, I'm going to appeal to Heron's formula for the area of a triangle in terms of its side lengths - quoting from the Wikipedia page https://en.wikipedia.org/wiki/Heron%27s_formula (deliberately not a clickable link to avoid problems with the spoiler protection), with reformatting for the limitations of this medium:
"Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
A = SquareRoot( s ( s − a ) ( s − b ) ( s − c ) ) ,
where s is the semi-perimeter of the triangle; that is,
s = (a + b + c) / 2."
The semi-perimeter is s = ((2n-m) + 2n + (2n+m))/2 = 3n, so we want whole numbers m < n for which:
SquareRoot ( 3n (3n - (2n-m)) (3n - 2n) (3n - (2n+m)) ) = 7446
Simplifying the easy parts of that formula, it becomes:
SquareRoot ( 3n (n+m) n (n-m) ) = 7446
Apply the identity (n+m)(n-m) = n^2-m^2, factorise 7446 into primes as 2 * 3 * 17 * 73, square both sides and take out the common factor of 3, and that becomes:
n^2 (n^2-m^2) = 2^2 * 3 * 17^2 * 73^2
Because n^2 appears on the left hand side of this, any prime factor of n must appear at least twice as many times on the right hand side. So prime factors of n can only be 2, 17 and 73, and each can appear at most once in its prime factorisation. So we only have 8 possible values for n, namely the numbers 2^x * 17^y * 73^z for each of x, y and z being either 0 or 1.
Furthermore, for each of those values, we can both determine what n^2-m^2 must be by dividing 2^2 * 3 * 17^2 * 73^2 = 18480972 by n^2, and place limits on what it can be, namely at most n^2-0^2 = n^2 and at least n^2-(n-1)^2 = 2n-1, since we know that m<n. This leads to the following results:
So we're left with the two possibilities n=73 and n=146, with n^2-m^2 = 3468 and 867 respectively. Now go back to regarding n^2-m^2 as (n-m)(n+m) and look at how the prime factorisations of 3468 = 2^2 * 3 * 17^2 and 867 = 3 * 17^2 can be split between n-m and n+m, bearing in mind that m >= 0 and so n-m <= n+m, and also that n must be their average, so they must either both be even or both odd:
n=73: (n-m)(n+m) = 2^2 * 3 * 17^2 is even, so at least one of n-m and n+m must be even, and therefore they both must be. So one factor of 2 must go to each, and the remaining factors 3, 17, and 17 be split between them, Taking n-m <= n+m into account, the possibilities are that (n-m,n+m) are:
(2, 2 * 3 * 17^2) = (2,1734), with average 868;
(2 * 3, 2 * 17*2) = (6,578), with average 292;
(2 * 17, 2 * 3 * 17) = (34,102), with average 68.
n=146: (n-m)(n+m) = 3 * 17^2 is odd, so both n-m and n+m must be odd, and we must split the factors 3, 17 and 17 between them. Again taking n-m <= n+m into account, the possibilities for (n-m,n+m) are:
(1, 3 * 17^2) = (1,867), with average 434;
(3, 17*2) = (3,289), with average 146;
(17, 3 * 17) = (17,51), with average 34.
The only one of those that correctly has n being the average of n-m and n+m is n=146, (n-m,n+m) = (3,289). So the unique solution is that the three sides of the triangle (2n-m, 2n, 2n+m) are (146+3, 2*146, 146+289) = (149, 292, 435), which are in arithmetic progression. And double-checking with Heron's formula to guard against arithmetic and copying mistakes:
s = (149+292+435)/2 = 438
A = SquareRoot(438 * (438-149) * (438-292) * (438-435)) = SquareRoot(438*289*146*3) = SquareRoot(55442916) = 7446.
One final observation that might be of interest is that some such 'Heronian triangles' with whole number sides and area are decomposable, in that they can be constructed straightforwardly from two Pythagorean triangles of equal height, as for instance a Heronian triangle of sides (13,14,15) can be, by gluing together Pythagorean triangles of width 9, height 12, hypotenuse 15 and width 5, height 12 and hypotenuse 13, or one of sides (4,13,15) can be by cutting the latter off the former, and some can't. This one can be - to find the construction first observe that the most usual formula A = bh/2 for the area of a triangle, where b is its base and h its height, implies h = 2A/b. The height for such a straightforward construction is the common side of the two Pythagorean triangles, so must be a whole number, so for such a construction to be possible, 2A must be divisible by the length of whichever side we use as the base. In this case, 2A = 2*7446 = 14892 is not divisible by either of the two sides 149 or 435, but is by the side 292: 14892 = 51 * 292. So such a construction must use the side of length 292 as the base, the common side of the two Pythagorean triangles must be 51 and their hypotenuses must be 149 and 435. The other sides of those triangles must be SquareRoot(149^2-51^2) and SquareRoot(435^2-51^2), and as it happens, those both turn out to be whole numbers, namely 140 and 432. The base 292 is the difference between them rather than their sum, so the solution triangle can be constructed as the result of chopping a (51, 140, 149) Pythagorean triangle off a (51, 432, 435) Pythagorean triangle - using somewhat crude (and very much not-to-scale!) ASCII graphics:
Not all Heronian triangles are decomposable - the Wikipedia page about them gives the example of the Heronian triangle with sides 5, 29 and 30, which is Heronian because it has semi-perimeter (5+29+30)/2 = 32 and so area SquareRoot(32 * (32-5) * (32-29) * (32-30)) = SquareRoot(32*27*3*2) = SquareRoot(5184) = 72, but not decomposable because its area is not divisible by any of its sides.
Gengulphus
First, the triangle is not a Pythagorean triangle (I rather suspect that the fact that the 3,4,5 Pythagorean triangle is a near match is a deliberate red herring by the puzzle-setter!). To prove that, note that three whole numbers in arithmetic progression with the middle term even must be 2n-m, 2n, 2n+m for whole numbers n and m. If they're also a Pythagorean triplet, then (2n-m)^2 + (2n)^2 = (2n+m)^2. Expanding the squares gives 4n^2 - 4nm + m^2 + 4n^2 = 4n^2 + 4nm + m^2, and cancelling equal terms in that reduces it to 4n^2 - 4nm = 4nm, or 4n(n-2m) = 0. So a Pythagorean triplet in arithmetic progression must either have n = 0, which isn't possible except for the totally degenerate triplet (0,0,0) because sides of triangles cannot be negative, or have n = 2m, which makes it a multiple of (2*2-1, 2*2, 2*2+1) = (3,4,5). Since (0,0,0) is a multiple of (3,4,5), it must therefore always be a multiple of (3,4,5) - but my previous post shows such triplets cannot be solutions to this problem.
The sides of the triangle must however still be of the form 2n-m, 2n, 2n+m to be in arithmetic progression, and the fact that they are sides of a triangle implies that 2n-m is non-negative, i.e. 2n >= m. Furthermore, the length 2n+m of the longest side must be at most the sum 4n-m of the lengths of the two shorter sides, so 2n+m <= 4n-m, which implies m <= n. Since the triangle has positive area, we know it is not degenerate, so that can actually be tightened up a bit, to m < n.
At this point, I'm going to appeal to Heron's formula for the area of a triangle in terms of its side lengths - quoting from the Wikipedia page https://en.wikipedia.org/wiki/Heron%27s_formula (deliberately not a clickable link to avoid problems with the spoiler protection), with reformatting for the limitations of this medium:
"Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
A = SquareRoot( s ( s − a ) ( s − b ) ( s − c ) ) ,
where s is the semi-perimeter of the triangle; that is,
s = (a + b + c) / 2."
The semi-perimeter is s = ((2n-m) + 2n + (2n+m))/2 = 3n, so we want whole numbers m < n for which:
SquareRoot ( 3n (3n - (2n-m)) (3n - 2n) (3n - (2n+m)) ) = 7446
Simplifying the easy parts of that formula, it becomes:
SquareRoot ( 3n (n+m) n (n-m) ) = 7446
Apply the identity (n+m)(n-m) = n^2-m^2, factorise 7446 into primes as 2 * 3 * 17 * 73, square both sides and take out the common factor of 3, and that becomes:
n^2 (n^2-m^2) = 2^2 * 3 * 17^2 * 73^2
Because n^2 appears on the left hand side of this, any prime factor of n must appear at least twice as many times on the right hand side. So prime factors of n can only be 2, 17 and 73, and each can appear at most once in its prime factorisation. So we only have 8 possible values for n, namely the numbers 2^x * 17^y * 73^z for each of x, y and z being either 0 or 1.
Furthermore, for each of those values, we can both determine what n^2-m^2 must be by dividing 2^2 * 3 * 17^2 * 73^2 = 18480972 by n^2, and place limits on what it can be, namely at most n^2-0^2 = n^2 and at least n^2-(n-1)^2 = 2n-1, since we know that m<n. This leads to the following results:
n n^2-m^2 Lower Upper Within
limit limit limits
============================================================
2^0 * 17^0 * 73^0 = 1 18480972 1 1 No
2^1 * 17^0 * 73^0 = 2 4620243 3 4 No
2^0 * 17^1 * 73^0 = 17 63948 33 289 No
2^1 * 17^1 * 73^0 = 34 15987 67 1156 No
2^0 * 17^0 * 73^1 = 73 3468 145 5329 Yes
2^1 * 17^0 * 73^1 = 146 867 291 21316 Yes
2^0 * 17^1 * 73^1 = 1241 12 2487 1540081 No
2^1 * 17^1 * 73^1 = 2482 3 4963 6160324 No
So we're left with the two possibilities n=73 and n=146, with n^2-m^2 = 3468 and 867 respectively. Now go back to regarding n^2-m^2 as (n-m)(n+m) and look at how the prime factorisations of 3468 = 2^2 * 3 * 17^2 and 867 = 3 * 17^2 can be split between n-m and n+m, bearing in mind that m >= 0 and so n-m <= n+m, and also that n must be their average, so they must either both be even or both odd:
n=73: (n-m)(n+m) = 2^2 * 3 * 17^2 is even, so at least one of n-m and n+m must be even, and therefore they both must be. So one factor of 2 must go to each, and the remaining factors 3, 17, and 17 be split between them, Taking n-m <= n+m into account, the possibilities are that (n-m,n+m) are:
(2, 2 * 3 * 17^2) = (2,1734), with average 868;
(2 * 3, 2 * 17*2) = (6,578), with average 292;
(2 * 17, 2 * 3 * 17) = (34,102), with average 68.
n=146: (n-m)(n+m) = 3 * 17^2 is odd, so both n-m and n+m must be odd, and we must split the factors 3, 17 and 17 between them. Again taking n-m <= n+m into account, the possibilities for (n-m,n+m) are:
(1, 3 * 17^2) = (1,867), with average 434;
(3, 17*2) = (3,289), with average 146;
(17, 3 * 17) = (17,51), with average 34.
The only one of those that correctly has n being the average of n-m and n+m is n=146, (n-m,n+m) = (3,289). So the unique solution is that the three sides of the triangle (2n-m, 2n, 2n+m) are (146+3, 2*146, 146+289) = (149, 292, 435), which are in arithmetic progression. And double-checking with Heron's formula to guard against arithmetic and copying mistakes:
s = (149+292+435)/2 = 438
A = SquareRoot(438 * (438-149) * (438-292) * (438-435)) = SquareRoot(438*289*146*3) = SquareRoot(55442916) = 7446.
One final observation that might be of interest is that some such 'Heronian triangles' with whole number sides and area are decomposable, in that they can be constructed straightforwardly from two Pythagorean triangles of equal height, as for instance a Heronian triangle of sides (13,14,15) can be, by gluing together Pythagorean triangles of width 9, height 12, hypotenuse 15 and width 5, height 12 and hypotenuse 13, or one of sides (4,13,15) can be by cutting the latter off the former, and some can't. This one can be - to find the construction first observe that the most usual formula A = bh/2 for the area of a triangle, where b is its base and h its height, implies h = 2A/b. The height for such a straightforward construction is the common side of the two Pythagorean triangles, so must be a whole number, so for such a construction to be possible, 2A must be divisible by the length of whichever side we use as the base. In this case, 2A = 2*7446 = 14892 is not divisible by either of the two sides 149 or 435, but is by the side 292: 14892 = 51 * 292. So such a construction must use the side of length 292 as the base, the common side of the two Pythagorean triangles must be 51 and their hypotenuses must be 149 and 435. The other sides of those triangles must be SquareRoot(149^2-51^2) and SquareRoot(435^2-51^2), and as it happens, those both turn out to be whole numbers, namely 140 and 432. The base 292 is the difference between them rather than their sum, so the solution triangle can be constructed as the result of chopping a (51, 140, 149) Pythagorean triangle off a (51, 432, 435) Pythagorean triangle - using somewhat crude (and very much not-to-scale!) ASCII graphics:
*-__
|\ --__
| \ --__
| \ --__
| \ --__
| \ --__
| 51 \ 149 --__435
| \ --__
| \ --__
| \ --__
| \ --__
|____________________\______________________--_
140 292
<--------------------432---------------------->
Not all Heronian triangles are decomposable - the Wikipedia page about them gives the example of the Heronian triangle with sides 5, 29 and 30, which is Heronian because it has semi-perimeter (5+29+30)/2 = 32 and so area SquareRoot(32 * (32-5) * (32-29) * (32-30)) = SquareRoot(32*27*3*2) = SquareRoot(5184) = 72, but not decomposable because its area is not divisible by any of its sides.
Gengulphus
-
cinelli
- Lemon Slice
- Posts: 553
- Joined: November 9th, 2016, 11:33 am
- Has thanked: 234 times
- Been thanked: 161 times
Re: Triangle
Reply to spoiler:
Well solved, Gengulphus - a masterful explanation. This problem uses my favourite of all mathematical formulae, Heron's formula with its beautiful symmetry.
Cinelli
Well solved, Gengulphus - a masterful explanation. This problem uses my favourite of all mathematical formulae, Heron's formula with its beautiful symmetry.
Cinelli
Re: Triangle
Sorry, I've not read other solutions, apologies if duplicate:
Denote triangle of sides: 2a-b, 2a, 2a+b, with semi-perimeter s = 3a (for integers a and b)
Note, unique factorisation of 7446 = 2.3.17.73 = A
By Heron's formula A^2 = 3a(a-b)a(a-b)
7446^2 = 3a^2(a-b)(a+b)
= 3a^2(a^2 - b^2)
Since LHS is a perfect square, RHS must be too.
Therefore for some integer c, with a>c, we have a^2 - b^2 = 3c^3
Taking prime factors we have
2.17.73 = ac, satisfying a^2 - 3c^2 = b^2
So either ac = 2482 x 1 or 1241 x 2 or 73 x 34 or 146 x 17
by inspection, we can discount the first 2 possibilities
check 73^2 - 3(34^2) = 5329 - 3468 = 1861 (this is prime and therefore not a square)
check 146^2 = 3(17^2) = 21316 - 867 = 20449 = 143^2 (used calculator here!)
therefore a = 146, c = 17, b = 143
so sides must be 149, 292, 435
-Dom
Denote triangle of sides: 2a-b, 2a, 2a+b, with semi-perimeter s = 3a (for integers a and b)
Note, unique factorisation of 7446 = 2.3.17.73 = A
By Heron's formula A^2 = 3a(a-b)a(a-b)
7446^2 = 3a^2(a-b)(a+b)
= 3a^2(a^2 - b^2)
Since LHS is a perfect square, RHS must be too.
Therefore for some integer c, with a>c, we have a^2 - b^2 = 3c^3
Taking prime factors we have
2.17.73 = ac, satisfying a^2 - 3c^2 = b^2
So either ac = 2482 x 1 or 1241 x 2 or 73 x 34 or 146 x 17
by inspection, we can discount the first 2 possibilities
check 73^2 - 3(34^2) = 5329 - 3468 = 1861 (this is prime and therefore not a square)
check 146^2 = 3(17^2) = 21316 - 867 = 20449 = 143^2 (used calculator here!)
therefore a = 146, c = 17, b = 143
so sides must be 149, 292, 435
-Dom
-
modellingman
- Lemon Slice
- Posts: 621
- Joined: November 4th, 2016, 3:46 pm
- Has thanked: 602 times
- Been thanked: 368 times
Re: Triangle
I must admit to prior ignorance of Heron's formula and so perhaps had a longer task than Gengulphus (and psychodom).
My approach was as follows:
The triangle has sides a, b and c in decreasing order of length. h is the perpendicular length from side a to the vertex of b and c. The perpendicular splits a into two lengths of xa and (1-x)a and, in effect, splits the triangle into two right-angled triangles: the larger has sides xa, h and b and the smaller (1-x)a, h and c. Applying Pythagoras' Theorem to each yields two equations involving h^2 which after a bit of re-arrangement allows a solution for x to be derived as
x = (a^2 + b^2 - c^2)/(2*(a^2))
Plugging this into the re-arranged Pythagorean equation h^2 = b^2 - (xa)^2 yields (within a couple of lines)
h^2 = [(2*a^2*b^2 + 2*a^2*c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4)]/[4*(a^2)] [1]
Since the triangle's area of 7446 is ah/2 it follows that (a^2)*(h^2) = 4 * 7446^2
and substituting for h^2 using [1] it follows that
(2*a^2*b^2 + 2*a^2+c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4) = 16 * 7446^2 [2]
Given that b is even and a, b and c are an arithmetic progression the substitions of 2n+m, 2n and 2n-m are respectively made for a, b and c and plugged into the LHS of [2]. After a bit of tedious work [2] simplifies to
48 * n^2 * (n^2 - m^2) = 16 * 7446^2
or
3 * (n^2) * (n^2 - m^2) = 7446^2
Splitting 7446 into its prime factors (of 2, 3, 17 and 73) allows further simplification to
(n^2) * (n^2 - m^2) = 2^2 * 3 * 17^2 * 73^2 [3]
which is the same equation that Gengulphus derives using a different approach.
Like Gengulphus I identified that [3] implies n can only be one or more of 2, 17 and 73 factored together, yielding 7 possibilities for n. In a slightly different approach to Gengulphus, I plugged each possibility for n into [3] to derive a value for m and found that only one of the 7 yielded a real integer solution - namely when n is 146 m is 143 corresponding to a triangle with sides 435, 292 and 149.
My approach was as follows:
The triangle has sides a, b and c in decreasing order of length. h is the perpendicular length from side a to the vertex of b and c. The perpendicular splits a into two lengths of xa and (1-x)a and, in effect, splits the triangle into two right-angled triangles: the larger has sides xa, h and b and the smaller (1-x)a, h and c. Applying Pythagoras' Theorem to each yields two equations involving h^2 which after a bit of re-arrangement allows a solution for x to be derived as
x = (a^2 + b^2 - c^2)/(2*(a^2))
Plugging this into the re-arranged Pythagorean equation h^2 = b^2 - (xa)^2 yields (within a couple of lines)
h^2 = [(2*a^2*b^2 + 2*a^2*c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4)]/[4*(a^2)] [1]
Since the triangle's area of 7446 is ah/2 it follows that (a^2)*(h^2) = 4 * 7446^2
and substituting for h^2 using [1] it follows that
(2*a^2*b^2 + 2*a^2+c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4) = 16 * 7446^2 [2]
Given that b is even and a, b and c are an arithmetic progression the substitions of 2n+m, 2n and 2n-m are respectively made for a, b and c and plugged into the LHS of [2]. After a bit of tedious work [2] simplifies to
48 * n^2 * (n^2 - m^2) = 16 * 7446^2
or
3 * (n^2) * (n^2 - m^2) = 7446^2
Splitting 7446 into its prime factors (of 2, 3, 17 and 73) allows further simplification to
(n^2) * (n^2 - m^2) = 2^2 * 3 * 17^2 * 73^2 [3]
which is the same equation that Gengulphus derives using a different approach.
Like Gengulphus I identified that [3] implies n can only be one or more of 2, 17 and 73 factored together, yielding 7 possibilities for n. In a slightly different approach to Gengulphus, I plugged each possibility for n into [3] to derive a value for m and found that only one of the 7 yielded a real integer solution - namely when n is 146 m is 143 corresponding to a triangle with sides 435, 292 and 149.
-
cinelli
- Lemon Slice
- Posts: 553
- Joined: November 9th, 2016, 11:33 am
- Has thanked: 234 times
- Been thanked: 161 times
Re: Triangle
I congratulate modellingman on his alternative approach. He clearly enjoys algebraic manipulation as much as I do. It just shows that there is more than one way to skin a cat. But in his working he has come very close to proving Heron’s formula The area of a triangle is (1/2)*a*h or
D^2 = (1/4) * h^2 * a^2
and from his equation [1] for h:
h^2 = [(2*a^2*b^2 + 2*a^2*c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4)]/[4*(a^2)]
we can write
D^2 = (1/4) * [(2*a^2*b^2 + 2*a^2*c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4)]/[4*(a^2)] * a^2
and this is an equation for D entirely in terms of the three sides. After a little simplification this becomes
D^2 = (1/16) * (-a^3 – b^3 – c^3 + 2*b^2*c^2 + 2*a^2*c^2 + 2*a^2*b^2)
and this factorises to
D^2 = (1/16) * (a+b+c) * (b+c-a) * (a+c-b) * (a+b-c)
[although this step does rather depend on knowing the answer].
This reduces to Heron’s formula
D^2 = s * (s-a) * (s-b) * (s-c)
Cinelli
D^2 = (1/4) * h^2 * a^2
and from his equation [1] for h:
h^2 = [(2*a^2*b^2 + 2*a^2*c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4)]/[4*(a^2)]
we can write
D^2 = (1/4) * [(2*a^2*b^2 + 2*a^2*c^2 + 2*b^2*c^2) - (a^4 + b^4 + c^4)]/[4*(a^2)] * a^2
and this is an equation for D entirely in terms of the three sides. After a little simplification this becomes
D^2 = (1/16) * (-a^3 – b^3 – c^3 + 2*b^2*c^2 + 2*a^2*c^2 + 2*a^2*b^2)
and this factorises to
D^2 = (1/16) * (a+b+c) * (b+c-a) * (a+c-b) * (a+b-c)
[although this step does rather depend on knowing the answer].
This reduces to Heron’s formula
D^2 = s * (s-a) * (s-b) * (s-c)
Cinelli
-
Gengulphus
- Lemon Quarter
- Posts: 4255
- Joined: November 4th, 2016, 1:17 am
- Been thanked: 2628 times
Re: Triangle
One minor point to add to modellingman's answer: he starts it with "The triangle has sides a, b and c in decreasing order of length." but it isn't immediately obvious where he uses that order. The following proof (which I must have first been told around 45 years ago, and it was clearly nowhere near new then) might provide some insight about that.
Consider a triangle ABC. Draw the bisector of the angle at A and the perpendicular bisector of the side BC. Assuming they're not parallel, they meet at a single point - call it X. In diagrammatic form (and please excuse the diagram not being to scale - ASCII diagrams are a somewhat limiting medium, and a proper embedded image is rather more trouble than it's worth, as well as not working well with spoiler protection if anyone wants to use it in a solution):
in which angle BAX = angle CAX (since the line AX is the bisector of angle BAC), D is the midpoint of line BC and angles ADX and CDX are both right angles (since the line DX is the perpendicular bisector of line BC).
Now drop perpendiculars from X to the sides AB and AC and label the points where they intersect those sides as Y and Z respectively:
in which angles AYX, BYX, AZX and CZX are all right angles, because the lines XY and XZ were constructed to be perpendicular to the lines AB and AC respectively.
Now we have angle YAX = angle ZAX (by the construction of X in the first step above), angle AYX = angle AZX (by the construction of Y and Z in the second step) and distance AX in the AXY triangle = distance AX in the AXZ triangle (obvious), from which it follows that triangles AYX and AZX are similar, and thus that distance AY = distance AZ and distance XY = distance XZ. (Yes, I know that those conclusions look blatantly wrong in the ASCII diagram, but as I said, please excuse that - it's very hard to draw this ASCII diagram to be even as roughly to scale as the pencil-and-paper sketch I was shown ~45 years ago...)
But now we have distance XB = distance XC (because X is on the perpendicular bisector of BC), distance XY = distance XZ (just proved) and angle XYB = angle XZC (since they were both constructed to be right angles in the second step above). So triangles XYB and XZC are also similar to each other, from which it follows that distance YB = distance ZC.
But now we have distance AB = distance AY + distance YB = distance AZ + distance ZC = distance AC.
There is the remaining possibility that the bisector of the angle at A and the perpendicular bisector of the side BC are parallel and so don't meet at a point (or meet at every point along them), so that the point X hasn't actually been defined. But in that case, the bisector of angle A meets the side BC perpendicularly at a point E, say, and we have the following diagram (still not to scale):
But we now have angle BEA = angle CEA (since AE is perpendicular to BC), angle BAE = angle CAE (since AE is the bisector of BC) and distance AE in triangle BAE = distance AE in triangle CAE. It follows that triangles BAE and CAE are similar, so that distance AB = distance AC. (And also that distance BE = distance CE, so in fact this case is when the bisector of the angle at A is the perpendicular bisector of BC and not merely parallel to it. So I could actually have drawn the diagram to be more to scale - but doing so would have surreptitiously assumed the answer, so I reckoned it wasn't a good idea.)
So in either case, I've proved that distance AB = distance AC - i.e. that the arbitrary triangle ABC I started with must have two equal sides. O more briefly, that all triangles are isosceles triangles.
Or have I? If not, where's the flaw in the proof?
Gengulphus
Consider a triangle ABC. Draw the bisector of the angle at A and the perpendicular bisector of the side BC. Assuming they're not parallel, they meet at a single point - call it X. In diagrammatic form (and please excuse the diagram not being to scale - ASCII diagrams are a somewhat limiting medium, and a proper embedded image is rather more trouble than it's worth, as well as not working well with spoiler protection if anyone wants to use it in a solution):
. \
\
_A__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ | --__
_- \ | --__
B------------------------------------------\-----D-------------------------------------C
\ |
\ |
\ |
\ |
\|
X
|\
| \
in which angle BAX = angle CAX (since the line AX is the bisector of angle BAC), D is the midpoint of line BC and angles ADX and CDX are both right angles (since the line DX is the perpendicular bisector of line BC).
Now drop perpendiculars from X to the sides AB and AC and label the points where they intersect those sides as Y and Z respectively:
. \
\
_A__
_- \ --__
_- \ --__
-_ _- \ --__
-_ _- \ --__
Y- \ --__
_- -_ \ --__ /
_- -_ \ --__ /
_- -_ \ -Z__
_- -_ \ / --__
_- -_ \ / --__
_- -_ \ / --__
_- -_ \ | / --__
_- -_ \ | / --__
B------------------------------------_-----\-----D-----/--------------------------------C
-_ \ | /
-_ \ | /
-_ \ | /
-_ \ | /
-_\|/
X_
/|\-_
/ | \ -
in which angles AYX, BYX, AZX and CZX are all right angles, because the lines XY and XZ were constructed to be perpendicular to the lines AB and AC respectively.
Now we have angle YAX = angle ZAX (by the construction of X in the first step above), angle AYX = angle AZX (by the construction of Y and Z in the second step) and distance AX in the AXY triangle = distance AX in the AXZ triangle (obvious), from which it follows that triangles AYX and AZX are similar, and thus that distance AY = distance AZ and distance XY = distance XZ. (Yes, I know that those conclusions look blatantly wrong in the ASCII diagram, but as I said, please excuse that - it's very hard to draw this ASCII diagram to be even as roughly to scale as the pencil-and-paper sketch I was shown ~45 years ago...)
But now we have distance XB = distance XC (because X is on the perpendicular bisector of BC), distance XY = distance XZ (just proved) and angle XYB = angle XZC (since they were both constructed to be right angles in the second step above). So triangles XYB and XZC are also similar to each other, from which it follows that distance YB = distance ZC.
But now we have distance AB = distance AY + distance YB = distance AZ + distance ZC = distance AC.
There is the remaining possibility that the bisector of the angle at A and the perpendicular bisector of the side BC are parallel and so don't meet at a point (or meet at every point along them), so that the point X hasn't actually been defined. But in that case, the bisector of angle A meets the side BC perpendicularly at a point E, say, and we have the following diagram (still not to scale):
. |
|
_A__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
B----------------------------E---------------------------------------------------------C
|
|
But we now have angle BEA = angle CEA (since AE is perpendicular to BC), angle BAE = angle CAE (since AE is the bisector of BC) and distance AE in triangle BAE = distance AE in triangle CAE. It follows that triangles BAE and CAE are similar, so that distance AB = distance AC. (And also that distance BE = distance CE, so in fact this case is when the bisector of the angle at A is the perpendicular bisector of BC and not merely parallel to it. So I could actually have drawn the diagram to be more to scale - but doing so would have surreptitiously assumed the answer, so I reckoned it wasn't a good idea.)
So in either case, I've proved that distance AB = distance AC - i.e. that the arbitrary triangle ABC I started with must have two equal sides. O more briefly, that all triangles are isosceles triangles.
Or have I? If not, where's the flaw in the proof?
Gengulphus
-
cinelli
- Lemon Slice
- Posts: 553
- Joined: November 9th, 2016, 11:33 am
- Has thanked: 234 times
- Been thanked: 161 times
Re: Triangle
I think Gengulphus is being unduly modest here. In the final diagram of the thread "hexagon of hexagons",
viewtopic.php?f=73&t=16916
he has demonstrated his mastery of the ASCII diagram. Not only does he show a convincing drawing of 19 interconnecting hexagons, he also places in those hexagons coloured numbers.
Representing the bisector at point A is hard, I admit. But representing distances shouldn't be too difficult.
On my screen BC measures 11 1/2 inches. D is supposed to be the midpoint of BC and yet BD is 6 1/2 inches - way off! I believe X should be much lower down. X must be outside the triangle, although I have been unable to prove this. And since X should be lower down, when perpendicular lines XY and XZ are drawn, one of Y or Z will be inside the triangle whereas the other will be outside. All the congruent triangles in the "proof" check out and BY does indeed equal CZ. Suppose the triangle is such that Y lies on AB whereas Z lies on AC produced. Then
AB = AY + YB
whereas
AC = AZ - CZ
and this is where the proof breaks down.
Cinelli
viewtopic.php?f=73&t=16916
he has demonstrated his mastery of the ASCII diagram. Not only does he show a convincing drawing of 19 interconnecting hexagons, he also places in those hexagons coloured numbers.
Representing the bisector at point A is hard, I admit. But representing distances shouldn't be too difficult.
On my screen BC measures 11 1/2 inches. D is supposed to be the midpoint of BC and yet BD is 6 1/2 inches - way off! I believe X should be much lower down. X must be outside the triangle, although I have been unable to prove this. And since X should be lower down, when perpendicular lines XY and XZ are drawn, one of Y or Z will be inside the triangle whereas the other will be outside. All the congruent triangles in the "proof" check out and BY does indeed equal CZ. Suppose the triangle is such that Y lies on AB whereas Z lies on AC produced. Then
AB = AY + YB
whereas
AC = AZ - CZ
and this is where the proof breaks down.
Cinelli
-
Gengulphus
- Lemon Quarter
- Posts: 4255
- Joined: November 4th, 2016, 1:17 am
- Been thanked: 2628 times
Re: Triangle
Well solved, cinelli! And that explains where modellingman uses "The triangle has sides a, b and c in decreasing order of length." in his proof - though in fact he only needs "The triangle has sides a, b and c with a the longest." to establish that the altitude he uses does divide the base of the triangle into two parts whose lengths sum to the base length, rather than breaking the projected base line into parts whose difference is the base length. Of course, the proof can be adapted to have separate cases for the two possibilities, but choosing a to be the longest side halves the amount of detailed working required...
Gengulphus
Gengulphus
Return to “Games, Puzzles and Riddles”
Who is online
Users browsing this forum: No registered users and 33 guests