Math challenge! (10 points to most elegant solution)?

we can do better.

if i don't wrong this question was on a recent China TST for IMO..the stronger statement (for a general triangle) is "let a triangle with integer sides;so if the inradious is 1 then the triangle is a right triangle".

about a year ago, when i still don't know this problem, i proposed a own(!!!) generalization of this one for a national competition: "let a triangle with integer sides; so if the inradious is integer then there exist only a finite number of triangles with this property".

however, i show the way for we get the result.

note. we define a,b,c the sides of the triangle and <a> the cyclic sum on this variables, for example <ab>=ab+bc+ca.

for hypotesis if the inradious is k then k=2S/p where S is the area and p the perimetre of the triangle. then by erone we can get: <a>*k= 1/2 sqrt( (<a^2>)^2 -2(<a^4>) ). we always use this notation,squares both member, and sostitute a=x+y, b=y+z, c=z+x, where x,y,z are unique positive rationals (in fact <x>=< (a+c-b)/2 > then (2x) and cyclic is integer). at the end we get a simple diofantea and we get the result, here there are all the main idea.

paolo

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if i must answer to your question (that is a very trivial corollary of mine) i can say for hypotesi: a+b+c=ab, where a and b are the catetes, and i can sobstitute a=m^2-n^2, b=2mn, c=m^2+n^2. we obtain 2m^2+2mn=2mn(m^2-n^2) implies 1=n(m-n) implies n=1 and m=2 so the triplet (3,4,5). :)))) bye

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@DrD, when you get b=2(a-1)/(a-2)>0 (so a>2) you can note that b=2+ 2/(a-2) implies directly a=3.

a² + b² = c²

a + b + c = ab

Multiply both sides by 2 and add a² + b² on both sides.

a² + b² + 2a + 2b + 2c = a² + 2ab + b²

a² + b² + 2a + 2b + 2c = (a + b)²

a² + b² + 2c = (a + b)(a + b - 2)

c² + 2c = (a + b)(a + b - 2)

c(c + 2) = (a + b)(a + b - 2)

Recall that a + b > c for any triangle.

Therefore c + 2 > a + b - 2.

c + 4 > a + b > c

c² + 8c + 16 > a² + b² + 2ab > c²

4c + 8 > ab > 0

8 > ab - 4√(a² + b²) > -4√(a² + b²)

With this constraint, we can guess all the possible values and see which are the only ones that work. It should be a = 3, b = 4 or a = 4, b = 3.

Of course, there is a lot of guessing involved here. But the best point here is that √(a² + b²) is an integer. We will also have to use rate of change as part of the things we'll use for the guessing.

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EDIT:

d = a + b

From our last equation:

(c + d) / (d - 2) = (c + d) / c

(c + d)(1 / (d - 2) - 1 / c) = 0

1 / (d - 2) = 1/c

c = a + b - 2

The only right triangle with this quality is (3, 4, 5).

EDIT:

Once I was in the frame of mind of looking for integer solutions, I found a much less algebraically intensive argument while I was out grocery shopping. It makes use of Heron's formula, which says that for any triangle with side lengths a, b, c, semi-perimeter s, and area A we have:

A^2 = s(s-a)(s-b)(s-c)

Setting s = A and assuming A is nonzero, this gives:

A = (A-a)(A-b)(A-c)

Now assume that A is a right triangle and c is the hypotenuse. Plugging in A = 1/2 ab and multiplying both sides by 4, we get:

2ab = (ab - 2a)(ab - 2b)(1/2ab - c)

2ab = ab(b - 2)(a - 2)(1/2ab - c)

2 = (a-2)(b-2)(1/2ab - c)

It is a standard fact that in a Pythagorean triple (a,b,c), a and b cannot both be odd. I'll supply a short proof of this below if you don't believe me. But granting this, we know that 1/2ab must be an integer because at least one of a,b is even. Certainly a-2 and b-2 are integers, so we have 3 integers whose product is 2. We have to be a little careful here. The only way to write 2 as the product of three positive integers is 2 = 1*1*2; however, there is no guarantee that all three factors are positive. Still, we know that the only possibilities for a-2 and b-2 are +/- 1 and +/- 2, and only one of them can be +/- 2. So since a and b are interchangeable, the only possibilities are:

a = 1, b = 1

a = 1, b = 3

a = 1, b = 4

a = 3, b = 3

a = 3, b = 4

We can compute the corresponding values for c using the Pythagorean theorem; they are sqrt(2), sqrt(10), sqrt(17), sqrt(18), and 5 respectively. The only case where c is an integer is a = 3, b = 4, c = 5.

Actually, looking back at Heron's formula, it wouldn't surprise me if one could use the cubic discriminant or something to find all possible integer sided triangles whose area is equal to its semi-perimeter; I'll leave that for another time. It's worth noting, however, that if you tweak this question even a little bit then it turns into a super hard algebraic geometry problem; for instance, the question of which rational numbers are the area of a right triangle with three rational sides rests on some deep conjectures in the theory of elliptic curves. But anyway.

Well, let's say the sides of the triangle are a,b,c with c the hypotenuse (all are positive). The semi-perimeter is 1/2(a+b+c) and the area is 1/2(ab), so for them to be equal we must have a + b + c = ab. On the other hand, for a right triangle we know that a^2 + b^2 = c^2. We would like to eliminate a variable; I'm not sure what the most elegant way to proceed is, but let's just follow our nose:

a + b = ab - c

(a+b)^2 = (ab-c)^2

a^2 + 2ab + b^2 = a^2 b^2 - 2abc + c^2

c^2 + 2ab = a^2 b^2 - 2abc + c^2

2ab = a^2b^2 - 2abc

a^2 b^2 = 2ab(1 + c)

We can assume that a and b are both nonzero (even if the degenerate triangle counts, its area is 0 and its semiperimeter is nonzero) and hence divide by ab to obtain:

ab = 2(1 + c)

c = 1/2(ab - 2)

Now we use the Pythagorean relation again:

a^2 + b^2 = c^2

a^2 + b^2 = 1/4(ab - 2)^2

a^2 + b^2 = 1/4(a^2 b^2 - 4ab + 4)

a^2 + 2ab + b^2 = 1/4(a^2 b^2 + 4ab + 4)

(a + b)^2 = 1/4(ab + 2)^2

a + b = 1/2(ab + 2)

2a - ab + 2b = 2

a(2-b) = 2 - 2b

a = 2* (b-1)/(b-2)

This relation gives a necessary and sufficient condition for two positive numbers a and b to define a right triangle whose semi-perimeter equals its area. So one comment is that this has tons of solutions; for example, a = 8/3, b = 5, c = 17/3 defines a right triangle whose area and semi-perimeter are both 20/3. Now that I think about it, given any right triangle it is possible to scale the side lengths until the area is the same as the semi-perimeter (not hard to show).

So I am guessing you are looking for integer solutions. Well, in order for a and b to both be integers, we should study the right-hand side of the expression for a given above. Note that b-1 and b-2 are always relatively prime (have no common factors), so if 2*(b-1)/(b-2) is to be an integer then we must have either that b-2 is 1 or b-2 is 2 (so that the 2's cancel). But this means that b is either 3 or 4, and hence a is either 4 or 3, respectively.

If a right triangle has the same area as its semiperimeter, then the radius of the inscribed circle = 1. If the tangent hypotenuse is at 45°, then the legs are 2 +√2. If the base of the triangle is 2, then the hypotenuse would be vertical. 3 is the only integer between 2 +√2 and 2. Hence, the only integer solution is 3, 4, 5.

Need more explanation why this is a proof? The area of any triangle is (1/2)rp, where r is the radius of the inscribed circle, and p is the perimeter. So, any hypotenuse of a right triangle that is tangent to the inscribed circle of radius 1 will result in a, b, c that meets the conditions. For symmetry reasons, we only need to consider hypotenuse angles 45° and greater. It's not hard to show that at 45° the base is 2+√2. At 90° angle, the base is 2. 3 is the only integer between the two. What else needs to be said?

OK you bailed yourself out by asking only for integer lengths.

sides are a, b, √(a^2 + b^2)

P = a + b + √(a^2 + b^2)

2A = ab

Set them equal and rearrange to get

b = 2*(a-1) / (a-2)

as anonymous dude has gotten

(a-1) and (a-2) are consecutive integers. So apart from a=3, (a-1) will never be divisible by (a-2).

Now can (a-1) be divisible by (a-2)/2?

Only if a = even, in which case a-1 is odd

So (a-2)/2 must also be odd

Further we get that a must be a multiple of 4 for this to work.

So our problem reduces to finding n such that

(4n-1) / (2n-1) is an integer

This requires that they have common factors, and if so their difference must also have those factors.

So we require that 2n-1, 4n-1 and 2n have common factors.

Since 2n and 2n-1 are consecutive integers, they do not have common factors, except for n=1 which corresponds to a=4.

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The only axiom I've relied on is that consecutive integers greater than 2 do not have common factors other than 1; and are not divisible by each other. Here's the proof.

If p is divisible by q, then you can write

p = k*q = q + (k-1)q

So the difference betwee p and q, which is

(k-1)*q must also be divisible by q.

In the case of consecutive integers, the difference between them is 1 and has no factor other than 1. Hence consecutive integers can not have common factors other than 1.

*EDIT*

bboy: Right, that's a much easier way.

There is no answer for this question. Because total of 5 odd numbers would be an odd number and it can't be 20. Therefore, its impossible that the butcher slaught 20 cows in 5 days if he cut the cows daily in odd numbers.

a^2 + b^2 = h^2

a=3 b=4 h=5

Therefore: 3^2 + 4^2 = 5^2 which is true.

Area is (b x h)/2 in the sum it's (a x b) /2 = (3 x 4)/2 = 6

3 + 4 + 5 = Its perimeter and if you divide it by 2 to find its "semi - perimeter" = 6

pythagorean theorem: a^2 + b^2 = c^2

Area of triangle is 1/2 b*h or 1/2 a*b

semiperimeter = 1/2(a+b+c)

1/2 ab=1/2 (a + b +c)

ab=a +b+c

ab= a + b + [(a^2+b^2)]^1/2

took a stab at it myself

semi parameter = (3+4+5)/2

= 6

area = 1/2 x base x height

= 0.5 x 3 x 4

= 6

hence proven

honestly i didnt cheated!!

HINT start with the Pythagorean Theorem