The length of a right triangle is an integer, and its area and perimeter are equal. Does such a right triangle exist? If so, determine its three sides. If not, explain the reason

The length of a right triangle is an integer, and its area and perimeter are equal. Does such a right triangle exist? If so, determine its three sides. If not, explain the reason

Find the three sides of a right triangle whose sides are integers and whose area is equal to the perimeter

Let the shortest side of the triangle be x (x is an integer), the other right side be KX (k is greater than or equal to 1), and the hypotenuse be x (1 + K ^ 2) ^ 1 / 2. Note: ^ 1 / 2 is 1 / 2 power, and K ^ 2 is the square of K
From the topic meaning: 0.5kx ^ 2 = [1 + K + (1 + K ^ 2) ^ 1 / 2] x
The solution is x = 2 + 2 [1 + (1 + K ^ 2) ^ 1 / 2] / K
Let [1 + (1 + K ^ 2) ^ 1 / 2] / k = t,
Because x is an integer, the value of T is 0,1 / 2,2 / 2,3 / 2,4 / 2,5 / 2
The solution is k = t / (T ^ 2-1),
That is, t / (T ^ 2-1) is greater than or equal to 1, the verification shows that t = 3 / 2, 4 / 2 satisfies the meaning of the problem, when t is greater than 4 / 2, K is less than 1, when t is less than 3 / 2, K has no solution,
By substituting the value of T, k = 12 / 5,4 / 3 can be obtained
When k = 12 / 5, the three sides of the triangle are x 12x / 5 13X / 5
It is concluded that the triangles are 5, 12 and 13, which satisfy the meaning of the question
Because the area is square with the side length and the perimeter is linear with the side length, the length of the three sides of the triangle is unique
Similarly, when k = 4 / 3, we can get a triangle 6,8,10. When the three sides of the triangle satisfy the proportional relation of K, the length of the three sides is also unique

The perimeter of the triangle is 10, where both sides are equal and the length is an integer, find the third side length

Let x 10-2x ＜ 2x 10-2x ＞ 0  2.5 ＜ x ＜ 5 ∵ X be positive integers 〉 x = 3,4 〉 10-2x = 4,2