Lesson Find a triangle with integer side lengths and integer area
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<H2>Find a triangle with integer side lengths and integer area</H2> <H3>Problem 1</H3>The lengths of the sides of a triangle are positive integers. One side has length 17 and the perimeter of the triangle is 54. If the area is also an integer, find the length of the longest side. <B>Solution</B> <pre> Use the Heron's formula for the area of the triangle area = {{{sqrt(s*(s-a)*(s-b)*(s-c))}}}. Here s = 54/2 = 27 is the semi-perimeter, a = 17, b and c are two other sides. Since the perimeter is 54 and side "a" is 17, we have b + c = 54 - a = 54 - 17 = 37. Let "b" be the longest side of the triangle. Then b >= 37/2 = 18.5 and since b is integer, we can write b >= 19. Also, b is less than semi-perimeter b < 54/2 = 27; c = 37-b. Then the formula takes the form area = {{{sqrt(27*10*(27-b)*(27-(37-b)))}}} = {{{3*sqrt(2*3*5*(27-b)*(b-10))}}}. (*) So, we seek for the integer value of "b" in the interval 19 <= b <= 26, which makes the right side of expression (*) integer number. +----------------------------------------------------------+ | Then one of the factors (27-b) or (b-10) should be 5, | | which gives b = 25. | +----------------------------------------------------------+ Indeed, then b = 25 is the sought side length, and the area (*) is area = {{{3*sqrt(2*3*5*2*15)}}} = {{{3*sqrt(2*3*5*2*3*5)}}} = {{{3*sqrt((2*3*5)^2)}}} = 3*2*3*5 = 90 square units. Thus the triangle sides are a= 17, b= 25 and c= 37-25 = 12 units; the longest side is 25 units. The triangle inequalities are held, so such triangle does exist. All requirements of the problem are held. <U>ANSWER</U>. Such a triangle does exist, and its longest side is 25 units long. 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