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Heronian Triangles CURRICULUM INSPIRATIONS www.maa.org/ci www.globalmathproject.org WHAT COOL MATH! CURIOUS MATHEMATICS FOR FUN AND JOY SUMMER 2021 BONUS HERON’S FORMULA BONUS ISSUE When recently looking into triangles As mentioned in my June 2021 essay, the posessing integer side lenghs and integer following formula for the area of a triangle area—Heronian triangles—I was led expressed solely in terms of the side lengths down a rabbit hole of trying to of the triangle is attributed to Hero of understand various claims stated on Alexandria (ca. 10 CE – 70 CE) websites and in the standard literature about them. Claims, it seems, are often Heron’s Formula: If a triangle has side made without proof or just with cryptic lengths a , b , and c , then its area A hints to (possible) proofs. is given by This essay is my attempt to put together abc++ −++ abcabcabc −+ +− A = ⋅ ⋅⋅ in one spot a mathematically robust 2 2 22 compendium of results. © James Tanton 2021 One way to obtain this formula is to make use of two results from trigonometry. If θ HERONIAN TRIANGLES is the angle between the sides labeled a and b , then we have A triangle with integer side lengths and 1 possessing integer area A is today called A= absin (θ ) 2 Heronian. Actually, people tend to broaden this definition a little. c2=+− a 22 b2 ab cos θ ( ) Definition: A triangle with side lengths and area each a rational value is called Heronian. Any Heronian triangle in this sense can be scaled to one with integer side lengths and integer area: simply scale the triangle by a common denominator of all four rational numbers a , b , c , and A . The relation We’ll call Heronian triangle with integer 22 cos(θθ) += sin( ) 1 side lengths and area integer Heronian. then gives As we saw in the June 2021 essay: 2 c2−− a 2 b 2 +16 A2 = 4 ab 22. ( ) Result 1: Every integer right triangle is integer Heronian. Thus 16A2 is a difference of two squares and we have Result 2: There are infinitely many examples of integer Heronian triangles. 16A2 = (We showed that there are infinitely-many almost-equilateral integer Heronian 222 222 (22ab−++ c a b)( ab +−− c a b ) triangles, in particular.) The first term in the product is Going further … 2 ((ab+−=) c2 ) Result 3: The perimeter of any integer (abcabc+−)( ++) Heronian triangle is an even integer. and the second term is Reason: Suppose, to the contrary, that 2 (c2 −−( ab) ) = abc++ is odd. Then each of the integers (−+abcabc +)( −+) −++=++−abc abc2 a Thus abc−+=++− abc2 b 2 +−=++− 16A=( abc ++)( −++ abcabcabc)( −+)( +−) abc abc2 c is also odd, and so, from Heron’s formula, giving Heron’s formula. 2 16A is the product of four odd numbers, and so is odd. This is patently not the case! www.jamestanton.com www.gdaymath.com www.globalmathproject.org © James Tanton 2021 Fun Fact 6: The semi-perimeter of any Corollary 4: For any integer Heronian integer Heronian triangle is a composite triangle with side lengths a , b , c the number. abc++ value s = is an integer. 2 Reason: We have 2 The value s is called the semi-perimeter of A=−−− ss( a)( s b)( s c). the triangle. Heron’s formula can be presented as In the prime factorization of A2 , any prime that appears must do so an even number of A= ss( −−− a)( s b)( s c) times since A2 is square. If s were prime, then s must consequently appear an even with the quantity under the square root a count of times in the right side of the product of four integers for integer expression above. But each of the values Heronian triangles. sa− , sb− , and sc− is less than s and so this cannot be the case. It must be then Corollary 5: There is no integer Heronian that s is composite. triangle with a side of length 1. Nor is there one with a side of length 2. Result 7: For each interior angle θ of a Heronian triangle sin (θ ), cos(θ ) , and Reason: Any two sides of a triangle must tan θ are rational numbers. have lengths that sum to more than the ( ) length of the third side. Thus, if an integer triangle has two sides of lengths 1 and a , Reason: Using the notation of the previous the third side must also have integer length section, if a , b , c and A have rational a . But then 1++aa is not even, and so 2A values, then so too do sin (θ ) = , the triangle cannot be Heronian. ab cab222−− If an integer triangle has two sides of length cos(θ ) = and 2ab 2 and a , the third side has length either a + sin (θ ) or a 1. tan (θ ) = . cos(θ ) In the second case, 21+++aa is not even and so the triangle cannot be Corollary 8: No Heronian triangle possesses Heronian. In the first case, the triangle has an angle of 30 , 45 , or 60 . area A satisfying 22 Aa=( +111)( )( )( a −= 1) a − 1. Result 9: Any triangle with the sine of each But no two non-zero square numbers differ angle in the triangle a rational number is by 1, so this too is impossible. similar to a Heronian triangle. (If one side length of the triangle is rational, then the triangle is Heronian.) Reason: Suppose a triangle has angles α , β , and γ opposite sides of lengths a , b , and c , respectively. Scale the triangle so that the length a is rational. Then the law of sines gives www.jamestanton.com www.gdaymath.com www.globalmathproject.org © James Tanton 2021 sin ()α sin ()γ ba= ⋅ and ca= ⋅ sin ()β sin ()α and the area of the triangle is 1 A= bcsin ()γ . 2 These are all rational values. Result 10: Each altitude in a Heronian triangle has rational length. If the altitude meets a side of the triangle, it divides its length into two sections, each of rational length. As there are infinitely many integer right Reason: Suppose an altitude of a Heronian triangles, there are infinitely many isosceles triangle with side lengths a , b , and c has integer Heronian triangles. length h and meets the side of length c , Pasting together the 6-8-10 and 8-15-17 dividing it into sections of lengths c1 and c2 , as shown. right triangles produces the 10-17-21 integer Heronian triangle. Let θ be the angle shown. From result 7, sin ()θ and cos()θ are rational. Thus Euclid (ca. 300BCE) established that every integer right triangle has side lengths given ha= sin ()θ and ca1 = cos()θ and by c= cc − are rational too. (Also, 21 kp22− q 1 () A= ch shows that h must be rational.) 2 k()2 pq This reasoning applies if the triangle is kp()22+ q oblique and the altitude of the triangle lies outside the triangle. for some integers k , p , q with p and q sharing no common factor. The area of such PASTING TOGETHER RIGHT TRIANGLES a triangle is the integer 1 ⋅22 −⋅ = If two integer right triangles share a k() p q k()2 pq 2 . common leg length (not hypotenuse), then 2 22− pasting together the two triangles along k pq() p q that common leg gives an integer Heronian If we scale each side length down by the triangle. For example, we create the 5-5-6 factor kpq , then we see that every integer and 5-5-8 isosceles integer Heronian right triangle is similar to one with rational triangles by pasting together two copies of side lengths the 3-4-5 right triangle. www.jamestanton.com www.gdaymath.com www.globalmathproject.org © James Tanton 2021 pq22− pq22+ Classification of Heronian Triangles 11 , 2 , Every Heronian triangle is a scaled version pq pq of an integer Heronian triangle with sides pq22− and rational area . pq a= nm()22 + h 22 Any two such rational right triangles can be b= mn() + h pasted together along the sides of length 2 2 c=+−() m n mn h and the figure can be rescaled to then () produce an integer Heronian triangle. where m , n , and h are positive integers One can also “subtract” two such triangles with mn> h2 . to obtain oblique integer Heronian triangles. This particular Heronian triangle has area A= hmn() m +− n() mn h2 . Reason: Consider a Herorian triangle with rational side lengths a , b , c , rational area A , and draw an interior altitude of length h . Suppose this altitude meets the side of length c dividing it into sections of length c and c as shown. Leonard Dickson in his volume History of 1 2 the Theory of Numbers, Vol II (New York: Dover 2005) suggests that Euler (1707- 1783) may have taken this approach to classify all Heronian triangles, and that before this, Brahmagupta (ca. 590 – ca. 668) may have conducted a similar approach. After all, result 10 shows that every Heronian triangle is composed to two right triangles with rational side lengths pasted 2A We have that h = , c1 and c2 are together along a common edge. c rational by result 10. Many standard websites, however, present a different classification of Heronian Let m denote the rational number ac+ 1 , triangles without clear attribution and n the rational number bc+ . (Wikipedia and MathWorld, for instance). 2 The stated classification appears in Robert We have that Carmichael’s 1915 volume Doiphante Analysis (New York: Whiley and Sons, 1915), 222 which is the earliest source of it I have been h=−= a c11 ma() − c able to trace.
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