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Grade School Triangles Written by: Jack S. Calcut Presented by: Ben Woodford (pay attention: there Will be a test at the end) Definitions • An angle is rational provided it is commensurable with a straight angle; equivalently, its degree measure is rational or its radian measure is a rational multiple of π. • A quadratic irrational is a number of the form r + s where r and s are rational, s = 0, and ∉ {0, 1} is a squarefree integer (i.e., 2 ∤ for all primes p ∈ Z). • A line segment is rational or quadratic irrational provided its length is rational or quadratic irrational respectively. More Facts • Fact 1. The only rational values of the circular trigonometric functions at rational multiples of π are the obvious ones. • Namely 0, ±1/2, and ±1 for cosine and sine, 0 and ±1 for tangent and cotangent, and ±1 and ±2 for secant and cosecant. • Corollary 1. The acute angles in each Pythagorean triple triangle are irrational. Fact 2. The acute angles in each Pythagorean triple triangle have transcendental radian measures and transcendental degree measures. Proof- Not very Enlightening. Example- Take the commonly seen 3-4-5 Triangle. Together with the law of cosines We can evaluate for the interior angles Here 2 = 2 + 2 − 2 cos 2 − 2 − 2 ) ( ⇒ α = cos −1 −2 Thus, α = 2 −52 −42 ) (3 cos −1 (−2∙5∙4) = −1 9 cos 10 = 0.4510268 … Main GST Theorem. The right triangles with rational angles and with rational or quadratic irrational sides are…? The (properly scaled) 45–45–90, 30–60–90, and 15–75–90 triangles. REDUCTION TO FINITELY MANY SIMILARITY TYPES. Suppose ΔABC is a right triangle whose acute angles are rational and whose sides are each rational or quadratic irrational as in Figure 4. Lemma 1. Each of the numbers cos α and cos β has degree 1, 2, or 4 over ℚ. Proof- As cos α = b/c ∈ ℚ(b, c), we have the tower of fields ℚ ⊆ ℚ(cos α) ⊆ ℚ(b, c). The degrees of these extensions satisfy [ℚ(b, c) : ℚ] = [ℚ(cos α) : ℚ] · [ℚ(b, c) : ℚ(cos α)] where [ℚ(b, c) : ℚ] equals 1, 2, or 4 since b and c each have degree 1 or 2 over ℚ. Lemma 2. If n > 2 and gcd(k, n) = 1, then 2 ℚ cos () = 2 Proof- Take (n) as the number of integers j such that 1 ≤ j ≤ n and gcd( j, n) = 1. So if ζ =cos(2kπ/n) + i sin(2kπ/n) is a primitive nth root of unity, then ℚ ℚ(ζ ) = (n) and ℚ(cos(2kπ/n)) = ℚ(ζ + ζ’) is the fixed field in ℚ(ζ ) of complex conjugation. Apparently, fact 1 follows from Lemma 2 with a bit of work. Recall: If p > 1 is prime and a ∈ N, then ϕ( ) = − −1 by direct inspection. Also, ϕ is multiplicative: if gcd(m, n) = 1, then ϕ(mn) = ϕ(m)ϕ(n) Lemma 3. ϕ(n) ≥ /2. The result is clear for n = 1, so let n =2 1 1 … be a prime factorization of n where a ≥ 0, the ’s are distinct positive odd primes, and ≥ 1 for each j . For ≥ 3, ∈ ℤ+ , For the prime then Lemma 3. ϕ(n) ≥√(/2). 2, we have ϕ(20 ) = 1 In either case, we have (2 )≥ (2 and if a ≥ 1, 2 )/ 2. Since is multiplicative we combine to obtain Lemma 4. The radian measures α and β both lie in the set Proof- By Lemma 1, cos α has degree 1, 2, or 4 over Q. Let α = 2kπ/n where gcd(k, n) = 1, k ∈ N, and n > 2 (since α < π/2). By Lemmas 2 and 3, we need only consider the cases 3 ≤ n ≤ 128. 2 By using a CAS we compute for these values of n and find that n lies in the set {3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30}. For each of these values of n, one simply produces the corresponding values of k with gcd(k, n) = 1 and 0 < α = 2kπ/n < π/2. As α and β are complementary and lie in S, we obtain our desired reduction to a finite set of possible similarity types. REDUCTION TO FINITELY MANY SIMILARITY TYPES. Proposition 1. The multiset {α, β} lies in the set T. EXPLICIT TRIANGLES Next we produce four explicit right triangles with algebraic side lengths. For each m ∈ N, define (x) = tan(m arctan x). These are the tangent analogues of the Chebyshev polynomials of the first kind for cosine. Let θ = arctan x; then The last equality defines the polynomials (x), (x)∈ ℤ[x]. Thus, each (x) is a rational function with integer coefficients. If tan(kπ/n) exists (i.e., k ≡ n/2 mod n), then tan(kπ/n) is a root of (x) and of (x). In other words, the minimal polynomial of tan(kπ/n) may be obtained by factoring (x) over ℤ[x] using a CAS and then choosing the correct irreducible factor. (let’s see an example) Let n = 10. Then 10 (x) = 10 9 − 120 7 + 252 5 − 120 3 + 10x factors over ℤ[x] into 10 (x) = 2x( 4 − 10 2 + 5)(5 4 − 10 2 + 1). Calculation shows that tan(π/10) ≠ 0 is not a root of ( 4 − 10 2 + 5) so it must be a root of ψ(x) = (5 4 − 10 2 + 1), why? Therefore tan = 10 ΔABC we have = 5−2 5 . 5 Recalling the set T and 5 − 2 5, = 5, and by Pythagoras’ theorem = 10 − 2 5. Repeating this process for π/12, π/8, and π/5, we obtain the four triangles I–IV described in Table 2. Table 2. Data for right triangles I–IV, namely the radian measure α of an acute angle, the minimal polynomial ψ(x) of tan α over ℚ, tanα in radical form, and the side lengths a, b, c as in Figure 4. Each triangle I–IV appears to contain at least one side whose length has degree 4 over ℚ. But looks can be deceiving… Observe, 6 − 2 5= ( 5 − 1)2 = 5 − 1 This suggests we need extra machinery to distinguish between the squares and nonsquares among the irrational side lengths. ALGEBRAIC TOOLS Recall: a number field is a subfield of C whose dimension as a vector space over ℚ is finite. Being a subfield of C, each number field is an integral domain. A quadratic number field K is a number field with [K : ℚ] = 2. So K = ℚ( ) for some squarefree d ∈ ℤ\{0,1}. i.e. If K is a number field, then the ring of integers of K is by definition: i.e. Where, ℤ[ ] = { + | , ∈ ℤ } and 1 ℤ[ 2 + 1 2 ] = { + 2 + 2 | , , ∈ ℤ } Define the norm of μ by: N(μ) = μμ’ = 2 − 2 d ∈ ℚ. The norm is multiplicative: N(μν) = N(μ)N(ν) for every μ, ν ∈ K. In particular, the restriction of N to is multiplicative and, takes integer values: N : → ℤ. Using these tools we obtain our sufficient condition to recognize nonsquares in . Proposition 2 Let K be a quadratic number field and let μ ∈ . If N(μ) is not a square in ℤ, then μ is not a square in . Proposition 3 Let K be a number field and R = . If α, β, γ ∈ R − { } and αβ2 = γ2 , then α is a square in R. Proof - If α, β, γ ∈ R − { } and αβ2 = γ2 , then 2 − α ∈ R[x] has γ/β as a root. Since is integrally closed γ/β ∈ R, thus, α = (γ /β)2 is a square in R. More Lemmas w/o Proof. Our Old example: 6 − 2 5= ( 5 − 1)2 = 5 − 1 Example: Consider D={1, 2, 3} which has elements that are linearly independent since 1 ≠ a 2 ≠ 3 , ∈ ℚ COMPLETION OF THE GST THEOREM The right triangles with rational angles and with rational or quadratic irrational sides are The (properly scaled) 45–45–90, 30–60–90, and 15– 75–90 triangles. In this section, we determine whether there exist triangles of the last four similarity types in T with rational or quadratic irrational sides. We begin with the last similarity type 36–54–90, which is represented by triangle IV with side lengths = 5 − 2 5, = 5, = 10 − 2 5. Let K = ℚ( 5) and recall = 1 ℤ[ 2 + 1 2 5] ⊇ ℤ[ 5] N(5-2 5) = 5 is not a square in ℤ, so 5-2 5 is not a square in by prop 2. Therefore ℚ = 4, By Lemma 6, and thus triangle IV is ruled out. It remains to rule out all triangles similar to triangle IV. Suppose, by way of contradiction, that a triangle, called IV’, is similar to triangle IV and satisfies the conditions in the theorem. All variables (except possibly λ) are rational integers and , 1, 2 > 1 are all squarefree. By (1), x and y are not both zero, further y ≠ 0. Otherwise (3) has degree 4 over ℚ. We suspect that equations (1)–(3) imply that d = 2 = 5. But squaring equation (3) yields ( + 5 )2 5 − 2 5 = ( + ℎ 5)2 This equation implies that 5 − 2 5 is a square in , but this is false, by prop 2. Thus, triangle IV does not exist. It remains to show that d = 2 = 5. Claim 1. d = 5. By eq. (2) we have (4) − + 5 + 5 − = + 1 . Since ≠ 0, if 5 ∤ , then (4) contradicts the linear independence of roots (Lemma 7). Therefore = 50 , with 5 ∤ 0 since is squarefree and we have, (5) − + 5 + 5 0 − 50 = + 1 . If 0 > 1, then eq. (5) contradicts Lemma 7. Thus, 0 = 1 = 5. Claim 2. ≠ 0. Proof- Otherwise y > 0 (since λ > 0) and Since L.H.S. has degree 1 or 2 over ℚ, lemma 6 implies 25 2 − 10 2 5 is a square in ℤ[ 5]. But the norm is 53 4 and this is not a square over ℤ, a contradiction to prop 2. (Remember prop 2 associates a square over Z with it’s norm.) Claim 3. 2 = 5. Proof. Otherwise, square both sides of (3) and conclude, by Lemma 7, that the coefficient −2 2 + 5 2 + 10 of 5 must equal zero. Setting this coefficient equal to zero and solving the resulting quadratic in x we obtain, 5 ± 5 = 2 This is a contradiction since x ∈ ℤ and y ≠ 0. Thus, no triangle similar to IV has rational or quadratic irrational sides. In an attempt to follow this argument for triangle 1 we find that we cannot reach any contradiction. Therefore we have our third similarity type of the Main Theorem. Who cares? Example- If , are standard values on the unit circle and , ∈ ℚ, then cos( − ) = a + b . Proof- By GST theorem there are only three such triangles down to similarity that cos( − ) assumes values for. Since each is quadratic irrational so is cos(−). Q.E.D. Alternate Proof- By the identity, cos( − ) = cos cos +sin sin Since the R.H.S. is the sum and product of rational or quadratic irrational values, so is the L.H.S.