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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 β’ 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 Ο. β’ A quadratic irrational is a number of the form r + s π where r and s are rational, s = 0, and π β {0, 1} is a squarefree integer (i.e., π2 β€ π for all primes p β Z). β’ A line segment is rational or quadratic irrational provided its length is rational or quadratic irrational respectively. More Facts β’ Fact 1. The only rational values of the circular trigonometric functions at rational multiples of Ο are the obvious ones. β’ 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. β’ 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 π2 = π 2 + π 2 β 2ππ cos πΌ 2 βπ 2 β π 2 ) (π β Ξ± = cos β1 β2ππ Thus, Ξ± = 2 β52 β42 ) (3 cos β1 (β2β5β4) = β1 9 cos 10 = 0.4510268 β¦ Main GST Theorem. The right triangles with rational angles and with rational or quadratic irrational sides areβ¦? The (properly scaled) 45β45β90, 30β60β90, and 15β75β90 triangles. REDUCTION TO FINITELY MANY SIMILARITY TYPES. Suppose ΞABC 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 Ξ± and cos Ξ² has degree 1, 2, or 4 over β. Proof- As cos Ξ± = b/c β β(b, c), we have the tower of fields β β β(cos Ξ±) β β(b, c). The degrees of these extensions satisfy [β(b, c) : β] = [β(cos Ξ±) : β] · [β(b, c) : β(cos Ξ±)] where [β(b, c) : β] equals 1, 2, or 4 since b and c each have degree 1 or 2 over β. Lemma 2. If n > 2 and gcd(k, n) = 1, then 2ππ πππβ cos π π(π) = 2 Proof- Take π(n) as the number of integers j such that 1 β€ j β€ n and gcd( j, n) = 1. So if ΞΆ =cos(2kΟ/n) + i sin(2kΟ/n) is a primitive nth root of unity, then πππβ β(ΞΆ ) = π(n) and β(cos(2kΟ/n)) = β(ΞΆ + ΞΆβ) is the fixed field in β(ΞΆ ) of complex conjugation. Apparently, fact 1 follows from Lemma 2 with a bit of work. Recall: If p > 1 is prime and a β N, then Ο(ππ ) = ππ β ππβ1 by direct inspection. Also, Ο is multiplicative: if gcd(m, n) = 1, then Ο(mn) = Ο(m)Ο(n) Lemma 3. Ο(n) β₯ π/2. The result is clear for n = 1, so let n =2π π1 π1 β¦ππ ππ be a prime factorization of n where a β₯ 0, the ππ βs are distinct positive odd primes, and ππ β₯ 1 for each j . For ππ β₯ 3, π β β€+ , For the prime then Lemma 3. Ο(n) β₯β(π/2). 2, we have Ο(20 ) = 1 and π In either case, we have π(2 )β₯ (2 π 2 if a β₯ 1, )/ 2. Since π is multiplicative we combine to obtain Lemma 4. The radian measures Ξ± and Ξ² both lie in the set Proof- By Lemma 1, cos Ξ± has degree 1, 2, or 4 over Q. Let Ξ± = 2kΟ/n where gcd(k, n) = 1, k β N, and n > 2 (since Ξ± < Ο/2). By Lemmas 2 and 3, we need only consider the cases 3 β€ n β€ 128. π π 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 < Ξ± = 2kΟ/n < Ο/2. As Ξ± and Ξ² 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 {Ξ±, Ξ²} lies in the set T. EXPLICIT TRIANGLES Next we produce four explicit right triangles with algebraic side lengths. For each m β N, define πΉπ (x) = tan(m arctan x). These are the tangent analogues of the Chebyshev polynomials of the first kind for cosine. Let ΞΈ = arctan x; then The last equality defines the polynomials ππ (x),ππ (x)β β€[x]. Thus, each πΉπ (x) is a rational function with integer coefficients. If tan(kΟ/n) exists (i.e., k β‘ n/2 mod n), then tan(kΟ/n) is a root of πΉπ (x) and of ππ (x). In other words, the minimal polynomial of tan(kΟ/n) may be obtained by factoring ππ (x) over β€[x] using a CAS and then choosing the correct irreducible factor. (letβs see an example) Let n = 10. Then π10 (x) = 10π₯ 9 β 120π₯ 7 + 252π₯ 5 β 120π₯ 3 + 10x factors over β€[x] into π10 (x) = 2x(π₯ 4 β 10π₯ 2 + 5)(5π₯ 4 β 10π₯ 2 + 1). Calculation shows that tan(Ο/10) β 0 is not a root of (π₯ 4 β 10π₯ 2 + 5) so it must be a root of Ο(x) = (5π₯ 4 β 10π₯ 2 + 1), why? π Therefore tan = 10 ΞABC we have π = 5β2 5 . 5 Recalling the set T and 5 β 2 5, π = 5, and by Pythagorasβ theorem π = 10 β 2 5. Repeating this process for Ο/12, Ο/8, and Ο/5, we obtain the four triangles IβIV described in Table 2. Table 2. Data for right triangles IβIV, namely the radian measure Ξ± of an acute angle, the minimal polynomial Ο(x) of tan Ξ± over β, tanΞ± in radical form, and the side lengths a, b, c as in Figure 4. Each triangle IβIV appears to contain at least one side whose length has degree 4 over β. But looks can be deceivingβ¦ Observe, 6 β 2 5= ( 5 β 1)2 = 5 β 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 β 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 : β] = 2. So K = β( π) for some squarefree d β β€\{0,1}. i.e. If K is a number field, then the ring of integers of K is by definition: i.e. Where, β€[ π] = {π + π π | π, π β β€ } and 1 β€[ 2 + 1 2 π] = {π + π 2 π + 2 π | π, π, π β β€ } Define the norm of ΞΌ by: N(ΞΌ) = ΞΌΞΌβ = π 2 β π 2 d β β. The norm is multiplicative: N(ΞΌΞ½) = N(ΞΌ)N(Ξ½) for every ΞΌ, Ξ½ β K. In particular, the restriction of N to ππ is multiplicative and, takes integer values: N : ππ β β€. Using these tools we obtain our sufficient condition to recognize nonsquares in ππ . Proposition 2 Let K be a quadratic number field and let ΞΌ β ππ . If N(ΞΌ) is not a square in β€, then ΞΌ is not a square in ππ . Proposition 3 Let K be a number field and R = ππ . If Ξ±, Ξ², Ξ³ β R β {ππ } and Ξ±Ξ²2 = Ξ³2 , then Ξ± is a square in R. Proof - If Ξ±, Ξ², Ξ³ β R β {ππ } and Ξ±Ξ²2 = Ξ³2 , then π₯ 2 β Ξ± β R[x] has Ξ³/Ξ² as a root. Since ππ is integrally closed Ξ³/Ξ² β R, 2 thus, Ξ± = (Ξ³ /Ξ²) is a square in R. More Lemmas w/o Proof. Our Old example: 6 β 2 5= ( 5 β 1)2 = 5 β 1 Example: Consider D={1, 2, 3} which has elements that are linearly independent since 1 β a 2 β π 3 πππ πππ¦ π, π β β COMPLETION OF THE GST THEOREM The right triangles with rational angles and with rational or quadratic irrational sides are The (properly scaled) 45β45β90, 30β60β90, and 15β 75β90 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 36β54β90, which is represented by triangle IV with side lengths π= 5 β 2 5, π = 5, π = 10 β 2 5. Let K = β( 5) and recall ππ = 1 β€[ 2 + 1 2 5] β β€[ 5] N(5-2 5) = 5 is not a square in β€, so 5-2 5 is not a square in ππ by prop 2. Therefore πππβ π = 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 IVβ, is similar to triangle IV and satisfies the conditions in the theorem. All variables (except possibly Ξ») are rational integers and π, π1, π2 > 1 are all squarefree. By (1), x and y are not both zero, further y β 0. Otherwise (3) has degree 4 over β. We suspect that equations (1)β(3) imply that d = π2 = 5. But squaring equation (3) yields (π₯ + π¦ 5 )2 5 β 2 5 = (π + β 5)2 This equation implies that 5 β 2 5 is a square in ππ , but this is false, by prop 2. Thus, triangle IV does not exist. It remains to show that d = π2 = 5. Claim 1. d = 5. By eq. (2) we have (4) βπ₯ + π₯ 5 + π¦ 5π β π¦ π = π + π π1 . Since π¦ β 0, if 5 β€ π, then (4) contradicts the linear independence of roots (Lemma 7). Therefore π = 5π0 , with 5 β€ π0 since π is squarefree and we have, (5) βπ₯ + π₯ 5 + 5π¦ π0 β π¦ 5π0 = π + π π1 . If π0 > 1, then eq. (5) contradicts Lemma 7. Thus, π0 = 1 πππ π = 5. Claim 2. π₯ β 0. Proof- Otherwise y > 0 (since Ξ» > 0) and Since L.H.S. has degree 1 or 2 over β, lemma 6 implies 25π¦ 2 β 10π¦ 2 5 is a square in β€[ 5]. But the norm is 53 π¦ 4 and this is not a square over β€, a contradiction to prop 2. (Remember prop 2 associates a square over Z with itβs norm.) Claim 3. π2 = 5. Proof. Otherwise, square both sides of (3) and conclude, by Lemma 7, that the coefficient β2 π₯ 2 + 5π¦ 2 + 10π₯π¦ of 5 must equal zero. Setting this coefficient equal to zero and solving the resulting quadratic in x we obtain, 5π¦ ± π¦ 5 π₯= 2 This is a contradiction since x β β€ and y β 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 πΌ, π½ are standard values on the unit circle and π, π β β, then cos(πΌ β π½) = a + b π. Proof- By GST theorem there are only three such triangles down to similarity that cos(πΌ β π½) assumes values for. Since each is quadratic irrational so is cos(πΌβπ½). Q.E.D. Alternate Proof- By the identity, cos(πΌ β π½) = cos πΌ cos π½ +sin πΌ sin π½ Since the R.H.S. is the sum and product of rational or quadratic irrational values, so is the L.H.S.