A little nticed right triangle Knstantine Hermes Zelatr Department f athematics Cllege f Arts and Sciences ail Stp 94 University f Tled Tled, OH 43606-3390 U.S.A.
A little nticed right triangle. Intrductin frm AB Γ Given a right triangle AB Γ, there is anther right triangle which naturally arises, and which seems t have escaped much attentin. In Figure, a nnissceles triangle (with right angle at the vertex A) is depicted. Let O be the midpint f the hyptenuse B Γ, which f curse is als the center f the triangles circumscribed circle. Cnsider the line segment AO ; it divides the right triangle issceles triangles ABO and AΓO, each having area AB Γ int tw nncngruent E, where E is the area f triangle AB Γ. Let and O be respectively the centers f the circumscribed circles f O triangles ABO and AΓO. It is triangle OO O, which is the subject matter f this wrk. Nte that since AB Γ is a nnissceles triangle, ne f the pints and O must lie in the exterir f O AB Γ, while the ther in its interir. Specifically if length A Γ > AB γ, then O lies in the exterir f AB Γ, while O in its interir (as in Figure while when < γ, it is the ther way arund). In Sectin 3, we easily shw by using a few simple gemetric arguments, that OO O is indeed a right triangle, similar t triangle AB Γ. In Sectin 4, we cmputatinally establish frmulas fr the lengths OO, OO, and O O ; in terms f triangle AB Γ s sidelengths,, γ. These are simple ratinal expressins. We als cmpute the area f OO O in terms f,, γ.
A little nticed right triangle In Sectin, we derive the frmula, fr any triangle AB Γ ; where R is the γ R 4E radius f the circumscribed circle, and E the triangle s area (see Figure ). We make use f this result in Sectin 4. In Sectin 5, we take a lk at the nearby trapezid, where is the O O midpint f the segment OB, and the midpint f O Γ. We calculate the trapezid s fur sidelengths, as well as its tw diagnal lengths d O and d O ; and als this trapezid s area. In Sectin 6, sme number thery enters the picture. We cnsider the case when AB Γ is a Pythagrean triangle, and we examine bth the triangle OO O and the trapezid in that case. Finally, in Sectin 7 we ffer three numerical examples. Figure
A little nticed right triangle 3. A derivatin f the frmula R 4E In Figure, a triangle γ ABΓ is inscribed in a circle f center O and radius R. If θ is the degree (r radian) measure f the triangle s internal angle at the vertex A; and als, with B Γ, BB R ( B is the pint n the circle, antidiametrical f B ), R OΓ OA OB ; we have frm the right triangle BΓ B, R sinθ. rever, let B be the ft f the perpendicular drawn frm B t the side A Γ ; and h BB. Then, h γ sinθ ; sin θ. And s we btain, γ h Rsinθ R h Rh γ. γ Als, E h, where γ AB, AΓ, and E stands fr the area f the triangle AB Γ. Slving fr h in terms f E gives E h ; and thus Rh γ further yields E γ R γ R () 4E Figure
A little nticed right triangle 4 3. The right triangle OO O Nw we g back t Figure. It is easy t see that O OO is a right triangle, in O fact similar t AB Γ. Since is the center f the circumscribed circle f the issceles triangle AOB, with O AO OB, it fllws that lies n the bisectr f the angle < AOB, since that angle bisectr cincides with the perpendicular bisectr f AB. In ther wrds, OO is perpendicular t AB. Thus, if ϕ is the degree measure f the angle < AΓB, then since OO is parallel t Γ A (bth being perpendicular t AB ), the angles < AΓB and < O OB must have the same degree measure ϕ, based n the relevant therem f Euclidean gemetry which pstulates that if tw parallel straight lines are cut by anther transversal line, then any pair f tw angles, ne internal the ther external, and lying n the same side f the transversal; must have the same degree measure. Nw, since < O OB has degree measureϕ, s des the angle < AOO, since as we have explained abve, the segment OO lies n the bisectr f the angle < AOB. rever, since the segment OO lies, by definitin, n the perpendicular bisectr f fllws that OO is parallel t AB (since AB is perpendicular t A Γ ; it A Γ as well). Applying again a fundamental therem f Euclidean gemetry, which pstulates that if tw straight parallel lines are cut by a transversal, then any pair f interir alternate angles (i.e. lying n ppsite sides f the transversal), must have the same degree measure. Let ω be the degree measure f angle < OAB. Since the degree measure f the angle < OAB is ω (same as the measure f the angle < OBA), it then fllws that the degree measure f the angle < AOO is alsω. Finally, since the line segment OA lies
A little nticed right triangle 5 in the interir f the angle < O OO, it fllws that the degree measure f the angle < O OO is equal t the sum f degree measures f the angles < O OA and < AOO ; hence it is equal t + ϕ 90, which prves that O OO is a right triangle with right ω angle at. The triangle OO is similar t the triangle O O AB Γ, since it is easily seen that the angle < O O O has measureω, while the angle < OO O has measureϕ. Immediate Observatins With ΓB, A Γ, AB γ ; we have Γ O O B AT 4 TO AO ΓO BO ΓK KA. The fllwing seven angles have the same degree measureϕ : < AΓB, < AOO, < O OB, < OAO, < ΓAO, < O O, and < O O. On the ther hand the fllwing angles have degree measureω : < ΓBA, OAB, < AO O, O O <, < OA. < O O O Als, we have R O O O A O B and O O O Γ O A, where is the R R radius f triangle AOB s circumscribed circle; R the radius f triangle AOΓ s circumscribed circle.
A little nticed right triangle 6 R O O 4. Cmputing the lengths, and, R If and E are the areas f the issceles triangles AOB and AOΤ respectively; E and E the area f the triangle Cmbining () and () we see that, similarly, AB Γ then, γ E E E () 4 ( ) ( ) OA OB AB γ OO R ; E γ 4 ( ) ( ) OA OΓ AΓ OO R. E γ 4 Altgether, R 4 R 4γ (3) Let x O. Frm the cngruent right triangles O TO and O O, we see that O O T ; and s O T x as well. Similarly we have y O OT. Cnsequently, O O OT + TO x + y (4) Frm the right triangle O O, x R by (3) ; 4 ( 4) 6 6 4 γ x 4 γ ; 4 x (since + γ ). Wrking similarly with the triangle O O we find that y 4γ. Putting these tw tgether,
A little nticed right triangle 7 γ x, y (5) 4 4γ Frm (4) and (5) O γ O 4 + 4γ ; which when cmbined with + γ gives, 3 O O (6) 4γ Als, the area f the right triangle OO O is equal t R R ; which when cmbined with (3) yields, Area f triangle 4 OO O (7) 3γ 5. The nearby trapezid OO We have already cmputed, in terms f,, andγ ; three f the fur sidelengths f this trapezid. Namely, the lengths O O, O x, and O y accrding t frmulas (5) and (6). The furth sidelength is O 4. rever, the area f the trapezid O O is given by, area f O O ( x+ y ) ( x + y), and by using (5) and + γ we arrive at, 4 4 Area f trapezid O O (8) 6γ
A little nticed right triangle 8 There are tw diagnals O and O f the trapezid OO. Let and d d be their lengths respectively. Frm the right triangles and O we btain O (since ) x + ( ) d and y + ( ) d. Implementing (5) further gives, 4 4γ d γ + 4, d + 4γ (9) 6. When AB Γ is a Pythagrean triangle In this sectin we venture a bit int number thery. If all three sidelengths,, γ are integers, then the triangle AB Γ is what is cmmnly referred t as a Pythagrean triangle. Clearly then, frmulas (3) and (6) tell us that the sidelengths R, O O f the triangle, while nt necessarily integers, are certainly ratinal R, numbers. Since {, γ } r [3]) we must have,, is a Pythagrean triple, as it is well knwn (may refer t [], [], ( m + n ), δ ( mn), γ ( m n ), δ δ (0) fr sme psitive integers δ, m, n such that ( m, n) (i.e. m and n are relatively prime), m > n, and m + n (md ) (i.e. m and n have different parities ne f them is dd, the ther even).
A little nticed right triangle 9 (Nte: Obviusly, δ ( m n ), γ δ ( mn) ; is the ther pssibility. But there is n need t distinguish between tw cases, since and γ are interchangeable, there is n additinal infrmatin abut r γ given here). That the abve parametric frmulas (invlving three parameters: δ, m, and n) generate the entire family f Pythagrean triples is a well-knwn fact in number thery. Almst every undergraduate text r bk in number thery, has sme material n Pythagrean triangles (typically the parametric frmulas and their derivatin and sme exercises n Pythagrean triples.) Let us apply frmulas (0) n (3), (6), and (7). After sme algebra we btain, R δ Area f ( m + n ) δ ( m + n ), R, 8mn 4( m n ) 4 δ ( m + n ) triangle OOO 64mn( m n ) O O δ 8mn 3 ( m + n ) ( m n ) () Likewise (5), (8), and (9) when cmbined with (0); they prduce δ x Area f And d ( m + n )( m n ) δmn( m + n ), y 8mn ( m n ) 4 δ ( m + n ) trapezid OO 3mn( m n ) 4 4 4 δ ( m + n ) m + 4m n + n δ ( m + n ) m m, d ( 8mn m n ) n + n 4 ()
A little nticed right triangle 0 Nw, let us cnsider the questin: fr what chice f the parameter δ, are all three R actually integers? First nte that since (, n) ratinals, R, OO m + n ( md), it fllws that, t t ( + n ), 8mn( m n ) ) m, m and fr any values f the nnnegative integer expnents and t. (This can be assigned as an exercise in an elementary thery curse). This then prves that, ( m n) 8mn) ( m + n ), 4( m n ) t 3 ( ) ( m + n ), 8mn( m ) +, n (3) Further, we als knw frm number thery that if an integer a divides the prduct bc f integers b and c; and ( ). Then a must be a divisr f c. Fr this, refer t [] r [3]. a, b Which shws that in view f (3) and frmulas (), and in rder fr all three R, O O t be integers, it is necessary and sufficient that δ be divisible by all three R, integers 8mn, 4( m n ), and 8 ( m n ) number thery. mn. We need ne mre basic result frm An integer δ is divisible by the psitive integers a, a, K, if, and nly if, δ is a r divisible by L, the least cmmn multiple f integer divisible by the r numbers). a,, a, K a r (L being the smallest psitive
A little nticed right triangle Applying the abve with a 8mn, a 4( m n ), and a3 8mn( m n ) Their least cmmn multiple is L 8mn( m n ) (the reader shuld verify).. The right triangle OO is a Pythagrean ne, precisely when ( δ K 8mn m n ), O where K can be any psitive integer. A Pythagrean triangle is called primitive, if its three sidelengths have n factr in cmmn, ther than. When δ, and nly then, the triangle AB Γ is primitive. In OO ( ) that case, all three sidelengths R, R, OO f triangle O are prper ratinal numbers, nne f them being an integer. On the ther hand, if δ K 8mn m n, bth right triangles AB Γ and OO O are nnprimitive; OO O is nnprimitive since ( R O O m + n ) is a cmmn factr in the case f, R, and. In cnclusin, if OO O is a Pythagrean triangle, it is always nnprimitive. Next, we apply similar cnsideratins t the trapezid O O, whse sidelengths are x, y, O O, and ( m + n ) δ. Using the same type f reasning (as in the case f the triangle OO O ), and by virtue f frmulas () and cnditins (3), we see nce again that, As in the case f triangle OO O, the trapezid OO will have all fur f its ( ) sidelengths integers if, and nly if, δ K 8mn m n, where K is a psitive integer. Als, in this case, the trapezid s area will als be an integer, as () clearly shws.
A little nticed right triangle And this brings us t the last questin in this paper. d d ( ) What abut the diagnal lengths and in (), fr δ K 8mn m n? Can either f them be an integer? The answer is n, neither nr d can be integers ( ) d when δ K 8mn m n. First, as we can easily see frm (), will be an integer d fr thse integer values f δ, if and nly if, the real number m + 4 4 + 4m n n is an integer. Similarly d will be an integer precisely when m + 4 4 m n n is an integer. Cnsider the fllwing result frm number thery. Let r and l be psitive integers. Then r l (the rth rt f l ) is a ratinal number if, and r nly if, l is the rth pwer f sme psitive integer i; that is l i. Equivalently, r l is either an integer r an irratinal number; the frmer ccurring precisely when r l i. In particular, the square rt (the case r ) f l will be either an integer r an irratinal number, the frmer ccurring exactly when interested reader may refer t [] r [3]. l i, fr sme psitive integer i. The Applying this t abve, we see that d will be integer precisely when d 4 4 m + 4m n + n i, fr sme psitive integer i. Which means that ( m, n, i wuld be 4 4 a slutin in psitive integers, t the Diphantine equatin x + 4x y + y z. Accrding t L.E. Dicksn s mnumental bk (see []), Euler was the first histrically knwn individual t prve that all the slutins t the abve equatins are given by, ) x y δ, z 4δ,
A little nticed right triangle 3 where δ can be any psitive integer. Thus, under the cnditin ( x, y), the abve equatin has nly ne slutin, ( ) namely x y, z 4. The integers m and n satisfy the same cnditin, m, n. But m and n have different parities (ne is even; the ther dd); thus, ( m, n) (,), which clearly shws that ( m, n, i) cannt really be a slutin t the Diphantine equatin 4 4 x + 4x y + y z, and s, m + 4 4 + 4m n n must be an irratinal number. We argue similarly in the case f : in rder fr d t be ratinal, it is d necessary and sufficient that m 4 4 m n + n i, fr sme psitive integer i; 4 4 m m n + n i, which means that ( m, n, i) wuld be a slutin, in psitive integers, 4 4 t the Diphantine equatin x x y + y z. L. Pcklingtn has prved (see [4]), that all the slutins in psitive integers, f the last equatin, are given by x y δ, z δ, where δ can be any psitive integer. x, y y z Under the cnditin ( ), there is nly ne slutin, x. Since (, n) m, and m + n ( md), nce mre we see that ( m, n, i) abve cannt be a slutin. Thus, m + 4 4 m n n must be an irratinal number. Pcklingtn s prf is als succinctly presented in W. Sierpinski s bk (see [3]). When ABΓ is a Pythagrean triangle, the tw diagnal lengths and d are always d irratinal numbers.
A little nticed right triangle 4 In a paper by S. Sastry (see [5]), a family f Hern Quadrilaterals was presented. These are quadrilaterals which have all fur sidelengths integers, integral area, and integral diagnal lengths. δ, the family f trapezids O btained, are never Hern Fr K8mn( m n ) O Quadrilaterals. They have all fur sidelengths OO, x, y, and being integers, integral area, but the tw diagnal lengths and d are always irratinal numbers. d Finally, fr δ K8mn( m n ),, and γ ; yield, the frmulas in () and (), and including ( )( ) R K mn( m + ) m n m, O O K ( m + ) 3 R +, K n n Area f triangle OO O K mn ( m n )( m + ) 4 n ( m n ) ( m + n ), y K 4 ( mn) ( m n ) x K + Area f trapezid O O K mn( m n )( m + ) 4 n 4 4 ( m n )( m + n ) m + m n d K 4 + n 4 4 ( m + n ) m m n d K 4mn + n ( m n )( m + n ), K 6( mn) ( m n ) γ K 8mn( m n ) K 4mn n
A little nticed right triangle 5 7. Numerical Examples Table (Fr triangle AB Γ ) γ K, m, n 40 9 44 K, m 3, n 30 880 00 K, m 4, n 860 3840 700 R R O O Table Area OO O x y Area O O d d K, m, n 75 00 5 3750 45 80 0 5 6 40 3 7500 K, m 3, n 845 08 97 856830 35 87 560 65 60 3 60 73660 K, m 4, n 4335 3 493 5060 385 088 4080 55 48 7 48 0050
A little nticed right triangle 6 8. Clsing Remarks Remark : It easily fllws frm (3) and + γ ; that R + R 4 4. (4) Equatin (4) implies the existence f a right triangle with leglengths R and R ; and hyptenuse length. Nte that each f the line segments Γ, O, O, and B, 4 in Figure, has length 4. Nw, given a line segment DE f length l, ne can cnstruct, in standard Euclidean fashin, a line segment having length l, as illustrated in Figure 3: Just draw the line perpendicular t DE (standard Euclidean cnstructin) and chse n it a pint F, s that FE. Then draw the perpendicular at F t the line segment DF, and let it intersect the straight line n which DE lies, at the pint G. The three right triangles EDF, EFG, and FDG, are all similar; and frm the similarity, it fllws that FE DE EG EG ; EG FE l. l Thus, given three line segments with lengths cnstruct three line segments having lengths; R, R, R 4 R,, and mentined abve; we can, and frm there the abvementined right triangle. One wnders what prperties this right triangle may have. Nte that it wuld be similar t R γ AB Γ ; since R, by (3).
A little nticed right triangle 7 Figure 3 Remark : Frm Figure, 0 < γ < 0 < ϕ < 45 < ω < 90 ; and ω + ϕ 90. Als (frm Figure r by (3)) nte that 0 < < R and 0 < γ < R. By (3), it is clear that R γ if, and nly if, R. A brief calculatin shws that this ccurs precisely when ( ) 4( ) + 0 + γ 4γ γ γ. Slving this quadratic equatin yields, tanω 3 r + 3. But tan ω > ; and s, γ tan ω + 3 tan 75 (see nte belw); ω 75 in this case. The fllwing can be readily established: ) 0 ϕ < 45 < ω < 60 0 < R < R < γ <. < ) 0 < ϕ < 45 < ω 60 ; then ϕ 30, γ, 3 ; 0 R < R γ <. In < this case, all fur pints B O, T, and O ; are aligned., 3) 0 ϕ < 45 < 60 < ω < 75 0 < R < γ < R <. < 4) 0 < ϕ < 45 < 60 < ω 75 ; ϕ 5. In this case R γ and R. The right triangles AB Γ and OO O are cngruent. 5) 0 < ϕ < 45 < 60 < 75 < ω < 90 0 < γ < R < < R.
A little nticed right triangle 8 Nte: With regard t the 75 angle. By using the duble-angle identity fr the tangent and tan 30 3 ; ne can establish tan5 3. Thus, tan 75 ct5 + 3 + 3 ( )( ) 4 3 + 3 tan5 3 3 + 3. References. Rsen, Kenneth H., Elementary Number Thery and its Applicatins, 993, Addisn- Wesley Publishing Cmpany (there is nw a furth editin as well), 544 p.p. ISBN 0-0-57889- a) Fur Pythagrean triples, see pages 436 44 b) The result which states that the nth rt f a psitive integer is either an integer r therwise irratinal; fllws frm Th.. in that bk, fund n page 96. a ( ) c) The result which states that if divides the prduct bc and a, b ; then a must divide c ; is stated n page 9 as Lemma.3.. Dicksn, L. E., Histry f Thery f Numbers, Vl. II, AS Chelsea Publishing, Rhde Island, 99. ISBN: 0-88-935-6; 803 p.p. (unaltered text reprint f the riginal bk, first published by Carnegie Institute f Washingtn in 99, 90, and 93) a) Fr material n Pythagrean triangles and ratinal right triangles, see pages 65 90. 4 4 b) Fr the Diphantine equatin x + 4x y + y z, see page 635.
A little nticed right triangle 9 3. Sierpinski, W., Elementary Thery f Numbers, Warsaw, Pland, 964, 480 p.p. (n ISBN number). re recent versin (988) published by Elsevier Publishing, and distributed by Nrth-Hlland. Nrth-Hlland athematical Library 3, Amsterdam (988). This bk is available by varius libraries, but it is nly printed upn demand. Specifically, UI Bks n Demand frm: Pr Quest Cmpany, 300 Nrth Zeeb Rad, Ann Arbr, ichigan, 4806-356 USA; ISBN: 0-598-5758-3 a) Fr a descriptin and derivatin f Pythagrean triples, see pages 38 4. a bc a, b b) Fr the result which states that if an integer is a divisr and ( ) ; then a must be a divisr f c ; see Therem 5 n page 4. c) Fr the result which states that a psitive integer is equal t the nth pwer f a ratinal number if, and nly if, that psitive integer is the nth pwer f an integer (which is equivalent t the statement that the nth rt f a psitive integer is either an integer r an irratinal number), see Th. 7 n page 6. 4 4 d) Fr the Diphantine equatin x + x y + y z, see pages 73 74. 4. Pcklingtn, H. C., Sme Diphantine impssibilities, Prc. Cambridge Philsphical Sciety, 7 (94), p.p. 08 5. K. R. S. Sastry, A descriptin f a family f Hern Quadrilaterals, athematics and Cmputer Educatin, Winter 005, p.p. 7-77