3-Isoincircles Problem. Trigonometric Analysis of a Hard Sangaku Chalenge.

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1 See discussions, stts, nd uthor profiles for this publiction t: 3-Isoincircles Problem. Trigonometric Anlysis of Hrd Sngku Chlenge. Article My 009 CITATIONS 0 READS uthor: Jesús Álvrez Lobo Profesores de Enseñnz Secundri 7 PUBLICATIONS 0 CITATIONS SEE PROFILE Some of the uthors of this publiction re lso working on these relted projects: Composition of sine wves: visul pproch to the Fourier theorem. View project Geometry Beyond Algebr View project All content following this pge ws uploded by Jesús Álvrez Lobo on 9 June 06. The user hs requested enhncement of the downloded file.

2 Scred Mthemtics : Jpnese Temple Geometry Fukgw Hidetoshi - Tony Rothmn Still Hrder Temple Geometry Problems Chpter 6 - Problem 3 A Remrkble properties in this construction. Let r be the inscribed circumferences inrdius (green color). Then: CH RS r. AMB perimeter squre of side BC perimeter. 3 OM BC. BH 3 r, CH r, BC 5r BHC is the Pythgoren r(3,,5). T O D S N H O O 3 B M R C

3 + é é α α + α ( ) r O P cos O M O N cos (tnα + tn α)cos α tnα cosα sin α tn α( cos α ) sin α sinα cosα + tnα sin α. A r Are of CHD Semiperimeter of CHD CH HD (sin )( cos ) CH + HD + DC sin + cos + sin cos sin. sin + cos + tn + + sec r (tn α sin α) r sin α - tn α sec α O r D P N r H O r 3 α O 3 B M C

4 By equling the clculted inrdius we obtin the following trigonometric eqution in the α ngle: or, sin α r r r tnα sin α, tn α secα tnα sin α tn α sec α, sin α tn α sec α tnα sin α tht cn be written s nd this is obtined, tn α sec α, i.e., sinα cosα sinα cosα tn α sec α, cos α + cos α ( tn α sec α) ( tn α sec α) ( + cos α) ( + cos α) ( + cos α) ( + cos α)( tn α sec α) cos α sin α tn α sec α + 0. By the vrible chnging tg α t, the bove trigonometric eqution becomes the lgebric eqution nd simplifying, t t t + t + 0, + t + t t t reduced cubic eqution, whose exct solutions could be obtined by the Crdno s formul, but it is not necessry since, by simple inspection, we see tht t is root of it. Moreover, this is the only permissible in the context of this problem. Indeed, eliminting this root, the other two, which re obtined from the resulting nd grde eqution, which re t, t, whose respective rcs tngent re α é 8 y α Pé 8, nd it is strightforwrd tht ABC cn not be isosceles for these vlues of α. tnα tn α 3 Thus, tnα, tn α nd tn α. tn α 3 tn α 7 3 Now it is esy to clculte the other prmeters ccording to this results: CH RS r t t ( + t) t 5t + 0 ( t) + t ( + t) t 8 CH sin sin α + t CH r tnα sin α 5 5 r 8 RS BR tn 3r tn( π α) 3r tn α 3r Without further difficulty cn be proved the remining properties:

5 perimeter of AMB BC 0r BM AM tn tn( π - ) tn[ π - ( π α ] tn α AM 7 α π t 5 AB sec sec( π - ) sec[ π - ( π α ] AB sin α t 7 α π + t Therefore, 5 BM + AM + AB ( ) BC 0 r OM BC BHC DHC re Pythgoren tringles (3,, 5) t 6 BH cos cos α BH 3r + t 5 t 8 HC sin sin α HC r ( BH, HC, BC) r(3,,5) + t 5 BC BC 5r In the first picture you cn see mny more reltionships; eg, AM bisects OH. Moreover, from the ngulr reltionships which re deducted directly from the nd figure, Consequently, O M tnα BC. + é α é. α + α tn tn( α π ) tn α, 7 vlue tht hd lredy been obtined to clculte the height of the tringle. Thus, the construction is only possible in isosceles tringles whose pir of equl ngles hs vlue such tht tn, 7 i.e., 73.7º To construct n isosceles tringle with these chrcteristics, we cn tke 7 units to the side nd on the uneven height, or 7 units for the inrdio, 35 for the side nd 60 on the uneven height.

6 Alterntive demonstrtion, chnging the reference system nd the wy we define the inrdius B( xb, yb) cos, 0 x y C( C, C) 0, sin x + x y + y M( xm, ym ), B C B C M( xm, ym ) cos, sin D( xd, yd) 0, sin A é é ADP + DOP O Q r DO y tn ( π DOP ) x + yd é rmo y tn ( x x M) + ym. O rbo y tn ( x x B) r HO y tn ( π P HO ) x y D O r P π π x O r H P O 3 B M C

7 Determintion of the incenter coordintes by intersections of pirs of bisectors: é rdo y tn x sin + Q rmo y cot x + cos sin rbo y tn x cos + r y x HO é y tn x sen + 3sin cos cot Q O ( x, y ) rdo r MO r x y cot x cos sin é + tn + cot Q cos tn y tn x + cos O ( x, y) rbo r HO r y x y + tn 3sin tn cos r tn tn + tn tn r + r 3 tn cot tn tn cos r cot + Performing now the chnge of vrible tn t, we hve: nd simplifying, t t t + 3 +, t t + t t t t t , eqution with no integer roots. By the vrible chnging z 3 t, the bove eqution becomes 3 z 7z 3z + 9 0, one of whose roots is z. Undoing the chnge of vrible is obtined t tn. The other 3 3 π two roots re t ±, both indmissible becuse they re relted to the rgument. Therefore, 3 tn tn 3 3 tn tn. tn tn 3 7 3

8 Problem 65 of REOIM Nº ISSN X. It is proposed generliztion of the problem studied bove, eliminting the importnt restriction tht implies the orthogonlity condition BH CD. It is therefore expected to be significntly hrder problem. After been prefixed the isosceles tringle ABC, there re degrees of freedom, tht cn be prmeterized by the ngles γ nd, elevtion ngles of CD nd BH, respectively. A r D O r O H h O 3 r 3 γ B BC 5 C For n nlyticl study, the choice of the reference system is criticl to the complexity of the resulting equtions. After severl ttempts, it is esy to be convinced tht the lloction shown in the figure below is the most convenient since it llows to define the equtions of two pirs of bisectors in the simplest possible wy, nd the inrdius r nd r 3 coincide in bsolute vlue with the ordinte of their incenters.

9 A D r O x P N O H r P y h P 3 r 3 O 3 γ B M C BH h H H( h, 0) rbo y tn x BHD γ + rho y tn ( x h) rbo y tn x 3 BHC γ + rho y cot ( x h) 3 NOP : r ON cos NOP (OM NM)cos NOP + γ OM tn MCO, NM tn NOP r tn cosγ sin γ NO + γ P γ, MCO

10 Coordintes of the incenters O y O 3 by intersection of bisectors nd the inrdius r nd r 3 : γ + tn x h γ + tn + tn h O ( x, y) rbo r HO r y r γ + tn tn γ + cot + cot y h γ + tn + tn γ + cot x 3 h γ + tn + cot h O3 ( x3, y3) rbo r 3 HO r 3 3 y3 r3 γ + tn cot γ + cot + tn y 3 h γ + tn + cot Identifying both inrdius, we obtin the equtions γ + γ + r r3 cot + cot cot + tn. + γ γ + h r r3 tn cosγ sin γ cot + tn. Eliminting the uxiliry prmeters nd h, h h sinγ BHC :, sinγ sin( γ + ) sin( γ + ) + γ γ + sin γ r r3 tn cosγ sinγ cot + tn, sin( γ + ) yields system of two trigonometric equtions in the prmeters γ nd, considering s known: or, γ + γ + cot + cot cot + tn + γ γ + tn cot γ cot + tn sin( γ + ) γ + γ + cot + cot cot + tn + γ γ + γ + γ + γ + tn cot cot tn tn cos

11 Developing, γ γ tn tn + tn tg tn + tn cot + + γ tn tn tn tn γ + tn tn γ γ γ γ tn + tn tn tn tn tn tn + + cot γ γ + tn tn tn γ γ tn tn tn tn γ tn + tn + γ tn tn γ γ γ tn tn tn tn tn tn tn + tg cot γ + + γ tn tn tn tn tn tn γ + + tn + tn γ γ γ γ γ tn + tn tn tn tn + cot tn + tn tn tn + + γ γ tn tn tn γ + γ γ tn tn tn tn tn tn γ γ tn tn tn tn tn + tg cot 0 γ + tn tn tn tn γ + tn + tn γ γ γ γ γ tn + tn tn tn tn tn tn + cot + tn tn + + γ γ tn tn tn γ 0 γ γ tn tn tn tn tn tn γ With the chnges of vribles, k tn, u tn, v tn, this trigonometric system becomes n lgebric system, + uv kv uv u + v k v v u + v k + u u u + v u + v u + v + 0 ku u v uv uv uv And by chnging uv z u + v in the eqution 3 we obtin the qudrtic eqution in z, + kv z + z 0. k v v

12 Solving, + kv + kv z. ± + v k v v k v + kv + kv In the context of this problem the solution z + v k v v k v is not cceptble becuse z < 0 uv >, but this is not possible s γ nd re cute ngles nd therefore γ 0 < tn <, 0 < tn <. So, we hve: + kv + kv z. + + v k v v k v Undoing the lst chnge of vrible, vlue: uv vz z u, u + v v + z nd replcing in this eqution z by its 5 + kv + kv v v k v + + v k v u + kv + kv v v k v v k v We obtined n eqution for u in terms of k (s known) nd v, but if we would eliminte u in replcing in this eqution the roots of 5, it would obviously led to n nlyticlly intrctble eqution for v: + kv + kv v v k v + + v k v u + kv + kv v + v? v k v + + v k v k + u u u + v u + v u + v + 0 ku u v uv uv uv Furthermore, it is trivil tht for ny isosceles tringle ABC there is unique solution becuse the three inrdius re monotonous nd continuous functions of the ngles γ nd (nyone cn be persuded of this through the empiricl demonstrtion tht this softwre provides). Therefore, we conclude tht the problem is solvble only by numericl methods, nd it is not possible to determine the eqution tht provides h r s explicit function of. View publiction stts