PROBLEM OF APOLLONIUS

Transcription

1 PROBLEM OF APOLLONIUS In the Jnury 010 issue of Amerin Sientist D. Mkenzie isusses the Apollonin Gsket whih involves fining the rius of the lrgest irle whih just fits into the spe etween three tngent irles of ifferent rii. The prolem ws first solve in etil y Rene Desrtes in Desrtes foun n the rtile sttes tht where,, re the rii of the three outer irles n is the rius of the irle just fitting into the gp forme y them when rought into ontt. A erivtion is not given sine the rtile is intene for the typil Sigm Xi reer n not ime iretly t mthemtiins. We remey this sitution here y eriving this result n lso etermine the re of the typil tringulr spe into whih the fourth irle just fits. Also some itionl vritions on the prolem re onsiere n worke out. Our strting point is the four irle igrm n the tringle forme y onneting the enter of the three outer irles y stright lines The semi-perimeter of the tringle is s so tht the Heron formul known to Desrtes yiels- AreTring le s s s s This re in turn must mth the re of the three tringles orere y the lue lines in the figure. These res gin n e otine vi the Heron formul n re-

2 ,, Are Are Are ACO BCO ABO Equting the res we otin the result- Although this result looks ifferent from tht given y Desrtes, it gives the sme nswer. For exmple, if 1 then we fin [/sqrt3] If -1,1/, n 1/, s in the exmple given in the rtile, the vlue of is foun to e 1/3. Next we look t the relte prolem of the re of the tringle with urve sies forme y the three outer irles. This re equls the re of tringle ABC minus the setors of the irles whose outer r form the sies of the inner tringle. One hs- / C B A AreInner where the ngles re otine from the lw of osines n re- ros ros ros C B A For the simplest se one tkes 1. This proues the ngles A B C ros0.5 π/3 so tht the re eomes AreInner sqrt3-π/ We hve use this result in some erlier investigtions on osilltory het trnsfer where one ws intereste in etermining the frtion of flui eing trnsporte through the interstiil spes in unle of open ene pillries.

3 One my generlize this prolem y sking for the imeter of the lrgest irle whih just fits into the str-shpe re forme y the surrouning of n irles n lso etermine the re of this str shpe re. By emning n/π rottionl symmetry of the str-shpe re, one hs the simplifition tht eh of the surrouning irles hs the sme rius R. A shemti of the prolem is s shown- One hs guiing regulr polygon with sies of length t whose vertexes we ple irles of rius R. Lines rwn inwr from the mile of eh of the polygon sies interset t point D. The length of these lines is h tn[π1/-1/n]. Next lines re rwn from the enter of eh irle to the sme point D. This results in n isoseles tringles with eh re equl to 1 1 AreIsoseles 1 h tn π n If one now sutrts the re of the two setors t the ottom of one of the isoseles tringles n multiplie the result y n, the re of the str region forme insie the surrouning irles will e- π π AreStr n tn[ 1 ] [1 ] n n n the rius of the lrgest irle entere t D will e- R π [se [1 ] n 1]

4 The irumferene of the str will e n. Working these numers out for n6, where the guiing polygon is hexgon, we finπ AreStr 6 3, Rius, Cirumferene 4π 3 We show you here onstrute figure of this n6 se- We point out tht the eges of the str o not orrespon the lssil steroi figure whih is generte y point on the periphery of smll yliner rolling within lrger yliner. The ifferene etween n stroi n one of our n4 strs is tht the sllops re somewht eeper for the str periphery s shown-

5 A prolem ssoite with these isussions els with the spe left when yliner of rius R is ple into right ngle orner. This spe n e looke s one fourth of the str re forme y four surrouning yliners of rius R eh. One fins- π π [ 4 ]tn π AreCorner n ovious result. Jnury 1, 010