Andrew Ryba Math Intel Research Final Paper 6/7/09 (revision 6/17/09)

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1 Andrew Ryb Mth ntel Reserch Finl Pper 6/7/09 (revision 6/17/09) Euler's formul tells us tht for every tringle, the squre of the distnce between its circumcenter nd incenter is R 2-2rR, where R is the circumrdius nd r is the inrdius. Not ll qudrilterls hve either circumcenter or n incenter, let lone both, but if one does, is there reltionship between the inrdius, the circumrdius, nd the distnce between the circumcenter nd incenter? Though this question is sked bout specil type of qudrilterl, in the end it is relly question bout system of two circles. Such system is determined by their rdii nd the distnce between their centers. The first thing we must think bout is how to go bout finding such formul relting the rdii to the distnce between the centers of the incircle nd circumcircle. We do not know for certin if such formul even exists. We know tht qudrilterl is cyclic if nd only if its opposite ngles re supplementry. The condition for qudrilterl to hve n inscribed circle is not so well known. There is n inscribed circle if the four ngle bisectors intersect t point. (This is becuse the center of ny inscribed circle must be equidistnt from the four sides. However point is equidistnt from two djcent sides if it is on the bisector of the ngle formed by the two sides. This shows tht n incenter must be point of intersection of the four ngle bisectors.) f we exmine n rbitrry cyclic qudrilterl whose ngle bisectors intersect (t the incenter of its inscribed circle), we would require formul relting R (the circumrdius), r (the inrdius), nd d (the distnce between the circumcenter nd the incenter). These quntities would depend on two djcent ngles of the cyclic qudrilterl. We could pln to express r nd d in terms of R nd the two djcent ngles nd then eliminte the ngle prmeters. However, we cn simplify the procedure by exmining cyclic qudrilterls where one of the ngles is fixed t 90 degrees. The other ngle will still be vilble s prmeter to eliminte in order to find reltionship between R, r, nd d. Now we re looking t kite, with two 90 degree ngles, nd nother pir of ngles which dd to 180 degrees. The kite A is shown in the digrm, nd hs incenter, circumcenter, both of which clerly lie long digonl. is obviously the midpoint of, nd thus hs length 2R.

2 y x A The line bisects both the ngles t nd. Therefore the incenter is t the point of intersection of nd the bisector of the ngle. y the ngle bisector theorem, / = y/x = tn. Also + = 2R Thus = 2R A And = (2R tn ) / ( 1 + tn ) = (2R sin ) / ( sin + cos ) And = R Let Z be rdius of the incircle, r, where Z intersects t Z (see the digrm below). Since Z is rdius, ngle Z = 90 degrees, nd Z is prllel to. So ngle Z =. 90 Z 90

3 Thus cos = r ( 1 + tn ) / (2R tn ) Therefore r = ( 2Rsin ) / ( 1 + tn ) = (2R sin cos ) / ( cos + sin ) = d = bs [ R ( (2 sin )/ (sin + cos ) 1)] = bs [ R ((sin cos )/ (sin + cos ))] Hence d 2 = R 2 ( ( 1 2sin cos )/ (1 + 2 sin cos )) so d 2 / R 2 = ( ( 1 2sin cos )/ ( 1 + 2sin cos )) nd (R 2 d 2 ) / (R 2 + d 2 ) = 2 sin cos So r 2 = 4 R 2 ( sin 2 cos 2 )/( sin cos )) nd r 2 = (R 2 ( (R 2 d 2 )/ (R 2 + d 2 ) ) 2 ) / ( 1 + (R 2 d 2 ) / ((R 2 + d 2 )) = (R 2 (R 2 d 2 ) 2 ) / ( (R 2 + d 2 ) 2 + (R 2 + d 2 )( R 2 d 2 ) ) Thus r 2 = (R 2 (R 2 d 2 ) 2 ) / (R 4 2 R 2 d 2 + d 4 + R 4 d 4 ) = (R 2 d 2 ) 2 / 2 (R 2 + d 2 ) Therefore 2r 2 ( R 2 + d 2 ) = (R 2 d 2 ) 2 And now we hve reltionship between R, r nd d for kite. Wht we need to do now is test whether this reltionship ppers to hold for ll cyclic qudrilterls tht hve n incircle. So my pln is to digrm generic cyclic qudrilterl tht hs n incircle with the help of Geometer s Sketchpd, nd test out the reltionship for it. n such qudrilterl A, opposite ngles must dd to 180 degrees (mking it cyclic), nd the ngle bisectors must meet t point (which is the incenter of its inscribed circle). To drw this qudrilterl, we cn strt with point s the incenter. Then we drw three rys through tht will contin the points A,, nd. The rys nd form ngles with the ry A tht we cll θ nd φ. We cn clculte tht the ngle of the qudrilterl t A must be 270 θ φ. (This is becuse the sum of the ngles A nd A is hlf the sum of pir of opposite ngles in cyclic qudrilterl which is 90. ut the sum of the ngles of qudrilterl A is 360 nd is 90 + θ + φ plus the ngle A.) The point A cn be nywhere on the ry A but the loctions of nd re then determined by the condition tht the lines A nd A mke ngles of 135 θ / 2 φ / 2 with the line A.

4 r = 2.27 cm R = 3.65 cm s A d = 1.05 cm ( R 2 -d 2 ) 2-2 r 2 ( R 2 +d 2 ) = 0.00 cm 4 q t K p r n the digrm, three lines A,, nd re shown s ordinry blue lines. A dotted blue ry (mrked p) tht mkes n ngle of 90 with is lso shown. The ngle between this ry nd the ry is 270 θ φ. A second dotted blue ry (mrked q) bisects this ngle. opies of the bisected ngle re plced t A to form the two bold lines A nd A. The condition tht nd bisect the ngles t nd determines the lines nd which re lso shown in bold. otted red lines (r nd s) mrk the perpendiculr bisectors of two sides of the qudrilterl A. They meet t the circumcenter of A. A third dotted red line (t) is perpendiculr from to A. This perpendiculr distnce gives the vlue of r. The distnce gives the vlue of R nd the distnce gives the vlue of d. A computer clcultion of (R 2 d 2 ) 2-2r 2 ( R 2 + d 2 ) shows result of 0. The result remins 0 s the point A nd the rys nd re moved in the Geometer s Sketchpd digrm. This suggests very strongly tht the reltionship must hold for ny cyclic qudrilterl tht hs n incircle. nline reserch now led me to result known s Poncelet s Porism tht cn use my clcultion for cyclic kites to prove the generl formul. lso found tht vrious forms of the formul were discovered by Steiner in (See ). The web reference indictes tht the problem hs been repetedly considered nd other erly solutions were found by Jcobi in 1823 nd Fuss in 1792.

5 Poncelet s Porism sttes tht if it is possible to find, for given n > 2, n n-sided polygon which is simultneously inscribed in circle nd circumscribed round circle, then it is possible to find exctly one such polygon with ny desired point belonging to s vertex. Q P d K Now let us ssume tht qudrilterl Q is inscribed in circle nd hs n inscribed circle. (n the digrm Q is the qudrilterl with bold blue edges.) Let be the center of circle nd be the center of circle. Let R nd r be the rdii of the circumcircle nd the incircle. Let d be the distnce between the centers of nd. Extend the line joining nd to meet the circle t point P. Poncelet s Porism tells us tht exctly one new qudrilterl K with P s vertex, s its circumcircle nd s its incircle cn be formed. (The qudrilterl K is shown with red sides in the digrm.) Since there is only such qudrilterl it must be the sme s its mirror imge cross the line. ut if qudrilterl is symmetric cross line through vertex it is kite. This proves tht K will hve to be kite. The circumcircle nd incircle of K re the sme circles nd. According to our formul for kites, which we proved erlier, 2r 2 ( R 2 + d 2 ) = (R 2 d 2 ) 2. Therefore this formul holds for the qudrilterl Q s well since Q nd K hve the sme system of two circles, nd thus R, r, nd d re the sme for Q nd for the kite K. This rgument shows tht the formul is true for ll cyclic qudrilterls with incircles.

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