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Question Number 17580 by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

Commented by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

triangle ABC,is given with: AB=c,BC=a,AC=b red lines,are perpendicular bisectors of sides. 1)find radius of a circle that passes trough midpoints of red lines. 2)if ′Z′,be the center of this circle, find: PZ (in terms of: a,b,c) trial case: a=15.93,b=15.76,c=12.51 ⇒r=2.16 ,PZ=0.81 .

triangleABC,isgivenwith:AB=c,BC=a,AC=bredlines,areperpendicularbisectorsofsides.1)findradiusofacirclethatpassestroughmidpointsofredlines.2)if′Z′,bethecenterofthiscircle,find:PZ(intermsof:a,b,c)trialcase:a=15.93,b=15.76,c=12.51⇒r=2.16,PZ=0.81.

Commented by mrW1 last updated on 08/Jul/17

(1) r=((abc)/(4(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))))) (2) d=((√(a^6 +b^6 +c^6 +3a^2 b^2 c^2 −a^2 (b^4 +c^4 )−b^2 (a^4 +c^4 )−c^2 (a^4 +b^4 )))/(4(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c)))))

(1)r=abc4(a+b+c)(a+b−c)(a−b+c)(−a+b+c)(2)d=a6+b6+c6+3a2b2c2−a2(b4+c4)−b2(a4+c4)−c2(a4+b4)4(a+b+c)(a+b−c)(a−b+c)(−a+b+c)

Commented by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

very right answers.beautiful relations. thank you master. please post the solution,if it possible.

veryrightanswers.beautifulrelations.thankyoumaster.pleasepostthesolution,ifitpossible.

Commented by mrW1 last updated on 09/Jul/17

Part 1: ΔDGE∼ΔACB with side length ratio 1:2 ΔKIJ∼ΔDGE with side length ratio 1:2 ⇒ΔKIJ∼ΔACB with side length ratio 1:4 let r=radius of circumcircle of ΔKIJ let R=radius of circumcircle of ΔABC ⇒r=(1/4)R R=PA=PB=PC diameter of circumcircle of ΔABC=((abc)/(2 A_(ΔABC) )) R=((abc)/(4 A_(ΔABC) ))=((abc)/(4(√(s(s−a)(s−b)(s−c))))) =((abc)/(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))) ⇒r=(1/4)R=((abc)/(4(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))))) Part 2: let x=distance between orthocenter (H) and circumcenter (P) x^2 =9R^2 −(a^2 +b^2 +c^2 ) =((9a^2 b^2 c^2 )/((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))−(a^2 +b^2 +c^2 ) =((9a^2 b^2 c^2 −[(a+b)+c][(a+b)−c][c+(a−b)][c−(a−b)](a^2 +b^2 +c^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((9a^2 b^2 c^2 −[(a+b)^2 −c^2 ][c^2 −(a−b)^2 ](a^2 +b^2 +c^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((9a^2 b^2 c^2 +[a^2 +b^2 −c^2 +2ab][a^2 +b^2 −c^2 −2ab](a^2 +b^2 +c^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((9a^2 b^2 c^2 +[(a^2 +b^2 −c^2 )^2 −4a^2 b^2 ](a^2 +b^2 +c^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((a^2 b^2 (−4a^2 −4b^2 +5c^2 )+(a^2 +b^2 −c^2 )^2 (a^2 +b^2 +c^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((a^2 b^2 (−4a^2 −4b^2 +5c^2 )+(a^2 +b^2 −c^2 )[(a^2 +b^2 )^2 −c^4 ])/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((a^2 b^2 (−4a^2 −4b^2 +5c^2 )+(a^2 +b^2 −c^2 )[a^4 +b^4 −c^4 +2a^2 b^2 ])/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =(((−4a^4 b^2 −4a^2 b^4 +5a^2 b^2 c^2 )+(a^6 +a^2 b^4 −a^2 c^4 +2a^4 b^2 +b^2 a^4 +b^6 −b^2 c^4 +2a^2 b^4 −c^2 a^4 −c^2 b^4 +c^6 −2a^2 b^2 c^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) =((a^6 +b^6 +c^6 +3a^2 b^2 c^2 −a^4 (b^2 +c^2 )−b^4 (a^2 +c^2 )−c^4 (a^2 +b^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) ⇒x=(√((a^6 +b^6 +c^6 +3a^2 b^2 c^2 −a^4 (b^2 +c^2 )−b^4 (a^2 +c^2 )−c^4 (a^2 +b^2 ))/((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))) distance between nine−point−circle center (N) and circumcenter (P) =(x/2) since circle Z is half as large as circle N, point Z is midpoint between nine−point−circle center and circumcenter ⇒d=(x/4)=((√(a^6 +b^6 +c^6 +3a^2 b^2 c^2 −a^4 (b^2 +c^2 )−b^4 (a^2 +c^2 )−c^4 (a^2 +b^2 )))/(4(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))))

Part1:ΔDGE∼ΔACBwithsidelengthratio1:2ΔKIJ∼ΔDGEwithsidelengthratio1:2⇒ΔKIJ∼ΔACBwithsidelengthratio1:4letr=radiusofcircumcircleofΔKIJletR=radiusofcircumcircleofΔABC⇒r=14RR=PA=PB=PCdiameterofcircumcircleofΔABC=abc2AΔABCR=abc4AΔABC=abc4s(s−a)(s−b)(s−c)=abc(a+b+c)(−a+b+c)(a−b+c)(a+b−c)⇒r=14R=abc4(a+b+c)(−a+b+c)(a−b+c)(a+b−c)Part2:letx=distancebetweenorthocenter(H)andcircumcenter(P)x2=9R2−(a2+b2+c2)=9a2b2c2(a+b+c)(−a+b+c)(a−b+c)(a+b−c)−(a2+b2+c2)=9a2b2c2−[(a+b)+c][(a+b)−c][c+(a−b)][c−(a−b)](a2+b2+c2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=9a2b2c2−[(a+b)2−c2][c2−(a−b)2](a2+b2+c2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=9a2b2c2+[a2+b2−c2+2ab][a2+b2−c2−2ab](a2+b2+c2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=9a2b2c2+[(a2+b2−c2)2−4a2b2](a2+b2+c2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=a2b2(−4a2−4b2+5c2)+(a2+b2−c2)2(a2+b2+c2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=a2b2(−4a2−4b2+5c2)+(a2+b2−c2)[(a2+b2)2−c4](a+b+c)(−a+b+c)(a−b+c)(a+b−c)=a2b2(−4a2−4b2+5c2)+(a2+b2−c2)[a4+b4−c4+2a2b2](a+b+c)(−a+b+c)(a−b+c)(a+b−c)=(−4a4b2−4a2b4+5a2b2c2)+(a6+a2b4−a2c4+2a4b2+b2a4+b6−b2c4+2a2b4−c2a4−c2b4+c6−2a2b2c2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)=a6+b6+c6+3a2b2c2−a4(b2+c2)−b4(a2+c2)−c4(a2+b2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)⇒x=a6+b6+c6+3a2b2c2−a4(b2+c2)−b4(a2+c2)−c4(a2+b2)(a+b+c)(−a+b+c)(a−b+c)(a+b−c)distancebetweennine−point−circlecenter(N)andcircumcenter(P)=x2sincecircleZishalfaslargeascircleN,pointZismidpointbetweennine−point−circlecenterandcircumcenter⇒d=x4=a6+b6+c6+3a2b2c2−a4(b2+c2)−b4(a2+c2)−c4(a2+b2)4(a+b+c)(−a+b+c)(a−b+c)(a+b−c)

Commented by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

thanks master.i am waiting for part2....

thanksmaster.iamwaitingforpart2....

Commented by mrW1 last updated on 09/Jul/17

Commented by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17

waiting is over. thank you very much dear master. so beautiful and smart. god blees you sir.thanks a lot again.

waitingisover.thankyouverymuchdearmaster.sobeautifulandsmart.godbleesyousir.thanksalotagain.

Commented by mrW1 last updated on 09/Jul/17

you′re welcome sir!

you′rewelcomesir!

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