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Question Number 80809 by mr W last updated on 06/Feb/20

Commented by mr W last updated on 06/Feb/20

I is a point inside the triangle ABC.  P,Q,R are the midpoint of AI,BI,CI  respectively.  What fraction of the triangle ABC  does the shaded hexagon present?

IisapointinsidethetriangleABC.P,Q,RarethemidpointofAI,BI,CIrespectively.WhatfractionofthetriangleABCdoestheshadedhexagonpresent?

Commented by ajfour last updated on 07/Feb/20

Commented by ajfour last updated on 07/Feb/20

△_(ABC) =Σ2qrsin α  eq. of AR:    v^� =r^� +λ(2p^� −r^� )  eq. of CP:     v^� =p^� +μ(2r^� −p^� )  v_E ^� = r^� +λ(2p^� −r^� )=p^� +μ(2r^� −p^� )  ⇒  1−λ=2μ  ,  2λ=1−μ  ⇒   1−λ=2−4λ   ⇒ λ=1/3  v_E ^� = r^� +((2p^� −r^� )/3) = (2/3)(p^� +r^� )  A_(IREP) =prsin β((4/9)+(1/9)+(1/9))               = (2/3)prsin β  A_(hexagon) =(2/3)Σprsin β  hence    (A_(hexagon) /△_(ABC) )=((2/3)/2) = (1/3) .

ABC=Σ2qrsinαeq.ofAR:v¯=r¯+λ(2p¯r¯)eq.ofCP:v¯=p¯+μ(2r¯p¯)v¯E=r¯+λ(2p¯r¯)=p¯+μ(2r¯p¯)1λ=2μ,2λ=1μ1λ=24λλ=1/3v¯E=r¯+2p¯r¯3=23(p¯+r¯)AIREP=prsinβ(49+19+19)=23prsinβAhexagon=23ΣprsinβhenceAhexagonABC=2/32=13.

Commented by mr W last updated on 07/Feb/20

answer is correct!  it′s a nice vector method sir!

answeriscorrect!itsanicevectormethodsir!

Commented by ajfour last updated on 07/Feb/20

Answered by mr W last updated on 07/Feb/20

Commented by mr W last updated on 07/Feb/20

an other way:  as example we treat only the part ΔIBC.  draw QN//BR  NR=IN=((IR)/2)=((RC)/2)  ⇒QD=((DC)/2)  ΔQDB=((ΔDCB)/2)=((ΔQCB)/3)=((ΔIBC)/6)  ΔRBC=((ΔIBC)/2)  A_(IQDR) =ΔIBC−ΔRBC−ΔQDB  A_(IQDR) =ΔIBC−((ΔIBC)/2)−((ΔIBC)/6)=((ΔIBC)/3)  ΣA_(IQDR) =((ΣΔIBC)/3)  ⇒hexagon =((ΔABC)/3)

anotherway:asexamplewetreatonlythepartΔIBC.drawQN//BRNR=IN=IR2=RC2QD=DC2ΔQDB=ΔDCB2=ΔQCB3=ΔIBC6ΔRBC=ΔIBC2AIQDR=ΔIBCΔRBCΔQDBAIQDR=ΔIBCΔIBC2ΔIBC6=ΔIBC3ΣAIQDR=ΣΔIBC3hexagon=ΔABC3

Commented by ajfour last updated on 07/Feb/20

Elementary and nice Sir.

ElementaryandniceSir.

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