Question Number 102836 by ajfour last updated on 11/Jul/20

Commented byajfour last updated on 11/Jul/20

△ABC is equilateral with side a.  If regions 1,2,3 have equal  perimeters, find x/a.

Answered by mr W last updated on 11/Jul/20

CD=p  CE=q  DE=(√(p^2 +q^2 −2pq cos 60°))=(√(p^2 +q^2 −pq))  BE=(√(a^2 +(a−q)^2 −2a(a−q) cos 60°))=(√(a^2 +(a−q)^2 −a(a−q)))  P_1 =p+q+(√(p^2 +q^2 −pq))  P_2 =a−p+(√(p^2 +q^2 −pq))+(√(a^2 +(a−q)^2 −a(a−q)))  P_3 =2a−q+(√(a^2 +(a−q)^2 −a(a−q)))  P_3 =P_2 :  2a−q=a−p+(√(p^2 +q^2 −pq))  a+p−q=(√(p^2 +q^2 −pq))  a^2 +2a(p−q)−pq=0  with α=(p/a), β=(q/a)  ⇒2(α−β)−αβ+1=0   ...(i)    P_1 =P_2 :  p+q=a−p+(√(a^2 +(a−q)^2 −a(a−q)))  2p+q−a=(√(a^2 +(a−q)^2 −a(a−q)))  4p^2 +4pq−4ap−aq=0  ⇒4α^2 +4αβ−4α−β=0   ...(ii)    from (i) and (ii):  β=((2α+1)/(α+2))=((4α(1−α))/(4α−1))  ⇒4α^3 +12α^2 −6α−1=0  we get:  α=0.5504, β=0.8237  (x/a)=1−α=0.4496

Commented byajfour last updated on 11/Jul/20

SUPERB!  Sir, Great solution,  thanks a lot.

Answered by 1549442205 last updated on 11/Jul/20

Commented by1549442205 last updated on 12/Jul/20

Putting BD=m,CE=n,BC=aWe need  find(x/a)= (m/a).From the cosine theorem we   get DE=(√(n^2 +(a−m)^2 −n(a−m)))  BE=(√(a^2 +(a−n)^2 −a(a−n))) .From   the hypothesis p(ΔABE)=p(ΔBDE)  we get AB+AE=BD+DE  ⇔a+a−n=m+(√((a−m)^2 +n^2 −n(a−m)))  ⇔(2a−n−m)^2 =(a−m)^2 +n^2 −n(a−m)  ⇔4a^2 +n^2 +m^2 −4an−4am+2mn=  a^2 −2am+m^2 +n^2 −an+mn  ⇔3a^2 −3an−2am+mn=0(1)  p(ΔBDE)=p(ΔCDE)⇔BD+BE=CD+CE  ⇔m+(√(a^2 +(a−n)^2 −a(a−n)))=a−m+n  ⇔a^2 +(a−n)^2 −a(a−n)=(a+n−2m)^2   ⇔a^2 −an+n^2 =a^2 +n^2 +4m^2 +2an−4am−4mn  ⇔4m^2 +3an−4am−4mn=0(2)  From (1) we get n=((3a^2 −2am)/(3a−m))(∗).Replace  into (2) we get  4m^2 +(3a−4m)×((3a^2 −2am)/(3a−m))−4am=0  ⇔12am^2 −4m^3 +9a^3 −18a^2 m+8am^2 −12a^2 m+4am^2 =0  ⇔4m^3 +30a^2 m−2am^2 −9a^3 =0  ⇔4((m/a))^3 −24((m/a))^2 +30((m/a))−9=0  ⇔(x/a)=(m/a)=0.4495847837  From(∗) we get:   y=(n/a)=2−(3/(3−(m/a)))=0.823720452

Commented byajfour last updated on 12/Jul/20

thanks Sir, excellent.

Commented by1549442205 last updated on 13/Jul/20

You are welcome sir.