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.