Let x=BP. Area(PQR)=0.5*PR*PQ*sin(<RPQ) But PR=2*x*sin(B/2), PQ=2([BC]-x)²cos(C/2) and <RPQ=180º-<RPB-<QPC=180-(180-B)/2-(180-C)/2=(B+C)/2=Pi/2-A/2 then Area(PQR)=2*x*([BC]-x)*sin(B/2)*sin(C/2)*cos(A/2) Then maximum area occurs when x=BC/2 and it is Maximum(A(PQR))=0.5*[BC]^2*sin(B/2)*sin(C/2)*cos(A/2) Then apply identities: 2*sin(u)*sin(v)=cos(u-v)-cos(u+v) 2*cos(u)*cos(v)=cos(u+v)+cos(u-v). and A,B,C relations to get the result.
Let x=BP.
ReplyDeleteArea(PQR)=0.5*PR*PQ*sin(<RPQ)
But
PR=2*x*sin(B/2), PQ=2([BC]-x)²cos(C/2)
and
<RPQ=180º-<RPB-<QPC=180-(180-B)/2-(180-C)/2=(B+C)/2=Pi/2-A/2
then
Area(PQR)=2*x*([BC]-x)*sin(B/2)*sin(C/2)*cos(A/2)
Then maximum area occurs when x=BC/2 and it is
Maximum(A(PQR))=0.5*[BC]^2*sin(B/2)*sin(C/2)*cos(A/2)
Then apply identities:
2*sin(u)*sin(v)=cos(u-v)-cos(u+v)
2*cos(u)*cos(v)=cos(u+v)+cos(u-v).
and A,B,C relations to get the result.
César Lozada
Venezuela
Please read
ReplyDeletePQ=2*[BC]-x)*cos(C/2)
instead of
PQ=2([BC]-x)²cos(C/2)
Thanks
César Lozada
Venezuela
César, thank you for confirming the result!
ReplyDelete