Annoying geometry problem

By David Loeffler (P865) on Monday, February 19, 2001 - 10:13 pm :

I am rather stuck trying to prove the following result:

''Suppose P is a point within an acute-angled triangle A B C. If angle A P B=C+g, B P C=A+a, C P A=B+b, show that a×A P/sin(a)=b×B P/sin(b)=g×C P/sin( g)''

The angles a, b and g seem to be the angles of the pedal triangle of P, but that doesn't seem to help a great deal! I have tried all sorts of approaches but they all get bogged down in messy algebra.

Can anybody help?

David Loeffler

By Kerwin Hui (Kwkh2) on Tuesday, February 20, 2001 - 03:38 pm :

OK, Construct the pedal triangle D E F of P, where the normal notation applies. Also note that A E P F are concyclic (similarly B D P F, C E P D).

First, apply sine rule to triangles D E F, we get

E F/(sina)=F D/(sinb)=D E/(sing) (*)

Now we apply sine rule to A E F to get

E F/(sinA)=A F/(cosÐF E B)=A F/(cosÐB A P)=A P

So, we have E F=A PsinA=(A P)[B C/(2R)]=a(A P)/(2 R)

Similarly, we get F D=b(B P)/(2R) and D E=c(C P)/(2 R).

Hence, when we substitute for D E, E F, F D into (*) and cancel the factor of 1/(2R), we obtain a(A P)/(sina = b(B P)/(sinb) = c(C P)/(sing)

as desired.

Kerwin

By David Loeffler (P865) on Thursday, February 22, 2001 - 11:14 pm :

Thanks Kerwin. The three terms must be equal to 4R_ABC R_DEF then, I suppose.

Thanks again.

David