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 $A P B = C + γ$ , $B P C = A + α$ , $C P A = B + β$ , show that $α \times A P / \sin (α) = β \times B P / \sin (β) = γ \times C P / \sin (γ)$ ''

The angles $α$ , $β$ and $γ$ 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

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

$E F / (\sin α) = F D / (\sin β) = D E / (\sin γ)$ (*)

Now we apply sine rule to $A E F$ to get

$E F / (\sin A) = A F / (\cos ∠ F E B) = A F / (\cos ∠ B A P) = A P$

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

Similarly, we get $F D = b (B P) / (2 R)$ 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 $1 / (2 R)$ , we obtain $a (A P) / (\sin α = b (B P) / (\sin β) = c (C P) / (\sin γ)$

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