Let $p-a=x$, $p-b=y$, $p-c=z$, $AP=n$, $PD=m$ then $x^2=n(n+m)$. Since the circumcircle of $\triangle{CPB}$ is tangent to $AC$ we can find that: $\displaystyle{\frac{n}{m}=\frac{2x+z}{y}}$.(1) Using this we conclude that $\displaystyle{(m+n)=\frac{x(2x+y+z)}{\sqrt{(2x+y+z)(2x+z)}}}$. But from Steward's Theorem: $\displaystyle{\frac{4xyz}{y+z}=m(n+m)}$. Therefore $\displaystyle{\frac{4xyz}{y+z}+x^2=(m+n)^2=\frac{{x^2}{(2x+y+z)}}{2x+z}}$ or $4z^2+7xz=xy$(2). Since $\angle{ACB}=90^\cdot$ it follows that $xy=z^2+zx+zy$ and (2) reduces to $3z+6x=y$. Using this and (1) we get that $\displaystyle{\frac{n}{m}=\frac{1}{3}}$. This implies $x=2n$, consequently $x+n=m$ and we are done.

Thank you, Shobber (for the your proposed problem) and Sailor (for the your solution) ! See PP12 from here a clearing up of the nice Sailor's solution.

CMO 2006= Chinese Mathmatical Olympiad 2006.

The claim is true regardless the fact that trangle ABC be rectangle. I will post later a sketch of a synthetic prove of that.

I presume some facts which can be proved using some well known projective results. Fact 1. If X is a generic point on AD and XB and XC cut (O) at Q and R one can prove that QR pass through S where S= FE.BC Fact 2. If RB cuts (O) in T one can prove that RQ, PT and DF concurr at one point. Fact 3. If U is the pole of RT, as B lies on RT the polar of B which is FD pass through U. Dim. From fact 1, as RQ pass trhough O it results that RQ is orthogonal to PD. Then from fact 2 we can deduce that D and T are the simmetrical points of P and F wrt diameter RQ. If UR intersect AB at V, as UV// AD we have that FR bisects VU, and from Fact 3 as F,D and U are colinears, FR bisects AD too at, let's say, M. Furthermore, as VR=VF and VRF~AMF, AM=AF. From Those it follows that AD = 2AF = 2 AP which is equivalent to the thesis.

sprmnt21 wrote: I presume some facts which can be proved using some well known projective results. Fact 1. If X is a generic point on AD and XB and XC cut (O) at Q and R one can prove that QR pass through S where S= FE.BC Fact 2. If RB cuts (O) in T one can prove that RQ, PT and DF concurr at one point. Fact 3. If U is the pole of RT, as B lies on RT the polar of B which is FD pass through U. Can someone show the proofs of these results using projective geometry?

I thought that Sangaku is more of a Japaneese specialty, rather than Chineese. I quess that, like most other things, it was smuggled into closed at that period Japan from the friendly neighbour China. We should look carefully into the true origins of Sangaku. Anyway, if one looks at the configuration, it is rather difficult not to notice the squares rotated around a common vertex and inscribed and circumscribed circles. Reminds of the famous Sangaku of folded square with inscribed circle in the triangle with the inradius equal to the overhang of the folded part. Billzhao is right, it would be interesting to get a projective solution for a Sangaku that traditionally was solved by algebraic methods. Dear Sprmnt21, please share with us the details of your projective solution. T.Y. M.T.

Sorry. Lately I didn't watched the forum. Here some more details about the "facts": Fact 1. If X is a generic point on AD and XB and XC cut (O) at Q and R one can prove that QR pass through S where S= FE.BC Proof 1. As S is harmonically conjugated with D wrt (B,C), then, if Y = AD/\QS and R’ = XC/\QS, {(Q,R’);(Y,S)} is a harmonic quadruple. If R” = (O)/\QS then, is well known, {(Q,R”);(Y,S)} )} is a harmonic quadruple too, i.e. R’ = R” = R, i.e. Q,R and S are collinear. Fact 2. If RB cuts (O) in T one can prove that RQ, PT and DF concur at one point. Proof 2. It is a well known fact that, if PQTR is an inscribed quadrilateral, the intersection of the diagonals PT, QR is the pole of the line through the points of intersection of the opposite sides. From this and the De la Hire theorem it follows what claimed. For this fact and other similar let have a look at http://www.maths.gla.ac.uk/~wws/cabripages/quad.html Fact 3. If U is the pole of RT, as B lies on RT the polar of B which is FD pass through U. Proof 3. This is simply the De La Hire theorem. ======== I use this post to correct a typo: The phrase:" From Those it follows that AD = 2AF = 2 AP which is equivalent to the thesis." should be: From Those it follows that AD = 2AF and AF = 2 AP which is equivalent to the thesis.

Thanks Sprmnt21. M.T.

anyone can post a solution with pure geometry

quiet easy for a cmo