#10 Mixed Problems II

By power of a point, \(PT^2=PA\cdot PB\). Hence \(100=5\cdot PB\), so \(PB=20\).

Since \(BD\perp CD\), \(\angle BDC=90^\circ\). Since \(BE\perp CE\), \(\angle BEC=90^\circ\). Thus \(D\) and \(E\) lie on the circle with diameter \(BC\), so \(B,C,D,E\) are concyclic.

Point \(P\) lies on the radical axis, so its powers with respect to \(\omega_1\) and \(\omega_2\) are equal. By the definition of power, \(PX\cdot PY=PU\cdot PV\).

\(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). By Ceva's theorem, lines \(AD,BE,CF\) are concurrent.

Triangles \(PAB\) and \(PCA\) are conveniently compared using the common base \(AP\). Their altitudes to line \(AP\), drawn from \(B\) and \(C\), are in the ratio \(BD:DC\), because \(B,D,C\) are collinear. Hence \(BD:DC=[PAB]:[PCA]=2:4=1:2\).

From point \(T\), tangent \(TA\) and secant \(TBC\) are drawn to the circle. Hence by power of a point, \(TA^2=TB\cdot TC\) in directed lengths.

Lines \(PAC\) and \(PBD\) are two secants of the same circle. By power of a point, \(PA\cdot PC=PB\cdot PD\).

Since \(X\in AB\), it lies on the radical axis of \(\omega_1,\omega_2\), so \(\operatorname{Pow}_{\omega_1}(X)=\operatorname{Pow}_{\omega_2}(X)\). Since \(X\in CD\), we get \(\operatorname{Pow}_{\omega_2}(X)=\operatorname{Pow}_{\omega_3}(X)\). Therefore \(\operatorname{Pow}_{\omega_1}(X)=\operatorname{Pow}_{\omega_3}(X)\), so \(X\) lies on the radical axis of \(\omega_1,\omega_3\), which is line \(EF\).

Since \(F\) lies on the extension beyond \(A\), \(CF:FA=3:1\). Check Menelaus' product: \(\frac{AD}{DB}\cdot\frac{BE}{EC}\cdot\frac{CF}{FA}=\frac{1}{2}\cdot\frac{2}{3}\cdot 3=1\). Hence \(D,E,F\) are collinear.

We have \(P=(6,0)+t(-6,8)=(6-6t,8t)\). The perpendicularity condition is \((6-6t,8t)\cdot(-6,8)=0\). Thus \(-36+36t+64t=0\), so \(t=\frac{9}{25}\). Hence \(BP:PC=t:(1-t)=9:16\).

Since \(AD:DB=2:3\), we have \(AD:AB=2:5\). From \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\) with scale factor \(\frac{2}{5}\). Areas of similar triangles are in the square of the scale factor, so \([ADE]:[ABC]=4:25\).

The four lines \(AB,AC,BE,CD\) form a complete quadrilateral. Its four circles are exactly \((ABE)\), \((ACD)\), \((BDP)\), \((CEP)\). By Miquel's theorem, they pass through one point.

Let \(A=(0,0)\), \(B=(d,0)\), and let the circle centres be \(O_1=(0,r)\), \(O_2=(d,s)\). Since the circles touch externally, \(O_1O_2=r+s\), so \(d^2+(s-r)^2=(r+s)^2\), hence \(d^2=4rs\). The tangency point \(T\) divides \(O_1O_2\) in the ratio \(r:s\), so \(T=\left(\frac{rd}{r+s},\frac{2rs}{r+s}\right)\). Then \(\overrightarrow{TA}\cdot\overrightarrow{TB}=\frac{rs(4rs-d^2)}{(r+s)^2}=0\). Therefore \(TA\perp TB\).

By Ceva's theorem, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Hence \(2\cdot\frac{3}{2}\cdot\frac{AF}{FB}=1\), so \(\frac{AF}{FB}=\frac{1}{3}\). Therefore \(AF:FB=1:3\).

Let the circle through \(A,C,D\) meet ray \(PB\) for the second time at \(X\). By power of point \(P\), \(PA\cdot PX=PC\cdot PD\). By the condition, \(PC\cdot PD=PA\cdot PB\). Since \(PA\ne 0\), \(PX=PB\). On one ray such a point is unique, so \(X=B\). Hence \(A,C,B,D\) are concyclic.

Line \(AB\) is the radical axis, so \(\operatorname{Pow}_{\omega_1}(P)=\operatorname{Pow}_{\omega_2}(P)\). These powers are \(PX^2\) and \(PY^2\), because \(PX\) and \(PY\) are tangents. Therefore \(PX^2=PY^2\), hence \(PX=PY\).

By Miquel's theorem, \(A,B,F,M\) lie on one circle. Therefore \(\angle AMB=\angle AFB\), since these angles subtend chord \(AB\). But \(FA\) is the same line as \(FD\), and \(FB\) is the same line as \(FC\). Hence \(\angle AFB=\angle DFC\). Therefore \(\angle AMB=\angle DFC\).

Tangents drawn from one point to a circle are equal, so \(TA=TC\). Also, the power of point \(T\) with respect to the circle equals both \(TA^2\) and \(TB\cdot TD\) along secant \(TBD\). Therefore \(TB\cdot TD=TA^2\).

Since each corresponding pair of triangles has a common height, we get \(BD:DC=2:3\), \(CE:EA=3:4\), \(AF:FB=2:1\). Then \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). By Ceva, lines \(AD,BE,CF\) are concurrent.

There is an external homothety sending the first circle to the second; its centre \(X\) lies on the line of the centres. A common external tangent maps to itself, so the tangency point \(A\) maps to \(C\), and \(B\) maps to \(D\). Therefore \(A,X,C\) are collinear and \(B,X,D\) are collinear. Hence lines \(AC\) and \(BD\) meet at \(X\) on the line of the centres.

Let circles \((AEF)\) and \((BFD)\) meet again at \(M\). We need to prove that \(C,D,E,M\) are cyclic. From \(M\in(BFD)\), \(\angle DMF=\angle DBF=\angle CBA\). From \(M\in(AEF)\), \(\angle FME=\angle FAE=\angle BAC\). Therefore \(\angle DME=\angle DMF+\angle FME=\angle CBA+\angle BAC=180^\circ-\angle BCA\). But \(\angle DCE=\angle BCA\). Thus \(\angle DME+\angle DCE=180^\circ\), and \(C,D,E,M\) lie on one circle.

For a point \(X\), the power with respect to the circle with diameter \(AB\) is \(\overrightarrow{XA}\cdot\overrightarrow{XB}\), and with respect to the circle with diameter \(AC\) it is \(\overrightarrow{XA}\cdot\overrightarrow{XC}\). Equality of powers gives \(\overrightarrow{XA}\cdot(\overrightarrow{XB}-\overrightarrow{XC})=0\), that is \(XA\perp BC\). Hence the radical axis of these two circles is the altitude from \(A\). Similarly, the other two radical axes are the altitudes from \(B\) and \(C\). Their common point is the orthocenter.

Since triangles \(ABD\) and \(ACD\) have the same altitude from \(A\), \(\frac{BD}{DC}=\frac{[ABD]}{[ACD]}\). Also, \(\frac{[ABD]}{[ACD]}=\frac{AB\cdot AD\sin\angle BAD}{AC\cdot AD\sin\angle CAD}=\frac{AB\sin\angle BAD}{AC\sin\angle CAD}\). The median \(AM\) divides the triangle into equal areas, so \(AB\sin\angle MAB=AC\sin\angle MAC\). By isogonality, \(\sin\angle BAD=\sin\angle MAC\), and \(\sin\angle CAD=\sin\angle MAB\). Therefore \(\frac{BD}{DC}=\frac{AB}{AC}\cdot\frac{\sin\angle MAC}{\sin\angle MAB}=\frac{AB}{AC}\cdot\frac{AB}{AC}=\frac{AB^2}{AC^2}\).

Since \(ABCD\) is cyclic, \(\angle PAB=\angle CAB=\angle CDB=\angle PDC\), and \(\angle PBA=\angle DBA=\angle DCA=\angle PCD\). Hence \(\triangle PAB\sim\triangle PDC\), so \(\frac{PB}{PC}=\frac{AB}{DC}\). Similarly, \(\triangle PAD\sim\triangle PBC\), hence \(\frac{PA}{PB}=\frac{AD}{BC}\). Multiplying these equalities gives \(\frac{PA}{PC}=\frac{AD}{BC}\cdot\frac{AB}{DC}=\frac{AB\cdot AD}{CB\cdot CD}\).