International Mathematical Olympiad

(i) Necessity.

Assume that $ABCD$ is a cyclic quadrilateral. Let the circle $\Gamma$ be the circumcircle of quadrilateral $ABCD$. Extend $BP$ and $DP$ beyond $P$ to meet the circle $\Gamma$ at $X$ and $Y$ respectively.

Since $DB$ does not bisect $\angle ABC$ and $P$ lies inside $ABCD$, it follows from $\angle PBC=\angle DBA$ that $\overline{CX}=\overline{AD}$, $D\ne X$, and the points $D$ and $X$ lie in the same half-plane bounded by the line $AC$. Hence $DX\parallel AC$. Similarly, $B\ne Y$ and $BY\parallel AC$.

The points $D$, $X$, $A$, $C$, $B$, $Y$ lie on the circle $\Gamma$, as mentioned above. So the points $D$ and $X$, the points $A$ and $C$, the points $B$ and $Y$ are symmetrical with respect to the perpendicular bisector $l$ of $AC$ respectively.

Since $P=DY\cap BX$, then $P$ is on the line $l$, or $AP=CP$. This completes the proof of the necessity.

(ii) Sufficiency.

Lemma. Let $l$ be a fixed line and $A,B,C$ be fixed points such that $A$ and $B$, $C$ lie in the different half-planes bounded by the line $l$. Assume that the point $X$ lies on the line $l$. Let $\alpha (X)$ be the smallest angle of rotation from the line $XA$ to the line $l$ anticlockwise. Let $\beta (X)$ be the smallest angle of rotation from the line $XC$ to the line $XB$ anticlockwise. If $\alpha (X)=\beta (X)$, then $X$ is called a good point. Prove that there are at most two good points.

Proof of the lemma

We set up a coordinate system with the line $l$ as the $x$-axis and the perpendicular of $l$ as the $y$-axis. Let $A(a,b)$, $B(c,d)$, $C(e,f)$, $X(x,0)$. Then $A=a+bi$, $B=c+di$, $C=e+fi$, and $X=x$. Hence $$\begin{array}{rl}A-X& =a-x+bi,\\ (a-x+bi)(\cos \alpha (X)+i\sin \alpha (X))& \\ & =(a-x)\cos \alpha (X)-b\sin \alpha (X)+[(a-x)\sin \alpha (X)+b\cos \alpha (X)]i.\end{array}$$

Rotate the line $XA$ by $\alpha (X)$ anticlockwise, coinciding with $l$. So $$(a-x)\cos \alpha (X)+b\cos \alpha (X)=0(1)$$

Moreover, since $\overrightarrow{XC}=e-x+fi$, $\overrightarrow{XB}=c-x+di$, then $\overrightarrow{XC}$ after being rotated by $\beta (X)$ anticlockwise becomes $$(e-x+fi)(\cos \beta (X)+i\sin \beta (X))=(e-x)\cos \beta (X)-f\sin \beta (X)+[(e-x)\sin \beta (X)+f\cos \beta (X)]i$$ which is parallel to $\overrightarrow{XB}$. Thus, $$(c-x)[(e-x)\sin \beta (X)+f\cos \beta (X)]-d[(e-x)\cos \beta (X)-f\sin \beta (X)]=0,$$ and $$[(c-x)(e-x)+df]\sin \beta (X)+[(c-x)f-d(e-x)]\cos \beta (X)=0.(2)$$

Since $\sin \theta$ and $\cos \theta$ are not 0 at the same time, so $X$ is a good point $\iff \alpha (X)=\beta (X)\iff$ the simultaneous equations in $u$ and $v$. $$\{\begin{array}{l}(a-x)u+bv=0,\\ [(c-x)(e-x)+df]u+[(c-x)f-d(e-x)]v=0\end{array}$$ have the non-zero solutions $$\begin{array}{rl}& \iff b[(c-x)(e-x)+df]+(a-x)[(c-x)f-d(e-x)]=0\\ & \iff (b+d-f)x^2+(af+cf-bc-be-ad-ed)x\\ & +bce+bdf+ade-acf=0.\end{array}$$

Let $g(x)=(b+d-f)x^2+(af+cf-bc-be-ad-ed)x+bce+bdf+ade-acf$. If $b+d-f=0$, it follows from $b\ne 0$ that $d\ne f$. Hence $BC$ and $l$ are not parallel. Let $BC$ intersect $l$ at $T(t,0)$. We have that $\beta (T)=0$ and $\alpha (T)>0$. Hence $T$ is not a good point. This implies $g(t)\ne 0$, so the equation $g(x)=0$ has at most two solutions.

Hence there are at most two good points.

This completes the proof of the lemma.

Let the points $A$, $B$, $C$, $D$ be arranged clockwise and $AP=CP$. Let $D^{\ast }$ be the point such that $A$, $D^{\ast }$, $C$, $B$ are concyclic, where $D^{\ast }$ lies on the ray $BD$. Since $AP=CP$ and $BD$ bisects neither $\angle ABC$ nor $\angle ADC$, $BP$ intersects the perpendicular bisector of $AC$ at the unique point $P$.

Let $D^{\ast }R$ be the ray satisfying $\angle C{D}^{\ast }R=\angle A{D}^{\ast }B$ and $P^{\ast }=D^{\ast }R\cap BP$. Together with (i), we have that $P^{\ast }A=P^{\ast }C$, or $P^{\ast }$ is the intersection point of $l$ and the perpendicular bisector of $AC$, so $P^{\ast }=P$. It follows from $\angle C{D}^{\ast }R=\angle A{D}^{\ast }B$ that $\angle A{D}^{\ast }B=\angle C{D}^{\ast }P$.

Replace $l$, $A$, $B$, $C$ by $BD$, $A$, $C$, $P$. Then $B$, $D$, $D^{\ast }$ are good points. Since $B\ne D$ and $B\ne D^{\ast }$, $D=D^{\ast }$. Hence $A$, $B$, $C$, $D$ is concyclic.

This completes the proof of the sufficiency.

Solution II

We may assume without loss of generality that $P$ lies in the rectangles $ABC$ and $BCD$.



Assume that the quadrilateral $ABCD$ is cyclic. Let the lines $BP$ and $DP$ meet $AC$ at $K$ and $L$ respectively. It follows from the given equalities and $\angle ACB=\angle ADB$, $\angle ABD=\angle ACD$ that the triangles $DAB$, $DLC$ and $CKB$ are similar. This implies $\angle DLC=\angle CKB$, so $\angle PLK=\angle PKL$. Hence $PK=PL$.

It follows from $\angle BDA=\angle PDC$ that $\angle ADL=\angle BDC$. Since $\angle DAL=\angle DBC$, the triangles $ADL$ and $BDC$ are similar. Hence $$\frac{AL}{BC}=\frac{AD}{BD}=\frac{KC}{BC},$$ yielding $AL=KC$. It follows from $\angle DLC=\angle CKB$ that $\angle ALP=\angle CKP$. Moreover, since $PK=PL$ and $AL=KC$, the triangles $ALP$ and $CKP$ are congruent. Hence $AP=CP$.