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 $\overline{CX} = \overline{AD}$, $D \neq X$, and the points $D$ and $X$ lie in the same half-plane bounded by the line $AC$. Hence $DX \parallel AC$. Similarly, $B \neq Y$ 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
$$\begin{aligned} A - X &= a - x + bi, \\ (a - x + bi)(\cos \alpha(X) + i\sin \alpha(X)) \\ &= (a - x)\cos \alpha(X) - b\sin \alpha(X) + [(a - x)\sin \alpha(X) + b\cos \alpha(X)]i. \end{aligned}$$

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

Moreover, since $\vec{XC} = e-x+fi$, $\vec{XB} = c-x+di$, then $\vec{XC}$ 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 $\vec{XB}$. Thus,
$$(c-x)[(e-x)\sin\beta(X) + f\cos\beta(X)] - d[(e-x)\cos\beta(X) - f\sin\beta(X)] = 0,$$
and
$$[(c-x)(e-x) + df]\sin\beta(X) + [(c-x)f - d(e-x)]\cos\beta(X) = 0. \quad (2)$$

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

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 $b \neq 0$ that $d \neq f$. 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 $g(t) \neq 0$, 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 $D^*$ be the point such that $A$, $D^*$, $C$, $B$ are concyclic, where $D^*$ 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 $D^*R$ be the ray satisfying $\angle CD^*R = \angle AD^*B$ and $P^* = D^*R \cap BP$.
Together with (i), we have that $P^*A = P^*C$, or $P^*$ is the intersection point of $l$ and the perpendicular bisector of $AC$, so $P^* = P$. It follows from $\angle CD^*R = \angle AD^*B$ that $\angle AD^*B = \angle CD^*P$.

Replace $l$, $A$, $B$, $C$ by $BD$, $A$, $C$, $P$. Then $B$, $D$, $D^*$ are good points. Since $B \neq D$ and $B \neq D^*$, $D = D^*$. 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 $AP = CP$.