, 2020

Solution:

A. The angle bisector of $\angle ECD$ intersects the circumcircle of $CDE$ (which is $\Gamma$) at the midpoint $M$ of arc $DBE$. It is well-known that the incentre is the intersection of the angle bisector segment $CM$ and the circle with centre at $M$ and passing through $D, E$. We will verify this property for $P$.

By the conditions we have $AP = OA = OB = OC = OD = OE = R$. Both lines $OM$ and $APB$ are perpendicular to $ED$, therefore $AP \parallel OM$; in the quadrilateral $AOMP$ we have $AP = OA = AM = R$ and $AP \parallel OM$, so $AOMP$ is a rhombus and its fourth side is $PM = R$.

In the convex quadrilateral $OMBP$ we have $OM \parallel PB$, so $OMBP$ is a symmetric trapezoid; the perpendicular bisector of its bases $AO$ and $PB$ coincide. From this symmetry we obtain $MD = OD = R$ and $ME = OE = R$. (Note that the triangles $OEM$ and $OMD$ are equilateral.) We already have $MP = MD = ME = R$, so $P$ indeed lies on the circle with center $M$ and passing through $D, E$. (Notice that this circle is the reflection of $\Gamma$ about $DE$.)

From $PB = PC$ and $OB = OC$ we know that $B$ and $C$ are symmetrical about $OP$; from the rhombus $AOMP$ we find that $A$ and $M$ are also symmetrical about $OP$. By reflecting the collinear points $B, P, A$ (with $P$ lying in the middle) we get that $C, P, M$ are collinear (and $P$ is in the middle). Hence, $P$ lies on the line segments $CM$.

B. Let $X$ be the second intersection of $CP$ with $\Gamma$. Using the power of the point $P$ in the circle $\Gamma$ and the fact that $PB = PC$, we find that $PX = PA = R$. The quadrilateral $AOXP$ has four sides of equal length, so it is a rhombus and in particular $OX$ is parallel to $AP$. This proves that $OXBP$ is a trapezoid, and because the diagonals $PX$ and $OB$ have equal length, this is even an isosceles trapezoid. Because of that, $DE$ is not only the perpendicular bisector of $PB$, but also of $OX$.

In particular we have $XD = XP = XO = XE = R$, which proves that $X$ is the middle of the arc $DE$ and $P$ belongs to the circle with center $X$ going through $D$ and $E$. These properties, together with the fact that $C, P, X$ are collinear, determine uniquely the incenter of $CDE$.

C. Let $Y$ be the circumcenter of triangle $BPC$. Then from $YB = YP$ it follows that $Y$ lies on $DE$ (we assume $D$ lies in between $Y$ and $E$), and from $YB = YC$ it follows that $Y$ lies on $OP$, where $O$ is the center of $\Gamma$.

From $\angle AOC = 2\angle ABC = \angle APC$ (because $\angle PBC = \angle PCB$) we deduce that $AOPC$ is a cyclic quadrilateral, and from $AP = R$ it follows that $AOPC$ is an isosceles trapezoid. We now find that $\angle YCP = \angle YPC = 180^{\circ} - \angle OPC = 180^{\circ} - \angle ACP$, so $Y$ lies on $AC$.

Power of a point gives $YO \cdot YP = YC \cdot YA = YD \cdot YE$, so $D, P, O$ and $E$ are concyclic. It follows that $2\angle DAE = \angle DOE = \angle DPE = \angle DBE = 180^{\circ} - \angle DAE$, so $\angle DAE = 60^{\circ}$. We can now finish the proof by angle chasing.

From $AB \perp DE$ we have $\angle AOD + \angle BOE = 180^{\circ}$ and from $\angle DOE = 2\angle DAE = 120^{\circ}$ it follows that $\angle BOD + \angle BOE = 120^{\circ}$. It follows that $\angle AOD - \angle BOD = 180^{\circ} - 120^{\circ} = 60^{\circ}$. Let $\angle OAB = \angle OBA = 2\beta$; then $\angle AOD + \angle BOD = \angle AOB = 180^{\circ} - 4\beta$. Together with $\angle AOD - \angle BOD = 60^{\circ}$, this yields $\angle AOD = 120^{\circ} - 2\beta$ and $\angle BOD = 60^{\circ} - 2\beta$. We now find $\angle AED = \frac{1}{2} \angle AOD = 60^{\circ} - \beta$, which together with $\angle DAE = 60^{\circ}$ yields $\angle ADE = 60^{\circ} + \beta$. From the isosceles trapezoid $AOPC$ we have $\angle CDA = \angle CBA = \frac{1}{2} \angle CPA = \frac{1}{2} \angle PAO = \beta$, so $\angle CDE = \angle CDA + \angle ADE = \beta + 60^{\circ} + \beta = 60^{\circ} + 2\beta$.

From $\angle BOD = 60^{\circ} - 2\beta$ we deduce that $\angle BED = 30^{\circ} - \beta$; together with $\angle DBE = 120^{\circ}$ this yields $\angle EDB = 30^{\circ} + \beta$. We now see that $\angle PDE = \angle BDE = 30^{\circ} + \beta = \frac{1}{2} \angle CDE$, so $P$ is on the angle bisector of $\angle CDE$. Similarly, $P$ lies on the angle bisector of $\angle CED$, so $P$ is the incenter of $CDE$.

D. We draw the lines $DP$ and $EP$ and let $D'$, resp. $E'$, be the second intersection point with $\Gamma$.

The triangles $APD'$ and $DPB$ are similar, and the triangles $APE'$ and $EPB$ are also similar, hence they are all isosceles and it follows that $E', O, P, D'$ lie on a circle with center $A$. In particular $AOD'$ and $AOE'$ are equilateral triangles. Angle chasing gives
$$\begin{gathered} \angle CDP = \angle CDD' = \frac{1}{2} \angle COD' = \frac{1}{2}(60^{\circ} + \angle COA) \\ \angle EDP = \angle EDD' = \angle EE'D' = \angle PE'D' = \frac{1}{2} \angle PAD' = \frac{1}{2}(60^{\circ} + \angle PAO) \end{gathered}$$
Similarly we prove $\angle CEP = \frac{1}{2}(60^{\circ} - \angle COA)$ and $\angle DEP = \frac{1}{2}(60^{\circ} - \angle PAO)$ so if we can prove that $\angle COA = \angle PAO$, we will have proven that $P$ belongs to the angle bisector of $\angle CED$ and to the angle bisector of $\angle CDE$, which is enough to prove that $P$ is the incenter of the triangle $CDE$.

Let $\beta = \angle ABC = \angle PCB$. We have $\angle APC = 2\beta$ and $\angle AOC = 2\beta$, so $AOPC$ is an inscribed quadrilateral. Moreover, since the diagonals $AP$ and $CO$ have equal length, this is actually an isosceles trapezoid, and hence $\angle PAO = \angle CPA = \angle COA$ which concludes the proof.

E. Without loss of generality we assume that $D$ and $C$ are in the same half-plane regarding line $AB$.

Since $PC = BP$ and $ABCD$ is inscribed quadrilateral we have $\angle PCB = \angle CBP = \angle CEA = \alpha$. As in the other solutions, $AOPC$ is an isosceles trapezoid and $2\alpha = \angle CPA = \angle PCO$.

Let $K$ be intersection of $EO$ and $\Gamma$. Then $\angle KDE = 90^{\circ}$, $AB \parallel DK$ and $KDBA$ is isosceles trapezoid. We obtain $DP = BD = AK$, which implies that $DPAK$ is a parallelogram and hence $DK = PA = R = OD = OK$. We see that $DOK$ is an equilateral triangle. Then $\angle ECD = \angle EKD = 60^{\circ}$.

Further we prove that $PC$ bisects $\angle ECD$ using $\angle DCB = \angle DKB = \angle KDA$ (from isosceles trapezoid $DKAB$) and that $\angle KEC = \angle OEC = \angle OCE$ (from isosceles triangle $OCE$):
$$\angle DCP = \angle DCB + \angle BCP = \angle KDA + \alpha = \angle KEC + \angle CEA + \alpha = \angle OCE + 2\alpha = \angle PCE$$
Further by $\angle DCP = \angle PCE = \frac{1}{2} \angle ECD = 30^{\circ}$ distances between $P$ and sides $\triangle CDE$ are $\frac{1}{2} PC = \frac{1}{2} PB$ (as ratio between cathetus and hypotenuse in right triangle with angles $60^{\circ}$ and $30^{\circ}$). So, we have found incentre.

F. Let $F, G, H$ be the projections of $P$ on the sides $DE, DC$ resp. $CE$. If $P$ is indeed the incenter, then the three line segments $PF, PG, PH$ have the same length. This means that the problem is equivalent to proving that $PG = PH = PF = \frac{1}{2} PB = \frac{1}{2} PC$ and thus trigonometry in the right-angled triangles $CPG$ and $CPH$ tells us that it is enough to prove that $\angle DCP = \angle ECP = 30^{\circ}$.

We introduce the point $X$ as the second intersection of the line $CP$ with $\Gamma$. Because $O$ is the center of $\Gamma$ we can reduce the problem to proving that $\angle XOD = \angle XOE = 60^{\circ}$, or equivalently that $XOD$ and $XOE$ are equilateral triangles. This last condition is equivalent to $X$ being the reflection of $O$ on the line $DE$. Following the chain of equivalences, we see therefore that in order to solve the problem it is enough to prove that $X$ is the reflection of $O$ on $DE$. We prove this property as in Solution B, using the fact that $OXBP$ is an isosceles trapezoid.

G. Assume $AB$ is parallel to the horizontal axis, and that $\Gamma$ is the unit circle. Write $f(\theta)$ for the point $(\cos(\theta), \sin(\theta))$ on $\Gamma$. As in Solution C, assume that $\angle AOB = 180^{\circ} - 4\beta$; then we can take $B = f(2\beta)$ and $A = f(180^{\circ} - 2\beta)$. As in Solution C, we observe that $AOPC$ is an isosceles trapezoid, which we use to deduce that $\angle ABC = \frac{1}{2} \angle APC = \frac{1}{2} \angle OAB = \beta$. We now know that $C = f(180^{\circ} - 4\beta)$.

The point $P$ lies on $AB$ with $AP = R = 1$, so $P = (\cos(180^{\circ} - 2\beta) + 1, \sin(2\beta)) = (1 - \cos(2\beta), \sin(2\beta))$. The midpoint of $BP$ therefore has coordinates $\left(\frac{1}{2}, \sin(2\beta)\right)$, so $D$ and $E$ have $x$-coordinate $\frac{1}{2}$. Without loss of generality, we take $D = f(60^{\circ})$ and $E = f(-60^{\circ})$.

We have now obtained coordinates for all points in the problem, with one free parameter $(\beta)$. To show that $P$ is the incenter of $CDE$, we will show that $P$ lies on the bisector of $\angle CDE$; analogously, one can show that $P$ lies on the bisector of angle $CED$. The bisector of angle $CDE$ passes through the midpoint $M$ of the arc $CE$ not containing $DE$; because $C = f(180^{\circ} - 4\beta)$ and $E = f(-60^{\circ})$, we have $M = f(240^{\circ} - 2\beta)$.

It remains to show that $P = (1 - \cos(2\beta), \sin(2\beta))$ lies on the line connecting the points $D = (\cos(60^{\circ}), \sin(60^{\circ}))$ and $M = (\cos(240^{\circ} - 2\beta), \sin(240^{\circ} - 2\beta)) = (-\cos(60^{\circ} - 2\beta), -\sin(60^{\circ} - 2\beta))$. The equation for the line $DM$ is
$$\left(Y + \sin(60^{\circ} - 2\beta)\right)\left(\cos(60^{\circ}) + \cos(60^{\circ} - 2\beta)\right) = \left(\sin(60^{\circ}) + \sin(60^{\circ} - 2\beta)\right)\left(X + \cos(60^{\circ} - 2\beta)\right)$$
which, using the fact that $\cos(60^{\circ}) + \cos(60^{\circ} - 2\beta) = 2\cos(\beta)\cos(60^{\circ} - \beta)$ and $\sin(60^{\circ}) + \sin(60^{\circ} - 2\beta) = 2\cos(\beta)\sin(60^{\circ} - \beta)$, simplifies to
$$\left(Y + \sin(60^{\circ} - 2\beta)\right)\cos(60^{\circ} - \beta) = \left(X + \cos(60^{\circ} - 2\beta)\right)\sin(60^{\circ} - \beta)$$
Because $\cos(60^{\circ} - 2\beta)\sin(60^{\circ} - \beta) - \sin(60^{\circ} - 2\beta)\cos(60^{\circ} - \beta) = \sin(\beta)$, this equation further simplifies to
$$Y\cos(60^{\circ} - \beta) - X\sin(60^{\circ} - \beta) = \sin(\beta).$$
Plugging in the coordinates of $P$, i.e., $X = 1 - \cos(2\beta)$ and $Y = \sin(2\beta)$, shows that $P$ is on this line: for this choice of $X$ and $Y$, the left hand side equals $\sin(60^{\circ} + \beta) - \sin(60^{\circ} - \beta) = 2\cos(60^{\circ})\sin(\beta)$, which is indeed equal to $\sin(\beta)$. So $P$ lies on the bisector $DM$ of $\angle CDE$, as desired.

H. Let $\Gamma$ be the complex unit circle and let $AB$ be parallel with the real line and $0 < \varphi = \arg b < \frac{\pi}{2}$. Then
$$|b| = 1, \quad a = -\bar{b}, \quad p = a + 1 = 1 - \bar{b}$$
From $\operatorname{Re} d = \operatorname{Re} e = \operatorname{Re} \frac{p + b}{2} = \frac{1}{2}$ we get that $d = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ and $e = \frac{1}{2} - \frac{\sqrt{3}}{2} i$ are conjugate 6th roots of unity; $d^3 = e^3 = -1, d + e = 1, d^2 = -e, e^2 = -d$ etc.

Point $C$ is the reflection of $B$ in line $OP$. From $\arg p = \arg(1 - \bar{b}) = \frac{1}{2}(\pi - \varphi)$, we can get $\arg c = 2\arg p - \arg b = \pi - 2\varphi$, so $c = -\bar{b}^2$.

Now we can verify that $EP$ bisects $\angle CED$. This happens if and only if $(p - e)^2(\bar{c} - \bar{e})(\bar{d} - \bar{e})$ is real. Since $\bar{d} - \bar{e} = -\sqrt{3} i$, this is equivalent with $\operatorname{Re}[(p - e)^2(\bar{c} - \bar{e})] = 0$. Here
$$\begin{aligned} (p - e)^2(\bar{c} - \bar{e}) & = (1 - \bar{b} - e)^2(-b^2 - d) = (d - \bar{b})^2(-b^2 - d) \\ & = -|b|^4 + 2d|b|^2 b - d^2 b^2 - d\bar{b}^2 + 2d^2\bar{b} - d^3 \\ & = -1 - 2db + \bar{d}b^2 - d\bar{b}^2 - 2\overline{db} + 1 \\ & = -2(db - \overline{db}) + (\bar{d}b^2 - d\bar{b}^2), \end{aligned}$$
whose real part is zero. It can be proved similarly that $DP$ bisects $\angle EDC$.