Let $\Gamma_1$ be a circle with center $O$ and $A$ a point on it. Consider the circle $\Gamma_2$ with center at $A$ and radius $AO$. Let $P$ and $Q$ be the intersection points of $\Gamma_1$and $\Gamma_2$. Consider the circle $\Gamma_3$ with center at $P$ and radius $PQ$. Let $C$ be the second intersection point of $\Gamma_3$ and $\Gamma_1$. The line $OP$ cuts $\Gamma_3$ at $R$ and $S$, with $R$ outside $\Gamma_1$. $RC$ cuts $\Gamma_1$ into $B$. $CS$ cuts $\Gamma_1$ into $D$. Show that $ABCD$ is a square.
Problem
Source: Mathematics Regional Olympiad of Mexico Center Zone 2020 P4
Tags: square, geometry, circles
x3yukari
14.11.2021 19:56
We use Cartesian coordinates.
Let $\Gamma_1$ be a circle of radius $1$ at the point $O=\left( -\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).$ Let $A$ be the point $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).$ Because $OAP$ is an equilateral triangle with side length $1, P$ is at $(0,0).$ Since the height of $OAP$ is $\frac{\sqrt{3}}{2}, PQ=\sqrt{3}.$ So $\Gamma_3$ has equation $x^2+y^2=3.$ Additionally, the line $OP$ has equation $y=\frac{\sqrt{3}}{2}x.$ So point $R,$ which is not in $\Gamma_1$ and thus in the first quadrant, is the point $\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right).$ Point $S$ has coordinates $\left(-\frac{\sqrt{3}}{2}, -\frac{3}{2}\right).$
We also know that $\Gamma_1$ has the equation $\left(x+\frac{1}{2}\right)^2+\left(y+\frac{\sqrt{3}}{2}\right)^2=1.$ Solving the system with the equation for $\Gamma_1,$ we obtain the two points $\left(-\frac{3}{2}, -\frac{\sqrt{3}}{2}\right)$ and $\left(0, -\sqrt{3}\right).$ However, the latter is the point $Q,$ so our desired point $C$ has coordinates $\left(-\frac{3}{2}, -\frac{\sqrt{3}}{2}\right).$ Combining this with point $R,$ the equation of line $CR$ is $y=x+\frac{3-\sqrt{3}}{2}.$ Solving the system with the equation for $\Gamma_1,$ we obtain the two points $\left(-\frac{3}{2}, -\frac{\sqrt{3}}{2}\right)$ and $\left(-\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right).$ But since the former is point $C,$ the latter is our desired point $B.$
Using our points $C$ and $S,$ the equation of line $CS$ is $y=-x-\frac{3+\sqrt{3}}{2}.$ Solving the system with the equation for $\Gamma_1,$ we obtain the two points $\left(-\frac{3}{2}, -\frac{\sqrt{3}}{2}\right)$ and $\left(-\frac{1}{2}, -1-\frac{\sqrt{3}}{2}\right).$ But again, since the former is point $C,$ the latter is our desired point $D.$
Now, since we have the coordinates of points $A, B, C, D,$ we can easily calculate the distance between each of them and verify that $AB=BC=CD=DA=\sqrt{2},$ implying $ABCD$ is indeed a square.