The circle $S_1$ intersects the hyperbola $y=\frac1x$ at four points $A$, $B$, $C$, and $D$, and the other circle $S_2$ intersects the same hyperbola at four points $A$, $B$, $F$, and $G$. It's known that the radii of circles $S_1$ and $S_2$ are equal. Prove that the points $C$, $D$, $F$, and $G$ are the vertices of the parallelogram.
Problem
Source: 2018 Belarusian National Olympiad 11.5
Tags: algebra, conics, hyperbola
RagvaloD
29.03.2019 16:48
Let center of first circle is $(x_1,y_1)$ then $a,b,c,d$ is solution of $(x-x_1)^2+(\frac{1}{x}-y_1)^2=r^2$ or $x^4-2x^3x_1+x^2(x_1^2+y_1^2-r^2)-2xy_1+1=0$ From here we can set two equalities: $abcd=1, 4r^2=(a+b+c+d)^2+(abc +abd+acd+bcd)^2-4(ab+ac+ad+bc+bd+cd)$ Same way we get $abfg=1,4r^2=(a+b+f+g)^2+(abf +abg+afg+bfg)^2-4(ab+af+ag+bf+bg+fg)$ Let $f(S)=(a+b+S)^2+(abS+\frac{a+b}{ab})^2-4(ab+(a+b)S+\frac{1}{ab})-4r^2$ We can note that $c+d$ and $f+g$ are solutions. $f(S)=a^2+b^2+S^2+2S(a+b)+2ab+a^2b^2S^2+(\frac{a+b}{ab})^2+2S(a+b)-4ab-4S(a+b)-\frac{4}{ab}-4r^2= a^2+b^2+S^2+a^2b^2S^2+(\frac{a+b}{ab})^2-2ab-\frac{4}{ab}-4r^2=S^2(1+a^2b^2)+a^2+b^2+(\frac{a+b}{ab})^2-2ab-\frac{4}{ab}-4r^2$ Obvious, that $f(S)$ has two solutions $s$ and $-s$ So we get that $c+d=s,f+g=-s,cd=fg=\frac{1}{ab} \to (c,d)=(-f,-g)$ or $(c,d)=(-g,-f)$ So it follows that there are two pairs of points $(C,G)$ and $(D,F)$ or $(C,F)$ and $(D,G)$ that are symmetric over point $(0,0)$
Vrangr
29.03.2019 20:34
Let $O$ be the centre of $S_1$. Note that $S_2$ is the reflection of $S_1$ over $AB$. It's well-known that the orthocenter of a triangle lies on the reflection of its circumcircle in a side. It's also known that any circumrectangular hyperbola of triangle passes through the orthocenter of that triangle. Hence, $\{F, G\}$ are the orthocenters of $\{\triangle CAB, \triangle DAB\}$. WLOG $F$ be the orthocenter of $\triangle CAB$. \[CF \perp AB, DG \perp AB \implies CF \parallel DG.\]\[CF = 2 \cdot \text{dist}(O, AB) = DG\]Hence, $CFGD$ is a parallelogram. [asy][asy] import geometry; size(7 cm); pair C = dir(108), B = dir(205), A = -1/B, D = dir(55); pair F = orthocenter(A, B, C), G = orthocenter(A, B, D); pair O = (0, 0); draw(C--F--G--D--cycle); draw(A--B); draw(F--foot(F, A, B)^^G--foot(G, A, B), gray+dashed); draw(conic((point) A, (point) B, (point) C, (point) D, (point) F), linewidth(1)); draw(circle(A, B, C)^^circle(G, A, B), linewidth(0.4)); dot(A^^B^^C^^D^^F^^G^^O); draw(O--foot(O, A, B), gray+linewidth(0.6)); label("$A$", A, dir(10)); label("$B$", B, -dir(0)); label("$C$", C, dir(80)); label("$D$", D, dir(95)); label("$F$", F, dir(140)); label("$G$", G, dir(30)); label("$O$", O, -dir(10)); [/asy][/asy]