Sharygin 2025 CR P12

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Gengar_in_Galar

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Gengar_in_Galar

#1 h Mar 10, 2025, 12:07 PM

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Circles $\omega_{1}$ and $\omega_{2}$ are given. Let $M$ be the midpoint of the segment joining their centers, $X$ , $Y$ be arbitrary points on $\omega_{1}$ , $\omega_{2}$ respectively such that $MX=MY$ . Find the locus of the midpoints of segments $XY$ .
Proposed by: L Shatunov

This post has been edited 1 time. Last edited by Gengar_in_Galar, Mar 11, 2025, 9:53 AM

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#2 h Mar 10, 2025, 12:18 PM

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sol

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#3 h Mar 10, 2025, 1:08 PM

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The answer is the radical axis of two circles ,we have many methods for proving it The problem in solving it synthetically is that we have to prove in both direction, First to prove every such point lie on the radical axis and second (construct) for every point on the radical axis such X and Y exists.

To overcome this we use coordinates Fixing one circle to be at origin with radius one and another circle at a distance of d with radius r, line joining centres to be the x-axis

Equation of the first circle $x^{2}+y^{2}=1$

Equation of the second circle $(x-d)^{2}+y^{2}=r^{2}$

the midpoint of their centres be $M:(\frac{d}{2},0)$

let X be a point on first circle and Y be a point on the second circle

$X:(x_{1},y_{1})$

$Y:(x_{2},y_{2})$

midpoint of XY say $F:(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$ given that $MX=MY$ equating using the coordinates gives

$\sqrt{(x_{1}-\frac{d}{2})^{2}+y_{1}^{2}}=\sqrt{(x_{2}-\frac{d}{2})^{2}+y_{2}^{2}}$

further simplifying gives (using the fact that

Equation of the first circle $x^{2}+y^{2}=$ 1Equation of the second circle

$\frac{x_{1}+x_{2}}{2}=\frac{1-r^{2}+d^{2}}{2d}$

$(x-d)^{2}+y^{2}=r^{2}$

which is constant and equal to the radical axis of the both circle we can also note that the condition $MX=MY$ does not depend on the y coordinate of X and Y

This post has been edited 1 time. Last edited by drago.7437, Mar 10, 2025, 1:09 PM

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cursed_tangent1434

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cursed_tangent1434

#4 h Mar 10, 2025, 3:28 PM

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We denote by $O_1$ and $O_2$ the centers of circles $\omega_1$ and $\omega_2$ respectively and $\ell$ the line $\overline{XY}$ .

We claim that the locus of the midpoints of segments $XY$ is simply the radical axis of circles $\omega_1$ and $\omega_2$ . Let $A$ and $B$ denote the second intersections of $\ell$ with $\omega_1$ and $\omega_2$ respectively. The following is the key claim.

Claim : $M$ is equidistant to points $A$ and $B$ \textit{i.e.} $MA=MB$ .

Proof : Let $P$ and $Q$ denote the midpoints of segments $AX$ and $BY$ respectively. Let $N$ denote the midpoint of segment $XY$ . Clearly $MN \perp \ell$ , $O_1P \perp \ell$ and $O_2Q \perp \ell$ due to perpendicular bisector reasons. Thus, by the Mean Geometry Theorem (or by considering the midline of trapezoid $O_1PQO_2$ ), since $M$ is the midpoint of segment $O_1O_2$ it follows that $N$ is the midpoint of segment $PQ$ . Further since clearly $NX=NY$ , we must have $XP=YQ$ . Thus,
$AX = 2PX = 2YQ = YB$ which implies that $\triangle AXM \cong \triangle BYM$ . Thus, $MA=MB$ as desired.

Thus as a result $NA=NB$ as well. But now note that,
$\text{Pow}_{\omega_1}(N) = NX \cdot NA = NY \cdot NB=\text{Pow}_{\omega_2}(N)$ so the midpoint of segment $XY$ lies on the radical axis of circles $\omega_1$ and $\omega_2$ , as desired.

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#5 h Mar 10, 2025, 3:29 PM

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Coord bash is the most reasonable here,
The trick of the question was the direct common tangent part
by varying the y coordinates of the two points X and Y we can only go till the direct common tangents so yes

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#6 h Mar 11, 2025, 2:01 PM

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CLAIM: ,Locus of midpoints of $XY$ is the radical axis $w_1$ and $w_2$ .

PROOF:Let the center of $w_1$ and $w_2$ be $O_1$ and $O_2$ .
Let $E$ is the midpoint of $XY$

$\star$ Drop $\perp$ from $O_1$ and $O_2$ to line $XY$ at $D_1$ and $D_2$ respectively. $\implies$ $D_1O_1$ is $\parallel$ to $D_2O_2$
Let $K= XY \cap w_1$ and $L= XY \cap w_2$

As $MX=MY$ $\implies$ $ME is \perp to XY$
Now as $M$ is the midpoint of $O_1O_2$ and $ME \perp D_1D_2$ $\implies$ $ME$ is $\parallel$ to $D_1O_1$ (mid-point theorem).
$\implies$ $E$ is the midpoint of $D_1D_2$ .

By above definition of $E$ we could say.:
$EX=EY=a$ and $ED_1=ED_2=b$ $\implies$ $EK=a-b=EL$
$\implies$ $EK.EX=EL.EY=(a-b).a$ $\implies$ $E$ indeed lies on the radical axis because of its converse theorem.

Is this correct.

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#7 h Mar 12, 2025, 10:11 AM

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Answer: The part of the radical axis between the 2 extreme common tangents, excluding that between the two interior common tangents (if they exist). Endpoints are included in the locus.

Image

Let $P$ be the midpoint of $XY$ , $XP \cap w_1 = G$ , $YP \cap w_2 = F$ .
Let $E$ & $H$ be the foot of projections of $A$ & $B$ on $XG$ & $FY$ .

Since $P$ is midpoint of $XY$ in isosceles triangle $XMY$ , we get $MP \perp XY$
Since $CM \parallel BI$ ( $\perp XY$ ) & $AM = MB$ , $\Rightarrow AC = CI \Rightarrow EP = PH$
$XP = PY \Rightarrow XP - PE = PY - PH \Rightarrow HY = XE \Rightarrow XG = FY$
$Pow_{w_1} P = PG \times PX = (PX - XG) \times PX = (PY - FY) \times PY = PF \times PY = Pow_{w_2} P$
$\Rightarrow P$ lies on radical axis of $w_1$ & $w_2$

For each $X$ , there are two possible positions for $Y$ : one below the center ( $Y_1$ ) & above it ( $Y_2$ ). We know that if $XY$ is tangent to $w_1$ , it is also to $w_2$ (by power of point). Also, the lowest position for both $X$ & $Y_1$ occurs when $XY$ is tangent to the circles. Hence, the lowest position for $P_1$ is the midpoint of their common tangent. Similarly, the highest position for $P_1$ is the midpoint of their second common tangent. We have similar results for $P_2$ , form which we get our result.

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maths_enthusiast_0001

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maths_enthusiast_0001

#8 h Mar 19, 2025, 7:02 AM

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Claim: Locus of the midpoints of segments $\color{blue}{XY}$ is the radical axis of circles $\color{blue}{\omega_{1}}$ and $\color{blue}{\omega_{2}}$ .
Proof: Denote the centers of the circles $\omega_{1}$ and $\omega_{2}$ by $O_{1}$ and $O_{2}$ respectively. Toss the figure on the coordinate plane with $O_{1} \equiv (-1,0)$ and $O_{2} \equiv (+1,0)$ . Thus, $M \equiv (0,0)$ . Let the radii of $\omega_{1}$ and $\omega_{2}$ be $r_{1}$ and $r_{2}$ respectively. Thus,
$\omega_{1} \equiv (x+1)^2+y^2=r_{1}^2$ $\omega_{2} \equiv (x-1)^2+y^2=r_{2}^2$ Now, let $X \equiv (h_1,k_1),Y \equiv (h_2,k_2)$ and $Z \equiv (h,k) \equiv \left(\frac{h_1+h_2}{2},\frac{k_1+k_2}{2}\right)$ where $Z$ is the midpoint of segment $XY$ . Then,
$h_{1}^2+k_{1}^2+2h_1+1-r_{1}^2=0 \cdots (1)$ $h_{2}^2+k_{2}^2-2h_2+1-r_{2}^2=0 \cdots (2)$ As $MX=MY$ ,
$h_{1}^2+h_{2}^2=k_{1}^2+k_{2}^2 \cdots (3)$ On subtracting $(2)$ from $(1)$ and using $(3)$ we get, $2(h_1+h_2)=(r_{1}^2-r_{2}^2)$ $\implies \boxed{h=\frac{r_{1}^2-r_{2}^2}{4}}$ Therefore, the locus of $Z$ is $\boxed{x=\frac{r_{1}^2-r_{2}^2}{4}}$ . Now, let $L$ denote the radical axis of $\omega_{1}$ and $\omega_{2}$ . Then,
$L \equiv \omega_{1}-\omega_{2}=0 \equiv \boxed{4x=(r_{1}^2-r_{2}^2)}$ Hence, the locus of $Z$ is radical axis of $\omega_{1}$ and $\omega_{2}$ . $\blacksquare$

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