Let $AB$ be a chord of the circle $(O)$. Denote M as the midpoint of the minor arc $AB$. A circle $(O')$ tangent to segment $AB$ and internally tangent to $(O)$. A line passes through $M$, perpendicular to $O'A$, $O'B$ and cuts $AB$ respectively at $C, D$. Prove that $AB = 2CD$.
Problem
Source: 2021 Saudi Arabia Training Lists p10 https://artofproblemsolving.com/community/c2758131_2021_saudi_arabia_training_tests
Tags: geometry, equal segments, tangent circles
Jalil_Huseynov
12.04.2022 10:10
Bump....
ythomashu
12.04.2022 10:47
The wording is ambiguous. I believe it should be specified that $O'$ is on the opposite side of $AB$ as $M.$
ythomashu
12.04.2022 11:33
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph;
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -3.224950463514445, xmax = 0.5503520430278914, ymin = 0.12438362533414005, ymax = 2.638430361753586; /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0);
draw((-2.321123646584331,1.533914914628198)--(-2.271901218751664,1.609002225658808)--(-2.3469885297822737,1.6582246534914749)--(-2.3962109576149406,1.583137342460865)--cycle, linewidth(.5) + qqwuqq);
/* draw figures */
draw(circle((-1.76,1.4636363636363636), 1));
draw((-2.70644348141704,1.7865058984072872)--(-0.8592204594698786,1.897913025871791), linewidth(.5));
draw((-2.70644348141704,1.7865058984072872)--(-1.7413153793856169,1.1538297817708227), linewidth(.5));
draw((xmin, 1.525469472693855*xmin + 5.2384840084369655)--(xmax, 1.525469472693855*xmax + 5.2384840084369655), linewidth(.5)); /* line */
draw((-1.820201211672225,2.461822625870621)--(-1.6997987883277748,0.46545010140210663), linewidth(.5));
/* dots and labels */
dot((-1.76,1.4636363636363636),dotstyle);
label("$O$", (-1.7436097939070259,1.5041466490256203), NE * labelscalefactor);
dot((-2.70644348141704,1.7865058984072872),dotstyle);
label("$A$", (-2.776315860719055,1.821576792512924), NE * labelscalefactor);
dot((-0.8592204594698786,1.897913025871791),dotstyle);
label("$B$", (-0.8421081864030824,1.9400840460815174), NE * labelscalefactor);
dot((-1.820201211672225,2.461822625870621),linewidth(4pt) + dotstyle);
label("$M$", (-1.908673468520424,2.494528696706008), NE * labelscalefactor);
dot((-1.6997987883277748,0.46545010140210663),linewidth(4pt) + dotstyle);
label("$F$", (-1.6843561671227292,0.5010673956057404), NE * labelscalefactor);
dot((-1.782831970443459,1.842209462139539),linewidth(4pt) + dotstyle);
label("$E$", (-1.7647718034728461,1.8765980173840566), NE * labelscalefactor);
dot((-1.7413153793856169,1.1538297817708227),linewidth(4pt) + dotstyle);
label("$O'$", (-1.7224477843412056,1.1867165055383166), NE * labelscalefactor);
dot((-2.2446377259302506,1.814357680273413),linewidth(4pt) + dotstyle);
label("$C$", (-2.3107516502710093,1.8554360078182364), NE * labelscalefactor);
dot((-2.3962109576149406,1.583137342460865),linewidth(4pt) + dotstyle);
label("$G$", (-2.501209736363392,1.5507030700704247), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy][/asy]
Let $E$ be the midpoint of $AB.$ It suffices to prove that $2AE=CE$ due to symmetry. Let the radius of $O,O'$ be $R,r,$ respectively.
Then $ME=2R-2r,$ $EO'=r,$ $EO=2r-R.$
And $AE^2+EO^2=AO^2\implies AE=\sqrt{R^2-(2r-R)^2}.$
The triangles $\triangle MCE$ and $\triangle AO'E$ are similar so \begin{align*}CE&=\frac{ME\cdot EO'}{AE}\\
\implies \frac{CE}{AE}&=\frac{(2R-2r)r}{R^2-(2r-R)^2}\\&=2.\end{align*}
Jalil_Huseynov
12.04.2022 11:52
ythomashu wrote:
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph;
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -3.224950463514445, xmax = 0.5503520430278914, ymin = 0.12438362533414005, ymax = 2.638430361753586; /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0);
draw((-2.321123646584331,1.533914914628198)--(-2.271901218751664,1.609002225658808)--(-2.3469885297822737,1.6582246534914749)--(-2.3962109576149406,1.583137342460865)--cycle, linewidth(.5) + qqwuqq);
/* draw figures */
draw(circle((-1.76,1.4636363636363636), 1));
draw((-2.70644348141704,1.7865058984072872)--(-0.8592204594698786,1.897913025871791), linewidth(.5));
draw((-2.70644348141704,1.7865058984072872)--(-1.7413153793856169,1.1538297817708227), linewidth(.5));
draw((xmin, 1.525469472693855*xmin + 5.2384840084369655)--(xmax, 1.525469472693855*xmax + 5.2384840084369655), linewidth(.5)); /* line */
draw((-1.820201211672225,2.461822625870621)--(-1.6997987883277748,0.46545010140210663), linewidth(.5));
/* dots and labels */
dot((-1.76,1.4636363636363636),dotstyle);
label("$O$", (-1.7436097939070259,1.5041466490256203), NE * labelscalefactor);
dot((-2.70644348141704,1.7865058984072872),dotstyle);
label("$A$", (-2.776315860719055,1.821576792512924), NE * labelscalefactor);
dot((-0.8592204594698786,1.897913025871791),dotstyle);
label("$B$", (-0.8421081864030824,1.9400840460815174), NE * labelscalefactor);
dot((-1.820201211672225,2.461822625870621),linewidth(4pt) + dotstyle);
label("$M$", (-1.908673468520424,2.494528696706008), NE * labelscalefactor);
dot((-1.6997987883277748,0.46545010140210663),linewidth(4pt) + dotstyle);
label("$F$", (-1.6843561671227292,0.5010673956057404), NE * labelscalefactor);
dot((-1.782831970443459,1.842209462139539),linewidth(4pt) + dotstyle);
label("$E$", (-1.7647718034728461,1.8765980173840566), NE * labelscalefactor);
dot((-1.7413153793856169,1.1538297817708227),linewidth(4pt) + dotstyle);
label("$O'$", (-1.7224477843412056,1.1867165055383166), NE * labelscalefactor);
dot((-2.2446377259302506,1.814357680273413),linewidth(4pt) + dotstyle);
label("$C$", (-2.3107516502710093,1.8554360078182364), NE * labelscalefactor);
dot((-2.3962109576149406,1.583137342460865),linewidth(4pt) + dotstyle);
label("$G$", (-2.501209736363392,1.5507030700704247), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy][/asy]
Let $E$ be the midpoint of $AB.$ It suffices to prove that $2AE=CE$ due to symmetry. Let the radius of $O,O'$ be $R,r,$ respectively.
Then $ME=2R-2r,$ $EO'=r,$ $EO=2r-R.$
And $AE^2+EO^2=AO^2\implies AE=\sqrt{R^2-(2r-R)^2}.$
The triangles $\triangle MCE$ and $\triangle AO'E$ are similar so \begin{align*}CE&=\frac{ME\cdot EO'}{AE}\\
\implies \frac{CE}{AE}&=\frac{(2R-2r)r}{R^2-(2r-R)^2}\\&=2.\end{align*}
In here you take $M-O-O'$ collinear, but it's only special case. It's not given that $O'$ lies on $MO$.
ythomashu
13.04.2022 02:28
I'm bad at math... i guess he wording of the problem threw me off
MathLover_ZJ
13.04.2022 03:57
I think it can be done with analytic geometry....
ythomashu
13.04.2022 04:00
I already solved it with distance bashing just need to put the solution together
ythomashu
13.04.2022 07:58
Let $E$ be where $(O')$ touches $AB$ and $F$ be the midpoint of $AB.$
It suffices to prove that $\frac{AC}{AE}=\frac12$ by symmetry.
Let $FE=b,$ $OM=R,$ $O'E=r.$
Let $G$ be on $OM$ such that $O'G||FE.$
Then: \begin{align*} &OG=\sqrt{(R-r)^2-b^2}\\
&OF=r+\sqrt{(R-r)^2-b^2}\\
&AF^2=R^2-\left(r+\sqrt{(R-r)^2-b^2}\right)^2\\
&MF=R-r-\sqrt{(R-r)^2-b^2}\\
&\triangle MCF \sim\triangle AO'E\implies CF=\frac{MF\cdot r}{AF+FE}\\
&EB=AF-FE\\
\end{align*}\begin{align*}\frac{CF}{EB}&=\frac{MF\cdot r}{AF^2-FE^2}\\
&=\frac{MF\cdot r}{R^2-\left(r+\sqrt{(R-r)^2-b^2}\right)^2-b^2}\\
&=\frac{MF\cdot r}{R^2-r^2-(R-r)^2+b^2-2r\sqrt{(R-r)^2-b^2}-b^2}\\
&=\frac{(R-r-\sqrt{(R-r)^2-b^2})r}{R^2-r^2-R^2-r^2+2Rr-2r\sqrt{(R-r)^2-b^2}}\\
&=\frac{Rr-r^2-r\sqrt{(R-r)^2-b^2}}{2Rr-2r^2-2r\sqrt{(R-r)^2-b^2}}\\
&=\frac12=\frac{AC+CF}{AE+EB}\\
&\implies \frac{AC}{AE}=\frac12.
\end{align*}
k12byda5h
13.04.2022 09:11
Let $(O')$ tangent to $AB$ and $(O)$ at $U,V$ respectively and $N,R$ be the midpoints of $AB,UV$ respectively. It's known by homothety that $U,V,M$ are collinear. Since $\angle RUO' = \angle VMN$. Hence, $\triangle UO'R \sim \triangle MNU$. Therefore, $\frac{UR}{MN} = \frac{UO'}{MU} \implies MU \cdot UV= 2UR \cdot MU = 2MN \cdot UO'$. Let circumcircle of $\triangle ABO'$ intersect line $O'U$ again at $X$. Note that $O'U \cdot UX = AU \cdot UB = MU \cdot UV = 2MN \cdot UO' \implies UX = 2MN$. Moreover, $\triangle XAB \cap \{U\} \sim \triangle MDC \cap \{N\}$. Therefore, $\frac{CD}{AB} = \frac{MN}{UX} = \frac{1}{2}$.
NN37252
14.04.2022 11:33
Let $(O’)$ be tangent to $AB$ and $(O)$ at $E$ and $F$ respectively, then $M,E,F$ are colinear and $\triangle MEA \sim\triangle MAF$, so $MA^2=MB^2=ME\cdot MF$. Then $M$ lies on the radical axis of $(O’)$ and the circle centered at $A$ with radius $0$. As $MC\perp AO’$, $C$ lies on the radical axis too, therefore $CA=CE$. Similarly, $DB=DE$, adding this up gives $AB=2CD$ as desired.