There are two circles with centers $O_1,O_2$ lie inside of circle $\omega$ and are tangent to it. Chord $AB$ of $\omega$ is tangent to these two circles such that they lie on opposite sides of this chord. Prove that $\angle O_1AO_2 + \angle O_1BO_2 > 90^\circ$. Proposed by Iman Maghsoudi
Problem
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Tags: IGO, 2018 igo, Iran, geometry, geometric ineq
InternetPerson10
20.09.2018 13:10
This problem has four tangents. The solution requires eight.
Please try to solve the problem first :DD If you're really stuck, here's the hintPlot points $C$ and $D$ so that triangle $ABC$ has incenter $O_1$ and triangle $ABD$ has incenter $O_2$.
Try the problem first :DD
SolutionPlot points $C$ and $D$ so that triangle $ABC$ has incenter $O_1$ and triangle $ABD$ has incenter $O_2$.
Let $P_1$ and $P_2$ be the points where $O_1$ and $O_2$ are tangent to $\omega$, respectively.
Notice that since $O_1$ and $O_2$ are where the angle bisectors meet, we have
$$2(\angle O_1AO_2 + \angle O_1BO_2) = \angle CAD + \angle CBD$$Thus, we may now prove that $\angle CAD + \angle CBD > 180^{\circ}$, or $\angle ACB + \angle ADB < 180^{\circ}$.
Since $C$ and $D$ lie outside $\omega$, we have $$\angle ACB<\angle AP_1B=90^{\circ}$$and $$\angle ADB<\angle AP_2B=90^{\circ}$$. Thus, we have $\angle ACB + \angle ADB < 180^{\circ}$, which implies $\angle CAD + \angle CBD > 180^{\circ}$, which finally means $\angle O_1AO_2 + \angle O_1BO_2 > 90^{\circ}$.
GeoMetrix
23.08.2019 09:09
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); 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 = -22.35475386748818, xmax = 30.950498831173114, ymin = -18.77190550787951, ymax = 16.43032955350907; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-1.4,0.21), 8.242912106773915), linewidth(2) + wrwrwr); draw(circle((3.5,-2.85), 2.4769335881286767), linewidth(2) + wrwrwr); draw(circle((-6.36,4.41), 1.7628386199536248), linewidth(2) + wrwrwr); draw((xmin, -0.2913966028869642*xmin + 0.7205606217214329)--(xmax, -0.2913966028869642*xmax + 0.7205606217214329), linewidth(2) + wrwrwr); /* line */ draw((xmin, 1.711061397854231*xmin + 18.786032958582997)--(xmax, 1.711061397854231*xmax + 18.786032958582997), linewidth(2) + wrwrwr); /* line */ draw((-6.36,4.41)--(-9.021648559008357,3.349438364256544), linewidth(2) + wrwrwr); draw((-9.021648559008357,3.349438364256544)--(3.5,-2.85), linewidth(2) + wrwrwr); draw((-6.36,4.41)--(-7.7356660737088045,5.483038536029795), linewidth(2) + wrwrwr); draw((-6.36,4.41)--(-6.46,2.65), linewidth(2) + wrwrwr); draw((3.5,-2.85)--(6.715057071168448,-1.23618419700914), linewidth(2) + wrwrwr); draw((-6.36,4.41)--(6.715057071168448,-1.23618419700914), linewidth(2) + wrwrwr); draw((3.827835662452707,-0.39485768672685784)--(3.5,-2.85), linewidth(2) + wrwrwr); draw((3.5,-2.85)--(5.44,-4.39), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.4,0.21),dotstyle); dot((5.44,-4.39),dotstyle); label("$B$", (5.578583098628483,-4.061040315067174), NE * labelscalefactor); dot((3.5,-2.85),dotstyle); label("$O_2$", (3.640210273222618,-2.50341929465175), NE * labelscalefactor); dot((-7.7356660737088045,5.483038536029795),dotstyle); label("$D$", (-7.6092748742221366,5.838506614684186), NE * labelscalefactor); dot((-6.36,4.41),dotstyle); label("$O_1$", (-6.22472285607509,4.765478800620227), NE * labelscalefactor); dot((-6.46,2.65),dotstyle); label("$K$", (-6.328564257436118,3.000174977482747), NE * labelscalefactor); dot((-9.021648559008357,3.349438364256544),dotstyle); label("A", (-8.889985491008154,3.6924509865562687), NE * labelscalefactor); dot((3.827835662452707,-0.39485768672685784),linewidth(4pt) + dotstyle); label("$H$", (3.9517344773057035,-0.11506706334810053), NE * labelscalefactor); dot((6.715057071168448,-1.23618419700914),linewidth(4pt) + dotstyle); label("$I$", (6.859293715414501,-0.9457982742363266), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] this is easy we have that $DK$ is perpendicular to $AO_1$ by simply joining $DK$ and then angle chasing. Now let $DK \cap AO_1=P$ now let $\angle ADK= \beta$ so we have $\angle AO_1K=\beta$ as $ADO_1K$ is cyclic now let $\angle O_1AD=\alpha$ and so we have $\angle DKO_1=\alpha$ now $\Delta APK \equiv \Delta KPO_1$ can be proved trivially using angle chasing we have that $AP=KP$ or that $2\alpha=90$ or $\alpha=45$ now so $\angle O_1AK=45$ we have $\angle O_1AO_2=45+x$ and similiarly performing these ssteps in circle with centre $O_2$ we have that $\angle O_2BO_1=45+y$ so we have $\angle O_1AO_2+\angle O_2BO_1=45+x+45+y=90+x+y>90$ and so we are done. $\blacksquare$
Afo
05.12.2020 06:49
InternetPerson10 wrote: Since $C$ and $D$ lie outside $\omega$, we have $$\angle ACB<\angle AP_1B=90^{\circ}$$and $$\angle ADB<\angle AP_2B=90^{\circ}$$. Thus, we have $\angle ACB + \angle ADB < 180^{\circ}$ There's an error here. It's not true in general that $\angle ACB<90^{\circ}$. However, we can prove it this way $$\angle ACB +\angle ADB<\frac1 2(\text{Minor Arc of AB} +\text{Major Arc of AB} )= 180^{\circ}$$
anurag27826
04.05.2023 16:50
GeoMetrix wrote: we have that $DK$ is perpendicular to $AO_1$ by simply joining $DK$ and then angle chasing. Actually this is wrong.