Let $ABC$ be an acute scalene triangle with circumcircle $\omega$ and incenter $I$. Suppose the orthocenter $H$ of $BIC$ lies inside $\omega$. Let $M$ be the midpoint of the longer arc $BC$ of $\omega$. Let $N$ be the midpoint of the shorter arc $AM$ of $\omega$. Prove that there exists a circle tangent to $\omega$ at $N$ and tangent to the circumcircles of $BHI$ and $CHI$.
Problem
Source: 2020 MEMO I-3
Tags: geometry, circumcircle, incenter, memo, MEMO 2020
timon92
30.08.2020 17:09
This problem was proposed by Burii.
Tintarn
30.08.2020 18:01
It seems that can be easily done by a bunch of angle chasing after observing that...
The touching points lie on $BN$ resp. $CN$.
Let $E$ be the intersection of $BN$ with $(BIH)$. Similarly, let $F$ be the intersection of $CN$ with $(CIH)$.
Let $O_1$ and $O_2$ be the centers of $(BIH)$ and $(CIH)$.
Let $D$ be the intersection of $EO_1$ and $FO_2$.
We claim that the desired circle is the one centered at $D$ passing through $E$ and $F$. Of course to this end we also need to show that $DE=DF$.
1) An easy angle chase shows that triangles $EHI$ and $FHI$ are congruent.
2) Another angle chase shows that the two circles $(BIH)$ and $(CIH)$ have the same radius.
3) Hence, by symmetry, we have that $DF=DE$ and $D$ lies on $HI$. In particular, the circle mentioned above exists.
4) Another angle chase shows that $\angle EDF=2\angle ENF$, hence $N$ lies on this circle.
5) Finally, to prove that the two circles touch, it suffices to compare the two angles over the chord $CN=FN$ i.e. we need to show that $\angle CBN=\angle FEN$ which is another (final) easy angle chase.
EagleEye
30.08.2020 18:52
Tintarn wrote: It seems that can be easily done by a bunch of angle chasing after observing that...
The touching points lie on $BN$, $CN$ resp.
Let $E$ be the intersection of $BN$ with $(BIH)$. Similarly, let $F$ be the intersection of $CN$ with $(CIH)$.
Let $O_1$ and $O_2$ be the centers of $(BIH)$ and $(CIH)$.
Let $D$ be the intersection of $EO_1$ and $FO_2$.
We claim that the desired circle is the one centered at $D$ passing through $E$ and $F$. Of course to this end we also need to show that $DE=DF$.
1) An easy angle chase shows that triangles $EHI$ and $FHI$ are congruent.
2) Another angle chase shows that the two circles $(BIH)$ and $(CIH)$ have the same radius.
3) Hence, by symmetry, we have that $DF=DE$ and $D$ lies on $HI$. In particular, the circle mentioned above exists.
4) Another angle chase shows that $\angle EDF=2\angle ENF$, hence $N$ lies on this circle.
5) Finally, to prove that the two circles touch, it suffices to compare the two angles over the chord $CN=FN$ i.e. we need to show that $\angle CBN=\angle FEN$ which is another (final) easy angle chase.
Good solution. It's almost similar to mine. I have slightly different proof for 4), 5).
1) $\angle IEH = \angle IBH=\angle ICH=\angle IFH$
$\angle HBC-\angle HCB = \angle IBC-\angle ICB$ and $\angle EBC-\angle FCB = \frac{1}{2}(\angle ABC-\angle ACB) =\angle IBC-\angle ICB$ implies $\angle EIH=\angle FIH$
2) $\odot(BIH) \equiv \odot(CIH)$ since $C$ is the orthocenter of $\triangle BIH$
Let $K$ be midpoint of small arc $BC$. Then, $K$ is the circumcenter of $\triangle IBC$. Note that $KBO_1I, KCO_2I$ are rhombus.
$\angle O_1 BA = \angle KBC = \angle MBO \implies \angle NBO = \angle O_1 BN =\angle O_1 EB \implies BO \parallel O_1 E$
Similarly, we get $CO \parallel O_2 F$. And $KO\parallel ID$. Therefore, $O_1 IO_2 D$ is a parallel transport of $BKCO$. Since $ON\parallel KA \parallel OD$, we get $O,D,N$ colinear and $ND = ON-OD=OB-O_1 B = O_1 D-OE=DE$.
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Kagebaka
31.08.2020 22:43
Here is a different solution: WLOG let $AB>AC$ because there are a lot of half angles so directing is kind of annoying. Let $BA,BM$ and $CA,CM$ meet $(BIH)$ and $(CIH)$ at $K,P$ and $L,Q,$ respectively. For shorthand, let $\angle BAC =\angle A,\angle ACB=\angle C,\angle CBA=\angle B.$ Observe that \begin{align*} \angle ACI&=\frac{1}{2}\angle C \\ \angle ACI&=\frac{1}{2}\angle B+\frac{1}{2}\angle C-90^\circ+\frac{1}{2}\angle A+\frac{1}{2}\angle C \\ \angle ACI&=\angle IBA-\angle MBA \\ \angle ACI&=\angle IBM \\ \angle LCH&=\angle HBP, \end{align*}so it follows that $R=MB\cap HC$ and $S=MC\cap HB$ lie on $(PLH)$ and $(KQH),$ respectively. Combined with $\angle CBM=90^\circ-\frac{1}{2}\angle A=180^\circ-\angle BIC=\angle BHC=\angle LHR,$ this implies that $\angle MQK=\angle SHK=\angle LPM\implies KQ||PL||BC.$ Thus, \[\angle HPL=\angle HRL=\angle CBL=\angle PLH\]and a very similar process also implies that $\angle HKQ=\angle KQH,$ so $KQ$ and $PL$ share a perpendicular bisector $IH$ and therefore are the parallel sides of an isosceles trapezoid. Finally, since \begin{align*} \angle MQL&=180^\circ-\angle LQB \\ &=\angle CHL \\ &=180^\circ-\angle LHR \\ &=\angle LPM, \end{align*}we find that $M$ also lies on $(LPKQ).$ This implies that $\angle KQL=\angle PKQ=\angle PMQ=\angle A$ so $A$ lies on $(LPKQ)$ as well and we're done by homothety (the desired circle circumscribes the midpoints of arcs $AM,KP,LQ$ on their respective circles). $\blacksquare$
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bora_olmez
06.09.2020 00:31
see my solution below (I can't delete this post)...
bora_olmez
06.09.2020 00:42
see my solution below (I can't delete this post)...
bora_olmez
25.02.2021 18:38
Probably one of my favourite geometry problems, here is the rather lengthy solution I gave at the competition Let $circ(BHI) = \omega_1$, $circ(CHI) = \omega_2$, centers of $\omega$, $\omega_1$, $\omega_2$ be $O, O_1, O_2$, respectively. Let $D$ be the midpoint of arc $BC$ not containing $A$. Also, let $\angle BAC$ = $\alpha$, $\angle ABC$ = $\beta$, $\angle ACB$ = $\gamma$ Let $HI$ $\cap$ $BC$ = $H_I$, $HB$ $\cap$ $CI$ = $H_B$ and $HC$ $\cap$ $BI$ = $H_C$. Now, $\angle HBI$ = $\angle H_BBI$ = $\frac{\alpha}{2}$ = $\angle ICH_C$ $ $ = $\angle HCI$ Let the radii of $\omega$, $\omega_1$, $\omega_2$ be $R$, $R_1$, $R_2$. We get: $HI$ = $2R_1 \sin \frac{\alpha}{2}$, $HI$ = $2R_2 \sin \frac{\alpha}{2}$ because of sine rule. $\implies$ $R_1$ = $R_2$ Claim: $AD$ $\parallel$ $ON$ Proof) Let $\angle ADM$ = $\phi$ Now, $\angle NDM$ = $\frac{1}{2}$ $\angle ADM$ because $N$ is the midpoint of arc AM. $\implies$ $\angle NDM $= $\frac{\phi}{2}$ Also, $\angle NOM$ = $2 \angle NDM$ because $O$ is the center of $\omega$. $\implies$ $\angle NOM$ = $\phi$. Therefore, $\angle NOM$ = $\angle ADM$ meaning that $AD$ is indeed parallel $ON$. $\square$ Moreover, $AD$ $\parallel$ $NO$ $\implies$ $AI$ $\parallel$ $NO$ (because $AI$ and $AD$ coincide). Claim: $BO_1$ $\parallel$ $AI$ (analagously, $CO_2$ $\parallel$ $AI$) Proof) From our initial angle chase, $\angle HBI$ = $\frac{\alpha}{2}$. Moreover, $\angle IBC$ = $\frac{\beta}{2}$. $\implies$ $\angle HBH_I$ = $\frac{\beta + \alpha}{2}$ Consequently, $\angle BHH_I$ = $90^{\circ}$ -$\frac{\beta +\alpha}{2}$ = $\frac{\gamma}{2}$ As $O_1$ is the center of $\omega_1$, $\angle O_1BI$ = $90^{\circ}$ -$\angle BHI$ = $90^{\circ}$ - $\angle BHH_I$ = $90^{\circ}$ - $\frac{\gamma}{2}$ = $\frac{\alpha + \beta}{2}$ Furthermore, $\angle BIA$ = $180^{\circ}$ - $\angle ABI$ - $\angle BAI$ = $180^{\circ}$ - $\frac{\alpha}{2}$ - $\frac{\beta}{2}$ from $ \triangle$$BIA$ $\implies$ $\angle BIA$ + $\angle O_1BI$ = $180^{\circ}$ Meaning that: $BO_1$ $\parallel$ $AI$ $\implies$ $BO_1$ $\parallel$ $CO_2$ $\parallel$ $AI$ $\parallel$ $ON$ (1) After these "preliminary steps" we can finally "construct" the circle which will be tangent to the three circles, $\omega, \omega_1, \omega_2$ Let $O_3$ be the point on $ON$ such that $BO_1$ = $OO_3$ and $D,O,O_3$ lie on $ON$ in this order. Now, (1)$ \implies$ $BO_1O_3O$ and $O_3OCO_2$ are parallelograms because $BO_1$ = $OO_3$. Let $R_3$ = $O_3N$ and let $\omega_3$ be the circle with center $O_3$ and radius $R_3$ (i.e $\omega_3$ is the circle with center $O_3$ through $N$). Then $R_3$ =$BO_1$ = $OO_3$ = $R-R_3$ $\implies$ $R$ = $R_1+R_3$. Also, $R_1+R_3$ = $R$ = $BO$ = $O_1O_3$. Let $K$ = $O_3O_1$ $\cap$ $\omega_1$. Then, $R_1+R_3$ = $O_1O_3$ = $O_1K + KO_3$ = $R_1$ + $KO3$. $\implies$ $KO_3$ = $O_3$. Therefore $K$ $\cap$ $\omega_1$ and $O_3$, $O_1$, $K$ are collinear meaning that $\omega_3$ is tangent to $\omega_1$ at $K$. Analogously, $\omega_3$ is tangent to $\omega_2$. Finally, $O_3$, $O$, $N$ are collinear. $\implies$ $\omega_3$ is tangent to $\omega$ at $N$ because $\omega$ has center $O$ and $\omega_3$ has center $O_3$. Essentially $\omega_3$ with center $O_3$ through $N$ satisfies all three conditions. $\square$
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lazizbek42
10.01.2022 08:11
I think very hard problem for p3
Mahdi_Mashayekhi
01.09.2022 07:45
Let $BN$ and $CN$ meet $BIH$ and $CIH$ at $X$ and $Y$. Let $M'$ be midpoint of arc $BC$. Claim $:$ circles $BIH$ and $CIH$ are symmetric wrt $HI$. Proof $:$ Note that $2R_{BHI} = \frac{BI}{\sin{BHI}} = \frac{BI}{\sin{ICB}} = \frac{CI}{\sin{IBC}} = \frac{CI}{\sin{CHI}} = 2R_{CHI}$ so circles are congruent and since $HI$ is their Radical Axis then Claim is proved. Now By this symmetry we have that $O_1X$ and $O_2Y$ intersect on Radical Axis. Let the intersection point be $K$. we claim $K$ is center of wanted circle. Claim $: BO_1IM'$ is Rhombus. Proof $:$ Note that $\angle BM'I = 2\angle BCI = 2\angle BHI = \angle BO_1I$ and $O_1B = O_1I$ and $M'B = M'I$. Claim $: O_1K || BO$ where $O$ is center of $ABC$. Proof $:$ Note that $\angle O_1BI = \angle M'BI$ so $\angle O_1BA = \angle CBM = \frac{\angle A}{2} = \angle MBO$ and since $\angle ABN = \angle NBM$ then $\angle O_1XB = \angle O_1BX = \angle O_1BN = \angle NBO = \angle XBO$ Note that with same approach $O_2Y || CO$ so $\angle XKY = \angle BOC = 2\angle A = 2\angle BNC = 2\angle XNY$ so $K$ is center of $XYN$ and we only need to prove that $N,K,O$ are collinear to prove that $XYN$ is tangent at $N$. Note that $ON \perp AM \perp AI || O_1B || KO$ so $KO || ON \implies O,K,N$ are collinear as wanted.