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Source: 2020 MEMO I-3

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XbenX

#1 h Aug 30, 2020, 11:19 AM

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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 $CHI$ .

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#2 h Aug 30, 2020, 2:09 PM

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This problem was proposed by Burii.

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#3 h Aug 30, 2020, 3:01 PM • 3 Y

Y by A-Thought-Of-God, Muaaz.SY, math_comb01

It seems that can be easily done by a bunch of angle chasing after observing that...
Hint
Solution

This post has been edited 1 time. Last edited by Tintarn, Aug 30, 2020, 3:59 PM

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#4 h Aug 30, 2020, 3:52 PM • 1 Y

Y by Tintarn

Tintarn wrote:

It seems that can be easily done by a bunch of angle chasing after observing that...
Hint
Solution

Good solution. It's almost similar to mine. I have slightly different proof for 4), 5).

Details for 1), 2)

Another ending instead of 4), 5)

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#5 h Aug 31, 2020, 7:43 PM

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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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#6 h Sep 5, 2020, 9:31 PM • 1 Y

Y by Mango247

see my solution below (I can't delete this post)...

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#7 h Sep 5, 2020, 9:42 PM • 1 Y

Y by Mango247

see my solution below (I can't delete this post)...

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#8 h Feb 25, 2021, 3:38 PM • 1 Y

Y by Muaaz.SY

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, $\angle HBI$ = $\angle H_BBI$ = $\frac{\alpha}{2}$ = $\angle ICH_C$ $\text{[math]}$ = $\angle HCI$

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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#9 h Jan 10, 2022, 5:11 AM

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I think very hard problem for p3

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#10 h Sep 1, 2022, 4:45 AM • 1 Y

Y by Mango247

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.

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