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Source: STEMS 2021/CAT B/P3

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1107 posts

#1 h Jan 24, 2021, 3:25 PM • 3 Y

Y by Severus, itslumi, Mathematicsislovely

Let $ABC$ be a triangle with $I$ as incenter.The incircle touches $BC$ at $D$ .Let $D'$ be the antipode of $D$ on the incircle.Make a tangent at $D'$ to incircle.Let it meet $(ABC)$ at $X,Y$ respectively.Let the other tangent from $X$ meet the other tangent from $Y$ at $Z$ .Prove that $(ZBD)$ meets $IB$ at the midpoint of $IB$

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1107 posts

#2 h Jan 24, 2021, 3:38 PM

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Solution
@below

This post has been edited 3 times. Last edited by Pluto1708, Jan 24, 2021, 3:50 PM

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#3 h Jan 24, 2021, 3:39 PM

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Let $T$ be the $A$ mixtillinear touchpoint. Let the tangents from $T$ to the incircle meet the circumcircle at $X'$ and $Y'$ . Then by Poncelet's Porism, $X'Y'$ is tangent to the incircle.

Let $M$ be the arc midpoint of $BAC$ . Then it is well known that $T-I-M$ are collinear. But $TI$ is the bisector of $\angle X'TY'$ which implies $M$ lies on the angle bisector of $\angle X'TY'$ . But $M$ lies on $(X'TY')$ , therefore $M$ lies on the perpendicular bisector of $X'Y'$ . However, by definition, $M$ lies on the perpendicular bisector of $BC$ . Thus $X'Y' \parallel BC \implies \{X', Y'\} = \{X, Y\}$ and $Z = T$ .

Now all that is left to show is $T, B, N, D$ are concyclic, where $N$ is the midpoint of $BI$ .
Note that because $IBD$ is a right angled triangle, $\angle IND = 2\angle IBD = \angle ABC$ .
Thus $T, B, N, D$ are concyclic $\iff \angle IND = \angle BTD \iff \angle ABC = \angle BTD \iff \angle ATC = \angle BTD \iff TA$ and $TD$ are isogonal with respect to $\angle BTC$ , but the latter is well known. Hence we are done. :-D

:-D

EDIT: Dang, sniped by a few seconds :/

This post has been edited 2 times. Last edited by gabrupro, Jan 24, 2021, 3:41 PM

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#4 h Jan 24, 2021, 5:55 PM • 1 Y

Y by PhysicsGirl123

This is also CAT A P5

Let $M_A$ denote the midpoint of $\widehat{BAC}$ and let $K$ be the midpoint of $\overline{BI}$ . Then we have the following claims.

Claim 1: $Z$ is the mixtillinear intouch point

Proof: Notice that by ponceletes porism $Z \in \odot(ABC)$ . Now notice that $\overline{XY} \parallel \overline{BC}$ and so $\overline{M_AX}=\overline{M_AY}$ which implies that $Z,I,M_A$ are collinear $\qquad \square$

Now its pretty well known that $\angle BZD= \angle ABC$ (reflect $Z$ across perpendicular bisector of $\overline{BC}$

) . This implies that $\angle BKD=180^\circ-\angle ABC$ and we're done $\qquad \blacksquare$

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595 posts

#5 h Jan 25, 2021, 8:44 AM

Y by

Proposed by Arindam Bhattacharya

Official Solution:
Let $\omega$ denote the incircle of $ABC$ , also let $AI$ meet $(ABC)$ at $N$ and let $M$ be the antipode of $N$ in $(ABC)$ . Let $T$ be the $A-$ mixtilinear incircle touchpoint.
Lemma 1:Let tangents from $T$ to $\omega$ meet $(ABC)$ at $B',C'$ . Then $B'C'$ is tangent to the $\omega$ at $D'$ .
Proof:
By Poncelet Porism, we know that $B'C'$ is tangent to the incircle.

Now, note that $\omega$ is also the incircle of $TB'C'$ . Since $T,I,M$ are collinear, we have that $M$ is the midpoint of arc $B'C'$ not containing $T$ .

Let $\Omega$ denote the circle centered at $M$ with radius $MI$ . Its easy to see that $B'C'$ is the radical axis of $(ABC)$ and $\Omega$ , so $B'C'$ is parallel to $BC$ and we're done.

Coming back to the problem, we have $Z$ lies on $(ABC)$ by Poncelet Porism, using our lemma we get that $Z$ is the $A-$ mixtilinear incircle touch point.

Now, consider $\Delta IBC$ . Its well known that $MI$ is the $I-$ symmedian of $IBC$ . Using the fact that $\angle IZN =\angle MZN = 90^\circ$ , we get that $Z$ is the midpoint of the $I-$ symmedian chord, a point known in folklore as the $Dumpty$ point.

Lemma 2:
Let $ABC$ be a triangle, let $E,F$ be the midpoints of the side $CA,AB$ respectively. Let the $A-$ symmedian meet $(ABC)$ at $K$ , let $Q$ be the midpoint of the $AK$ . Let $D$ be the foot of altitude from $A$ onto $BC$ . Then $QDBF$ is cyclic.
Proof
Let $O$ be the circumcenter of $ABC$ . Since $\angle OQA = 90^\circ$ , we have that $Q$ lies on $(AOEF)$ . Suppose $Q'$ lies on $(DBF)$ and $(AOEF)$ , by miquel's pivot theorem, it lies on $(DCE)$ . Now using the fact that $FB=FD$ and $EC=ED$ , we get
$\angle BQ'C = \angle BQ'D + \angle CQ'D = \angle BFD + \angle DEC = 180-2\angle B + 180 - 2\angle C = 2 \angle A = \angle BOC$ Thus $Q=Q'$ .

Coming back to the problem, use Lemma 2 on $\Delta IBC$ to get the required result.

This post has been edited 1 time. Last edited by kapilpavase, Jan 25, 2021, 9:06 AM

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Mathematicsislovely

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Mathematicsislovely

#6 h Feb 1, 2021, 8:39 PM • 2 Y

Y by amar_04, Mango247

Claim: $Z$ is the point where $A$ -mixtilinear incircle touches $\odot(ABC)$
$\emph{proof.}$ Let $Z_1$ be the point wehere $A$ -mixtilinear incircle touches $\odot(ABC)$ .Let $Z_1S,Z_1T$ are the tangents from $Z_1$ to the incircle of $ABC$ .Let $Z_1S$ cut $\odot(ABC)$ again at $Y_1$ and $Z_1T$ cut $\odot(ABC)$ again at $X_1$ .By Poncelet Porism, $X_1Y_1$ must be tangent with the incircle.We will prove that $X_1\equiv X$ and $Y_1\equiv Y$ which will in turn prove $Z_1\equiv Z$ .
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Let $A'$ be the point where $Z_1D$ cut $\odot(ABC)$ .It is well known that $AA'||BC$ .

By DDIT, $(Z_1B,Z_1C);(Z_1A,Z_D);(Z_1S,Z_1T)$ are reciprocal pairs of some involution at the pencil of lines through $Z_1$ .Projecting in $\odot(ABC)$ we get that $(B,C);(X_1,Y_1);(A,A')$ are reciprocal pairs of a involution.

Since, $BC||AA'$ and as in any involuton that maps a circle to the same circle the line joining points reciprocal pair must passes through a same line hence $X_1Y_1||BC$ . But as $X_1Y_1$ is tangent to incircle it must passes through $D'$ .We are done $\blacksquare$

So $Z\equiv Z_1$ .Then $\angle BZD=\angle BZC-\angle CZD=\angle BZC-\angle CZA'=180^{\circ}-\angle A-\angle C=\angle B$ Also since midpoint of $IB$ say $M$ is circumcircle of $\odot IBD$ so $\angle BMD=2\angle BIM=180^{\circ}-\angle B$ So $Z,B,M,B$ are cyclic $\blacksquare$

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L13832

#7 h Aug 23, 2024, 6:04 PM • 1 Y

Y by radian_51

Refer to the solution @below by Eka01
Incomplete Solution

This post has been edited 6 times. Last edited by L13832, Aug 28, 2024, 7:30 AM

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Eka01

#8 h Aug 28, 2024, 7:03 AM • 2 Y

Y by Sammy27, L13832

@above your solution seems a bit vague and incomplete. In particular, you didn't define $M_a$ nor did you show that $X'Y'$ is parallel to $BC$ and instead used the properties of the original $X$ and $Y$ .

Here's my solution.
We claim that $Z$ is the $A$ mixtilinear touch point. Let the tangents from $Z'$ (The mixtilinear touch point) to the incircle meet $(ABC)$ again at $X'$ and $Y'$ respectively. By Poncelet's Porism, $X'Y'$ is tangent to the incircle. Since $I$ is also the incenter of $\Delta Z'X'Y'$ , it follows that the midpoint of major arc $\overarc{BAC}$ which we call $L$ is the midpoint of minor arc $\overarc{X'Y'}$ since it is well known that $Z',I,L$ are collinear. This implies that $L$ lies on the perpendicular bisector of both $X'Y'$ as well as $BC$ . Similarly the center $O$ of $(ABC)$ must lie on the perpendiuclar bisector of $X'Y'$ as well as $BC$ so it follows that both lines share the same perpendicular bisector and are hence parallel. Now it is an $NCERT$ fact that if two parallel lines are tangent to a circle, then they must cut the circle at diametrically opposite points implying $X'Y'$ is tangent to the circle at diamterically opposite point of $D$ implying that $X',Y',Z' \equiv X,Y,Z$ respectively.

Now we can delete $X$ and $Y$ from the picture. Inverting and restating in terms of the intouch triangle, we now have the following equivalent problem

Quote:

In $\Delta ABC$ , midpoint of $AB$ is $M$ , orthocenter is $H$ and center is $O$ . Midpoint of $AH$ is $T$ and reflection of $O$ across $AB$ is $O'$ . Prove that $(OAO'T)$ are cyclic.

Note that the result is equivalent to proving that $\angle MTX= \angle BCA$ where $X$ is foot of $A$ onto the $A$ medial line which follows from trivial angle chasing since $T$ is the orthocenter in triangle $\Delta AMN$ ~ $\Delta ABC$ where $N$ is midpoint of $AC$ .

This post has been edited 2 times. Last edited by Eka01, Aug 28, 2024, 7:24 AM

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Eka01

#9 h Aug 28, 2024, 8:00 AM

Y by

Pluto1708 wrote:

Solution
@below

Uhh is it well known though, given the crux of the solution is to actually show that $Z$ is the mixtilinear touch point. Also you didn't define $J$ . The original poster(much orz than me) won't see this since im replying to a 3 year old solution but the purpose of this reply is to assure the people trying this that the geo is definitely not easy unless you start out knowing $Z$ from a previous problem.

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Eka01

#10 h Aug 28, 2024, 8:05 AM • 1 Y

Y by Sammy27

I have fat thumbs

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#11 h Apr 15, 2025, 8:46 AM • 1 Y

Y by GeoKing

Let $A'$ be point on $(ABC)$ such that $\overline{AA'} \parallel \overline{BC}$ .

Claim: $Z$ is $A$ -Mixtilinear intouch point.
Proof: We will phantom point this. Redefine $Z$ be the above and let $X$ , $Y$ $\in (ABC)$ such that $\overline{ZX}$ and $\overline{ZY}$ are tangents to incircle.

By Poncelet's Porism, $\overline{XY}$ is tangent to the incircle and by DDIT from $Z$ to $ABDC$ with the incircle as inconic; we get involution pairs $(\overline{ZA},\overline{ZD})$ ; $(\overline{ZB},\overline{ZC})$ ; $(\overline{ZX},\overline{ZY})$ and now project from $Z$ to $(ABC)$ to get pairs $(A,A')$ ; $(B,C)$ ; $(X,Y)$ and hence $\overline{XY} \parallel \overline{BC}$ as required because this means $D' \in \overline{XY}$ . $\square$

Now it is just an angle chase. Let $N$ be midpoint of $\overline{BI}$ and note that this is the center of $(BID)$ . And so $\measuredangle BND=2\measuredangle BID=2 \measuredangle IBD=\measuredangle ABC=\measuredangle BCA'=\measuredangle BZA'=\measuredangle BZD$ And done.

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