The Dumpty Parabola (So, sad to know :( That it's already discovered :( )

by AlastorMoody, Sep 11, 2019, 4:15 PM

This Parabola is already discovered

and is also known as Artzt Parabola, But anyway, I'll like to call this "Dumpty Parabola"

...

Let's try to motivate this parabola!

uraharakisuke_hsgs wrote:

Let $d_1 , d_2$ be two abitrary lines that are not perpendicular . Let $A,B$ be abitrary points lie on $d_1,d_2$ . How to construct a parabola that is tangent to $d_1,d_2$ at $A,B$

Well, *coincidence* I was also working on the same construction, 'coz It is almost impossible to construct a parabola tangent to two lines at given point on GEOGEBRA...

While searching for some Lemmas, in SL Loney's Coordinate Geometry book, I found the following Lemma:

Similarity on Parabola Lemma wrote:

Let $\mathcal{P}$ be a Parabola with $X,Y \in \mathcal{P}$ . Let the tangents at $X,Y$ meet at $A$ . Then, $\Delta TAY \sim \Delta TXA$

Now, this Lemma clearly motivates the construction of a parabola with focus at Dumpty Point and so we have the following Problem

Problem: Let $\Delta ABC$ be a triangle. Let $D$ be the midpoint of $A-$ symmedian Chord. Let $M$ be midpoint of $BC$ . Let $AM$ intersect Nine-Point Circle at $M'$ . Let $\ell$ be perpendicular to $AM$ at $M'$ . Prove that, There Exists a Parabola, $\mathcal{P}$ with Focus at $D$ tangent to $AB,AC$ at $B,C$ and Directrix as $\ell$

Proof: (By Supercali)

Supercali wrote:

Let $B'$ and $C'$ be midpoints of $AC$ and $AB$ respectively. Let $\ell$ meet $AC,AB$ at $E,F$ respectively. Let $AM$ meet $BD$ at $P$ and $CD$ at $Q$ . By angle chasing (using property of symmedians, midpoint theorem and the fact that $M',B',C',M$ are concyclic), we can easily see that $E,M',D,Q,B'$ are concyclic. Hence $\angle EDC= \angle EM'Q = 90$ . Similarly $\angle FDB = 90$ . Hence it will be sufficient to show that $B,C$ lie on the parabola with focus $D$ and directrix $\ell$ .

Let $G$ be foot from $C$ on $\ell$ . $\angle EDC=\angle EGC = 90$ . By angle chasing, we can easily see that $\angle ECD=\angle ECG$ , hence $EGCD$ is a kite, which implies $CG=CD$ , hence $C$ lies on the parabola. Similarly, $B$ also lies on the parabola, and we are done.

Proof: (By MathPassionForever)

MathPassionForever wrote:

Consider an inparabola of $AB'C'$ with directrix as perpendicular bisector of $AM'$ . Note that since the foci of the parabola are isogonal conjugates, if the axis is parallel to the $A$ median, the other focus is on the $A$ symmedian. Also, since the isogonal conjugate of a point at infinity is on the circumcircle, the focus lies on $\odot(AB'C')$ , which is just a homothety of ratio $\frac{1}{2}$ of $\odot(ABC)$ from $A$ . So clearly $D$ lies on the former. So $D$ is the focus of this inparabola!

This discussion could have ended here, but AoPS guys are just too amazing and they found many more properties which I couldn't have managed to find out nor could I have proved them!! >_< - https://artofproblemsolving.com/community/q1h1912112p13117117

Here's a Property that can be killed using the SL Loney Similarity Lemma

Aryan-23 wrote:

Prove that the circumcircle of the triangle formed by tangents at any 3 points on the parabola passes through its focus??

These are complete Supercali's efforts:

Supercali wrote:

Another interesting property: $B'C'$ is tangent to the parabola and point of intersection lies on $AM$ .

This gives another interesting problem:
In $\Delta ABC$ , Let $X$ be a point such that $ABXC$ is harmonic. Let $\ell$ be the line passing through the orthocentre of the triangle which is perpendicular to the $A$ -median. Prove that the parabola with focus $X$ and directrix $\ell$ is tangent to the sides of $\Delta ABC$

Proof: Let $Q$ be the midpoint of $AM$ . Using angle chasing, we can get $QF=QD$ (you will have to use concyclicities used in my original solution). Let $\ell$ meet $B'C'$ at $K$ . Then, by angle chasing, $FQDK$ is a kite. Hence $\angle QDK=90$ and we are done.

Supercali wrote:

The niceness continues:
1) Let $BB'$ meet the parabola at $X$ Then $CX$ passes through the centroid of $\Delta AB'C'$
2) Let $X_a$ be the $A$ -Humpty point and let $H$ be the orthocentre. Then $Q,X_a$ , foot from $A$ onto $BC$ , $K$ defined above and midpoint of $AH$ are concyclic.

Supercali wrote:

Another thing:
Let $AX$ meet the parabola again at $X'$ . Then $BX'$ is parallel to $AC$ .
Also, if $CX'$ meets $AM$ at $N$ , then $A$ is the midpoint of $MN$ .

Supercali wrote:

Let us call it the Artzt parabola from now on. Here are some properties of its tangents and tangency points:
1) (Well known) Let $L$ be any point on $BC$ and let $M,N$ be points on $AC,AB$ respectively such that $AMLN$ is a parallelogram. Then $MN$ is tangent to the parabola.
2) If $S$ is the point of tangency, and $CS$ meets $LN$ at $N'$ , then $N$ is the midpoint of $LN'$ .
3) (Unsolved, but my favourite) Apparently $B'$ (midpoint of $AC$ ), $Q$ (midpoint of $B'C'$ ), $S$ , $B$ , $C$ , $N$ , $Q_B$ (intersection of $LN$ and $AQ$ ) lie on a conic.
4) (Unsolved) Apparently the locus of the centre of the above conic as $L$ varies on $BC$ is a line passing through the midpoint of $BC$ (not shown in figure).

These are math_pi_rate's efforts at providing such amazing proofs to these properties:

math_pi_rate wrote:

Here are (brief) proofs of some of the properties (Assuming that the properties in post #1 and post #3 are solved; Also, I'll use the diagram in post #10):

Supercali wrote:

Let $BB'$ meet the parabola at $T$ Then $CT$ passes through the centroid of $\triangle AB'C'$

Let $\mathcal{P}$ denote the parabola from now on. Let $AQ \cap \mathcal{P}=Z$ and $BQ \cap A \infty_{BC}=U$ . It's easy to see that $AUMB$ is a parallelogram, which directly gives that $AUCM$ is also a parallelogram (i.e. $CU \parallel AM$ ). By Pascal on $QZCBBQ$ , we get that $BQ \cap CZ$ lies on $AU$ , which gives $CZ \parallel AM$ , or that $Z=\infty_{AM}$ .

Now, Let $B \infty_{AM} \cap MC'=L$ . It's easy to show that $AMBL,ACML$ and $AB'C'L$ are all parallelograms. This also implies that $L \in CQ$ and that $B'L$ bisects $AC'$ (i.e. $G' \in B'L$ ). Then, Pascal on $QZCCTQ$ and $QTBZCC$ gives that $CT \cap AM,B',L$ are collinear. As $B'L \cap AM=G'$ , so we get $G' \in CT$ , as desired.

Supercali wrote:

Let $X_a$ be the $A$ -Humpty point and let $H$ be the orthocentre. Then $Q,X_a$ , foot from $A$ onto $BC$ (say $A_H$ ), $K$ defined above and midpoint of $AH$ (say $H_A$ ) are concyclic.

Note that $A_H,H,M,X_a$ are concyclic. Then POP, and dividing both sides by half, easily gives that $A_H,X_a,Q,H_A$ are concyclic. Now, $\angle KQX_a=90^{\circ}+\angle H_AAX_a=90^{\circ}+\angle H_AX_aA=\angle KH_AX_a \Rightarrow K \in \odot (QH_AX_a)$

Supercali wrote:

Let $AT$ meet the parabola again at $T'$ . Then $BT'$ is parallel to $AC$ .

By Pascal on $BT'TCCB$ , it suffices to show that the line joining $AT \cap BC$ and $CT \cap AB$ is parallel to $AC$ . This is easy by some cross ratio chasing (I am seriously not in the mood of writing that up

Supercali wrote:

Let $L$ be any point on $BC$ and let $M,N$ be points on $AC,AB$ respectively such that $AMLN$ is a parallelogram. Then $MN$ is tangent to the parabola at a point $S$ .

Animate $S$ on the parabola $\mathcal{P}$ . One can easily show that $S \mapsto M,N$ are projective maps (Try to use the fact that focus of the the parabola lies on the circle passing through the point of intersection of the tangents at any two points, and the point where these tangents meet the tangent at the vertex). And, $M \mapsto M \infty_{AB} \cap BC$ and $N \mapsto N \infty_{AC} \cap BC$ are also projective. Thus it suffices to prove the problem for three positions of $S$ . Take $S=Q,B,C$ for this.

Supercali wrote:

$B'$ (midpoint of $AC$ ), $Q$ (midpoint of $B'C'$ ), $S$ , $B$ , $C$ , $N$ , $Q_B$ (intersection of $LN$ and $AQ$ ) lie on a conic.

Apply Pascal on $SQQCBB$ and $SSQCCB$ to get that $MC',SQ,BC$ are concurrent. Then, the converse of Pascal's Theorem on $B'QQ_BNBC$ and $NSQB'CB$ gives the desired conclusion.

This is from MathPassionForever's Blog,

Quote:

I am extremely indebted to AlastorMoody, Supercali and math_pi_rate for this one. I admittedly have just shifted an entire thread to my blog post, but this was just too beautiful so........

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So here is the description of some of the objects:
$\Delta ABC$ is the reference triangle
$B',C'$ are the midpoints of sides $AB,AC$
$Q$ is the midpoint of $B'C'$ and $M$ is the midpoint of $BC$
$X_a$ is the $A$ -Humpty point
$H$ is the orthocenter of $\Delta ABC$ , $H_a$ is the foot of $A$ -altitude, $H_A$ is the orthocenter of $\Delta AB'C'$ and hence the
midpoint of $AH$ .
$D$ is the $A$ -Dumpty point
$M'$ is the reflection of $D$ about $B'C'$ and hence the midpoint of $AX_a$
$l$ is the line through $M'$ perpendicular to $AM$
$E,F$ are the intersections of $l$ with $AC,AB$ respectively
$D'_1,D'_2$ are the feet of perpendiculars from $B,C$ on $l$
$K$ is the intersection of $B'C'$ and $l$
$S$ and $T$ will be defined later.
$P,Q$ are $BD \cap AM$ and $CD \cap AM$ respectively.

Here's the first part:
EXISTENCE OF ARTZT PARABOLA: Prove that there exists a parabola tangent to $AB, AC, B'C'$ at $B,C,Q$ respectively. Prove that its focus is $D$ and its directrix is $l$ . It will be denoted by $\mathcal{P}$ from now on.

My partial proof : Consider the inparabola of $\Delta AB'C'$ with its axis parallel to the $A$ -median. Then its focus must lie on the $A$ -symmedian as well as on $\odot(AB'C')$ and hence is $D$ . Since its directrix is perpendicular to $AM$ and passes through $H_a$ (Orthocenter always lies on directrix of inparabola), the directrix must be $l$ . From here, I have only Supercali's proof for tangency to $B,C$ . For tangency at $Q$ , note that $B'C$ is the perpendicular bisector of $DM'$ and that $QM \perp l$

Supercali's pure angle chase proof:
(i) $B,R,D,M',E$ are cyclic and so are $C,P,D,M',F$
Proof: $\measuredangle M'EB = \measuredangle M'EA = \dfrac{\pi}{2} - \measuredangle EAM' = \dfrac{\pi}{2} - \measuredangle DBQ = \measuredangle M'DB$ and we have $B,D,M',E$ cyclic.
$\measuredangle M'RD = \measuredangle DAC + \measuredangle ACD = \measuredangle M'BQ + \measuredangle QBD = \measuredangle M'BD$ and we have $B,R,D,M'$ cyclic. Similarly for the other one.
(ii) $\measuredangle CDE = \measuredangle RDE = \measuredangle RM'E = \dfrac{\pi}{2}$ so it suffices to show $B,C$ lie on the parabola.
(iii) $\measuredangle ECD = \measuredangle EAM' = \measuredangle ECD'_2$ which makes $DED'_2C$ a kite $\implies ED=ED'_2$ and we're done!

Corollary 1: $\Delta QB'C, \Delta QC'B, \Delta ABC$ share an Artzt Parabola and hence a Dumpty point!
Proof 1: By degrees of freedom, a unique parabola is tangent to $AB,AC$ at $B,C$ . But since the parabola is tangent to $B'Q,B'C$ at $Q,C$ and $C'Q,C'B$ at $Q,B$ we're done!

And then math_pi_rate reminds us that Pascal might be a little more than a unit of pressure at times:

Such a Clean report, I guess I'll ever be able to present this so neatly

This post has been edited 9 times. Last edited by AlastorMoody, Oct 14, 2019, 6:50 AM

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Beautiful.

by kn07, Nov 11, 2023, 6:06 PM

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a lot of these properties follow from properties of inparabola of degen quad. (AA)(BC)

by qwerty123456asdfgzxcvb, Mar 4, 2025, 2:39 PM

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what a goat, u used to be friends with my brother
by bookstuffthanks, Jul 31, 2024, 12:05 PM
hello fellow moody!!
by crazyeyemoody907, Oct 31, 2023, 1:55 AM
@below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read
by kamatadu, Jan 3, 2023, 1:25 PM
Lots of good stuffs here.
by amar_04, Dec 30, 2022, 2:31 PM
But even if he went to jee he could continue with this.

Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows
by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM
Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
Btw @below finally everyone falls to the monopoly of JEE Coz IIT's are the best in India.
by BVKRB-, Feb 1, 2022, 12:57 PM
Kukuku first shout of 2022,why did this guy left this and went for trashy JEE
by Commander_Anta78, Jan 27, 2022, 3:42 PM
When are you going to br alive again ,we miss you
by HoRI_DA_GRe8, Aug 11, 2021, 5:10 PM
kukuku first shout o 2021
by leafwhisker, Mar 6, 2021, 5:10 AM
wow I completely forgot this blog
by Math-wiz, Dec 25, 2020, 6:49 PM
buuuuuujmmmmpppp
by DuoDuoling0, Dec 22, 2020, 10:54 PM
This site would work faster if not all diagrams were displayed on the initial page. Anyway I like your problem selection taste.
by WolfusA, Sep 24, 2020, 7:58 PM
nice blog
by Orestis_Lignos, Sep 15, 2020, 9:09 AM
Hello everyone, nice blog
by Functional_equation, Sep 12, 2020, 6:22 PM
pro blogo
by Aritra12, Sep 8, 2020, 11:17 AM
Nice !
by Synthetic_Potato, Mar 10, 2019, 8:30 AM
Thanks dude!
by AlastorMoody, Mar 1, 2019, 1:39 PM
Amazing work bro!!
by Night_Witch123, Mar 1, 2019, 1:33 PM
@below Thanks a lot
by AlastorMoody, Feb 26, 2019, 2:48 PM
Nice Blog, loved it.
by Delta0001, Feb 26, 2019, 6:07 AM
@below Thanks a lot!
by AlastorMoody, Feb 16, 2019, 3:11 PM
Nice work.
by TheDarkPrince, Feb 16, 2019, 2:39 PM
@below Thanks buddy!
by AlastorMoody, Jan 29, 2019, 9:31 AM
Nice blog! 6th shout!
by PhysicsMonster_01, Jan 29, 2019, 9:21 AM
@below I am noob right now I hope to develop myself within a year or so
by AlastorMoody, Jan 28, 2019, 7:24 PM
Nice blog. But the properties are a bit 'basic'.
by ayan.nmath, Jan 28, 2019, 7:14 PM
Nice!!!!!!!!
by enthusiast101, Jan 13, 2019, 4:44 PM
can I have contrib?
by Kagebaka, Dec 23, 2018, 8:24 PM
Math is fun!
by karolina.newgard, Nov 21, 2018, 6:37 PM

120 shouts

Magic of Hogwarts

The Dumpty Parabola (So, sad to know :( That it's already discovered :( )

by AlastorMoody, Sep 11, 2019, 4:15 PM

2 Comments