Angle chasing with ellipses [Part 1]

by p_square, May 24, 2021, 4:48 PM

Let's start with a classic puzzle. Imagine the following scenario. You stand at a point $F_1$ on the Euclidean plane. A picturesque straight river lies within your view, its nearer shore some line $PQ$. A thirsty cat longs for water at some point $F_2$ on the same side of the river as you. What's the shortest path you can take to aid the cat with some water? A bit more mathematically, what's the shortest path from $F_1$ to $F_2$ which touches some line $PQ$?
[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(60); pair C = dir(30), P = dir(180), Q = dir(0), F2 = dir(-30), F1 = dir(-110); pair A = extension(F1, C, P, Q); pair A1 = A + (-0.3, 0); pair P1 = P + (0,0.1), Q1 = Q + (0,0.1), P2 = P+(0,0.05), Q2 = Q+(0,0.05); draw(P--Q, blue); draw(P1--Q1, blue); draw(P2--Q2, blue); dot("$F_1$",F1,N); dot("$F_2$",F2,N); label("You",F1 - (0,0.2)); label("Cat",F2 - (0,0.2)); dot("$P$",P,W); dot("$Q$",Q,E); [/asy]
First observe that the path can be broken down into two straight lines, your trip to some point on the river (let's call that point $A$), and then the rush back towards the cat.
There are two very fundamentally different ways to approach this question.
Let's start with the more common one, and the one that provides the more satisfactory answer. Assume that there is another cat which stands at the reflection of $F_2$ across $PQ$. (on the other side of the river). Let's denote this point by $C$. There's an interesting correspondence between paths that bring the original cat water, and paths to this new cat $C$: The correspondence being to reflect the portion of the path where you're carrying water across $PQ$. Any path from $F_1$ to $F_2$ which crosses $PQ$ is the same length as a path from $F_1$ to $C$.This implies that the length of your path is at least the distance between $F_1$ and $C$. Conversely, it's easy to come up with a path that has a length $F_1C$. Take the straight line $F_1C$, and simply reflect the section of it lying on the other side of the river across $PQ$.
[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); pair C = dir(30), P = dir(180), Q = dir(0), F2 = dir(-30), F1 = dir(-110); pair A = extension(F1, C, P, Q); pair A1 = A + (-0.3, 0); pair P1 = P + (0,0.1), Q1 = Q + (0,0.1); draw(P--Q, blue); draw(F1--A--F2, fuchsia); draw(A--C, fuchsia + dashed); draw(F1--A1--F2, springgreen); draw(A1--C, springgreen + dashed); dot("$F_1$",F1,dir(F1)); dot("$F_2$",F2,dir(F2)); dot("$C$",C,dir(C)); dot("$P$",P,N); dot("$Q$",Q,N); [/asy]
As a side, note that since $AF_2$ is the reflection of $AF_1$ in $PQ$, that $PQ$ is the external angle bisector of $F_1AF_2$, that is $\angle PAF_1 = \angle QAF_2$.

There's a completely different way of trying to find what the point $A$ on line $PQ$ should be. Let's pick a real number $d$, and see if there is a point $A$ on line $PQ$, such that $F_1A + F_2A = d$ (ie if your path length could possibly have distance $d$). Or, slightly rephrased, whether the ellipse with focii $F_1, F_2$ and focal distance $d$, intersect $PQ$. The smallest $d$ for which it does intersect line $PQ$, is the shortest path length possible. Notice however, that for the smallest such $d$, the ellipse will intersect the path at precisely one point, i.e. $PQ$ will be tangent to the ellipse!
This gives us an interesting setup. An ellipse with focii $F_1, F_2$ is tangent to $PQ$ at some point $A$. And since we know that $A$ is the point on $PQ$ with distance $F_1A + AF_2$ minimal, from the first part, we know that $PQ$ is the external angle bisector of $F_1AF_2$. And since by choosing $F_1, F_2$, the ellipse could be arbitrary, this angle bisector property is true for every ellipse and tangent to that ellipse! Some people might recognize this as the so-called 'optical property'. It's pretty amazing how this tiny co-incidence, that just offers one pair of unexpected equal angles in an ellipse, paves way for many more.
[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.3, min = 1, a = 0.8, b; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a); pair P =A+1.3*dir(el, a), Q = A-1.3*dir(el,a); pair G = 2*foot(F1, P, Q)-F1; draw(P--Q, blue); draw(A--G, dashed+brown); draw(F1--G, orange); draw(F2--A--F1, brown); draw(anglemark(P,A,F2)); draw(anglemark(F1,A,Q)); dot("$F_2$",F1,S); dot("$F_1$", F2, S); dot("$Q$", Q, dir(Q)); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$G$", G, dir(G)); [/asy]

If there's one thing to take away from the previous thought experiment, it's that thirsty cats will themselves walk towards the river to drink water, if you delay too long lost in mathematical thought. Another, equally profound, more mathematical take away could be that reflections of the focii over tangents to an ellipse are nice and deserve further investigation. Let's try a similar trick, this time with two tangents.
In the diagram below, the two tangents are $AP$ and $BP$, and $G,H$ are the reflections of $F_1$ over these lines respectively. From the previous setup we see that $F_2AG$ and $F_2BH$ are collinear.
There's a lot of stuff going on in this diagram. Since $G$ and $H$ were reflections, we know that $PG = PF_1 = PH$. Also, again from the previous diagram we had that $F_2G = F_2A + AF_1 = F_2B + BF_1 = F_2H$. these together imply that $PF_2$ is the angle bisector of $GH$!! The neat angle condition that arises from this, is that $\angle AF_2P = \angle BF_2P$! The result also of course holds for the other focus as well, that is $\angle AF_1P = \angle BF_1P$.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.3, min = 1, a = 1.3, b=0.6; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a), B = point(el,b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); pair G = 2*foot(F2, P, A) - F2, H = 2*foot(F2, P, B) - F2; draw(H--F1--G, brown); draw(H--F2--G, royalblue); draw(F1--P, purple); draw(P--((G+H)/2), purple+dashed); draw(G--H, springgreen); draw(P--foot(F2,A,P), fuchsia); draw(B--foot(F2,B,P), fuchsia); draw(G--P--H, orange); draw(F2--P, orange); draw(anglemark(P,F1,A)); draw(anglemark(B,F1,P)); dot("$F_2$",F1,S); dot("$F_1$", F2, S); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$H$", H, dir(H)); dot("$G$", G, dir(G)); [/asy]

A brief side before we look at the third angle property, we see that $F_1P, AP, BP, F_2P$ are respectively the angle bisectors of $BF_1A, F_1AF_2, AF_2B, F_2BF_1$ respectively, or $P$ is the incentre (excentre?) of self intersecting quadrilateral $F_1AF_2B$. That's in itself quite a cool result.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(80); real maj = 1.3, min = 1, a = 1.3, b=0.6; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a), B = point(el,b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); draw(A--F2--B--F1--A, royalblue); draw(A--foot(P,A,F2), royalblue); draw(B--foot(P,B,F1), royalblue); draw(circumcircle(foot(P,A,F1),foot(P,A,F2),foot(P,B,F1)), brown); dot("$F_2$",F1,S); dot("$F_1$", F2, S); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); [/asy]

This also gives a very rich configuration when $AB$ passes though $F_1$! The excentre result translates to $PF_1 \perp AB$, and to $P$ being the excentre of $F_2AB$.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.3, min = 1, a = 1.1, b=0.6; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a); real ipt[] = intersect(el, (F2--(6F2-5A))); b = ipt[0]; pair B = point(el, b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); draw(circumcircle(foot(P,A,F1),foot(P,A,F2),foot(P,B,F1)), brown); draw(A--B--F1--A, royalblue); draw(P--F2, orange); dot("$F_1$",F1,S); dot("$F_2$", F2, S); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); [/asy]

Last time, we reflected the same focus across both the tangents. This time we will reflect $F_1$ across one tangent ($AP$) to get $G$, and $F_2$ across the other ($BP$) to get $H$. The surprise lying in store for us this time, is a pair of congruent triangles. In particular, it turns out that $GPF_2$ is congruent to $F_1PH$. [the respective sides can easily be shown to be equal]
What's even more interesting is the pair of equal angles that emerges from this. Since $\angle F_1PH = \angle GPF_2$, we know that $\angle GPF_1 = \angle F_2PH$, and dividing by $2$, $\angle APF_1 = \angle F_2PB$!

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.3, min = 1, a = 1.4, b=0.3; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a), B = point(el,b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); pair G = 2*foot(F2, P, A) - F2, H = 2*foot(F1, P, B) - F1; draw(F1--P--G--F1, royalblue); draw(F2--P--H--F2, purple); draw(A--P--B, brown); draw(anglemark(G,P,A)); draw(anglemark(B,P,H)); draw(anglemark(A,P,F2)); draw(anglemark(F1,P,B)); dot("$F_2$",F1,S); dot("$F_1$", F2, S); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$H$", H, dir(H)); dot("$G$", G, dir(G)); [/asy]

Here's all three angles conditions in a master diagram. Make sure that you're fairly comfortable with the diagram; For the rest of the post, I'll freely use all the three angle properties.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.3, min = 1, a = 1.4, b=0.3; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el,a), B = point(el,b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); pair Pa = 2*A - P, Pb = 2*B - P; draw((1.4B-0.4P)--P--(1.5A-0.5P), fuchsia); draw(A--F2--B--F1--A, royalblue); draw(F2--P--F1, brown); draw(anglemark(A,P,F2), royalblue); draw(anglemark(F1,P,B), royalblue); draw(anglemark(P,F2,A), springgreen); draw(anglemark(B,F1,P), fuchsia); draw(anglemark(P,F1,A), fuchsia); draw(anglemark(B,F2,P), springgreen); draw(anglemark(Pa, A, F2), brown); draw(anglemark(F1, A, P), brown); draw(anglemark(P, B, F2)); draw(anglemark(F1, B, Pb)); dot("$F_2$",F1,S); dot("$F_1$", F2, S); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); [/asy]

The last property in particular, helps immensely when we seek characteristics of in-ellipses of polygons. Before we dive into what in-ellipses of triangles look like, a few definitions. Consider an arbitrary angle, say $\angle BAC$. Two points $F_1$ and $F_2$ are said to be isogonal in $\angle BAC$, iff $F_2$ lies on the reflection of $AF_1$ over the angle bisector of $BAC$. The isogonal of $AF_1$ in $\angle BAC$ is the the reflection of $AF_1$ across the bisector of $\angle BAC$.
A point $F_2$ the isogonal conjugate of $F_1$ with respect to triangle $ABC$, if it lies on all three isogonals i.e. $F_2$ is the unique point which lies on the
  1. reflection of $AF_1$ over bisector of $\angle BAC$,
  2. reflection of $BF_1$ over bisector of $\angle CBA$, and
  3. reflection of $CF_1$ over bisector of $\angle ACB$.
The slick thing is that in a triangle, isogonal conjugates always exist - the above three lines always concur. Why do they concur? You guessed it - Ellipses.

Let the $B$-isogonal of $BF_1$, and $C-$isogonal of $CF_1$ meet at $F_2$. We'll show that $F_2$ also lies on the third isogonal. For this, consider the ellipse with foci $F_1$ and $F_2$ tangent to $BC$ at some point $X$. Let the other tangent from $B$ to the ellipse touch the ellipse at $Y$, and let the other tangent from $C$ to the ellipse touch the ellipse at $Z$. Since we have that $\angle ABF_1 = \angle XBF_2 = \angle YBF_1$, we see that $AYB$ are collinear, and similarly $AZC$ are collinear. $A$ is the intersection of tangent from $Y$ to the ellipse, and $Z$ to the ellipse. This literally spells out $\angle YAF_1 = \angle ZAF_2$, this is exactly the third isogonality which we wanted to prove!

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.2, min = 1, a = 2.8, b=0.3, c = 1.5; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair X = point(el,a), Y = point(el,b), Z = point(el, c); pair A = extension(Y, dir(el, b)+Y, dir(el, c) + Z, Z); pair B = extension(Z, dir(el, c) + Z, X+dir(el, a), X); pair C = extension(X, X+dir(el, a), Y+dir(el, b), Y); draw(A--B--C--A, royalblue); draw(F2--A--F1, orange); draw(F2--B--F1, brown); draw(F2--C--F1, brown); dot("$F_1$",F1,S); dot("$F_2$", F2, S); dot("$C$",C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$X$",X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); [/asy]

The above also gives a nicer definition of the isogonal conjugate. The isogonal conjugate of a point $F_1$ in triangle $BAC$, is the unique point $F_2$, such that there is an inellipse of $ABC$ with foci $F_1$ and $F_2$.

We'll also now claim, that the 6 feet of altitudes, from the two foci to the three the sides of a triangle lie on a circle. (This is called the pedal circle). We'll further show that this circle is centered at the centre of the ellipse, $O$.
[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(70); real maj = 1.4, min = 1, a = 2.9, b=0.3, c = 1.5; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair X = point(el,a), Y = point(el,b), Z = point(el, c); pair A = extension(Y, dir(el, b)+Y, dir(el, c) + Z, Z); pair B = extension(Z, dir(el, c) + Z, X+dir(el, a), X); pair C = extension(X, X+dir(el, a), Y+dir(el, b), Y); pair U = foot(F1,B,C); pair V = 2*U - F1; pair O = (0,0); draw(A--B--C--A, royalblue); draw(F2--A--F1, orange); draw(F2--B--F1, orange); draw(F2--C--F1, orange); draw(O--U, brown); draw(F2--V, purple); draw(circumcircle(U, foot(F1,A,B), foot(F1,A,C)), brown); draw(F1--foot(F1,C,A), dotted); draw(F1--foot(F1,B,A), dotted); draw(F1--V, dotted); draw((-maj,0)--(maj,0), brown); draw(F2--foot(F2,C,A), dotted); draw(F2--foot(F2,B,A), dotted); draw(F2--foot(F2,C,B), dotted); dot("$F_1$",F1,S); dot("$F_2$", F2, S); dot("$C$",C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$X$",X, dir(X)); dot("$Y$", Y, dir(-Y)); dot("$Z$", Z, dir(-Z)); dot("$O$",O, dir(90)); dot("$U$", U, dir(U)); dot("$V$", V, dir(V)); [/asy]
Let $V$ be the reflection of $F_1$ in $BC$ and let $U$ be the foot of altitude from $F_1$ onto $BC$. Remember that $F_2VX$ are collinear, and that $F_2X = F_2X + XV = F_2X + F_1X$. Dilating by $\frac12$ from $F_1$, we see that $OU = \frac{F_1X + F_2X}{2}$. Since this quantity will be constant whichever focus we take, and whichever side we drop the altitude to, we get that all six points fomed by dropping a perpendicular from the focii to the sides are concyclic.

Isogonal conjugates and inellipses become even more interesting in quadrilaterals (and I'll definitely write a post on those too soon!), but let's come back to those later.
Let's for now move on to a few more elementary properties. Suppose that we have two parallel chords $AB$ and $CD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$. Then, $MN$ passes through the centre of the ellipse!
To see this, we first need a new way of thinking about ellipses. It's popularly known that an ellipse can be formed by cutting a cone with a plane. [this is probably the most standard definition] What's slightly less commonly heard of, but as neat, is that the intersection of a cylinder and plane is also an ellipse! This intuitively ought to be true, because in a sense, an infinite cylinder is the limiting case of a cone, just imagine taking the top vertex of a cone and sliding it to infinity. The proof that the figure that emerges is an ellipse is also exactly the same.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(60); real maj = 1.2, min = 1, a = 2.85, b=1.35, c = 0.4, d; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el, a), B = point(el, b), C = point(el, c); real ipt[] = intersect(el, (C+0.1A-0.1B)--(C+5A-5B)); d = ipt[0]; pair D = point(el, d), O = (0,0); pair M = (A+B)/2; pair N = (C+D)/2; draw(A--B, brown); draw(C--D, brown); draw(M--N, royalblue); dot("$F_1$",F1,S); dot("$F_2$", F2, S); dot("$C$",C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$",D, dir(D)); dot("$O$",O,S); [/asy]

Now that we have that, let's think of our ellipse as being such an intersection, and let's consider projecting the ellipse down to a plane perpendicular to the cylinder. The ellipse gets projected down to a circle, and the parallel chords remain parallel chords. It's also easy to check that midpoints go to midpoints, and so does the line through the midpoints. To finish, we just need to show the theorem for a circle. This luckily is really easy. The line joining the midpoints of parallel chords in a circle is just the perpendicular bisector and that definitely passes through the centre of the circle.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(60); real maj = 1, min = 1, a = 2.85, b=1.35, c = 0.4, d; path el = ellipse((0,0),maj,min); draw(el, fuchsia); pair F1 = (sqrt(maj*maj - min*min),0); pair F2 = -F1; pair A = point(el, a), B = point(el, b), C = point(el, c); real ipt[] = intersect(el, (C+0.1A-0.1B)--(C+5A-5B)); d = ipt[0]; pair D = point(el, d), O = (0,0); pair M = (A+B)/2; pair N = (C+D)/2; draw(A--B, brown); draw(C--D, brown); draw(M--N, royalblue); dot("$C$",C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$",D, dir(D)); dot("$O$",O,S); [/asy]
In case the 3d-geo above was difficult to follow, there's a nice alternative way of thinking about the transformation we just did. Take the ellipse and it's chords, and forcefully squish it down into a circle. More formally, first ensure that the major axis of the ellipse is parallel to the $x-$axis. We'll pick some constant $c$, and send the point with co-ordinates $(x, y)$ to the point with co-ordinates $(cx, y)$. Wisely picking $c$ changes the ellipse to a circle, and the definition of every other point in the figure remains the same.

Leaving ellipses aside, let's now consider parabolas.
Formally, a parabola is the locus of points which are equidistant from some point $P$ and some line $l$. In an informal non-rigorous sense, a parabola is an ellipse, with one focus at infinity Don't quote me on this, ok?. If we take $Q$ to be the point at infinity along the line perpendicular to $l$, and let $Z$ be a variable point on the parabola, speaking informally, $QZ + PZ$ should be the fixed distance from $Q$ to line $l$. [As an interesting side note, this intuition about parabolas being the same as ellipses can actually be realized by shifting to the projective plane]
Formalities set aside however, the various angle conditions and various properties we discussed for ellipses also hold true for parabolas too!
Here is a figure with all three angle properties. Convince yourself that all marked pairs of angles are the same.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(80); import graph; real f(real x) { return x*x; } path el = graph(f, -1.2, 1.2); pair F1 = (0, 0.25); pair D1 = (-1.4, -0.25), D2 = (1.4, -0.25); path d = D1--D2; draw(el, fuchsia); draw(d, fuchsia); real a = 20, b = 95; pair A = point(el,a), B = point(el,b); pair P = extension(A+dir(el,a), A, B, B+dir(el, b)); pair P0 = 1.4*F1 - 0.4*P; pair Af = foot(A,D1,D2), Bf = foot(B, D1, D2), Ff = foot(F1, D1, D2), Pf = foot(P, D1, D2); draw(A--Af, dashed); draw(P--Pf, dashed); draw(B--Bf, dashed); draw(A--P--B, royalblue); draw(A--F1--B, orange); draw(P--P0, brown); draw(anglemark(P,A,F1), brown); draw(anglemark(P,B,Bf), brown); draw(anglemark(P0,F1,A), springgreen); draw(anglemark(B,F1,P0), springgreen); draw(anglemark(F1,P,A), fuchsia); draw(anglemark(B,P,Pf), fuchsia); dot("$F_1$", F1, N); dot("$P$",P, dir(P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); [/asy]

In fact, since a parabola in some sense is a very special ellipse, some additional properties are true too! [and some previously stated properties become even nicer]
Let $X, Y, Z$ be arbitrary points on some parabola with focus $F_1$, and directrix $l$, and let the tangents create some triangle $ABC$. For convenience, let the point at infinity along line perpendicular to $l$ be $F_2$. Remember that $F_1$ is the isogonal conjugate of $F_2$ in triangle $ABC$? Since in a parabola $F_2$ is at infinity, $F_1ABC$ are concyclic! Further, by Simson line, the reflections of F_1 across $AB, BC, CA$ and the orthocenter of $ABC$ are collinear. Since by optical property and definition of a parabola, the first three points lie on the directrix, we see that the orthocenter of $ABC$ lies on the directrix.

[asy] defaultpen(fontsize(10pt)); import math; import olympiad; unitsize(80); import graph; real f(real x) { return x*x; } path el = graph(f, -1.3, 1.3); pair F1 = (0, 0.25); pair D1 = (-1.5, -0.25), D2 = (1.5, -0.25); path d = D1--D2; draw(el, fuchsia); draw(d, fuchsia); real a = 55, b = 94, c = 10; pair X = point(el,a), Y = point(el,b), Z = point(el, c); pair A = extension(Y, dir(el, b)+Y, Z, dir(el, c) + Z); pair B = extension(Z, dir(el, c) + Z, X+dir(el, a), X); pair C = extension(X, X+dir(el, a), Y, Y+dir(el, b)); draw(circumcircle(A,B,C), brown); pair H = orthocenter(A,B,C); draw(A--foot(A,B,C),dotted); draw(B--foot(B,A,C),dotted); draw(C--foot(C,B,A),dotted); draw(Z--A--Y, royalblue); draw(B--C, royalblue); dot("$F_1$", F1, N); dot("$C$",C, dir(C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$Z$",Z, dir(Z)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$H$", H, dir(H)); [/asy]

To be truthful, this topic isn't very relevant to olympiad geometry and readers will at best stumble across 3-4 questions which have a solution using the theory discussed here. I still think that the topic is quite fun and interesting though, hence the blog post. :) To me, when I first saw all of this, it was surprising that angle conditions could be true in conics at all, since I used to feel that projective geometry was the only way of dealing with conics. There might be few applications of this apart from maybe giving thirsty cats water in an optimal way, but it's still a pretty beautiful piece of math.
That said, it's not completely useless, and in the following post, I'll definitely showcase a few oly geo problems where knowing angle chasing in conics helps. (Including IMO 18/6!)

To end with, a couple of exercises/theorems on ellipses/parabolas that will lead into part 2. There are some fancy names in this part which I have not bothered to define, just look those up if necessary.
  1. Show that any in-ellipse of a quadrilateral has its centre on the Newton-Gauss line
    Hint
  2. Prove that the focus of the parabola tangent to four sides of a quadrilateral, is the quadrilateral's Miquel point.
    Hint
  3. Prove that the directrix of a parabola tangent to four sides of a quadrilateral is the Gauss-Bodenmiller line.
    Hint
This post has been edited 2 times. Last edited by p_square, Jun 13, 2021, 3:05 PM
geometry
ellipse
conic

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9 Comments

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Wow this is really cool! :coolspeak:

by L567, May 24, 2021, 5:00 PM

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This is super cool ! Will you post more content on this blog?

Thanks! Hopefully, yes. I'm planning on reviving the blog.
This post has been edited 1 time. Last edited by p_square, May 25, 2021, 4:47 PM

by Aryan27, May 25, 2021, 2:20 AM

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As expected from an IMO Gold Medalist. Anyways , It was brilliant just like your other posts

by Bratin_Dasgupta, May 25, 2021, 1:57 PM

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If anyone found this blog post interesting and prefers seeing things in video format, @rg230403 covered some similar things in this video, Do check it out!

by p_square, May 25, 2021, 4:52 PM

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Pro gawd stuff

by 554183, May 30, 2021, 10:37 AM

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It's so cool :coolspeak:

by MatBoy-123, Sep 20, 2021, 12:41 PM

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Just asking Idk if I am sensible but what happens to an ellipse when you invert around a point outside of it/on it?

Also what happens to an ellipse when you invert around the point at infinity?

by Project_Donkey_into_M4, Jan 9, 2022, 10:44 AM

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I think in the isogonal conjugate-cum-ellipse diagram, $Y$ and $Z$ are labelled incorrected (switched). This means that people first come to comment and not to read :P Whether I do the same is left as an exercise for the reader (but fancy geo is certainly intidimating for those who aren't familiar with it, including me). Good Luck for INMO whoever is a contestant $\cap$ reading this!

by hellomath010118, Mar 3, 2022, 1:30 PM

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@Project_Donkey you get a limacon

by qwerty123456asdfgzxcvb, Dec 26, 2024, 8:29 AM

Alive again :D

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p_square
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  • thank you for teaching me the humpty point

    by ihatemath123, Jun 27, 2024, 2:43 PM

  • ORZZZZZZZZZZ

    by avisioner, Jun 19, 2024, 12:02 AM

  • He never had motivation to multiply out random polynomials

    by the_mathmagician, Oct 25, 2023, 1:49 AM

  • Getting a part 3 is unlikely... anyways, if it ever gets released, it would be great if you listed some problems where this methods are useful!

    by Alfombra12, Aug 13, 2023, 10:36 AM

  • still waiting for part 3

    by kn07, Jun 14, 2023, 10:55 PM

  • Orz orz :) :omighty:

    by aansc1729, Mar 13, 2023, 1:13 PM

  • waiting for part 3 since 1 year :(

    by anurag27826, Jul 18, 2022, 11:19 AM

  • pro kid moment

    by the_mathmagician, Feb 15, 2022, 2:21 AM

  • Ded blog

    by SPHS1234, Jan 31, 2022, 7:43 AM

  • orzity orz orz

    by tigerzhang, Oct 15, 2021, 10:08 PM

  • congrats on IMO gold medal!

    by mathlearner2357, Jul 25, 2021, 5:04 AM

  • orzorzorz

    by PEKKA, Jul 21, 2021, 3:55 PM

  • Wow this is awesome!!

    Waiting for part3

    by lneis1, Jun 24, 2021, 5:42 PM

  • Part 3 hype

    by NJOY, Jun 15, 2021, 9:29 AM

  • Ooh niceee! Waiting for part 3!

    by L567, Jun 14, 2021, 4:43 AM

63 shouts
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