Desargues’ Involution Theorem in Olympiad Geometry

Malay Mahajan - MBL 2024

Abstract

DIT and DDIT are “modern-fads” in Olympiad Geometry. This is a powerful tool for solving
olympiad problems and it comes under projective geometry. DIT stands for Desargues’
Involution Theorem and DDIT stands for Dual of DIT. In this course, we will discuss DIT,
DDIT and solve problems based on them.

1. Building the projective world

Euclidean geometry deals with tools such as angles, lengths, cyclicities, etc. However, in projective
geometry we deal with conditions about collinearity, concurrency, tangencies and cross ratio.
In this section we will define the real projective plane, and then the important tool in projective
geometry, namely the cross ratio.

1.1. Real Projective Plane

In general, we seam to care a lot about intersections. Generally any two lines intersect; except
if they are parallel. However we would want it that any two lines in the plane intersect. The
natural intuitive way of making parallel lines intersect, is at infinity.

Definition 1.1 (Constructing RP2 ). Start with Euclidean plane R2 . Then,

1. Along every direction, we add a “point at infinity”.

2. All points along infinity lie on the “line at infinity”.

Because of point 1, any line has a point at infinity (which we constructed along that direction),
and naturally any two parallel lines intersect at a point at infinity (the one along their common
direction).

1.2. Cross Ratio

Definition 1.2. Cross Ratio of 3 types is defined as follows :

• (Points on a line) The cross ratio of four points A, B, C, D on a projective line ℓ is defined
as
XA Y A
(A, B, X, Y ) = ÷
XB Y B
• (Lines through a point). The cross of four lines a, b, x, y ∈ LP (set of lines passing through
point P ) is defined as
sin ∡(x, a) sin ∡(y, a)
(a, b, x, y) = ÷
sin ∡(x, b) sin ∡(y, b)

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 • (Points on a conic) The cross ratio of four points A, B, X, Y on a conic γ is defined as

(A, B, X, Y )γ = (P A, P B, P X, P Y )

where P is any point on γ.

Remark 1.3. The cross ratio is actually signed. The orientation of points on line determines
the sign of each distance, and orientation of the angle determines the sign of each sin-ratio. But
don’t bother too much with this.

Remark 1.4. In particular, one definition of “conic” is the locus of points P for such the
aforementioned cross ratio of lines is constant. If you don’t know what a conic is, it’s alright.
All you need to remember is there is a unique conic passing through 5 points, and cross ratio
definition.
We will mostly be dealing only with circles in this course. For a circle, you can actually check
that (A, B, X, Y ) using Sine rule turns out to be ± XB
XA
÷ YY B
A
where the sign is negative if segments
AB and XY intersect and positive otherwise.

1.3. Perspectivity

Theorem 1.5 (Perspectivity). Suppose A, B, X, Y lie on a line, and P be any point not on the
line. Then
(A, B, X, Y ) = (P A, P B, P X, P Y )

Proof. This is really just applying sine-rule multiple times.

This is a powerful theorem since it basically connects the cross ratios of the 3 types! A natural
corollary of perspecitvity is the following lemma.

Lemma 1.6. (Projecting cross ratios) We can actually “project” as follows

• (Line to line) Let ℓ1 , ℓ2 be lines and let A1 , B1 , C1 , D1 be points lying on ℓ1 . Pick any point
P not lying on l1 or l2 . Let A2 , B2 , C2 , D2 be intersections of P A1 , P B1 , P C1 , P D1 with ℓ2 .
Then
P
(A1 , B1 , C1 , D1 ) = (A2 , B2 , C2 , D2 )

• (Conics to line) Let γ, ℓ be a conic and a line and let A1 , B1 , C1 , D1 be points on γ. Pick any
point P lying on γ. Let Let A2 , B2 , C2 , D2 be intersections of P A1 , P B1 , P C1 , P D1 with ℓ.
Then
P
(A1 , B1 , C1 , D1 )γ = (A2 , B2 , C2 , D2 )

• (Conic to conic) Let γ be a conic and let A1 , B1 , C1 , D1 be points on γ. Pick any point P .
Let A2 , B2 , C2 , D2 be second intersections of P A1 , P B1 , P C1 , P D1 with γ. Then
P
(A1 , B1 , C1 , D1 )γ = (A2 , B2 , C2 , D2 )γ

Proof. For the first two cases, (A1 , B1 , C1 , D1 )=(P A1 , P B1 , P C1 , P D1 )=(P A2 , P B2 , P C2 , P D2 )=

(A2 , B2 , C2 , D2 ).

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 1.4. Pole-polars

Definition 1.7. Let ω be a circle with center O and radius r. Let P be any point in the plane.
Let P ′ be the point on ray OP such that OP · OP ′ = r2 . if ℓ is the line through P ′ which is
perpendicular to OP , then ℓ is called the polar of P , and P is called pole of ℓ.

Theorem 1.8 (La Hire). P lies on polar of Q iff Q lies on polar of P .

Lemma 1.9. If A, B, C, D are points on a line ℓ. Suppose ℓA is polar of A wrt ω and so on.
Prove that
(A, B, C, D) = (ℓA , ℓB , ℓC , ℓD )

2. Involution

2.1. Definition of Involution

We start with definition of an involution. As you might guess, we want some function f such that
f ◦ f =id. However we can’t forget the most important tool in projective geometry, cross ratio!

Definition 2.1. Let P be a line or a conic or LX . Then f : P → P is an involution if

• f ◦ f is identity.

• f preserves cross ratio on P.

• f is not identity.

As a consequence of first condition i.e. f (f (P )) = P ∀P ∈ P, we call (P, f (P )) a reciprocal pair.

Exercise 2.1. Suppose f : P → P preserves cross ratio and swaps A, B ∈ P. Prove that f is an
involution.

2.2. Involutions on Line

Let’s start off with listing some examples of involutions.

1. Reflection about a point on the line.

2. Isogonal conjugation about an angle.

3. Inversion about a point on the line.

Exercise 2.2. Prove that inversion preserves cross ratio, using lengths.

We will now classify all involutions on a line. Turns out any involution is an inversion!

Theorem 2.2 (Involutions are inversions). Any involution f on a line takes one of the following
three forms :

• Inversion about a Euclidean point. (Possibly negative inversion)

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 • Reflection about a Euclidean point. (Should be though as inversion with center at infinity)

Wait, are reflections inversions as well? Yes they can be thought of as inversions about circle
with diameter P ∞.

Proof. Let f swap ∞ with O and have reciprocal pairs (X1 , X2 ), (Y1 , Y2 ). Then

OX1 OY2
(∞O, X1 Y1 ) = (O∞, X2 Y2 ) =⇒ =
OY2 OX2

Hence OX1 ·OX2 = OY1 ·OY2 . Due to directed lengths and arbitrary choice of X, Y we infer that
f must be an inversion centered at O! If O ̸= ∞, we fall in the 1st case.
f (A)f (B)
If O = ∞. Since (∞, A, B, C) = (∞, f (A), f (B), f (C)) we infer that AB
AC = f (A)f (C) . Hence f is
a homothety on the line. Since f 2 =id, the scaling factor must be ±1 and we fall in the 2nd
case.

Lemma 2.3. Let P, Q, R, S are points on a line ℓ such that P ̸= S, R and Q ̸= S, R. Prove that
there is a unique involution with reciprocal pairs (P, Q), (R, S)

Proof. If suffices to prove that there is exactly one point X such that XP · XQ = XR · XS.
Let A be a random point not on ℓ. Let the radical axis of the 2 distinct circles (AP Q) and (ARS)
intersects ℓ at X. Then X is precisely the only point on ℓ that satisfies XP · XQ = XR · XS, so
we are done!

2.3. Involutions on Pencil of lines

Suppose f : LX → LX is an involution and let m be a line not passing through X.

Project every line ℓ ∈ LX onto m. We get an involution on m!

Exercise 2.3. Check that it is indeed an involution. To be precise, consider g : LX → m given

by g(ℓ) = ℓ ∩ m. Prove that there is an involution on m with reciprocal pairs (g(ℓ), g(f (ℓ))).

2.4. Involutions on conic

Theorem 2.4. For any involution f on a conic C, there exists a fixed point P such that P ∈
Af (A) ∀A ∈ C.

Proof. Since the statement is purely projective, consider a projective transformation sending C
to a circle. Let (A1 , A2 ), (B1 , B2 ), (C1 , C2 ) be reciprocal pairs of f on C. Pick X ∈ C and invert
about X and suppose C inverts to line ℓ.
Now (A′1 , A′2 ), (B1′ , B2′ ), (C1′ , C2′ ) become reciprocal pairs of an involution f ′ on ℓ. By Theorem 2.2,
∃O ∈ ℓ such that OA′1 · OA′2 = OB1′ · OB2′ = OC1′ · OC2′ Since P ̸= O, (P A′1 A′2 ), (P B1′ B2′ ), (P C1′ C2′ )
are coaxial and hence have a second intersection. Hence after inverting A1 A2 , B1 B2 , C1 C2 concur
as desired!

Note that we can project involutions from lines and conics. This is useful to use concurrency of
chords, as well as prove concurrency of chords.

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 3. DIT and DDIT

Finally, it’s time to present the nukes :)

3.1. DIT

Theorem 3.1. Let ABCD be a quadrilateral with circumconic C. A line ℓ intersects AB,
CD, AD, BC, AC, BD at points X1 , X2 , Y1 , Y2 , Z1 , Z2 respectively; and suppose ℓ intersects
C at W1 , W2 . Then (X1 , X2 ), (Y1 , Y2 ), (Z1 , Z2 ), (W1 , W2 ) are reciprocal pairs of some (unique)
involution on ℓ.

Proof. Let f be the unique involution that swaps X1 , X2 and W1 , W2 (by Lemma 2.3). By
symmetry, it suffices to show that this involution swaps Y1 , Y2 due to symmetry.
A C
(X1 , Y1 , W1 , W2 ) = (B, D, W1 , W2 ) = (Y2 , X2 , W1 , W2 ) = (X2 , Y2 , W2 W1 )

Hence f swaps Y1 , Y2 as desired.

There are other degenerate forms of DIT, obtained by taking the quadrilateral as ABCA and
ABAB respectively :

Corollary 3.2 (3 point DIT). Let ABC be a triangle with circumconic C. A line ℓ intersects
AB, AC, BC at X1 , X2 , Y2 and the tangent to C at A at Y1 ; ℓ intersects C at W1 , W2 . Then
(X1 , X2 ), (Y1 , Y2 ), (W1 , W2 ) are reciprocal pairs of some (unique) involution on ℓ.

Corollary 3.3 (2 point DIT). Let A, B be points on conic C. A line ℓ intersects AB at X, tan-
gent to C at A, B at Y1 , Y2 respectively; ℓ intersects C at W1 , W2 . Then (X, X), (Y1 , Y2 ), (W1 , W2 )
are reciprocal pairs of some (unique) involution on ℓ.

3.2. DDIT

Theorem 3.4. Let ABCD be a quadrilateral with inconic C. Let E = AB ∩ CD and F =

AD ∩ BC. Let P be a point not lying on any side of the quadrilateral. Let P X, P Y be the
tangents from P to C. Then there exists a (unique) involution on LP that with reciprocal pairs
(P A, P C), (P B, P D), (P E, P F ) (P X, P Y ).

Again as before, degenerate cases of DDIT are as follows :

Corollary 3.5. (3 point DDIT) Let ABC be a triangle with inconic C. Let C touch BC at P .
If P X, P Y are tangents from P to C, then there is an involution with reciprocal pairs (P A, P D),
(P B, P C), (P X, P Y ).

Corollary 3.6. (2 point DDIT) Let A, B lie on conic C and P be a point. If tangents to C
at A, B intersect at Z, and P X, P Y are tangents from P to C. Then there is an involution with
reciprocal pairs (P X, P Y ), (P A, P B), (P Z, P Z).

Proof. Start with DIT and take pole-polar transformation wrt C. Apply DDIT on ABDC and
AZBZ to get last 2 forms.

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 4. Example Problems

Example 4.1 (Butterfly Theorem). Let AB, CD, P Q be chords of a circle ω concurrent at M .
Let X = P Q ∩ AD and Y = P Q ∩ BC. Suppose M P = M Q. Show that M X = M Y .
Proof. Apply DIT on the line P XM QY intersecting quadrilateral ABCD with circumconic ω.
We have (M, M ), (P, Q), (X, Y ). The unique involutiom swapping first two pairs is reflection
about M . Hence M X = M Y .

Example 4.2 (USAMO 2012/5). Let P be a point in the plane of △ABC, and γ a line passing
through P . Let A′ , B ′ , C ′ be the points where the reflections of lines P A, P B, P C with respect
to γ intersect lines BC, AC, AB respectively. Prove that A′ , B ′ , C ′ are collinear.
Proof. Let X = AB ∩ A′ B ′ . We wish to show X = C ′ . Applying DDIT on the quadrilateral
ABA′ B ′ , we have an involution with reciprocal pairs (P A, P A′ ), (P B, P B ′ ), (P X, P C).
However reflection about γ is the unique involution swapping first two pairs. Hence P Z is
reflection of P C over γ. Since Z ∈ AB, we must have Z = C ′ and we are done!

Example 4.3 (China TST 2017/2/3). Let ABCD be a quadrilateral and let ℓ be a line. Let
ℓ intersect the lines AB, CD, BC, DA, AC, BD at points X, X ′ , Y, Y ′ , Z, Z ′ respectively. Given
that these six points on l are in the order X, Y, Z, X ′ , Y ′ , Z ′ , show that the circles with diameter
XX ′ , Y Y ′ , ZZ ′ are coaxial.
Proof. Apply DIT on the line ℓ intersecting quadrilateral ABCD, there is an involution with re-
ciprocal pairs (X, X ′ ), (Y, Y ′ ), (Z, Z ′ ). Now recollect that any involution is an inversion/reflection.
Due to the ordering, it can’t be a reflection, and hence it is an inversion about some point which
we call O.
Hence OX · OX ′ = OY · OY ′ = OZ · OZ ′ . Hence O lies on axis of pairwise radical axes of the 3
circles (XX ′ ), (Y Y ′ ), (ZZ ′ ). But all the pairwise radical axes are perpendicular to ℓ, hence the
3 circles have a radical axis and we are done!

Example 4.4 (Lemma from 2023 BMoEG). Let ABCD be a cyclic quadrilateral with cir-
cumcirle ω. Let ℓ be a lint that intersects ω at X, Y , and meets lines AB and AD in points
S1 , S2 respectively. Define U = CX ∩ AD and V = CY ∩ AB. Further define T = U S1 ∩ V S2 and
R = CT ∩ XY . Finally, define Z = BD ∩ XY . Then prove that (ZR; S1 S2 ) = (AC; DB)ω .
Proof. By DDIT on quadrilateral U S1 V S2 about point C, we have an involution with pairs
(CU, CV ), (CS1 , CS2 ), (CA, CT ). Projecting on line XY , we get an involution Φ1 with pairs
(X, Y ), (S1 , S2 ), (R, C ′ ) where C ′ = AC ∩ XY .
By 3 point DIT on ABD with circumconic ω on line XY , we get that an involution Φ2 with
reciprocal pairs (X, Y ), (S1 , S2 ), (Z, A′ ) where A′ = AA ∩ XY .
But let Φ be the unique involution swapping (X, Y ), (S1 , S2 ). Since Φ shares two pairs with Φ1
and Φ2 , we have Φ = Φ1 = Φ2 .
Hence Φ swaps (X, Y ), (S1 , S2 ), (R, C ′ ), (Z, A′ ). Hence,
Φ
(ZR; S1 S2 ) = (A′ C ′ , S2 S1 ) = (AC; DB)ω

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The following is a really hard problem from Taiwan TST, to solve with euclidean geometry. It is
also of my favourite statements.

Example 4.5 (Taiwan TST 2014/3/3). Let M be any point on the circumcircle of triangle
ABC. Suppose the tangents from M to the incircle meet BC at two points X1 and X2 . Prove
that the circumcircle of triangle M X1 X2 intersects the circumcircle of ABC again at the tangency
point of the A-mixtilinear incircle.

Proof. Let ω be the incircle of ABC which touches BC at D, and let T be the touch point of
A-mixtilinear incircle with (ABC).
By 3 point DDIT on ABC with inconic ω, there is a involution with reciprocal pairs (M B, M C),
(M A, M D), (M X1 , M X2 ). Projecting onto BC, we have an involution with reciprocal pairs
(B, C), (M ′ , D), (X1 , X2 ) where M ′ = M A ∩ BC.
Let S be the centre of inversion of the aforementioned involution on BC, and we have SB ·SC =
SM ′ · SD = SX1 · SX2 . Hence S lies on the radical axis of (ABC), (M M ′ D), (M X1 X2 ). Since
they intersect at a point M , M K is the radical axis of the three circles. Hence they meet at a
second point.
We want to prove that T lies on (M X1 X2 ) and hence it’s equivalent to show that T lies on
(M M ′ D) is cyclic.
First note that if A′ ∈ (ABC) such that AA′ ∥ BC, then T , D, A′ are collinear. (This can be
seen by reflecting these three points along the perpendicular bisector of BC) Hence if we let
T D ∩ AM = Z, then
ZD ZA ZM
= =
ZM ′ ZA′ ZT
Hence ZD · ZT = ZM ′ · ZM which implies M M ′ DT is cyclic, and we are done!

Example 4.6 (Serbia 2017/6). Let k be the circumcircle of △ABC and let ka be A-excircle.
Let the two common tangents of k, ka cut BC in P, Q. Prove that ∡P AB = ∡CAQ.

Proof. Let ka touch BC at E and let the common tangents meet at Z. By 3 point DDIT on
ABC with inconic ka , there is an involution with reciprocal pairs (ZB, ZC), (ZA, ZE), (ZP, ZQ).
Projecting onto BC, we infer that there is an involution on BC with pairs (B, C), (P, Q), (D, E)
where D = ZA ∩ BC.
The problem asks us to prove that the unique involution swapping first two pairs is isogonal
conjugation in ∠BAC. Hence it’s equivalent to show that ∠BAD = ∠EAC.
If T is touch point of A-mixtilinear incircle with (ABC), it’s equivalent to show that X, A, T are
collinear. But this follows by Monge on (ABC), ka and the A- mixtilinear incircle.

5. Finale

In this section we will address two problems. These problems are much more involved. Unlike
most of the example problems, we wouldn’t be directly done after applying one DIT/DDIT. These
problems (second one to be more precise) involve a mix of ideas and much more than a plain
DIT/DDIT. In it’s glory we first present HMIC 2024/5.

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Problem 5.1. Let ABC be an acute, scalene triangle with circumcenter O and symmedian point
K. Let X be the point on the circumcircle of triangle BOC such that ∠AXO = 90◦ . Assume
that X ̸= K. The hyperbola passing through B, C, O, K, and X intersects the circumcircle of
triangle ABC at points U and V , distinct from B and C. Prove that U V is the perpendicular
bisector of AX.
Proof. It’s well known that A, K, X are collinear. Let H be the unique conic passing through
B, C, O, K, X and ω be (ABC). What is a way to use the conic? Treat it as the circumonic of
a quadrilateral and apply DIT!!
Claim 5.1. U V bisects AX.

Proof. Let Φ be the unique involution on line ℓ = OX applied on quadrilateral U V BC (which

swaps (U B ∩ ℓ, V C ∩ ℓ), (U C ∩ ℓ, V B ∩ ℓ))

• Φ swaps (U V ∩ OX, T ) where T = OX ∩ BC.

• By considering ω as the circumonic of U V BC, Φ swaps (Y1 , Y2 ) where Y1 , Y2 = OX ∩ ω.
• By considering H as the circumconic of U V BC, Φ swaps (X, O).

Since OXBC is cylic, T X · T O = T B · T C = T Y1 · T Y2 . Hence T is the involution centre of Φ!

Since Φ swaps (U V ∩OX, T ) we infer that U V ∩OX = ∞OX hence U V ∥ OX =⇒ U V ⊥ AX.
Claim 5.2. U V bisects AX.

Proof. As we saw considering same quadrilateral and line, but amusingly 2 different circumonics
was a magical trick. Who said we do it once?! Do it again!!
Let Φ be the unique involution on line ℓ′ = AX applied on quadrilateral U V BC (which swaps
(U B ∩ ℓ′ , V C ∩ ℓ′ ), (U C ∩ ℓ′ , V B ∩ ℓ′ ))

• Φ swaps (M, D) where M = U V ∩ AX and D = AX ∩ BC.

• By considering ω as the circumonic of U V BC, Φ swaps (A, S) where (AS; BC)ω = −1.
• By considering H as the circumconic of U V BC, Φ swaps (X, K).

Let Z be the centre of inversion which swaps the involution Φ. Since we want to show M is the
Want Φ
midpoint of AX, −1 = (AX; M ∞AX ) = (SK; DZ). It suffices to show that (SK; DZ) = −1.
Φ
Note that −1 = (AS, X∞AX ) = (SA, KZ). We will now characterize Z.
Let B ′ = BK ∩ ω and C ′ = CK ∩ ω.
B
−1 = (AB, CC ′ ) = (A, D; K, BC ′ ∩ AD)
By symmetry (A, D; K, CB ′ ∩ AD) = −1 hence BC ′ , CB ′ , AD concur say at Z ′ . By brocard on
BCB ′ C ′ , Z ′ lies on polar of K wrt ω. But since (SA, KZ) = −1, Z also lies on the polar of K
wrt ω. Since both Z, Z ′ lie on AD and polar of K, Z = Z ′ . Hence Z lies on BC ′ . Now
B K
(SA, KZ) = (SB ′ , B ′ C ′ ) = (AS, BC) = −1
and we are done!

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Combining the 2 claims, the problem is proven.
And now its time for Turkey TST 2024 P9.
Problem 5.2. In a scalene triangle ABC, I is the incenter and O is the circumcenter. The line
IO intersects the lines BC, CA, AB at points D, E, F respectively. Let A1 be the intersection of
BE and CF . The points B1 and C1 are defined similarly. The incircle of ABC is tangent to
sides BC, CA, AB at points X, Y, Z respectively. Let the lines XA1 , Y B1 and ZC1 intersect IO
at points A2 , B2 , C2 respectively. Prove that the circles with diameters AA2 , BB2 and CC2 have
a common point.
Proof. The first key step is to understand the point A2 .
Claim 5.3. Lines BA2 and CA2 are reflections of each other in line OI.

Proof. Begin with the following observations.

• By DIT on the ℓ = A1 A2 applied on quadrilateral BEF C we infer that there is a involution

with reciprocal pairs (AB ∩ ℓ, AC ∩ ℓ), (A1 , A1 ), (X, A2 ). Projecting from A, we get an
involution Φ1 on BC with pairs (B, C), (D′ , D′ ), (X, T ) where D′ = AA1 ∩ BC and T =
AA2 ∩ BC.
A A
• Observe that (BC; D′ D) = (F, E; AA1 ∩ EF, D) =1 (CB; D′ , D). Let Φ2 be the unique invo-
lution on BC that preserves D′ and swaps (B, C). Because of the cross ratio we have Φ2
preserves D as well.

Since Φ1 , Φ2 share two pairs, we have Φ1 = Φ2 . Hence we have an involution on BC with pairs

(B, C); (X, T ); (D, D) (1)

We still haven’t used the fact that X is the touch point of incircle ω with BC. In order to use
this, we take pole-polars wrt ω. Recollect that taking pole polars preserves cross ratios and hence
involutions. Let the dual of T be XP for some point P on ω. Let X ′ be the other tangent from
D and it follows that X ′ is the reflection of X over OI.
Taking duals of the involution in (1), we get an involution on LX with reciprocal pairs

(XY, XZ); (XX, XP ); (XX ′ , XX ′ )

Projecting through X onto ω we get an involution on ω with reciprocal pairs

(Y, Z); (X, P ); (X ′ , X ′ )

Hence by theorem 2.4 Y Z, XP, X ′ X ′ concur. Using La Hire we infer that their duals wrt ω must
be collinear. Thus A, T and X ′ are collinear.
Hence A2 , T, X ′ are collinear. Hence the unique involution on BC with pairs (X, T ) and (D, D) is
isogonal conjugation in reflection about OI followed by projecting from A2 back onto BC. Hence
from (1), BA2 , CA2 are reflections over OI, proving the the claim.
Since OB = OC and that A2 O bisects ∠BA2 C, from Fact 5, we infer that OBCA2 is cyclic.
Inverting about (ABC), we see that points A2 and D are inverses in (ABC). Hence

OA2 = OD · OA2

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Now it’s time to guess the concurrency point. Since the problem is completely symmetric in
A, B, C and we have this super useful line OI, the best guess is the miquel of the lines AB, AC,
BC, OI which we call S.
Since OS 2 = OA2 = OD · OA2 , we see that △OSA2 ∼ △ODS.
We intend to show ∠ASA2 = 90◦ , so it is equivalent to show that if SA2 intersects (ABC) again
at A′ , then A′ is the A- antipode in (ABC). We have

∠SA3 O = ∠OSA2 = ∠SDO = ∠SDF = ∠SBF = ∠SA3 A

implying that A, O, A3 are collinear and we are done!!

6. Conics

We present a different proof to DIT. Suppose given a quadrilateral ABCD and a line ℓ which is
not one of the sides of ABCD.
Define the function f : ℓ 7→ ℓ given by f (X) is the second intersection of the unique conic passing
through A, B, C, D, X and ℓ. We will show that f is an involution. f 2 = id is obvious and it
suffices to prove that cross ratios are preserved.
Consider Isogonal Conjugation in △ABC. This takes any point P not lying on the sides and
sends it to P ∗ which is isogonal conjugate of P wrt ABC. Check that isogonal conjugation is an
involution and it swaps lines with circumconics.
Suppose given X1 , X2 , X3 , X4 on ℓ and let Yi = f (Xi ).

• Since A, B, C, D, Xi , Yi lie on a conic, D∗ , Xi∗ , Yi∗ are collinear.

• Since Xi , Yi lie on a line for all i, Xi∗ , Yi∗ lie on a fixed conic γ that passes through A, B, C.

(X1 , X2 ; X3 , X4 ) = A(X1 , X2 ; X3 , X4 ) = A(X1∗ , X2∗ ; X3∗ , X4∗ ) = (X1∗ , X2∗ ; X3∗ , X4∗ )γ

Similarly, (Y1 , Y2 ; Y3 , Y4 ) = (Y1∗ , Y2∗ ; Y3∗ , Y4∗ )γ . But

D∗
(X1∗ , X2∗ ; X3∗ , X4∗ )γ = (Y1∗ , Y2∗ ; Y3∗ , Y4∗ )γ

Hence we get (X1 , X2 ; X3 , X4 ) = (Y1 , Y2 ; Y3 , Y4 ). Proving that f preserves cross ratio.

In the statement of DIT, f swaps AB ∩ ℓ with CD ∩ ℓ by taking the conic to be the degenerate
conic AB ∪ CD and so on.

7. Problems

Click on the problem source to visit the AoPS thread of the problem.

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 7.1. Problem Set

Problem 7.1 (Isogonal Line Lemma). Let P, Q be isogonal points in ∠BAC. Let X = BP ∩CQ
and Y = BQ ∩ CP . Prove that X and Y are also isogonal in ∠BAC.1
Problem 7.2 (CGMO 2017). Let the ABCD be a cyclic quadrilateral with circumcircle ω1 .Lines
AC and BD intersect at point E,and lines AD,BC intersect at point F .Circle ω2 is tangent to
segments EB, EC at points M, N respectively,and intersects with circle ω1 at points Q, R.Lines
BC, AD intersect line M N at S, T respectively. Show that Q, R, S, T are concyclic.
Problem 7.3 (AoPS Thread). Let ABC and DEF be two triangles which share an incircle
ω and circumcircle γ such that D lies on minor arc AB in γ. Let L be the tangency point of
EF on ω and define K similarly on BC. Select N ≡ AL ∩ γ and M ≡ DK ∩ γ. Show that lines
AM, EF, BC, N D are concurrent.
Problem 7.4 (ARMO 2024 11.4). A quadrilateral ABCD without parallel sides is inscribed in
a circle ω. We draw a line ℓa ∥ BC through the point A, a line ℓb ∥ CD through the point B,
a line ℓc ∥ DA through the point C, and a line ℓd ∥ AB through the point D. Suppose that the
quadrilateral whose successive sides lie on these four straight lines is inscribed in a circle γ and
that ω and γ intersect in points E and F . Show that the lines AC, BD and EF intersect in one
point.
Problem 7.5 (Sharygin 2017 9.8). Let AK and BL be the altitudes of an acute-angled triangle
ABC, and let ω be the excircle of ABC touching side AB. The common internal tangents to
circles CKL and ω meet AB at points P and Q. Prove that AP = BQ.
Problem 7.6 (Sharygin 2017 9.8). Let AK and BL be the altitudes of an acute-angled triangle
ABC, and let ω be the excircle of ABC touching side AB. The common internal tangents to
circles CKL and ω meet AB at points P and Q. Prove that AP = BQ.
Problem 7.7 (USMCA 2020/7). Let ABCD be a convex quadrilateral, and let ωA and ωB be
the incircles of △ACD and △BCD, with centers I and J. The second common external tangent
to ωA and ωB touches ωA at K and ωB at L. Prove that lines AK, BL, IJ are concurrent.
Problem 7.8 (Sharygin 2023 P17). A common external tangent to circles ω1 and ω2 touches
them at points T1 , T2 respectively. Let A be an arbitrary point on the extension of T1 T2 beyond
T1 , and B be a point on the extension of T1 T2 beyond T2 such that AT1 = BT2 . The tangents
from A to ω1 and from B to ω2 distinct from T1 T2 meet at point C. Prove that all nagelians of
triangles ABC from C have a common point.
Problem 7.9. In a triangle ABC, let H, O be the orthocenter and circumcenter respectively.
Let P = BH ∩CO and let Q = CH ∩BO. Prove that A, the circumcenter of (AP Q), and HO∩BC
are concurrent.
Problem 7.10 (China TST 2017/5). In the non-isosceles triangle ABC,D is the midpoint of side
BC,E is the midpoint of side CA,F is the midpoint of side AB.The line(different from line BC)
that is tangent to the inscribed circle of triangle ABC and passing through point D intersect line
EF at X.Define Y, Z similarly.Prove that X, Y, Z are collinear.
Problem 7.11 (Generalization of India 2023 TST 3 P3, by i3435). In triangle ABC, with
orthocenter H and circumcircle Γ, K is an arbitrary point on BC. Point Q lies on Γ such that
1 This lemma is super useful. I can think of numerous problems that use this form of DDIT. I haven’t included

them in the handout though.

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AQ ⊥ QK. Circumcircle of △AQH meets AC at Y and AB at Z. Let BY and CZ meet at T .
Prove that T H is perpendicular to the isogonal of AK in ∠BAC.

Problem 7.12 (Sharygin Final Round 2024 9.3). Let (P, P ′ ) and (Q, Q′ ) be two pairs of points
isogonally conjugated with respect to a triangle ABC, and R be the common point of lines P Q
and P ′ Q′ . Prove that the pedal circles of points P , Q, and R are coaxial.

Problem 7.13 (Infinity Dots MO 3 P3). In a scalene triangle ABC, the incircle ω has center I
and touches side BC at D. A circle Ω passes through B and C and intersects ω at two distinct
points. The common tangents to ω and Ω intersect at T , and line AT intersects Ω at two distinct
points K and L. Prove that either KI bisects ∠AKD or LI bisects ∠ALD.

Problem 7.14 (IMOSL 2015 G7). Let ABCD be a convex quadrilateral, and let P , Q, R, and
S be points on the sides AB, BC, CD, and DA, respectively. Let the line segment P R and QS
meet at O. Suppose that each of the quadrilaterals AP OS, BQOP , CROQ, and DSOR has an
incircle. Prove that the lines AC, P Q, and RS are either concurrent or parallel to each other.

Problem 7.15 (Tevl Cohl). Given two triangles △ABC and △DEF with common orthocenter
H such that A, B, C, D, E, F lie on a conic C that is not a rectangular hyperbola. Prove that H
lies on the radical axis of (ABC) and (DEF ).

Problem 7.16 (IMOSL 2022 G8). Let AA′ BCC ′ B ′ be a convex cyclic hexagon such that AC is
tangent to the incircle of the triangle A′ B ′ C ′ , and A′ C ′ is tangent to the incircle of the triangle
ABC. Let the lines AB and A′ B ′ meet at X and let the lines BC and B ′ C ′ meet at Y . Prove
that if XBY B ′ is a convex quadrilateral, then it has an incircle.

Problem 7.17 (IMOSL 2021 G8). Let ABC be a triangle with circumcircle ω and let ΩA be
the A-excircle. Let X and Y be the intersection points of ω and ΩA . Let P and Q be the
projections of A onto the tangent lines to ΩA at X and Y respectively. The tangent line at P to
the circumcircle of the triangle AP X intersects the tangent line at Q to the circumcircle of the
triangle AQY at a point R. Prove that AR ⊥ BC.

7.2. Special Section

For the problems in the following section, solutions involving DIT/DDIT are rare, as compared
to the ones not involving them. But the solutions involving DIT/DDIT are instructive in some
sense, or might even showcase the power of DIT/DDIT.

Problem 7.18 (Sharygin Correspondence Round 2024 P11). Let M, N be the midpoints of sides
AB, AC respectively of a triangle ABC. The perpendicular bisector to the bisectrix AL meets
the bisectrixes of angles B and C at points P and Q respectively. Prove that the common point
of lines P M and QN lies on the tangent to the circumcircle of ABC at A.

Problem 7.19 (USAMO 2024/5). Point D is selected inside acute △ABC so that ∠DAC =
∠ACB and ∠BDC = 90◦ + ∠BAC. Point E is chosen on ray BD so that AE = EC. Let M be
the midpoint of BC. Show that line AB is tangent to the circumcircle of triangle BEM .

Problem 7.20 (IMOSL 2022 G7). Two triangles ABC, A′ B ′ C ′ have the same orthocenter H
and the same circumcircle with center O. Letting P QR be the triangle formed by AA′ , BB ′ , CC ′ ,
prove that the circumcenter of P QR lies on OH.

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Problem 7.21 (IMO 2021/3). Let D be an interior point of the acute triangle ABC with
AB > AC so that ∠DAB = ∠CAD. The point E on the segment AC satisfies ∠ADE = ∠BCD,
the point F on the segment AB satisfies ∠F DA=∠DBC, and the point X on the line AC satisfies
CX = BX. Let O1 and O2 be the circumcenters of the triangles ADC and EXD, respectively.
Prove that the lines BC, EF, and O1 O2 are concurrent.

8. Resources

This handout was a part of a course I taught at MBL 2024. I am grateful that I got the opportunity
to teach at this amazing camp. The LATEXtemplate has been provided by them.
Special thanks to Archit Manas, Kanav Talwar and Siddharth Choppara for sharing countless
new ideas, and sharing lots of problems on projective geometry. Thanks to Sidharth Suresh and
Rushil Mathur for helping me prepare the content of the course.
Thanks to the countless users on AoPS. I have linked AoPS solutions to every problem. All
credits for those solutions go to the respective user.

References

[1] http://ericshen.net/handouts/ZG-nuclear.pdf

[2] https://services.artofproblemsolving.com/download.php?id=
YXR0YWNobWVudHMvZS9jLzYzYWY3MmU0MmJjNjhiZmYwMDVhMjEzOWQzYmZjMGVmODVlOTZjLnBkZg=
=&rn=ZGVzYXJndWVzLWludm9sdXRpb24tdGhlb3JlbS5wZGY=

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