Hey guys! I know this is many months late
, but I finally got something worthy to post on (after the massive success of the Queue and Ex points post
). The main content of the discussion ahead is based on two extremely well-known problems, namely IMOSL 2002 G7 and IMOSL 2011 G4. The basic agenda is to unravel the relation between these two seemingly unrelated configurations, and then to pick up some problems around this. Hopefully you guys will enjoy this one too. Anyway, let's begin by first showing the two problems about which this whole post is (with solutions included for completeness' sake):
IMOSL 2002 G7: The incircle
of the acute-angled triangle
is tangent to its side
at a point
. Let
be an altitude of triangle
, and let
be the midpoint of the segment
. If
is the common point of the circle
and the line
(distinct from
), then prove that the incircle
and the circumcircle of triangle
are tangent to each other at the point
.
Solution: Suppose the tangent to
at
meets
(which is the tangent to
at
) at some point
. Let
be the incenter
,
-excenter of
, and
be the midpoint of
. It is well known that
are collinear, so we get
This gives
(since
is the inverse of
in
), and hence the result.
IMOSL 2011 G4: Let
be an acute triangle with circumcircle
. Let
be the midpoint of
and let
be the midpoint of
. Let
be the foot of the altitude from
and let
be the centroid of the triangle
. Let
be a circle through
and
that is tangent to the circle
at a point
. Prove that the points
and
are collinear.
Solution: Let
be the midpoint of
, and
be the foot of perpendicular from
to
. Notice that the negative homothety centered at
that takes
to
, also takes
to
, and so
are collinear. Thus it suffices to show that
lies on
. Let
be the circumcenter of
, and suppose the tangent to
at
meets
at
. As
is also tangent to
, from the Radical Axes Theorem on
we get that
is tangent to
at
. Now,
is cyclic. As
, we have that
, where the last equality follows from the fact that
lies on
, which is the perpendicular bisector of
. Thus,
lies on
, hence proving what was required.
Now that we are done with the formalities, let's move onto the more important part, i.e. the relation between these two problems.
The Whole Configuration: Let
have incircle
, incenter
and intouch triangle
. Let
be the medial triangle of
. Let
be the midpoint of the
-altitude, and
be the point where
meets
again. By IMOSL 2002 G7,
is tangent to
. Let
be the point on
such that
is tangent to
. Then
.
Proof: Invert about
. Then
and
are swapped.. Thus,
and
get swapped. But the inverse of
(i.e.
) must also be tangent to
. This gives
directly.
![[asy]
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[/asy]](//latex.artofproblemsolving.com/b/8/f/b8f37135a01bc9b0009a6dd2b94032b481b4c289.png)
Well having seen the relation, I guess it's time for some properties (From now on we will work with the above configuration only, i.e. same notations).
PROPERTY-1 Let
be the antipode of
in
, and
be the foot of perpendicular from
to
(i.e.
is the isotomic point of the foot of the
-altitude). Then
are isogonal in
. Also lines
are concurrent.
Proof: Seeing isogonal lines with one foot on the circumcircle and the other on the corresponding side inspires us to consider
inversion (If it doesn't, then make sure it does the next time
). So we invert about
with radius
, followed by reflection in
(denoting images under this transformation by
). Then
are the reflections of
in
respectively, while
is a point such that
is tangent to
. Thus, we have
Also, as
is the center of
, so we have that
is the perpendicular bisector of
. Combining this with
, we get that
, as desired.
For the second part, let
be the reflection of
in line
(i.e.
lies on
such that
). Note that
passes through the foot of the
-altitude. Then the same inversion gives that
passes through
. Combined with
and
(which has been shown above), we have the desired result
PROPERTY-2 (Probably my favourite property related to this configuration): Replace
by
, and define
similarly (We'll continue this notation from now on). Then
concur at the homothety center
of
and the excentral triangle of
. Finally,
also lies on the Euler line of
. (NOTE:
is known as the Isogonal Mittenpunkt Point of
, and is also the homothety center of
and the orthic triangle of
.)
Proof: Using Property-1, lines
concur at the isogonal conjugate of the isotomic conjugate of the orthocenter
of
. Now, let
be the excentral triangle of
, and
be the Bevan Point of
(i.e. the center of
). By IMOSL 2002 G7's proof, we have
. This means that the concurrency point of
is the homothety center
of
and
.
This gives that
also lies on the line joining the circumcenters of
and
, i.e.
. As
is the orthocenter of
, so line
is the Euler line of
. But, it is well known that the Euler lines of
and
are the same. Thus, we get that
lies on the Euler line of
also, as desired.
PROPERTY-3 Lines
are concurrent.
Proof: We prove the following stronger result (which is known as the Steinbart theorem).
Steinbart wrote:
Let
be points on
such that
are concurrent at a point
. Then
are also concurrent.
This can easily be proved using Trig Ceva, but I'll be modest, and just use moving points
. Fix
, and animate
on line
. Then
are projective. Thus, it suffices to prove that
for three distinct positions of
on line
. The result trivially follows on taking
, as desired.
REMARK (IMPORTANT): Unlike most of the famous theorems, the converse of Steinbart Theorem is not completely true. The extended Steinbart theorem deals with this anomaly. You can find it
here (written by Darij Grinberg).
Well, it's time to move on to some problems (wait, what's the distinction between problems and properties
). Anyway, let's hope I add enough motivational content for each one
.
PROBLEM-1 (Generalization of IMOSL 2011 G4; given by babu2001
here)
Let
. Let
be the reflection of
in the midpoint of
. Let the isogonal ray to
intersect
at
. Let
be the reflection of
in the perpendicular bisector of
. Then
are collinear.
Solution 1: Let
be the midpoint of
, and
such that
is the
-symmedian . We will do a cross ratio chase (don't get afraid if you haven't done one before; it is a wonderful method definitely worth knowing about). The basic inspiration is to note that isogonal rays and isotomic points are related to each other projectively (i.e. they preserve cross ratios). We start with the ratio
and try to prove it is equal to
, after which we are obviously done. First, by isotomic conjugation,
. Then we have
Thus, to show the original problem it suffices to prove that
are collinear. This can equivalently be formulated as
This equivalent statement gives that if we show the problem for even one finite position of
(which is not necessarily
), we can deduce that the problem is true for all positions of
. Taking
as the isotomic point of the foot of the
-altitude (i.e. the foot of perpendicular from the
-antipode on
) quite easily works. Hence, done.
NOTE (A possible trap): We cannot just show the problem to be true for
or
, and say it is always true, after we have deduced the equivalence relation. This is because, when
, the cross ratio
becomes invariant of point
. A similar thing happens when
is the point at infinity, which is why we must ensure that
is a finite point distinct from
.
Solution 2: There is an easy alternative moving points solution (those who don't know moving points may skip this). Animate
on
. Then, since isogonal and isotomic conjugation are projective maps, so the map
must also be projective. Thus, it suffices to prove the problem for three distinct positions of
. Taking
and the point at infinity on
trivially give the conclusion.
PROBLEM-2 (
EMMO 2016 Junior P5)
Let
be a triangle with circumcenter
and circumcircle
The point
lies on
such that
is the
- symmedian of triangle
The line through
perpendicular to
intersects
in
respectively. Denote by
the nine-point circle of triangle
and let
and
intersect again in
Further, let the tangent to
at
meet the line
in
and let
be the antipode of
with respect to circle
Prove that the points
are collinear.
Solution: Redefine
as the point where
meets
again, in which case we have to show that
lies on the nine-point circle of
. Note that, by Property-1 (second part),
is in fact the 2011 G4 point of
. Suppose the perpendicular bisector of
(i.e. line
) meets
at
. Then we wish to show that
. Consider
inversion. Then, using Property-1, and with the help of the inversive distance formula, we get the following problem:-
Inverted problem wrote:
Let
be the midpoint and the isotomic point of the
-altitude in
. Suppose
meets
at
. Then show that
is cyclic.
As we do with almost all problems, we will assume that the problem is true, and try to deduce whatever we can. Note that, if we denote the reflection of
in
by
, then
and
. The reason we consider
is that it also lets us characterize point
; it is simply the foot of perpendicular from
to
. Thus,
is nothing but the pedal circle of
wrt
. Now, whenever pedal circles are involved, I have a tendency to go towards the isogonal conjugate (and so should you, if you don't have such an urge already). Anyway, the main point here is that isogonal conjugates share their pedal circles (which is also called the
six point circle theorem). So we consider the isogonal conjugate of
, which is nothing but the point
where the tangent to
at
and
meet. Note that
, and so using the fact given above, we get that
, as desired.
REMARK: The trick of considering the common pedal circle of isogonal conjugates is a wonderful tool for tackling problems involving these.
Here's another example of this trick (the geometry one, not the combi one
).
PROBLEM-3 (Proposed by AlastorMoody
here)
In an acute triangle
, Let
be the isotomic conjugate of the orthocenter
of
. Let
and
be the midpoint of
. Let
be the orthic triangle WRT
and
be the centroid of
. Define
and Let
is the midpoint of
(i) Let
be the midpoint of
. Show that the intersection of
and
lies on
. Moreover, show that this intersection point lies on
(ii) Suppose that the intersection point in (i) is
, then prove,
lies on
(iii) Let
at
, then prove,
Solution: Redefine
as the 2011 G4 point of
, since Part (ii) and (iii) kind of give away that this should be
. In that case, (ii) and (iii) are true by solution of IMOSL 2011 G4. So we'll just show that
and
. Using
as given in (iii), we notice that
directly gives
by converse of Reim's Theorem. Again using converse of Reim's Theorem, we get
from
(since
and
).
For the final part, i.e. showing
, we bring our old friend, the
-Ex point
, into picture (You'll see soon why). Seeing so many circles, for obvious reasons, we try to use radical axes theorem. Applying it to
, we see that
lies on radical axis of
and
. But, by Property-4
here, we know that point is nothing but
, which in turn means that
are concurrent at
. Then
gives the result.
REMARK: I believe the problem would have been much harder (especially in exam conditions), if the last 2 parts weren't given. They are sort of the giveaways. Apart from that, the part of getting
concurrent could be observed even by a good eye. Like in my diagram, I had a good hunch that it might be true (and these hunches go a long way with geometry).
PROBLEM-4 (
STEMS 2020 Category B Subjective P4)
In triangle
with incenter
, the incircle
touches sides
and
at points
and
, respectively. A circle passing through
and
touches
at point
. The circumcircle of
meets
at
. Prove that
is parallel to
.
Solution: Let
be the
-intouch point of
, and note that
is the 2011 G4 point of
. Redefine
as the point such that
. Then
combined with
gives that
is the center of
, i.e.
is the reflection of
in
. Now, we consider
inversion followed by reflection in angle bisector of
(denoting inverse of point
by
). Let
be the line through
parallel to
. Then
such that
and
. Since,
is the midpoint of
, so we get that
is the foot of perpendicular from
to
. Note that
is equal to the distance of
from the
-altitude, and so we have
. Combined with
, we get that
is a parallelogram. Thus, we have
, and so
must be an isosceles trapezoid, or equivalently,
is cyclic, as desired.
REMARKS: Wonderfully easy if you know the above theory
.
With that, we come to an end of this section. For a bonus problem, try
this one using what we have done till now (Don't see my solution
). After this I will probably be posting on this blog after May only. So enjoy
P.S. Do write in the comments section if you wish to share something (like a problem or another property). Also, I am extremely sorry for any typos. You can notify me about those too in the comments section. Thanks
This post has been edited 4 times. Last edited by math_pi_rate, Jan 31, 2020, 6:04 AM
by sqing, Mar 26, 2025, 3:49 AM 
and 
. Prove that
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](http://latex.artofproblemsolving.com/2/2/e/22e6ec2f98777db80e39333542664e903663bf81.png)
![$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$](http://latex.artofproblemsolving.com/f/c/a/fca343ee701eeff4c309b31beb3426ec96b49b55.png)
Let 
. Prove that
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，and any rectangular strip of width 1 covers at most two points of S.
. Let 
be a point on side 
, and 
be a point on the extension of 
such that 
. Let 
be the circumcenter of 
, and 
be a point on the side extension of 
satisfying 
. Line BP intersects AC at point Q. Prove that 
be a finite set of real numbers, 
be a real number, and 
be 2025 non-zero real numbers. Define![\[A =
\left\{
(x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d
\right\},\]](http://latex.artofproblemsolving.com/2/e/1/2e18b6d25bfaa3e3d52f7efe5e4cba7a9cb87db9.png)
![\[B =
\left\{
(x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0
\right\},\]](http://latex.artofproblemsolving.com/4/1/f/41f6b126cc3cf24757b37147906c767d49a0fdd9.png)
![\[C =
\left\{
(x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0
\right\}.\]](http://latex.artofproblemsolving.com/d/e/6/de6586b88004e3f5aa7774525b00ee975c072ff9.png)
Show that 
.
, ![$P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$](http://latex.artofproblemsolving.com/0/5/1/0514f85850830af9a9e1a4dd5e7a4e8e68b2dff9.png)
.
polynomial with only one root.
?
lie in a square of area 
. Prove that among the 
that intersect in a region of area not less than 
.
over all pairs of positive integers 
.
of the acute-angled triangle 
is tangent to its side 
at a point 
. Let 
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
at 
meets 
) at some point 
. Let 
be the incenter,
-excenter of 
, and 
be the midpoint of 
. It is well known that 
are collinear, so we get 
This gives 
(since 
be an acute triangle with circumcircle 
be the midpoint of 
and let 
be the midpoint of 
. Let 
be the foot of the altitude from 
be the centroid of the triangle 
be a circle through 
. Prove that the points 
and 
are collinear.
be the midpoint of 
be the foot of perpendicular from 
. Notice that the negative homothety centered at 
, also takes 
are collinear. Thus it suffices to show that 
. Let 
is also tangent to 
we get that 
is tangent to 
is cyclic. As 
, we have that 
, where the last equality follows from the fact that 
. Thus, 
and intouch triangle 
. Let 
be the midpoint of the 
meets 
is tangent to 
be the point on 
is tangent to 
.
and 
are swapped.. Thus, 
get swapped. But the inverse of ![[asy]
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/b/8/f/b8f37135a01bc9b0009a6dd2b94032b481b4c289.png)
be the antipode of 
(i.e. 
are isogonal in 
. Also lines 
are concurrent.
inversion (If it doesn't, then make sure it does the next time 
, followed by reflection in 
). Then 
are the reflections of 
respectively, while 
is a point such that 
is tangent to 
Also, as 
, so we have that 
is the perpendicular bisector of 
. Combining this with 
, we get that 
, as desired.
be the reflection of 
(i.e. 
). Note that 
passes through the foot of the 
passes through 
and 
(which has been shown above), we have the desired result 
, and define 
similarly (We'll continue this notation from now on). Then 
concur at the homothety center 
of 
be the excentral triangle of 
be the Bevan Point of 
). By IMOSL 2002 G7's proof, we have 
. This means that the concurrency point of 
. As 
is the Euler line of 
are concurrent.
be points on 
are concurrent at a point 
are also concurrent.
, and animate 
. Then 
are projective. Thus, it suffices to prove that 
for three distinct positions of 
. The result trivially follows on taking 
, as desired.
. Let 
be the reflection of 
intersect 
at 
be the reflection of 
are collinear.
such that 
is the 
and try to prove it is equal to 
, after which we are obviously done. First, by isotomic conjugation, 
. Then we have 
Thus, to show the original problem it suffices to prove that 
are collinear. This can equivalently be formulated as 
This equivalent statement gives that if we show the problem for even one finite position of 
(which is not necessarily 
or 
, and say it is always true, after we have deduced the equivalence relation. This is because, when 
, the cross ratio 
becomes invariant of point 
.
must also be projective. Thus, it suffices to prove the problem for three distinct positions of 
and the point at infinity on 
The point 
such that 
is the 
The line through 
in 
respectively. Denote by 
the nine-point circle of triangle 
and let 
Further, let the tangent to 
and let 
are collinear.
meets 
. Note that, by Property-1 (second part), 
) meets 
. Then we wish to show that 
. Consider 
be the midpoint and the isotomic point of the 
meets 
. Then show that 
is cyclic.
and 
. The reason we consider 
is nothing but the pedal circle of 
and 
meet. Note that 
, and so using the fact given above, we get that 
, as desired. 
, Let 
be the isotomic conjugate of the orthocenter 
and 
be the orthic triangle WRT 
and Let 
is the midpoint of 
and 
lies on 
at 
and 
directly gives 
by converse of Reim's Theorem. Again using converse of Reim's Theorem, we get 
from 
(since 
and 
).
, we bring our old friend, the 
, into picture (You'll see soon why). Seeing so many circles, for obvious reasons, we try to use radical axes theorem. Applying it to 
, we see that 
lies on radical axis of 
. But, by Property-4 here, we know that point is nothing but 
are concurrent at 
gives the result. 
concurrent could be observed even by a good eye. Like in my diagram, I had a good hunch that it might be true (and these hunches go a long way with geometry).
with incenter 
and 
meets 
. Prove that 
is parallel to 
.
as the point such that 
. Then 
combined with 
gives that 
, i.e. 
). Let 
be the line through 
such that 
and 
. Since, 
is the midpoint of 
, so we get that 
is equal to the distance of 
. Combined with 
, we get that 
is a parallelogram. Thus, we have 
, and so 
must be an isosceles trapezoid, or equivalently, 
). So, without further ado, let's start this 'safar' (I have tried to arrange the problems in order of difficulty):-
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by 
the orthocenter of 
be the point of intersection of the circumcircle 
of the triangle 
, and 
be the point of intersection (other than 
and the circle 
must hold.
from now on). Seeing this point, we add in the 
is harmonic, for which we will require some sort of perspectivity. As all points are closely linked to the point 
is harmonic (I guess this has been given in the proof of Property-3), and with the help of Property-1, we get that 
be its circumcircle, and 
be midpoints of 
, respectively. 
are the feet of the altitudes from 
and 
at 
intersects 
at 
intersects smaller arc 
are concyclic.
intersect 
(
), respectively. Prove that 
are concurent.
, to find that 
are in fact collinear. At first glance, the given conditions do not give any more significant information (atleast not to me 
, we get that 
passes through the 
So we arrive at the most crucial part of the solution. We just need to show that the line 
is the 
. This is nice, as we know that 
, and so we get that 
. Now, it is some manipulations before finding a way out. What I did was take notice of the fact that 
) must lie on 
is antiparallel wrt 
(as 
). But, 
and 
, after which one can finish using the fact that 
are concurrent.
let 
and 
and 
. Prove that 
is cyclic. Our solution is motivated by this well-known lemma and its converse (Lemma 9.17 in EGMO). Without knowing that lemma, it might be difficult to produce this proof (Basically I am just telling you to learn it ASAP 
(as 
, so we get 
. Combining these two equalities, we see that 
is cyclic. As 
, so we get that in fact 
bisects 
externally. This is the crucial part of the solution. After this it is just some angle chasing (unfortunately there's not much motivation for that; you just try to find nice angles). The full proof can be found here.
, 
. Suppose the circle 
, and meets line 
at a point 
. The tangent to 
at 
and 
intersect at a point 
.
is the 
. So 
, which means that 
is also tangent to 
, 
and 
(all tangents taken wrt 
and 
, we get that 
are collinear (this is true for all cyclic quadrilaterals; learn this result as a fact as it is quite useful). Now, these two new intersection points move us closer towards a complete quadrilateral. So let's "complete" this complete quadrilateral configuration (pun intended 
. As discussed in Problem-3 of Part One, this should motivate us to look for Brokard's Theorem. Then using Brokard (and the collinearity discussed earlier), we get that 
is the polar of 
, which means that 
. Suppose what we are supposed to prove is true, i.e. 
is the polar of 
(as 
). So we get that 
be the inverse of 
, so 
. Thus it suffices to show that 
, which is equivalent to proving that 
(since 
are collinear). However, 
is nothing but the power of 
are tangent to each of the circle 
, with 
interior to 
exterior to 
and the common tangent of 
intersect at the point 
be the diameter of 
, and 
. Show that 
is cyclic.
. But, applying Property-3 to 
is harmonic (See the Remark given after Problem-3 above). Now, we again turn our attention back to 
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
. The length of the first jump is 
, the second 
, the third 
, and so on. The length of 
jump is equal to 
. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?
January 2020
September 2018
