markan (19:30:43)
In this math jam we will discuss inversion, which is a geometric transformation of the plane like reflection or translation. Many complicated-looking geometric theorems can be proved easily after inversion.

markan (19:30:56)
are there any general questions before we dive in to the definition?

markan (19:31:21)
ok

markan (19:31:23)
Let's start with the definition.

markan (19:31:28)


markan (19:31:34)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/inversion.asy.png

markan (19:31:39)


markan (19:31:53)
So inversion exchanges the interior and exterior of the circle. Points closer to the center before inversion end up further away after inversion, and vice versa.

markan (19:32:02)
Also, note that inverting about the circle a second time is like undoing the original inversion.

markan (19:32:09)
There's actually a simple straightedge-and-compass construction that can be used to invert points.

markan (19:32:26)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/construction.asy.png

markan (19:32:28)


markan (19:32:33)
Let A be a point inside the circle gamma centered at O, and let l be the line through A perpendicular to OA. Let K be a point where l meets gamma, and m be the line tangent to gamma at K. The point where m meets OA is the inverse of A.

markan (19:32:51)
Let's prove this before we go on. What do we want to show?

markan (19:33:05)
(and by the way, please speak up if i'm going too fast)

markan (19:33:45)
i'm going to give you a minute or so to catch up reading the above

PI-Dimension (19:33:04)
OA*OA'=r^2

laxatives (19:33:08)
oa(oa')=r2

Karth (19:33:26)
Prove that OA * OA' = OK^2

tjhance (19:33:30)
that OA*OA'=r^2

Ivan Zhang (19:33:59)
OA*OA'=r^2

markan (19:34:42)
yes, we want to show that

markan (19:34:44)


toadoncart (19:34:17)
what's the subject of this math jams? i just got in

markan (19:34:54)
it's about geometric inversion

markan (19:34:59)
If we divide we can rewrite this as

markan (19:35:11)


markan (19:35:20)
This looks like something that might arise from similar triangles. What similar triangles do we want?

indianamath (19:34:57)
and we show that by similarity

Teki-Teki (19:35:18)
which immediately suggests similar triangles.

Ivan Zhang (19:36:14)
My bad. OAK and OKA'.

indianamath (19:36:28)
the 2 triangles with the radius as a side length

tjhance (19:36:36)
OAK is similar to OKA'

markan (19:37:22)
yep

markan (19:37:37)
remember, when you're giving similar triangles, you should order the points so that they correspond

markan (19:37:44)
in other words, OAK is similar to OKA'

markan (19:37:52)
but OAK is not similar to KA'O

markan (19:37:59)
alright, so how do we know that these two triangles are similar?

PI-Dimension (19:38:13)
same angle at O and one right triangle

kevinlang (19:38:15)
2 congruent angles

not_trig (19:38:18)
they share an angle and both have right angles

laxatives (19:38:20)
AA~

dingzhou (19:38:31)
they are both right triangles, and they share another angle

markan (19:38:42)
The triangles share two angles: a right angle and angle KOA = A'OK, so they are similar.

laxatives (19:38:37)
Could you repost the picture?

markan (19:39:07)


markan (19:39:09)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/construction.asy.png

markan (19:39:29)
Ok, so now we have OA/OK = OK/OA'

markan (19:39:34)
and OA/OK is what?

boogiepop90 (19:39:49)
OA/r

markan (19:40:05)
whereas OK/OA' is.....

tjhance (19:40:12)
r/OA'

Teki-Teki (19:40:13)
r/OA'

not_trig (19:40:12)
so OA*OA' = r^2

markan (19:40:23)
yep

markan (19:40:29)
this is what we wanted to prove

markan (19:40:37)
so A' is indeed the inverse of A through the circle

markan (19:40:42)
From now on, we will denote the inverse of any point X by X'.

markan (19:40:53)
Ok, next question: given the diagram we've been using, where is K'?

not_trig (19:41:02)
K' = K

kevinlang (19:41:04)
same point

indianamath (19:41:04)
k

PI-Dimension (19:41:05)
on K

Teki-Teki (19:41:05)
K'=K (they're the same point)

Ivan Zhang (19:41:14)
K itself.

Karth (19:41:16)
K' would be on the circle itself

markan (19:41:35)
It's the same as K, because

markan (19:41:37)


markan (19:41:42)
So you'll notice that if we invert A and K about O, we form a new triangle similar to OAK, namely OK'A'. Does this happen for any triangle, not just the ones where we construct K in a particular way?

markan (19:42:16)
Here's a picture:

markan (19:42:19)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/test.asy.png

markan (19:42:24)


markan (19:42:32)
What do you think, does it look like we have similar triangles?

Teki-Teki (19:42:00)
Yes.

indianamath (19:42:09)
yes

laxatives (19:42:12)
yes

not_trig (19:42:48)
yes?

PI-Dimension (19:42:56)
OBA and OA'B?

Ivan Zhang (19:43:00)
OBA and OA'B' are similar.

123s (19:43:09)
it looks like it

Karth (19:43:10)
BOA ~ A'OB'?

algebra3000 (19:43:12)
well, if you flip B'OA' then it looks similiar to BOA

markan (19:43:28)
It does turn out that triangle OAB is similar to triangle OB'A'. Why?

naitixuy (19:43:27)
side-angle-side

Teki-Teki (19:43:56)
OB*OB' = OA*OA' = r^2 can be rearranged so that OB/OA = OA'/OB'. Hence OAB is similar to OB'A'.

tjhance (19:44:01)
OB * OB' = r^2 = OA * OA' so OB / OA' = OA / OB'

markan (19:44:42)
We have

markan (19:44:44)


markan (19:45:00)
where we used the definition of inversion twice

markan (19:45:05)
this ratio and the shared angle O is enough to make the triangles similar.

markan (19:45:17)
alright, what is the inverse of O?

Teki-Teki (19:45:27)
a point at infinity.

kevinlang (19:45:28)
doesnt exist

darkprince (19:45:30)
infinity?

not_trig (19:45:31)
infinity

tjhance (19:45:35)
the point that is infinitely far away from the circle

markan (19:45:48)
yeah, trick question

markan (19:45:53)
No point in the plane could be its inverse, so we will define an ""extra"" point known as the ""point at infinity."" It and O are inverses of each other. Also, we declare that any line goes through the point at infinity, since lines go off to infinity.

markan (19:46:06)
this is just a choice we get to make

markan (19:46:16)
the only reason we make it is because things turn out conveniently later

markan (19:46:24)
With this convention, inversion is a one-to-one mapping of the plane. This fact will prove useful later.

markan (19:46:29)
Also keep in mind that undoing an inversion is the same thing as repeating the inversion.

Teki-Teki (19:46:37)
but not an isometry, right?

markan (19:46:47)
right

markan (19:47:08)
next question:

markan (19:47:21)
ah, many of you want to know what an isometry is

markan (19:47:31)
it's a transformation that doesn't change distances

markan (19:47:44)
so rotations, translations, and mirror images are examples

markan (19:48:25)
(in case you're curious, iso means same and meter means distance)

markan (19:48:40)
anyway

markan (19:48:42)
mirro

markan (19:48:45)
oops..

markan (19:48:47)
What is the inverse of a line through O?

kevinlang (19:48:56)
the same line

not_trig (19:48:56)
itself

Teki-Teki (19:49:00)
The same line

laxatives (19:49:04)
line through O'?

SorcererofDM (19:49:04)
itself?

indianamath (19:49:06)
the same line

tjhance (19:49:07)
the same line

markan (19:49:13)
why?

Teki-Teki (19:49:28)
Because X' is on the ray OX.

markan (19:49:49)
yeah, it just comes from the definition

markan (19:50:17)
any point r away from O will just go to one 1/r away from O along the same line

markan (19:50:22)
and as for O, it will go to the point at infinity

markan (19:50:27)
but we already declared that the point at infinity is on every line

markan (19:50:35)
Ok, now it gets interesting. We're going to start inverting whole figures, instead of just points.

Teki-Teki (19:50:34)
and the point at infinity maps to O

junezyu (19:50:39)
could you please repost the definition of inversion?

markan (19:50:50)
sure

markan (19:50:58)


markan (19:51:03)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/inversion.asy.png

markan (19:51:05)


markan (19:51:21)


markan (19:51:28)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/invertline.asy.png

markan (19:51:33)


markan (19:51:43)
We've drawn in M' and P'. Ignore the second circle for the moment.

markan (19:51:54)
i'll give you a minute to read

markan (19:53:08)
so basically we want to see where P' could possibly be, given some P on the line

markan (19:53:16)
what is angle OP'M'?

Teki-Teki (19:53:26)
90 degrees

boogiepop90 (19:53:26)
90

kevinlang (19:53:27)
90

Ivan Zhang (19:53:29)
90

markan (19:53:37)
how do you know?

Teki-Teki (19:53:44)
because OMP is similar to OP'M'

indianamath (19:53:45)
similar triangles

boogiepop90 (19:53:52)
similar

markan (19:54:03)
It's 90 degrees from the rule about similar triangles.

markan (19:54:13)
so what can we say about P'?

Teki-Teki (19:54:28)
It must lie on the circle with diameter OM'

Karth (19:54:42)
its locus is a circle

tjhance (19:54:44)
it is on the circle with diameter OM'; this will be true for any point on the line

markan (19:54:52)
P' must lie on the circle with diameter M'. (Which is the second circle we drew in.)

markan (19:55:00)
It's not too hard to see that the inverse of l covers all of this circle. Why?

not_trig (19:55:15)
lines are continuous?

markan (19:56:12)
that's an interesting idea but you'd have to do some more work to turn it into a proof

markan (19:56:15)
there's a simpler answer....

tjhance (19:56:59)
show that any point on the circle maps to a point on the line

markan (19:57:13)
good

markan (19:57:26)
what's an easy way to do that?

markan (19:58:06)
For any point Q on the circle other than O, just construct the chord to O. That chord will intersect l at some point L, and then the inverse of L will be Q. As for O, we decided by convention that l includes the ""point at infinity,"" which inverts to O.

markan (19:58:32)
make sense?

not_trig (19:58:36)
could you please repost the diagram?

markan (19:58:48)
yeah

markan (19:58:53)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/invertline.asy.png

markan (19:58:59)


markan (19:59:44)
so indeed, a line that is not through O inverts into a full circle through O

markan (20:00:10)
Note that the circle with diameter OM' in our figure is the same as the circumcircle of triangle OAB. So the inverse of any line through gamma is the circle containing O and the two points where the line intersects gamma.

markan (20:00:13)
So in conclusion, we have the following important rule:

markan (20:00:18)
Under inversion around O, lines not through O invert to circles through O, and vice versa.

markan (20:00:26)
Our next step will be to determine what happens when we invert a circle that does _not_ pass through the center of inversion.

markan (20:00:37)
here's a diagram:

Karth (20:00:37)
so then does a circle map into a line then? (is the converse true?)

markan (20:00:58)
a circle through the center of inversion maps into a line

markan (20:01:09)
so yes, the converse is true

markan (20:01:23)
we're about to see what happens if the circle doesn't go through O

markan (20:01:39)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/invertcircle.asy.png

markan (20:01:44)


markan (20:01:47)
Assume O is our center of inversion. We won't draw in the circle we're inverting about. gamma is the circle we want to invert.

markan (20:01:58)


markan (20:02:24)
how can we proceed?

markan (20:03:28)
hmm....it's very quiet

markan (20:03:31)
are people confused?

markan (20:04:21)
there's a little bit of confusion, so let me try to explain what's going on

markan (20:04:31)
we're trying to learn what happens to circles when we invert

markan (20:04:39)
so we draw a circle gamma

markan (20:04:44)
and we're going to invert it around O

markan (20:04:52)
(we didn't bother to draw the circle of inversion, since it won't matter)

markan (20:05:05)
but we did draw in the line from O to the center of gamma

markan (20:05:10)
which meets gamma at A and B

markan (20:05:23)
then P is arbitrary, and we drew in the various inverses

markan (20:05:29)
and now we want to find the angle A'P'B'

markan (20:05:36)
so many of you have good ideas:

Teki-Teki (20:03:33)
Similar triangles again.

indianamath (20:03:42)
triangles OAP and OA'P' are similar i think

Ivan Zhang (20:03:44)
similar triangle.

tjhance (20:03:54)
i think APB and A'P'B' are similar

ColbertCo (20:04:07)
OAP is similar to OP`A`

Karth (20:04:33)
or what's our radius of inversion?

markan (20:06:10)
it turns out it doesn't matter

markan (20:06:15)
you can assume it's 1 inch, if you like

Teki-Teki (20:04:23)
OAP is similar to OP'A'. Hence angle OA'P' is equal to angle OPA. Similarly angle OB'P' is equal to angle OPB.

markan (20:07:00)
this is good - we've got two angles being equal

markan (20:07:15)
this is all from the rule about similar triangles that we found a while back

markan (20:07:29)
how can we use this to get A'P'B'?

algebra3000 (20:06:15)
can you post the image again?

markan (20:08:01)


markan (20:08:03)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/invertcircle.asy.png

indianamath (20:09:16)
well....isn't A'P'B' just 90 degrees?

markan (20:09:39)
it sure looks that way

markan (20:09:47)
and in fact you're right, but we haven't proven it yet

not_trig (20:09:48)
but that's [i]if[/i] it's on the circle

markan (20:10:09)
let me suggest an idea

markan (20:10:33)
let's try to rewrite A'P'B' in terms of the angles of some of our similar triangles

markan (20:10:46)
because similar triangles tend to be powerful

tjhance_2 (20:11:15)
A'P'B' = OP'A' - OP'B' = OAP - OBP = (180 - PAB) - (90 - PAB) = 90

markan (20:12:27)
many of you have good ideas now, i just picked one to run with so we don't get confused

markan (20:13:14)
i believe tjhance is exactly right

markan (20:13:20)
do people see where this string of equalities is coming from?

not_trig (20:13:26)
why is OBP = 90-PAB

markan (20:13:57)
this is because PAB is a right triangle

markan (20:14:05)
angle A and angle B within it must add to 90

not_trig (20:14:03)
OH! SORRY... thought it was P'

markan (20:14:32)
So indeed, P' lies on gamma'. This means gamma inverts to gamma'.

markan (20:14:41)
(where gamma' is the circle with radius A'B')

markan (20:14:50)
We drew the diagram with O outside gamma, but we'd get the same result regardless.

markan (20:14:52)
Circles not through the center of inversion invert to other circles not through the center of inversion.

markan (20:15:03)
Before we get on to some problems, let's talk about tangency for a second. What's the definition of tangency for two circles, or a circle and a line?

PI-Dimension (20:15:24)
touch at one point only

not_trig (20:15:36)
they ""touch at one point""

markan (20:15:49)
Two circles/lines are tangent if they have exactly one point in common.

markan (20:15:51)
When working with inversion, we're also going to say that two lines are tangent when they are parallel, because they share the point at infinity.

markan (20:16:02)
Now, since inversion is a one-to-one mapping, if two figures have exactly one point in common before inversion, they will have exactly one point in common after inversion. That means that two tangent figures will invert to two other tangent figures. This will come in handy later.

markan (20:16:29)
actually, i'll give you one example of this now:

markan (20:16:37)
suppose we have two parallel lines, and then we invert them

markan (20:16:42)
we'll get two circles

markan (20:16:53)
what will be special about those circles, based on our discussion of tangency?

kevinlang (20:17:04)
tangent at O

not_trig (20:17:04)
we get two tangent circles, with centers on the perpendicular from O to the lines

Ivan Zhang (20:17:09)
tangent to each other.

laxatives (20:17:12)
They become tangent?

indianamath (20:17:12)
they're tangent

markan (20:17:22)
yeah

markan (20:17:27)
ok, good

markan (20:17:35)
Now we are going to turn to applications of inversion. Many complicated-looking theorems are actually very simple once one inverts.

markan (20:17:41)
We'll start with a problem from Turkey's 1998 national olympiad. Here is the statement and a diagram:

markan (20:17:49)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/turkey98_5.asy.png

markan (20:17:54)


markan (20:17:59)
Let ABC be a triangle. Suppose that the circle through C tangent to AB at A and the circle through B tangent to AC at A have different radii, and let D be their second intersection. Let E be the point on the ray AB such that AB=BE. Let F be the second intersection of the ray CA with the circle through A, D, and E. Prove that AF = AC.

markan (20:18:04)
i'll let you read for a minute

markan (20:18:20)
and feel free to ask questions on anything that came before, if you're not quite caught up

darkprince (20:19:10)
so, if a line l passes through the center of inversion O, then would it be a circle of infinite radius?

tjhance_2 (20:19:23)
what happens when we invert a line that does not intersect the circle of inversion?

markan (20:20:03)
darkprince: yes, you can think of a line as a circle of infinite radius

markan (20:20:22)
that allows you to state some of the theorems about inversion more simply

markan (20:20:53)
tjhance: it becomes a circle through the center of inversion

Teki-Teki (20:19:52)
It becomes a circle through O that does not intersect the circle we are inverting about.

markan (20:21:09)
(in other words, it will be entirely inside the circle of inversion)

markan (20:21:20)
ok, let's get back to our problem now

markan (20:21:22)
any ideas?

indianamath (20:19:22)
i think we'll need to invert everything from A

Karth (20:21:43)
panic?

tjhance_2 (20:21:55)
invert about the circle containing E, D, A, and F?

not_trig (20:22:13)
invert about the circle through A, D, C?

123s (20:22:18)
invert around D?

markan (20:22:43)
many interesting ideas

markan (20:23:20)
let me make some comments

markan (20:23:35)
very often the circle of inversion doesn't matter very much, it's only the center that matters

markan (20:23:51)
changing the radius of the circle is just going to scale the diagram up or down

Teki-Teki (20:22:30)
Well, we're going to have to do something with AB=BE because, since inversions are not isometries, the information will be almost useless after we invert.

Teki-Teki (20:24:09)
Invert around B: Then we can use AB = BE.

markan (20:24:28)
this is another interesting point

markan (20:24:51)
if we don't invert around a, b, or e, the information that ab = be won't be preserved in any especially simple form

indianamath (20:22:13)
we need to invert all the points and circles around A so the 3 circles become 3 lines

markan (20:25:22)
but the latest observation is an important one

markan (20:25:27)
things become really nice if we invert about A

markan (20:25:32)
This problem is crying out for inversion around A, because there are three circles and two tangent lines going through it. Anytime you have multiple circles going through a point, it's worth considering inversion, because it will turn them into lines, which are often easier to work with.

markan (20:25:54)
so let me invert about A and draw the inverted points

markan (20:26:05)
i'll repost the original for comparison

markan (20:26:10)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/turkey98_5.asy.png

markan (20:26:13)


markan (20:26:18)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/turkey98_5_inverted_points.asy.png

markan (20:26:29)


markan (20:26:36)
ok

markan (20:26:38)
now....

markan (20:26:45)
we had 5 objects (lines or circles) going through A.

markan (20:26:54)
three circles and two lines

PI-Dimension (20:26:41)
woah...

markan (20:27:05)
In terms of A and primed points, what circles/lines do these five objects invert to?

markan (20:27:25)
many people are making good observations already

markan (20:27:36)
let's systematically go through the objects though:

markan (20:27:42)
what does circle ACD turn into?

markan (20:28:19)
(i'll draw them in once we know what they are)

kevinlang (20:28:06)
line D'C' ??

markan (20:28:30)
yeah

markan (20:28:46)
remember, circles through the center of inversion become lines

markan (20:28:54)
how about circle ABD?

indianamath (20:29:09)
line B'D'

not_trig (20:29:09)
line through B', D'

Karth (20:29:10)
line B'D'

Ivan Zhang (20:29:10)
line B'D'

markan (20:29:16)
line FAC?

Teki-Teki (20:29:24)
Line F'AC'

indianamath (20:29:25)
line F'C'

Karth (20:29:29)
stays the same

not_trig (20:29:30)
F'C'

markan (20:29:40)
line ABE?

tjhance_2 (20:29:47)
stays the same

not_trig (20:29:49)
same

Karth (20:29:50)
stays the same - B'E'

markan (20:30:06)
and finally, circle FADE?

not_trig (20:30:15)
line through F', D', E'

kevinlang (20:30:23)
F'E'D'

markan (20:30:33)
yep

markan (20:30:46)
so we just proved that these three points are collinear

not_trig (20:30:39)
(they all inverted to lines...)

markan (20:30:54)
convenient!

markan (20:30:56)
to summarize:

markan (20:31:02)
circle ACD --> line C'D'
circle ABD --> line B'D'
line FAC --> line F'AC'
line ABE --> line AE'B'
circle FADE --> line F'E'D'

markan (20:31:18)
ok, i'll draw in the lines now

markan (20:31:26)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/turkey98_5_inverted.asy.png

markan (20:31:28)


darkprince (20:31:51)
so lines NOT through the center become circles, but if not, then they stay the same? sorry, this is the first time

markan (20:32:17)
that's correct

markan (20:32:23)
lines through the center of inversion stay the same

markan (20:32:38)
alright, now let's start noticing things about our inverted diagram

markan (20:32:41)
what do you see?

tjhance_2 (20:32:14)
AC' || B'D' since AC and circle ADB where tangent, similarily AB' || C'D'

not_trig (20:32:31)
since FC is tangent to circle ABD, F'C' || B'D'; similarly, AB' || C'D'

Karth (20:33:01)
please say AB'D'C' is a parallellogram...

kevinlang (20:33:05)
AB' looks parallel to C'D'

markan (20:33:23)
yup

markan (20:33:43)
we've got two pairs of parallel lines, which gives us a parallelogram

markan (20:34:01)
(this is because the two pairs each came from two tangent objects in the original figure)

markan (20:34:27)
and since they're lines, the only way they can now be tangent is if they meet at infinity

markan (20:34:32)
(in other words they must be parallel)

PI-Dimension (20:32:56)
ABDC is a parallelogram, and E is the midpoint of A'B', so AF'=AC'

naitixuy (20:32:56)
a parallelogram and B'E'=E'A'

PI-Dimension (20:34:01)
E' is the midpoint of AB', thus proving AF'=AC'

markan (20:34:58)
how do we know E' is the midpoint here?

Teki-Teki (20:35:26)
Because AE = 2 AB, so AE' = AB'/2

indianamath (20:35:48)
AB=AE, so A'E'=E'B'

markan (20:36:11)
good

markan (20:37:08)
i see a lot of you trying to use similar triangles to get this result

markan (20:37:11)
i don't think that will work

markan (20:37:21)
you have to refer to the original diagram and use that information somehow

markan (20:37:29)
be careful not to assume things about the diagram that haven't been proven

markan (20:37:42)
in any case, many of you have observed that once we have a parallelogram

markan (20:37:52)
and once we know E' is the midpoint of AB', we have some congruent triangles

not_trig (20:34:41)
now, tri. D'E'B' is congruent to tri. F'E'A and we're done

andrewmath (20:36:38)
AF'E' is congruent to B'D'E'

markan (20:38:04)
AE'F' and B'E'D' are congruent, so AF'=B'D'=AC'. Thus AF=AC!

indianamath (20:37:35)
with that, Because A'B'=C'D"", so F'A'=1/2*F'C'

PI-Dimension (20:38:19)
yay =)

Ivan Zhang (20:38:32)
It becomes so easy!

markan (20:38:43)
Tangency was very important here, because it gave us our parallel lines. One of the nice things about inversion is that it preserves tangency.

markan (20:38:50)
alright

markan (20:38:55)
we've got one more part to work through

markan (20:39:06)
which is the famous Ptolemy's inequality

markan (20:39:11)
it says:

Karth (20:39:05)
when we use inversion on an olympiad, would it suffice to have a table and say BLAH inverts to BLAH?

markan (20:39:39)
most likely yes, as long as you're using basic rules like the ones we've mentioned

markan (20:39:49)
ptolemy:

markan (20:39:54)
Given a quadrilateral ABCD which does not intersect itself,

markan (20:40:02)


ThAzN1 (20:40:11)
(flipped)

markan (20:40:27)
oops

markan (20:40:30)
thanks

markan (20:40:35)


markan (20:40:53)
with equality exactly when ABCD is cyclic.

markan (20:41:29)
does anyone have any ideas how inversion might help with this?

markan (20:42:03)
(just curious....there's no reason to see the answer, because itcomes out of nowhere)

PI-Dimension (20:41:49)
we are trying to prove it right?

markan (20:42:10)
yes

Teki-Teki (20:41:49)
Well, since equality occurs when ABCD is cyclic, we might want to consider inverting about the circumcircle of ABC because then we have a special case when D lies on said circle.

not_trig (20:41:53)
lengths are tricky with inversion...

kevinlang (20:42:01)
invert with center at the intersection of diagonals

kshweh_2 (20:42:14)
make a circle through three points and invert through one of them

darkprince (20:41:58)
are we trying to prove the inequality using inversion?

markan (20:42:39)
yes, we're trying to prove the inequality using inversion

markan (20:43:00)
nobody's guessed the trick yet, though those are good ideas :)

markan (20:43:21)
it turns out we can invert about one of the points of the quadrilateral _itself_

markan (20:43:26)
and things become simple

markan (20:43:44)
we make the inspired guess to invert around one of the vertices of the quadrilateral (say B), and then apply the triangle inequality to triangle A'D'C'.

markan (20:43:49)
here's a picture

markan (20:44:00)
//s3.amazonaws.com/classroom.artofproblemsolving.com/Community/MJImages/inversion_mathjam/ptolemy.asy

markan (20:44:02)


markan (20:44:16)
what does the triangle inequality tell us?

kevinlang (20:44:37)
each side is less than the sum of the other two sides

Teki-Teki (20:44:42)
Well, we have A'C' < A'D' + D'C' as one of the inequalities.

ColbertCo (20:44:42)
A`D` + D`C` > A`C`

not_trig (20:44:43)
A'D' + D'C > A'C' (if it's not degenerate =P)

Karth (20:44:44)
A'C' < A'D' + C'D'

PI-Dimension (20:44:47)
A'C'<A'D'+D'C'

markan (20:44:55)


markan (20:45:11)
now we're going to write each term here using only lengths of the original quadrilateral

markan (20:45:19)
let's start with A'C'

indianamath (20:45:19)
why less than or equal to? it should aways be less than

markan (20:45:52)
if D' were on the line between A' and C', we'd have equality

not_trig (20:45:56)
well, BAC is similar to BC'A'

markan (20:46:18)
good

markan (20:46:28)
where can we go with this to get A'C'?

Teki-Teki (20:46:22)
Well we have A'C'/BC' = CA/AB and we can rewrite BC' = r^2/BC

tjhance_2 (20:46:32)
A'C' = AC * BA' / BC

markan (20:46:46)
Using similar triangles BAC and BC'A', we can write

markan (20:46:48)


markan (20:46:56)
which means that

markan (20:47:01)


markan (20:47:11)
Transforming the other terms in the triangle inequality in the same manner and cancelling r^2, we get

markan (20:47:14)
....what?

markan (20:48:15)
you're guessing we get ptolemy, which is right of course :)

markan (20:48:17)
but before that.....

Teki-Teki (20:48:17)
AC/AB*CB < AD/AB*DB + DC/DB*CB

markan (20:48:33)
yeah

markan (20:48:38)
the other terms behave like the first one

Teki-Teki (20:48:40)
Should be less than or equal to - sorry.

markan (20:49:01)
we change X'Y' to XY*r^2 / XB*YB for whatever X' and Y' we have

markan (20:49:03)
so the result is

markan (20:49:19)


Teki-Teki (20:49:01)
and then we multiply through by AB*CB*DB

Karth (20:49:06)
multiply it all by AB * BC * BD

markan (20:49:26)
Now we cross multiply and get

markan (20:49:31)


markan (20:49:33)
as desired.

markan (20:49:43)
now let's talk about when equality occurs

markan (20:49:50)
equality holds if and only if D' is collinear with, and between, A' and C'.

markan (20:50:15)
Let's suppose equality does hold. Then, by the above statement, the inverse of the line containing A',D', and C' would be what?

indianamath (20:50:13)
which means the quadrilater ABCD is cyclic

Teki-Teki (20:50:20)
Which means that D must be on a circle through A,B,C

not_trig (20:50:26)
CIRCLE

indianamath (20:50:31)
CIRCLE!!!!

markan (20:50:45)
a circle through B which would also have to hit A, D, and C in that order. Hence ABCD would be cyclic.

markan (20:51:02)
Conversely, if ABCD were cyclic in that order, inverting the circle would yield a line containing A', D', and C' in that order, and equality would hold.

tjhance_2 (20:50:02)
If ABCD is cyclic then A', C', and D' are collinear, so equality occurs in the triangle inequality

markan (20:51:24)
And we're done!

markan (20:51:36)
This proof is rather magical. I don't know of any particular reason why one would try inversion on this problem, if one didn't already know it would work out. So don't worry if it wasn't your intuition either.

markan (20:51:51)
that's all we have planned for today

not_trig (20:51:52)
is there a purely euclidean wAY?

markan (20:52:20)
i believe so

markan (20:52:25)
not sure what it is though

Teki-Teki (20:52:29)
Yes - use law of sines and areas and whatnot on the point of intersection.

Karth (20:51:48)
is there a good source for inversive geometry problems?

markan (20:52:58)
richard suggests the berkeley math circle archives

ThAzN1 (20:53:03)
there's a book by Prasolov as well

markan (20:53:03)
(google for it)

ThAzN1 (20:53:12)
http://students.imsa.edu/~tliu/Math/

not_trig (20:52:59)
geometry revisited!

Teki-Teki (20:53:00)
Geometry Revisited by Coxeter and Greitzer

ThAzN1 (20:53:19)
that has a lot of problems with solutions

markan (20:53:32)
before we go let me also mention one more fact about inversion

markan (20:53:37)
which is that it preserves angles

markan (20:53:44)
so if you start with two lines at angle alpha to each other

markan (20:53:46)
and then invert them

markan (20:53:58)
you'll get two circles whose tangents at the point of intersection are angle alpha from each other

markan (20:54:08)
(you can try to prove this on your own :))

markan (20:54:25)
i'm going to open the room up to discussion

markan (20:54:29)
feel free to ask more questions

PI-Dimension (20:54:41)
[img id=em-12]

markan (20:54:46)
thanks for coming everyone!