| Transcript
for the Math
Jam "Inversion"
on Aug 1. |
| Math Jam hosted by markan
(Sean Markan ). |
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)
http://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)
http://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)
http://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)
http://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)
http://artofproblemsolving.com/Community/MJImages/inversion_mathjam/inversion.asy.png
markan (19:51:05)
markan (19:51:21)
markan (19:51:28)
http://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)
http://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)
http://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)
http://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)
http://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)
http://artofproblemsolving.com/Community/MJImages/inversion_mathjam/turkey98_5.asy.png
markan (20:26:13)
markan (20:26:18)
http://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)
http://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)
http://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!
