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darij
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Jan-11-04, 08:15 PM (EST)
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"Synthetic proof of CB's Conjecture"
 
  

Synthetic proof of
Christopher Bradley's Conjecture

Version 2: 8 renewed.

Many thanks to Rafi for initiating a very
animate and interesting discussion thread in the Hyacinthos
newsgroup, starting with the message #8910 "Quadrilateral
problem."; many thanks also to Jean-Pierre Ehrmann, Charles
Worrall, Alexey Zaslavsky, Peter Scales, Marcello Tarquini,
Alexander Bogomolny, Juan Carlos Salazar, Eric Danneels and
Nikolaos Dergiades for taking part and contributing many
interesting details to this thread. By making use of your
contributions, I have succeeded in finding a synthetic proof of
the result which is known as "Christopher Bradley's
Conjecture" in this forum. However, let me apologize for not
having read a great part of the discussion done before I
subscribed to the CTK Exchange, and hence, I may just repeat what
was already said. Let me also excuse for my notations (they are
completely different from those used in this discussion).

1. Triangle lemmata

Consider a triangle ABC and a point P on its
side AC. Let the incircles of triangles PAB and PBC have centers
X and Y, respectively; the incircle of triangle ABC touches AC at
U. Then:


Theorem 1. The points P
and U lie on the circle with diameter XY.



Theorem 2. The point U
lies on the internal common tangent of the incircles of
triangles PAB and PBC different from the line BP.


For Theorem 1, Juan Carlos Salazar refers to
the American Mathematical Monthly, but he doesn't remember the
exact citation.

Proofs. Here is a restated proof of
Theorem 2 by Rafi:

Let U be the point where the internal common
tangent of the incircles centered at X and Y different from BP
meets AC. We will show that the incircle of triangle ABC touches
AC at U.

Let the internal common tangent of the
incircles of triangles PAB and PBC touch these circles at X' and
Y', respectively. Then, since the internal common tangents to two
circles have equal length, we get X'Y' = X"Y", where
X" and Y" are the points where the incircles of
triangles PAB and PBC touch BP.

Now, this yields


PX" - PY" = X"Y" = X'Y'
= UY' - UX'.


If J and K are the points where the incircles
of triangles PAB and PBC touch AC, then the well-known fact that
the two tangent segments from a point to a circle are equal in
length yields that PX" = PJ, PY" = PK, UX' = UJ, UY' =
UK; hence,


PJ - PK = UK - UJ, i. e.

KJ - 2 PK = KJ - 2 UJ,


and PK = UJ. Thus,


AU = AJ + UJ = AJ + PK

= (AB + AP - BP)/2 + (BP + CP - BC)/2

= (AB + AP + CP - BC)/2

= (AB + AC - BC)/2,


and U is the point of tangency of the incircle
of triangle ABC with the side AC. This proves Theorem 2.

Now, the points X and Y lie on the two angle
bisectors of the angle formed by the lines AC and UX'Y' (because
both of these lines touch the two incircles!). Hence, the angle
XUY is 90, and U lies on the circle with diameter XY. The same
reasoning (using the line BP instead of UX'Y') shows that P lies
on the circle with diameter XY, too. This proves Theorem 1.

2. The configuration
of Christopher Bradley

A circumscribed quadrilateral will
mean a quadrilateral with an incircle.

Now let ABCD be a circumscribed quadrilateral
with the incenter O. The diagonals AC and BD meet at P. Let X, Y,
Z, W be the incenters of triangles PAB, PBC, PCD, PDA,
respectively, and (X), (Y), (Z), (W) be the respective incircles.
Then, Christopher Bradley's conjecture states:


Theorem 3. The points X,
Y, Z, W lie on one circle.


This fact was proven, together with its
converse, in <1> and <2>. Here, we don't worry about the converse
and seek a synthetic proof.

3. Incircles and
exsimilicenters

Call (O) the incircle of quadrilateral ABCD,
its center being O. Let also (A'), (B'), (C'), (D') be the
incircles of triangles DAB, ABC, BCD, CDA, where A', B', C', D'
are their respective centers.

Since incenters lie on angle bisectors, the
following triples of points are collinear:


(A; O; A'), (B; O; B'), (C; O; C'), (D; O;
D');

(A; B'; X), (B; C'; Y), (C; D'; Z), (D; A'; W);

(A; D'; W), (B; A'; X), (C; B'; Y), (D; C'; Z).


We will use the abbreviation exsimilicenter
for the external center of similtude (= external homothetic
center) of two circles.

The 1st Monge theorem states that the
pairwise exsimilicenters of three circles are collinear.

Let R be the exsimilicenter of the circles (A')
and (C'). Of course, R lies on A'C'. Now, the exsimilicenter of
(O) and (A') is A, and the exsimilicenter of (O) and (C') is C.
Hence, the 1st Monge theorem yields that R lies on the line AC.

The lines AX and CY meet at B'; the lines XA'
and YC' meet at B; the lines AA' and CC' meet at O. Since B, O,
B' are collinear, we can apply the Desargues theorem to triangles
AXA' and CYC', and infer that the lines AC, XY and A'C' concur.
Since AC and A'C' meet at R, we see that R lies on XY.

Now, what have we proven? R lies on the line XY
- joining the centers of the circles (X) and (Y) - and on their
common tangent AC. Hence, R is the exsimilicenter of (X) and (Y).
Similarly, R is the exsimilicenter of (Z) and (W) and lies on the
line ZW. Moreover, there will be a similar point Q on BD with
analogous properties. We have obtained:


Theorem 4. The
exsimilicenter R of the circles (A') and (C') is
simultaneously the exsimilicenter of (X) and (Y) and the
exsimilicenter of (Z) and (W). It lies on the lines AC, A'C',
XY and ZW.



The exsimilicenter Q of the circles (B')
and (D') is simultaneously the exsimilicenter of (Y) and (Z)
and the exsimilicenter of (W) and (X). It lies on the lines
BD, B'D', YZ and WX.


Finally, we state a trivial fact which hardly
merits the name "theorem":


Theorem 5. The lines XZ
and YW are the angle bisectors of the angles APB = CPD and
DPA = BPC and pass through P.


This is because incenters of triangles lie on
angle bisectors.

4. U and T; Proof of
Theorem 3

We will make use of a well-known result about
circumscribed quadrilaterals:


Theorem 6. The incircles
(A') and (C') of triangles DAB and BCD touch BD at one point
T.


The incircles (B') and (D') of triangles
ABC and CDA touch AC at one point U.


Doubters can perform an easy computation of
lengths (making use of the fact that AB + CD = BC + DA in the
circumscribed quadrilateral ABCD).

Of course, the point T lies on A'C'; moreover,
the points T and R divide the segment A'C' harmonically, since T
is the internal center of similtude of (A') and (C') and R is
their external center of similtude.

Now, we can apply Theorem 1 to triangle ABC and
the point P on its side AC. Consequently, the points P and U lie
on the circle with diameter XY. Hence, the points X, Y, P, U are
concylic, and RX * RY = RP * RU. Similarly, RZ * RW = RP * RU,
and thus RX * RY = RZ * RW, and the points X, Y, Z, W are
concyclic. This proves Theorem 3. The last part of this elegant
proof is due to Juan Carlos Salazar.

5. The center of the
circle XYZW

Now regard the center of the circle through X,
Y, Z, W. How could it be else:


Theorem 7. The center O'
of the circle through X, Y, Z, W is the point of intersection
of the lines A'C' and B'D'.


This was found by Charles Worrall. Here is,
again, my proof, based on a lemma found by Rafi:


Theorem 8. The triangle
PQR is autopolar with respect to the circle through X, Y, Z,
W.


Proof. The diagonal triangle
of a quadrilateral is the triangle formed by the intersection of
the diagonals and the two intersections of opposite sidelines.

Now, the triangle PQR is the diagonal triangle
of the cyclic quadrilateral XYZW; hence, it is autopolar with
respect to the circumcircle of this cyclic quadrilateral. (In
fact, it is well-known that the diagonal triangle of a cyclic
quadrilateral is autopolar with respect to the circumcircle.)
Theorem 8 is established.

Hence, the line PR - i. e., the line AC - is
the polar of Q with respect to the circle through X, Y, Z, W.
Consequently, the lines AC and O'Q are perpendicular (O' is the
center of the circle through X, Y, Z, W). On the other hand, the
line B'D' is perpendicular to AC, too (since the circles (B') and
(D') touch AC at one point U), and it also passes through Q.
Hence, the line O'Q coincides with the line B'D', i. e. the point
O' lies on B'D'. Similarly, O' lies on A'C'. This proves Theorem
7.

6. Another property
of A'C' and B'D'

There is more to say about the lines A'C' and
B'D':


Theorem 9. The line B'D'
is the reflection of the line BD in the angle bisector of the
angle formed by the lines WX and YZ.


The line A'C' is the reflection of the line
AC in the angle bisector of the angle formed by the lines XY
and ZW.


Instead of proving this directly, we will show
a stronger result:


Theorem 10. The points P
and O' are isogonal conjugates with respect to any of the
four triangles XQY, YRZ, ZQW, WRX.


Proof. Since O' is the circumcenter of
triangle ZWX, we have angle O'WZ = 90 - angle WXZ. On the other
hand, the right-angled triangle WPX gives 90 - angle WXZ = 90
- angle WXP = angle XWP. Hence, angle O'WZ = angle XWP, and the
line WO' is the reflection of the line WP in the angle bisector
of the angle QWZ. Similarly, the line ZO' is the reflection of
the line ZP in the angle bisector of the angle QZW. In other
words, the reflections of two cevians of P in triangle ZQW in the
corresponding angle bisectors pass through O'. Hence, O' is the
isogonal conjugate of P with respect to triangle ZQW. Similarly
for the triangles XQY, YRZ, WRX.

7. Four internal
common tangents

The geometry related to our configuration
doesn't seem to have an end. If we consider our triangle ABC with
the point P on AC again, and apply Theorem 2, then we see that
the point U lies on the internal common tangent of the incircles
of triangles PAB and PBC different from the line BP. In our
terminology:


Theorem 11. The point U
lies on the internal common tangent of the circles (X) and
(Y) different from the line BD.


Similarly we find three other internal
common tangents of (Y) and (Z), of (Z) and (W), and of (W)
and (X).


Let the internal common tangent of the circles
(X) and (Y) different from BD be denoted by b', and the three
other ones by c', d', a' (in the natural order). Then, U lies on
b' and d', and T lies on a' and c'. Now, I conjecture that


Theorem 12. The lines a',
b', c', d' are equidistant from O'. In other words, there
exists a circle centered at O' and touching the lines a', b',
c', d'.


Moreover, Peter Scales suspects the following:


Theorem 13. The point of
intersection of the lines a' and b' lies on O'X.


Similarly for b' and c', c' and d', d' and
a'.


Proofs of Theorem 12 and 13 would be greatly
appreciated.

8. Alexey Zaslavsky's
tangential quadrangle

In <2>, the editors mention an additional
result proved by Alexey Zaslavsky:


Theorem 14. The tangents
to the circle through X, Y, Z, W at X, Y, Z, W form a
quadrilateral with two vertices on AC and two vertices on BD.


Zaslavsky's proof is analytic; here is a
synthetic proof based upon an idea of Rafi:

Call (O') the circle through X, Y, Z, W. The
tangents to (O') at W and at X meet at a point A" which is
the pole of the line WX with respect to (O'). Similarly, the
tangents to (O') at X and at Y meet at the pole B" of XY,
the tangents to (O') at Y and at Z meet at the pole C" of
YZ, and the tangents to (O') at Z and at W meet at the pole
D" of ZW.

For establishing Theorem 14, we have to show
that A" and C" lie on AC, and B" and D" lie
on BD.

Since XY passes through R, the pole B" of
XY lies on the polar of R, i. e. on the line PQ (since triangle
PQR is autopolar), i. e. on the line BD. Similarly, D" lies
on BD, and A" and C" on AC. Theorem 14 is proven.

Rafi has lately found out more about these
points A", B", C", D":


Theorem 14a. The
quadrilateral A"B"C"D" is inscribed; its
circumcenter lies on the line PO'.


Proof. If A1, B1,
C1, D1 are the midpoints of the segments
WX, XY, YZ, ZW, then, since A", B", C", D"
are the poles of the lines WX, XY, YZ, ZW with respect to (O'),
the points A", B", C", D" are the images of A1,
B1, C1, D1 in the inversion with
respect to (O').

Now, since A1, B1, C1,
D1 are the midpoints of the sides of quadrilateral
XYZW, the quadrilateral A1B1C1D1
is a parallelogram with sidelines parallel to the diagonals of
XYZW (Varignon parallelogram), i. e. a rectangle (since the
diagonals XZ and YW of XYZW are perpendicular to each other). The
center of this rectangle A1B1C1D1
is the centroid of the quadrilateral XYZW.

Now, remember that in any inscribed
quadrilateral with perpendicular diagonals, the intersection of
the diagonals is the anticenter (the point of concurrence of the
so-called maltitudes, the perpendiculars from the
midpoints of the sides to the opposite sides). Hence, in our
inscribed quadrilateral XYZW with perpendicular diagonals, the
point P is the anticenter. Now, the anticenter of a cyclic
quadrilateral is the reflection of the circumcenter in the
centroid; hence, the centroid of the quadrilateral is the
midpoint between the anticenter and the circumcenter. In our
case, the centroid of the quadrilateral XYZW is the midpoint
between the anticenter P and the circumcenter O'.

In other words, the midpoint O1 of
PO' is the center of the rectangle A1B1C1D1.
But any rectangle has a circumcircle centered at the rectangle's
center; hence, the points A1, B1, C1,
D1 lie on a circle centered at O1.

Now I will use the following fact: Inversions
map circles to circles, but (in general) centers not to centers.
However, the center of the image circle always lies on the line
joining the center of the original circle with the center of
inversion.

Applied to the inversion with respect to the
circle (O'), and remembering that A", B", C",
D" are the inversive images of A1, B1,
C1, D1, our result that the points A1,
B1, C1, D1 lie on a circle
centered at O1 implies that the points A",
B", C", D" lie on one circle with a center on the
line O1O', i. e. on the line PO'. This proves Theorem
14a.

9. Metric relations

Last, but not least, there is a series of
metric relations in the configuration. If r, x, y, z, w, a', b',
c', d' are the radii of the circles (O), (X), (Y), (Z), (W),
(A'), (B'), (C'), (D'), respectively, then <3> states that


Theorem 15. We have 1/x +
1/z = 1/y + 1/w.


The two proofs given in <3> are not really easy
and little delightful; the following one I have found seems to be
rather nice:

We will establish more than Theorem 15; in
fact, we will show that


Theorem 16. We have 1/x -
1/a' = 1/y - 1/c'.


This entails 1/x - 1/y = 1/a' - 1/c', and
similarly, 1/w - 1/z = 1/a' - 1/c'. Therefore, 1/x - 1/y = 1/w -
1/z, proving Theorem 15.

Proof of Theorem 16. The points X, Y,
R being collinear, the Menelaos theorem in triangle BC'A' entails


-1 = BX / XA' * A'R / RC' * C'Y / YB (with
signed lengths).


Now, it is easy to prove that BX / XA' = x /
(a' - x) and C'Y / YB = (c' - y) / y; further, A'R / R'C = - a' /
c' (since R is the exsimilicenter of (A') and (C')). Hence,


-1 = x / (a' - x) * (- a' / c') * (c' - y)
/ y,


and


x a' (c' - y) = (a' - x) c' y.


This can be easily seen equivalent to 1/x -
1/a' = 1/y - 1/c', proving our Theorem 15.

10. Note about
excenters

What is true for an incenter also holds for
every excenter - this is the extraversion principle for triangles
(applicable to equations, of course, not to inequalities).
However, whether the incenters of triangles PAB, PBC, PCD, PDA
can be replaced by the excenters corresponding to P in our
configuration without the need of replacing the circumscribed
quadrilateral ABCD by a quadrilateral with an excircle
is
not clear to me.

Anyway, the analogue of Theorem 3 for excenters
is valid:


Theorem 17. The excenters
of triangles PAB, PBC, PCD, PDA corresponding to P lie on one
circle.


This was established by Nikolaos Dergiades.

References

<1> Toshio Seimiya, Peter Y. Woo, Solution
of Problem 2338
, Crux Mathematicorum 4/25 (1999) pp.
243-245.

See also https://www-students.biola.edu/~woopy/math/cr2338.htm

<2> I. Vaynshtejn, Reshenie zadachi M1524,
Kvant 3/1996 pp. 25-26.

https://kvant.mccme.ru/1996/03/resheniya_zadachnika_kvanta_ma.htm";

<3> I. Vaynshtejn, N. Vasiljev, V. Senderov, Reshenie
zadachi M1495
, Kvant 6/1995 pp. 27-28.

https://kvant.mccme.ru/1995/06/resheniya_zadachnika_kvanta_ma.htm

Darij Grinberg

11 Jan 2004



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  Subject     Author     Message Date     ID  
Synthetic proof of CB's Conjecture darij Jan-11-04 TOP
  Autopolar ? Bractals Feb-09-04 1
     RE: Autopolar ? darij Feb-10-04 2
         RE: Autopolar ? Bractals Feb-10-04 3
  Three more cases Bractals Feb-13-04 4
     RE: Three more cases Bractals Mar-19-04 5
         RE: Three more cases darij Apr-06-04 6

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Bractals
Member since Jun-9-03
Feb-09-04, 01:28 PM (EST)
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1. "Autopolar ?"
In response to message #0
 
   Hi darij,

Do you know a place on the internet where I can find a proof (in english) that the diagonal triangle of a cyclic quadrilateral is autopolar?

Thanks,

Bractals


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darij
Member since Jan-9-04
Feb-10-04, 08:40 AM (EST)
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2. "RE: Autopolar ?"
In response to message #1
 
   Dear Bractals,

You wrote:

>> Do you know a place on the internet where I can find a
>> proof (in english) that the diagonal triangle of a
>> cyclic quadrilateral is autopolar?

Well, it is just necessary to show that if we have four
points A, B, C, D on a circle k, and K is the
intersection of AB and CD, L is the intersection of BC
and DA, and M is the intersection of AC and BD, then
triangle KLM is autopolar with respect to k.

Theorem 1 of Vladimir's "Quadrilateral inversions" post
in this forum (College Math), applied to the cyclic
quadrilateral ABCD, yields that L lies on the polar of
K with respect to k. The same fact, applied to the
cyclic quadrilateral ABDC, shows that M lies on the
polar of K. Hence, the line LM is the polar of K with
respect to k. Similarly, MK is the polar of L, and KL
is the polar of M, so that the triangle KLM is indeed
autopolar.

Sincerely,
Darij Grinberg


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Bractals
Member since Jun-9-03
Feb-10-04, 08:39 PM (EST)
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3. "RE: Autopolar ?"
In response to message #2
 
   Hi Darij,

Thanks for the pointer to Vladimir's Theorem #1.

Bractals


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Bractals
Member since Jun-9-03
Feb-13-04, 01:34 PM (EST)
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4. "Three more cases"
In response to message #0
 
   Hi Darij,

I have followed your proof through paragraph 5 for Case I (see below) the configuration of Christopher Bradley. Using the following notation:


i(XYZ) denotes the incenter of triangle XYZ
e(XYZ) denotes the excenter of triangle XYZ
on the bisector of angle XYZ
e(YZX) denotes the excenter of triangle XYZ
on the bisector of angle YZX
e(ZXY) denotes the excenter of triangle XYZ
on the bisector of angle ZXY

The proof also holds for cases II - IV (replacing 'in' with 'ex' where required).

CASE
I II III IV

A' i(DAB) e(DAB) i(DAB) e(DAB)
B' i(ABC) e(ABC) e(ABC) i(ABC)
C' i(BCD) e(BCD) i(BCD) e(BCD)
D' i(CDA) e(CDA) e(CDA) i(CDA)

X i(PAB) e(BPA) e(ABP) e(PAB)
Y i(PBC) e(CPB) e(PBC) e(BCP)
Z i(PCD) e(DPC) e(CDP) e(PCD)
W i(PDA) e(APD) e(PDA) e(DAP)


Note: The four cases exhaust the sixteen "centers" of the triangles PAB, PBC, PCD, and PDA.

Bractals


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Bractals
Member since Jun-9-03
Mar-19-04, 03:25 PM (EST)
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5. "RE: Three more cases"
In response to message #4
 
   Hi Darij,

Here is a
proof for Cases I - IV in the previous post. If you have a chance to look it over, you might tell me why I did not need Monge's theorem. Is it only because the results of Monge's theorem are needed in your proof from paragraph 6 and beyond?

Thanks,

Bractals


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darij
Member since Jan-9-04
Apr-06-04, 03:32 PM (EST)
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6. "RE: Three more cases"
In response to message #5
 
   Dear Bractals,

You wrote:

>> Here is a proof for Cases I - IV in the previous
>> post. If you have a chance to look it over, you
>> might tell me why I did not need Monge's theorem.

Thanks - your proof is indeed a bit shorter than
mine, since I use Monge and Desargues, while you used
Desargues only. Once we know that the lines AC, A'C',
XY (your A*B*) and ZW (your C*D*) concur, it is
trivial that their point of concurrence is the
exsimilicenter of the circles (A') and (C'), the one
of the circles (X) and (Y), and the one of (Z) and
(W), since AC is a common tangent of these pairs of
circles and A'C', XY and ZW are their respective
central lines. Hence, we don't need Monge at all!

Sincerely,
Darij Grinberg


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