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A Mathematical Droodle
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A conjecture by Christopher Bradley has been posted to one of the old CTK Exchange forums by Nathan Bowler. Mr. Bradley claimed that, given an inscriptable convex quadrilateral, the incenters of the four triangles formed by its diagonals are concyclic.
Many people tried to solve the problem, mostly unsuccessfully until it has migrated to the Hyacinthos geometry list where additional information and the problem's history have surfaced. Eventually, Darij Grinberg came up with a proof that he posted to the old CTK Exchange. The proof is presented below.
Synthetic proof of Christopher Bradley's Conjecture
by Darij Grinberg
11 Jan 2004
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.
§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, the n, 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,
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.
Toshio Seimiya, Peter Y. Woo, Solution of Problem 2338, Crux Mathematicorum 4/25 (1999) pp. 243-245.
I. Vaynshtejn, Reshenie zadachi M1524, Kvant 3/1996 pp. 25-26.
I. Vaynshtejn, N. Vasiljev, V. Senderov, Reshenie zadachi M1495, Kvant 6/1995 pp. 27-28.