Bicentric quadrilateral

yetti

Bicentric quadrilateral

**Posted:** Mon Jan 10, 2005 10:41 am

_________________
Carthage must be destroyed.

yetti

Nobody bites into it, so here is the solution:

Let $P, Q, R, S$ b the tangency points of the incircle $(I)$ with the quadrilateral sides $AB, BC, CD, DA$ . There are numerous proofs that the incenters $I_P, I_Q, I_R, I_S$ of the triangles $\triangle ABX, \triangle BCX, \triangle CDX, \triangle DAX$ and the excenters $E_P, E_Q, E_R, E_S$ of these triangles against their common vertex $X$ are concyclic iff the quadrilateral $ABCD$ is tangential. Let $I_A, I_B, I_C, I_D$ be the incenters of the triangles $\triangle DAB, \triangle ABC, \triangle BCD, \triangle CDA$ and $E_A, E_B, E_C, E_D$ the excenters of these triangles against the vertices $A, B, C, D$ . It is also fairly well known that in a tangential quadrilateral $ABCD$ , the lines $I_AI_B, I_PI_Q, I_RI_S$ concur at a point $K_I$ on the diagonal $AC$ and the lines $I_BI_C, I_QI_R, I_SI_P$ concur at a point $L_I$ on the diagonal $BD$ . Similarly, the lines $E_AE_B, E_PE_Q, E_RE_S$ concur at a point $K_E$ on the diagonal $AC$ and the lines $E_BE_C, E_QE_R, E_SE_P$ concur at a point $L_E$ on the diagonal $BD$ . In addition, since the incircles $(I_A), (I_C)$ and the excircles $(E_A), (E_C)$ touch each other on the diagonal $BD$ , the lines $I_AI_C$ and $E_AE_C$ are perpendicular to this diagonal. Similarly, the the incircles $(I_B), (I_D)$ and the excircles $(E_B, (E_D)$ touch each other on the diagonal $AC$ and the lines $I_BI_D$ and $E_BE_D$ are therefore perpendicular to this diagonal. It is also well known that the lines $I_AI_C$ , $I_BI_D$ intesect at the circumcenter $O_I$ of the cyclic quadrilateral $I_PI_QI_RI_S$ and similarly, the lines $E_AE_C$ , $E_BE_D$ intesect at the circumcenter $O_E$ of the cyclic quadrilateral $E_PE_QE_RE_S$ .

Again, there are numerous proofs that in a cyclic quadrilateral $ABCD$ , the incenters $I_A, I_B, I_C, I_D$ and the excenters $E_A, E_B, E_C, E_D$ form rectangles and as such, they are also concyclic. The rectangle diagonals $I_AI_C, I_BI_D$ intersect at the circumcenter $O_I$ of the rectangle $I_AI_BI_CI_D$ and the rectangle diagonals $E_AE_C, E_BE_D$ intersect at its circumcenter $O_E$ of the rectangle $E_AE_BE_CE_D$ . It follows that in a bicentric quadrilateral $ABCD$ , the circumcircle of the cyclic quadrilateral $I_PI_QI_RI_S$ and the circumcircle of the rectangle $I_AI_CI_BI_D$ have the common circumcenter $O_I$ . Likewise, the circumcircle of the cyclic quadrilateral $E_PE_QE_RE_S$ and the circumcircle of the rectangle $E_AE_BE_CE_D$ have the common circumcenter $O_E$ . The whole point is to show that the first pair of circumcircles has the same radius and the second pair of circumcircles also has the same radius. One way to do this is to prove the incenters $I_P, I_Q, I_R, I_S$ and the excenters $E_P, E_Q, E_R, E_S$ to be concyclic in a bicentric quadrilateral $ABCD$ in a special way, which yields similarity of these to quadrilaterals with the cyclic quadrilateral $PQRS$ of the tangency points.

Step 1: The opposite sides $PQ, RS$ and the opposite sides $QR, SP$ of the cyclic quadrilateral $PQRS$ meet at points $K, L$ on the diagonals $AC, BD$ and the diagonals $PR, QS$ of $PQRS$ meet at the same point $X$ as the diagonals $AC, BD$ of $ABCD$ . The proof is by the repeated use of Pacsal's theorem for the 2 diagonals and 2 tangents of the cyclic quadrilateral $PQRS$ , showing the 4 points $A, X, C, K$ and the 4 points $B, X, D, L$ to be collinear. Since the vertices $A, B, C, D$ are the poles of the lines $SP, PQ, QR, RS$ , the points $K, L$ are poles of the lines $BD, AC$ .

Step 2: In an inversion in the incircle $(I)$ , the vertices $A, B, C, D$ of the bicentric quadrilateral $ABCD$ are carried into the midpoints $A', B', C', D'$ of the sides $SP, PQ, QR, RS$ of $PQRS$ , thus forming a parallelogram with the sides $B'C' \parallel D'A' \parallel PR$ and $A'B' \parallel C'D' \parallel QS$ parallel to the diagonals $PR, QS$ of $PQRS$ . Since this parallelogram is inscribed in the inverted circumcircle $(O)$ of the bicentric quadrilateral $ABCD$ , it is a rectangle and consequently, the diagonals $PR \perp QS$ are perpendicular to each other.

Step 3: Let $E, G$ be the intersections of the line $KX \equiv AC$ with the opposite sides $SP, QR$ and $F, H$ the intersections of the line $LX \equiv BD$ with the opposite sides $PQ, RS$ , forming another quadrilateral $EFGH$ with the diagonals $EG \equiv, AC, FH \equiv BD$ intersecting at the point $X$ . In a polarity transformation WRT to an arbitrary circle $(X)$ centered at the diagonal intersection $X$ , the quadrilaterals $PQRS$ and $EFGH$ are carried into parallelograms $(p, q, r, s) \equiv P'Q'R'S$ and $(e, f, g, h) \equiv E'F'G'H'$ , where $p \equiv P'Q', q \equiv Q'R', r \equiv R'S', s \equiv S'P'$ and $e \equiv E'F', f \equiv F'G', g \equiv G'H', h \equiv H'E'$ . Since the diagonals $PR \perp QS$ are perpendicular to each other, the paralleogram $P'Q'R'S'$ is a rectangle. Since $EFGH$ is inscribed in $PQRS$ , the rectangle $P'Q'R'S'$ is inscribed in the parallelogram $E'F'G'H'$ . Since $K \equiv PQ \cap RS$ , $L \equiv QR \cap RS$ , the lines $k = Q'S', l = P'R'$ are polars of the points $K, L$ WRT $(X)$ . Since the points $K, E, G$ and the points $L, F, H$ are collinear with the center of the polar circle $(X)$ , their polars $k \equiv Q'S' \parallel e \equiv E'F' \parallel g \equiv G'H'$ and $l \equiv P'R' \parallel f = F'G' \parallel h \equiv H'E'$ are parallel. The rectangle diagonals $l \equiv P'R', k \equiv Q'S'$ are thus the middle lines of the parallelogram $E'F'G'H'$ and conversely, the parallelogram diagonals $n \equiv E'G', m = F'H'$ are the middle of the rectangle $P'Q'R'S'$ . Hence, the parallelogram $E'F'G'H'$ has perpendicular diagonals $E'G' \perpendicular F'H'$ and it is a rhombus. The opposite sides $EF, GH$ and $FG, HE$ meet at the poles $M, N$ of the rhombus diagonals $m \equiv F'H', n = E'G'$ . Since these rhombus diagonals are parallel to the rectangle opposite sides $m \equiv F'H' \parallel q \equiv Q'R' \parallel s \equiv S'P'$ , $n \equiv E'G' \parallel p \equiv P'Q' \parallel r \equiv R'S'$ , the points $M, Q, S$ and the points $N, P, R$ are collinear with the center of the polar circle $(X)$ . Hence, the diagonals $QS, PR$ pass through the points $M, N$ . The rectangle diagonals $k = Q'S', l = P'R'$ are perpendicular to the lines $KX, LX$ , identical with the diagonal lines $EG, FH$ , connecting their poles $L, K$ with the center of the polar circle $(X)$ . Similarly, the rhombus diagonals $n = E'G', m = F'H'$ are perpendicular to the lines $NX, MX$ , identical with the diagonal lines $PR, QS$ , connecting their poles $N, M$ with the center of the polar circle $(X)$ . Because of the symmetry of the transformed configuration, the rhombus diagonals are bisectors of the angles between the rectangle diagonals. But this also means that the diagonals of the quadrilateral $PQRS$ are bisectors of the angles between the diagonals of the quadrilateral $EFGH$ . Since the diagonal lines $EG, FH$ are identical with the diagonal lines $AC, BD$ and since the diagonals of the quadrilaterals $I_PI_QI_RI_S$ , $E_PE_QE_RE_S$ are bisectors of the angles between the diagonals $AC, BD$ by definition, the diagonal lines of the quadrilaterals $PQRS$ and $I_PI_QI_RI_S$ , $E_PE_QE_RE_S$ are identical.

Step 4: The triangles $\triangle AXS, \triangle AXP$ have the side $AX$ is common and the sides $AS = AP$ equal, being the tangent lengths to the incircle $(I)$ from the vertex $A$ . The sides $XS, XP$ are therefore internally cut by the internal bisectors of the angles $\measuredangle AXS, \measuredangle AXP$ in the same ratio. Using the same argument for the triangle pairs $(\triangle BXP, \triangle BXQ)$ , $(\triangle CXQ, \triangle CXR)$ , $(\triangle DXR, \triangle DXS)$ ,

$\frac{XI_P}{PI_P} = \frac{XI_Q}{QI_Q} = \frac{XI_R}{RI_R} = \frac{XI_S}{SI_S}$

$\eta_I = \frac{XI_P}{XP} = \frac{XI_Q}{XQ} = \frac{XI_R}{XR} = \frac{XI_S}{XS} < 1$

Accordingly, the quadrilateral $I_PI_QI_RI_S$ is centrally similar to the cyclic quadrilateral $PQRS$ with homothety center $X$ and coefficient $0 < \eta_I < 1$ . The circumcircle $(O_I)$ of the quadrilateral $I_PI_QI_RI_S$ is centrally similar to the circumcircle $(I)$ of the quadrilateral $PQRS$ and the poles $K_I, L_I$ of the diagonal lines $BD, AC$ WRT the circle $(O_I)$ are centrally similar to their poles $K, L$ WRT the circle $(I)$ with the same homothety center and coefficient. The center $O_I$ of the circumcircle $(O_I)$ is then the intersection of normals dropped from the poles $K_I, L_I$ to their polars $BD, AC$ . Using the fact that an external angle bisector cuts the opposite side of a triangle externally in the ratio of the two adjacent sides, we can similarly show that the triangle excenters quadrilateral $E_PE_QE_RE_S$ is centrally similar to the cyclic quadrilateral $PQRS$ with homothety center $X$ and coefficient $\eta_E > 1$ and that the center $O_E$ of its circumcircle $(O_E)$ is the intersection of normals dropped from the poles $K_E, L_E$ to their polars $BD, AC$ .

Step 5: Let $(O_I, r_I)$ , $(O_E, r_E)$ be the circumcircles of the rectangles $I_AI_BI_CI_D$ and $E_AE_BE_CE_D$ , with the same centers $O_I, O_E$ as the circumcircles of the cyclic quadrilaterals $I_PI_QI_RI_S$ and $E_PE_QE_RE_S$ , but possibly different radii. The centers $I_A, I_C, O_I$ of the circles $(I_A), (I_C), (O_I, r_I)$ are collinear. Let $r_A, r_C, r_I$ be the radii of these circles and assume that $r_A > r_C$ Since the circles $(I_A), (I_C)$ touch each other on the diagonal $BD$ and since the circle $(O_I)$ passes through their centers $I_A, I_C$ , the radius $r_I$ is equal to

$r_I = \frac{r_A + r_C}{2}$

The common external tangents of the circles $(I_A), (I_C)$ meet at the point $K_I$ , their external homothety center. Accordingly,

$\frac{K_II_C}{K_II_A} = \frac{r_C}{r_A}$

$K_II_A = K_II_C + 2r_I = K_II_A \frac{r_C}{r_A} + r_A + r_C$

$K_II_A = \frac{r_A + r_C}{r_A - r_C}\ r_A$ . and . $K_II_C = \frac{r_A + r_C}{r_A - r_C}\ r_C$

$K_IO_I = K_II_C + r_I = \frac{r_A + r_C}{r_A - r_C}\ r_C + \frac{r_A + r_C}{2} = \frac{r_A + r_C}{r_A - r_C}\ r_I$

$\frac{K_IO_I}{K_II_A} = \frac{r_I}{r_A}$ . and . $\frac{K_IO_I}{K_II_C} = \frac{r_I}{r_C}$

This implies that the point $K_I$ is the external homothety center of the circle pairs $(I_A), (O_I, r_I)$ and $(I_C), (O_I, r_I)$ as well; the 3 circles $(I_A), (I_C), (O_I, r_I)$ have a single pair of the common external tangents meeting at the point $K_I$ . In exactly the same way, we can show that the 3 circles $(I_B), (I_D), (O_I, r_I)$ also have a single pair of the common external tangents meeting at the point $L_I$ . If the circles $(I_A), (I_C)$ or $(I_B), (I_D)$ happen to have the same radii (i.e., the quadrilateral $ABCD$ is a bicentric kite), the circle $(O_I, r_I)$ also has the same radius and the corresponding homothety center $K_I$ or $L_I$ moves to infinity on the diagonal line $AC$ or $BD$ .

Step 6: Since the common external tangents of the circle pairs $(I_A), (I)$ and $(I_A), (O_I, r_I)$ meet at the points $A, K_I$ , the external homothety center of the circles $(I), (O_I, r_I)$ is on the diagonal $AC$ . Since the common external tangents of the circle pairs $(I_B), (I)$ and $(I_B), (O_I, r_I)$ meet at the points $B, L_I$ , the external homothety center of the circles $(I), (O_I, r_I)$ is also on the diagonal $BD$ . This homothety center is then identical with the intersection $X$ of these two diagonals. But the external homothety center of the circles $(I), (O_I)$ is also at the point $X$ . In addition, the circumcircles $(O_I), (O_I, r_I)$ of the cyclic quadrilateral $I_PI_QI_RI_S$ and of the rectangle $I_AI_BI_CI_D$ have the common center $O_I$ . Consequently, these two circles are identical, $(O_I) \equiv (O_I, r_I)$ . In exactly the same way, we can show that the circumcircle $(O_E)$ of the cyclic quadrilateral $E_PE_QE_RE_S$ is identical with the circumcircle $(O_E, r_E)$ of the rectangle $E_AE_BE_CE_D$ .

It remains to be shown that if the quadrilateral $ABCD$ is not bicentric, the incenters $I_P, I_Q, I_R, I_S, I_A, I_B, I_C, I_D$ and the excenters $E_P, E_Q, E_R, E_S, E_A, E_B, E_C, E_D$ are not concyclic. The proof that the incenters $I_P, I_Q, I_R, I_S$ and the excenters $E_P, E_Q, E_R, E_S$ cannot be concyclic in a quadrilateral $ABCD$ , which is not tangential, is well known. Perhaps, it is necessary to show that the quadrilaterals $I_AI_BI_CI_D$ , $E_AE_BE_CE_D$ cannot be cyclic unless the quadrilateral $ABCD$ is also cyclic (and then they are necessarily rectangles). The proofs I have seen demonstrated only that $I_AI_BI_CI_D$ , $E_AE_BE_CE_D$ are rectangles in a cyclic quadrilateral, but not that they cannot be cyclic in some other quadrilateral.

Step 7: Assume that the quadrilateral $ABCD$ is not cyclic, for example, the sum of the opposite angles $\measuredangle BAD + \measuredangle BCD < 180^o$ . The vertex $A$ then lies outside of the circumcircle $(O_C)$ of the triangle $\triangle BCD$ and the vertex $C$ lies outside of the circumcircle $(O_A)$ of the triangle $\triangle BAD$ . Let the vertex $A$ move along the diagonal $AC$ into the intersection $A'$ of this diagonal with the circle $(O_C)$ . The incenter $I_C$ is fixed and the incenters $I_B, I_D$ move toward the vertex $C$ along the bisectors of the fixed angles $\measuredangle BCA, \measuredangle DCA$ into new positions $I_B', I_D'$ . Since the quadrilateral $A'BCD$ is now cyclic, the angle $\measuredangle I_B'I_CI_D' = 90^o$ is necessarily right. But since the incenter $I_C$ lies on the common bisector of the angles $\measuredangle BCD, \measuredangle I_BCI_D$ inside of the angle $\measuredangle I_BCI_D = \frac{BCD}{2} < 90^o$ , the original incenters $I_B, I_D$ both lie inside of the right angle $\measuredangle I_B'I_CI_D'$ and consequently, the original angle $\measuredangle I_BI_CI_D < 90^o$ . For exactly the same reasons, the angle $\measuredangle I_BI_AI_D < 90^o$ as well. The sum of the opposite angles of the quadrilateral $I_AI_BI_CI_D$ is therefore less than $180^o$ , which means that this quadrilateral cannot be cyclic.

**Posted:** Sun Mar 06, 2005 1:46 pm

_________________
Carthage must be destroyed.

Beat · Yang-Mills Theory Offline Joined: 12 Nov 2005 Posts: 662

#3

tangential, excentre, concyclic. Can somebody explain me these words, please.
The problem looks hard and interesting and i cannot translete a bit of it. Thank you

**Posted:** Mon Dec 05, 2005 9:29 am

Virgil Nicula

#4

Another property

**Posted:** Mon Dec 05, 2005 5:59 pm

mecrazywong

#5

Re: bicentric quadrilateral

**Posted:** Mon Dec 05, 2005 6:24 pm

yetti

#6

Re: another property

_________________
Carthage must be destroyed.

yetti

#7

**Posted:** Tue Dec 06, 2005 5:44 am

_________________
Carthage must be destroyed.