geometry incenter rectangle circumcircle parallelogram rhombus symmetry

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 $O_E$. [color=blue] 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.[/color]  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.

[color=red][b]Step 1:[/b][/color] 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 $BD, AC$.

[color=red][b]Step 2:[/b][/color] 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, [color=blue]the diagonals $PR \perp QS$ are perpendicular to each other[/color].

[color=red][b]Step 3:[/b][/color] 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 $E'G' \perp endicular F'H'$ 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, [color=blue]the diagonal lines of the quadrilaterals $PQRS$ and $I_PI_QI_RI_S$, $E_PE_QE_RE_S$ are identical[/color].

[color=red][b]Step 4:[/b][/color] 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, [color=blue]the quadrilateral $I_PI_QI_RI_S$ is centrally similar to the cyclic quadrilateral $PQRS$ with homothety center $X$ and coefficient $0 < \eta_I < 1$[/color]. 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 $BD, AC$.

[color=red][b]Step 5:[/b][/color] 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,

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

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

$K_II_A = \frac{r_A + r_C}{r_A - r_C}\ r_A$ [color=white].[/color] and [color=white].[/color] $K_II_C = \frac{r_A + r_C}{r_A - r_C}\ r_C$

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

$\frac{K_IO_I}{K_II_A} = \frac{r_I}{r_A}$ [color=white].[/color] and [color=white].[/color] $\frac{K_IO_I}{K_II_C} = \frac{r_I}{r_C}$

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; [color=blue]the 3 circles $(I_A), (I_C), (O_I, r_I)$ have a single pair of the common external tangents meeting at the point $K_I$[/color]. In exactly the same way, we can show that [color=blue]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 $L_I$[/color]. 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 $BD$.

[color=red][b]Step 6:[/b][/color] 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, [color=blue]these two circles are identical, $(O_I) \equiv (O_I, r_I)$.[/color] 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.

[color=red][b]Step 7:[/b][/color] 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.

[quote="yetti"]A quadrilateral $ABCD$ with the diagonal intersection $X$ is tangential iff the incenters of the triangles $\triangle ABX, \triangle BCX, \triangle CDX, \triangle DAX$ are concyclic. If the quadrilateral $ABCD$ is cyclic, the incenters of the triangles $\triangle DAB, \triangle ABC, \triangle BCD, \triangle CDA$ make a rectangle, i.e., they are also concyclic. Prove that the quadrilateral $ABCD$ is bicentric iff the incenters of all 8 triangles are concyclic and the iff the excenters against the vertex $X$ for the first 4 triangles and the excenters against the vertices $A, B, C, D$ for the last 4 triangles are concyclic.[/quote]
The first two results are actually well known. As long as I remember, they appeared in Chinese TST, APMO, Crux, and much more...
Anyway, nice results for the sufficient and necessary condition of a quadrilateral being bicentric ;)

[quote="Virgil Nicula"]Let $ABCD$ be a convex tangential (circumscribed) quadrilateral to the circle $C=C(O,R)$. I note the intersection $X\in AC\cap BD$ and the incircles $C_a=(I_a,r_a)$, $C_b=C(I_b,r_b)$, $C_c=C(I_c,r_c)$, $C_d=C(I_d,r_d)$ of the triangles $XAB$, $XBC$, $XCD$, $XDA$. We know, from the[i] Yetti's posted problem[/i], that the quadrilateral $I_aI_bI_cI_d$ is cyclic. [u]Prove[/u] that the tangents in the vertexes of the quadrilateral $I_aI_bI_cI_d$ to the its circumcircle form a quadrilateral with two vertexis on the line $AC$ and with another vertexis on the line $BD$ ([b]Alexei Zaslawsky[/b]).[/quote]

I have seen a simple proof (by Darij) using poles and polars, but I prefer my original proof of the fact that the incenters $I_1, I_2, I_3, I_4$ are concyclic, because it is so illuminating. It also yields the above proposition as a side effect. Only later I figured out the simple angle chase leading to the cyclic quadrilateral condition $\angle I_4I_1I_2 + \angle I_2I_3I_4 = 180^\circ$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21768]Quadrilateral (incenters form a cyclic quadrilateral)[/url]).

Let ABCD is an arbitrary tangential quadrilateral with the diagonal intersection X and $(I_1), (I_2), (I_3), (I_4)$ the incircles of the triangles  $\triangle XAB, \triangle XBC, \triangle XCD, \triangle XDA$. Assume that the incircle $(I_1)$ is the largest one, let $A_0$ be the intersection of the other tangent to $(I_1)$ parallel to the diagonal BD (other than BD) with the diagonal AC and let $B_0$ be the intersection of the other tangent to $(I_1)$ parallel to to the diagonal AC (other than AC) with the diagonal BD. If A' is an arbitrary point on the diagonal ray $[A_0A$, then we can draw tangents from A' to the circles $(I_1), (I_4)$ (other than AC) intersecting BD at points B', D'. Then we can draw tangents from B', D' to the circles $(I_2), (I_3)$, respectively, and these tangents will intersect at a point C' on the diagonal AC, forming a tangential quadrilateral A'B'C'D' with the same diagonal lines $A'C' \equiv AC, B'D' \equiv BD$ and with the same incircles $(I_1), (I_2), (I_3), (I_4)$ of the triangles $\triangle XA'B', \triangle XB'C', \triangle XC'D', \triangle XD'A'$ as the original quadrilateral ABCD.

Starting with the point $A' \equiv A_0$, the vertex B' is at infinity on BD (on the ray $[B_0B$). If we move A' to infinity on AC (on the ray $[A_0A$), then B' moves to $B' \equiv B_0$. Assuming that B' is at infinity, the angle $\angle B'C'A' > \angle B'D'A'$, while assuming that A' is at infinity, the opposite is true - the angle $\angle B'C'A' < \angle B'D'A'$. Therefore, a position of the point $A' \in [A_0A$ exists, such that $\angle B'C'A' = \angle B'D'A'$. But then these 2 angles subtend a common arc on the chord A'B', which means that the tangential quadrilateral A'B'C'D' is also cyclic, i.e., bicentric. But if A'B'C'D' is bicentric, its contact quadrilateral E'F'G'H' is centrally similar to the fixed quadrilateral $I_1I_2I_3I_4$ with the homothety center X (the diagonal intersection). Since E'F'G'H' is cyclic, $I_1I_2I_3I_4$ is also cyclic. Moreover, the tangential quadrilateral $PQRS$ of the cyclic quadrilateral $I_1I_2I_3I_4$ is centrally similar to the bicentric quadrilateral A'B'C'D' with the same homothety center X, hence, $P, R \in A'C' \equiv AC$ and $Q, S \in B'D' \equiv BD$.

All that remains to be shown is the claim that the quadrilateral A'B'C'D' has the same diagonals as the quadrilateral ABCD (and remains tangential) when moving the starting point A' in any way on the ray $[A_0A$ - this is quite long. Since ABCD is tangential with an incircle (I), the incircles $(J_1), (J_3)$ of the triangles $\triangle s ABC, \triangle CDA$ touch at a point $M \in AC$ and the incircles $(J_2), (J_4)$ of the triangles $\triangle BCD, \triangle DAB$ touch at a point $N \in BD$. Using the collinearity of 3 external homothety centers of the circle triples $(J_2), (J_4), (I)$, the external homothety center K of $(J_2), (J_4)$ lies on AC and similarly, the external homothety center L of $(J_1), (J_3)$ lies on BD. Similarly, BK is the external homothety axis of the both circle triples $(J_2), (J_4), (I_1)$ and $(J_2), (J_4), (I_2)$, hence, the external homothety center of $(I_1), (I_2)$ lies on BK and since AC is their common external tangent, this homothety center is the intersection $K \equiv BK \cap AC$. Likewise, the external homothety center of the circles $(I_3), (I_4)$ is $K \in AC$ and the extenal homothety center of both $(I_2), (I_3)$ and $(I_4), (I_1)$ is $L \in BD$. The center lines $J_2J_4 \perp BD$ passes through K and the center line $J_1J_3$ passes through L. Moreover, XM is equal to the common internal tangent lengths of $(I_2), (I_3)$ and $(I_4), (I_1)$, while XN is equal to the common internal tangent lengths of $(I_1), (I_2)$ and $(I_3), (I_4)$ (the proof is virtually the same as in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=62239]A very nice problem: constant sum[/url], where XM = 0).

Now erase the tangential quadrilateral ABCD with its incircle (I), as well as the triangle incircles $(J_1), (J_2), (J_3), (J_4)$, but leave the diagonal lines $p \equiv AC, q \equiv BD$ with the homothety centers K, L, the normals $KM \perp q, LN \perp p$ and the circles $(I_1), (I_2), (I_3), (I_4)$. Let $A' \in [A_0A$ be arbitrary, let the tangents from A' to the circles $(I_1), (I_4)$ other than p meet the line q at B' D', respectively, and let the tangents from B', D' to the circles $(I_2), (I_3)$ other than q meet the line p at $C_1, C_2$, respectively. Let $(J_4')$ be the incircle of the triangle $\triangle D'A'B'$ touching the line q at the point M (because XM equal to the common internal tangent lengths of $(I_4), (I_1)$ and $(I_2), (I_3)$ is fixed). Let $(J_1'), (J_3')$ be the incircles of the triangles $\triangle A'B'C_1, \triangle C_2D'A'$. The circlles $(J_1'), (J_3')$ touch each other at the point $N \in p$ (again, because XN equal to the common internal tangent lengths of $(I_1), (I_2)$ and $(I_3, (I_4)$ is fixed). Let $(J_2')$ be a circle touching the line q at the point M and also touching the line $B'C_1$. Similarly, let $(H_2')$ be a circle touching the line q at the same point M and also touching the line $D'C_2$.

The external homothety axis of both the circle triples $(I_1), (I_2), (J_4')$ and $(I_1), (I_2), (J_2')$ is the line B'K, hence, the external homothety center of the circles $(J_4'), (J_2')$ is the intersection $K \equiv B'K \cap KM$ of this homothety axis and their center line $KM \equiv J_2'J_4'$. Likewise, the external homothety axis of both the circle triples $(I_3), (I_4), (J_4')$ and $(I_3), (I_4), (H_2')$ is the line D'K, hence, the external homothety center of the circles $(J_4'), (H_2')$ is the intersection $K \equiv D'K \cap KM$ of this homothety axis and their center line $KM \equiv H_2'J_4'$. But this implies that [color=blue]the circles $(J_2') \equiv (H_2')$ are identical[/color], both touching the circle $(J_4')$ at the same point M and both having the same external homothety center K with this circle. Keep the label $(J_2')$ for this circle. The common external tangents $q, B'C_1$ of the circles $(I_2), (J_2')$ meet at the point $B' \in q$, the common external tangents $q, D'C_2$ of the circles $(I_3), (J_2')$ meet at the point $D' \in q$ and the common internal tangent length of the circles $(I_2), (I_3)$ on the line p is equal to XM. It follows that the common external tangents $B'C_1, D'C_2$ meet on the line p, i.e., the points $C_1 \equiv C_2 \equiv C'$ are identical and the quadrilateral A'B'C'D has the same diagonal lines p, q as the original tangential quadrilateral ABCD. Since the triangle incircles $(J_2'), (J_4')$ touch each other at the point $M \in q \equiv B'D'$ and the triangle incircles $(J_1'), (J_3')$ at the point $N \in p \equiv A'C'$, the new quadrilateral A'B'C'D' remains tangential.

[quote="Beat"]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[/quote]

Concyclic points are points lying on a single circle. Every 3 points lie on a single circle, hence, the term is used only if 4 or more points lie on a single circle.

Tangential quadrilateral is a quadrilateral (convex of concave), whose side lines touch a single circle. This single circle can be centered inside the quadrilateral (incircle with an incenter) or outside the quadrilateral (excircle with an excenter). Tangential quadrilateral is also called inscriptable. Compare with a triangle, which always has one incircle and 3 excircles.

Any other terms ?

Yetti


Putnam
Harvard
college
MIT
college contests

Hey,

So, what motivates you guys to prepare for the Putnam?

Thanks,

-APSStudent

The prospect of getting into a good grad school.

I don't think we do

nobody at harvard does at least

dudes at MIT do just 'cause they're crazy like that

Contests are fun, and doing well on them is more fun. We're in it because we like solving problems.

I actually dislike competitions.  I always found that the competitive aspect took away some of the fun.

Well, it is always nice to match yourself against both difficult problems and other strong people. If nothing else, it corrects your self-esteem a bit ;) I certainly liked math. competitions as a kid...


Quadrilateral $ ABCD$ is a parallelogram. If the measure of angle $ A$ is 62 degrees and the measure of angle $ ADB$ is 75 degrees, what is the measure of angle $ ADC$, in degrees?

Note that $ m\angle ADC\equal{}m\angle D\equal{}180^\circ\minus{}62^\circ\equal{}\boxed{118^\circ}$.

you don't need the information that angle ADB is 75 degrees


analytic geometry
geometry
linear algebra
matrix
real analysis
real analysis unsolved

Assume $ S \subset \mathbb{R}^2$, can we get the conclusion that:If under the rectangular coordinate $ S$ is Jordan measurable, and its' Jordan area $ \equal{}a$.

Consider coordinate transformation:$ \left\{\begin{matrix}
x'\equal{}\varphi\left(x \right ) \\
y'\equal{}\psi \left(y \right )
\end{matrix}\right.$

Is $ S$ Jordan measurable under the new coordinate? and has the same area? :roll:

Not unless $ \phi,\psi$ have special properties.
For example:  $ x/2, 2y$ yes.  But $ x, 2y$ no, this doubles the area.


[quote="chien than"]Prove that:$64xyz(x+2t)(y+2t)(z+2t) \leq 27(t^{2}+xy+yz+xz)^{3}$
where x;y;z;t are non-negative[/quote]

I think it's false because if we take $x=y=z$ we get $4x(x+2t) \leq 3(t^{2}+3x^{2}) \Longrightarrow 3t^{2}+5x^{2}\geq 8xt$
Which is not true for $\frac{5x}{3}>t>x$

[b]cefer wrote[/b]:
[quote]Which is not true for $\frac{5x}{3}>t>x$[/quote]
Why?

AFAIK this is true
$27xyz(x+2t)(y+2t)(z+2t)\le (3t^{2}+2(xy+yz+zx))^{3}$
 :wink:


Let $a,b,c$ be $0<a\leqq 2,0<b\leqq 2,0<c\leqq 2$.

Prove that $\frac{\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}}{\sqrt{3}}+\frac{9abc}{ab+bc+ca}\leqq 4\sqrt{3}$.

\[
\begin{gathered}
  \frac{{\sum {\sqrt {4 - a^2 } } }}
{{\sqrt 3 }} + \frac{{9abc}}
{{ab + bc + ca}} \hfill \\
   \leqslant \frac{{\sum {\sqrt {4 - a^2 } } }}
{{\sqrt 3 }} + a + b + c \hfill \\
   = \frac{{\sum {\sqrt {4 - a^2 } }  + \sqrt 3 a}}
{{\sqrt 3 }} \hfill \\
   \leqslant \frac{{\sum {\sqrt {\left( {4 - a^2  + a^2 } \right)\left( {1 + 3} \right)} } }}
{{\sqrt 3 }} \hfill \\
   = 4\sqrt 3  \hfill \\
\end{gathered}
\]

Nice solution,siuhochung.

Here is the back ground of making this problem.

$\sqrt{4-x^2}\leqq -\sqrt{3}x+4\ (0<x\leqq 2)$


geometry
parallelogram
trapezoid
rectangle
quadratics
power of a point
radical axis

[b]School round in Slovakia 2004/2005[/b]
1) Find number of infinite arithmetic progressions of integers $a_1,a_2,a_3,\ldots$ such that there exists two distinct integers $i,j$ for which $1\leq i,j \leq 10$ such that $a_i=1, a_j=2005$.

2) In parallelogram ABCD, |AB|>|BC|  let K,L,M,N be points of touch of incircles of triangles ACD, BCD, ABC, ABD (respectively) to one of diagonals (so K is the point where incircle of ACD touches AC). Prove that KLMN is rectangle.

3) For what positive integers $k$ has $k(k-2)\leq(k+\frac{1}{k})x\leq k^2(k+3)$ exactly $(k+1)^2$ soluitions in integers?

[b]District round in Slovakia 2004/2005[/b]
4) For positive reals $a,b,s$ with product 1 prove and find when it attains equality:
$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq \frac{3}{4}$

5) Solve this system of equalities in integers:
$x(y+z+1)=y^2+z^2-5$
$y(z+x+1)=z^2+x^2-5$
$z(x+y+1)=x^2+y^2-5$

6) In a plane there's triangle KLM with |KM|=|LM|. Let circle k be tangent to line KM in point K and let l be tangent to line LM in point L. Let k and l be externally tangent in point T. Find locus of all points T.

7) Find all pairs of positive integers $a,b$ such that $a+b$ has last digit 3, $|a-b|$ is prime and $ab$ is square.

[b]National round in Slovakia 2004/2005[/b]
8) Let's have arithmetic progressions of real numbers $\{x_1\}_{i=1}^\infty$, $\{y_1\}_{i=1}^\infty$ with $x_1=y_1$ and for some $k>1$ holds:
$x_{k-1}y_{k-1}=42\qquad x_ky_k=30\qquad x_{k+1}y_{k+1}=16$
Find all such progressions for which $k$ is biggest (or in another words, find the biggest possible $k$ and then all progressions that fullfil the conditions).

9) Find for which $m$  there are exactly $2^{15}$ subsets $X$ of $\{1,2,\ldots,47\}$ for which holds: $m$ is smallest term of $X$ and for each $x\in X$ is $x+m\in X$ or $x+m>47$.

10) In trapezoid $ABCD$ is $AB$ parallel to $CD$ and let $E$ be the midpoint of $BC$. Suppose we can inscribe circle in $ABED$ and also into $AECD$. Then if $|AB|=a,|BC|=b,|CD|=c,|DA|=d$ prove:
\[a+c=\frac{b}{3}+d\qquad \frac{1}{a}+\frac{1}{c}=\frac{3}{b}\]

11) In a plane there is a triangle $AKL$. Let $ABCD$ be rectangle (can be also square) such that $K$ lies on side $BC$ and $L$ on side $CD$. Find locus of intersections of diagonals of $ABCD$.

12) Prove that for all real numbers $p,q,r,s$ with $q\neq -1, s\neq -1$ holds:
Quadratic equations
\[x^2+px+q=0\qquad x^2+rx+s=0\]
have in real numbers one common root and the product of the other two is 1 iff for $p,q,r,s$ holds:
\[pr=(q+1)(s+1)\qquad p(q+1)s=r(s+1)q\]

13) Find out whether it's possible to arrange numbers $1,2,\ldots,15$ in such way that you cannot color them using four colors so that numbers of one color makes monoticaly increasing or decreasing progression.

[b]Discussions about them:[/b]
[hide]
1)[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=22456[/url]
2)[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=22455[/url]
3)[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=22454[/url]
5)[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24201]http://www.mathlinks.ro/Forum/viewtopic.php?t=24201[/url]
6) here  :D
7)[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=153806]http://www.mathlinks.ro/Forum/viewtopic.php?p=153806[/url]
10)[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=205697#p205697]www.mathlinks.ro/Forum/viewtopic.php?p=205697#p205697[/url]
[/hide]

please also post the problems in separate topics and post the links back in this post :) thanks!

I've done it. Now I see, that most of this download post has problems to download, but it isn't necessary if it's just three problems. Or is it? After Slovakia selection test I'll post it all in pdf.

Thanks. It's easier to have links to all the problems (because that's how we work on the forum) but have all the problems in the same page in case someone wants to print them and work on them at the same time.

The single common internal tangent of the circles k, l passing through their tangency point T is their radical axis. The tangent lengths from any point on the radical axis (and from such points only) to both circles are equal. Therefore, the radical axis passes through the point M and MT = MK = ML. If we insist that the circles k, l have an external tangency with each other, the locus of the tangency points T is the shorter circular arc KL of the circle (M) with radius MK = ML. If both external and internal tangencies of k, l are allowed, the radical axis may be the single common tangent (internal or external) of the circles k, l and the locus of the points T is the entire circle (M). When the tangency point T is diametrally opposite to K or to L, the circle k or l degenerates into a line (circle of infinite radius).

...

Your approach is fine, but I think this section is just for posting problems, not solutions...


anyone here from New Mexico?
I want to see if anyone else from the land of enchantment is on AoPs.

my friend's dad's cousin's son's cousin's friend's grandpa's divorced wive's son's daughter is from New Mexico!

lizz...

it's okay not to answer those kind of questions...

[quote="Valentin Vornicu"]I'm not sure if you're aware that there is a [url=http://www.mathlinks.ro/Forum/index.php?f=294]New Mexico local community[/url] on the site :) (it's not very active though ... :unsure:).[/quote]
really I never knew that :P  thank u!

I have the distinct pride of living in the state with [b]2nd[/b]  to worst education system (Mississippi just beats us out, though we gain on them every year).  Yeah, we should get some posts going on the local forum.


there was this problem, that went something like this:

if we start with the number 8^1989, then take the sum of its digits, the sum of the digits of the sum of the digits, and so on, we arrive at a single digit. what is that digit?

i went mod 9, saw that all of these numbers mod 9 have the same remainder as 8^1989, so the answer was 8. thats right, but when i went mod 3, i got 2. how do we know that 8 is right, but that 2 is not?

Ok, start out with the number 8^1989. This has remainder 8 when we divide it by 9, which you probably can check. You know that the divisibility test for 9 is to take the sum of the digits and see whether that value is a multiple of 9 or not. Well, when we take the sum of the digits of that number, the number is 8(mod 9). When we take the sum of the digits of the previous number to see what it's remainder is, we again get 8(mod 9). This will continue on and on, since (mod 9) the sum of the digits is invariant. This stops when we finally get down to our single digit, 8(mod 9).

But when you do this (mod 3), it ends up at 2(mod 3).

The reason is that with 2 mod 3, we wont' actually arrive at 5, since 5 is too small a number, and it's possible that we just stop at 8 and keep going 8, 8, 8, ... without decreasing, since there are no other digits.  While it may be true that it's 2 mod 3, this is inconclusive.  8 mod 9 is, because summing digits will always decrease the number until we arrive at a single-digit number, at which it all stops.  There is only one single-digit number  :equiv:  8 (mod 9)

Mod 3 doesn't give you the sum of the digits of a number.  Mod 3 tells you whether or not a number is divisible by 3.  It happens that the easiest way to do that is to sum the digits.

In other words (sum of digits) tells you everything you want to know about (mod 3).

But (mod 3) doesn't tell you very much information about (sum of digits).

(mod 9) does tell you everything you want to know about the sum of digits, if you go all the way down to a single digit.


A set of balls is given. Each ball is coloured red or blue, and there is at least one of each colour. Each ball weighs either 1 pound or 2 pounds, and there is at least one of each weight. Prove that there are two balls having different weights and different colours.

[hide]
There is a red ball. There is a blue ball. If they are of different weights, the problem is solved. If they have the same weight, there is another ball of the opposite weight. But this ball is red or blue, which means it is different from one of the original balls. So the problem is solved again.[/hide]


[quote="zhou_qiang"]"D lies on the same side of AB as C". What does this meant?[/quote]
Draw AB. If C is above AB then so is D. If C is below AB then so is D.

above and below...
i think more rigorously, you can say that
there does not exist a point E on AB such that angle CED=180


Inequality is reversed

$ x^8\plus{}y^8\plus{}z^8\equal{}3$

Show that $ x^7\plus{}y^7\plus{}z^7 \le 3$

$ (x^7\plus{}y^7\plus{}z^7)^8 \le 3(x^8\plus{}y^8\plus{}z^8)^7$

$ \sqrt[7]{\frac{x^7\plus{}y^7\plus{}z^7}{3}}\le\sqrt[8]{\frac{x^8\plus{}y^8\plus{}z^8}{3}}$

Which is power mean.


If I'm in 9th grade, and I want to make MOSP, what do I have to get on the USAMO?

Any ideas?? :huuh:

The cutoff has been 9 the last two years.

On another note, any news of an official date? I recall diophantient asking this a few months ago, and 'unofficial' dates were provided by AMC Director, but has anything been finalized?

[quote="CatalystOfNostalgia"]On another note, any news of an official date? I recall diophantient asking this a few months ago, and 'unofficial' dates were provided by AMC Director, but has anything been finalized?[/quote]

The website says June 10-July 3.

12 anybody, 12 non-seniors, 24ish freshmen

Is MOSP only for excactly 9th grade, or all 6th-8th?

[quote="distracted523"]Can 8th graders make the Red Group too? Or is it just freshmen?[/quote]
Red is for freshmen only I think, 8th graders must make it through blue :O My friend scored 10 last year in 8th grade so I know.

Red = Ninth Only
Black = Top 12
Blue = Top 12 non-seniors after that

[quote="distracted523"]Can 8th graders make the Red Group too? Or is it just freshmen?[/quote]

really really smart 8th graders can make black/blue MOP

like evan o'dorney whos like totally awesomeness (i think there have been like 5 or so in the last 6 or 7 years-dont mark my words though)

anyways, but Red is only 9th graders-or too many people.

Huh? the future isn't a mid schooler...
"or too many people" (I'm assuming that's what sounded wierd) I think means if there were more than just 9th graders, there would be too many people.

I was under the impression that MOP was something like 12/18/30 [b]last year[/b] instead of 12/12/24 for Black/Blue/Red, no? Any chance that will happen again?

I believe that it is always 12/18/30.

EDIT: The AMC website (assuming this section isn't out of date) says they try for 15-16 blue moppers (although they don't use the terminology blue mop)

[url]http://www.unl.edu/amc/a-activities/a6-mosp/factoidsmosp.shtml[/url]

[quote="samath"][quote="aspiringchesslete"][quote="the future"]
anyways, but Red is only 9th graders-or too many people.[/quote]

what

why are middle schoolers pretending they know things about MOP[/quote]
what
why are people with 9 posts pretending they know things about other aopsers?[/quote]

well excuse me

I have been to MOP

and he has clearly not

[quote="aspiringchesslete"][/quote][quote="samath"][quote="aspiringchesslete"][quote="the future"]
anyways, but Red is only 9th graders-or too many people.[/quote]

what

why are middle schoolers pretending they know things about MOP[/quote]
what
why are people with 9 posts pretending they know things about other aopsers?[/quote][quote="aspiringchesslete"]

well excuse me

I have been to MOP

and he has clearly not[/quote]

uhh what did i say that was wrong...oh well dont have to be so arrogant about it...
and yea ive never been to MOP, im obviously inferior to you in life (since you know I havent memorized the 100th digit of pi yet...)----or i could create another account....

... wow ... totally uncalled for there, chesslete. honestly

OK I apologize. Sorry for that, it didn't seem so bad when I posted it, but now it does look mean. I was getting defensive.


By AM-GM \[ A_1 = (x^2 + yz)(y^2 + zx)(z^2 + xy) \leq \frac{(2x^2 + y^2 + z^2)(x^2 + 2y^2 + z^2)(x^2 + y^2 + 2z^2)}{8}\]
Another AM-GM application yields that this expression does not exceed \[\frac{1}{8} \cdot \left(\frac{(2x^2 + y^2 + z^2) + (x^2 + 2y^2 + z^2) + (x^2 + y^2 + 2z^2)}{3}\right)^3 = \frac{8(x^2 + y^2 + z^2)^3}{27}\]
Finally by the Power Mean inequality the latter expression does not exceed \[8\left(\frac{x^2 + y^2 + z^2}{3}\right)^3 \leq 8\left(\frac{x^3 + y^3 + z^3}{3}\right)^2 = \frac{8}{9}(x^3 + y^3 + z^3)^2.\]

in my mind, those exercices like this one where you just have to compute with knowing inequalities are boring.
don't you think so ?

Yes they are boring, but your comment is as boring as the above exercise.

l apologize for my boring comment. after all, maybe l am the one that found these problems boring.


function
inequalities
algebra
functional equation

$ 1$.Find all functions $ f: R\rightarrow R$ such that for every $ x,y$ from real numbers we have:
$ f(xy)\equal{}yf(x)\plus{}xf(y)$
$ 2$.find all functions $ f: R\rightarrow R$ satisfying the following functional equation:
$ f(x)\plus{}(x\plus{}\frac {1}{2})f(1\minus{}x)\equal{}1$   (for every real number $ x$)

mamnoon misham age javabhashoono begid.

Fek konam soorate avallio kamel naneveshti   :P (shayad man eshtebah karde basham :!: )
[b]1[/b].$ f(0) \equal{} 0$ be rahati bedast miad.hala age do taraf moadela ro bar $ xy$ taghsim koni va tarif koni $ g(x) \equal{} \frac {f(x)}{x}$
darim : $ g(xy) \equal{} g(x) \plus{} g(y)$ .Pas $ g$ mitoone ye tabeye Logarithmi bashe ke fek nakonam manzooret in joor tabe ha bashe...

[b]2[/b].Kafie be jaye $ x$ bezari $ 1\minus{}x$ ke ye 2 moadele 2 majhool ba moadele ye asli tashkil bedi ke javabesh mishe :
$ f(x)\equal{}\frac {2}{1\minus{}2x}$ (tabe be ezaye $ x\equal{}\frac{1}{2}$ mishe $ \frac{1}{2}$ )

[quote="amparvardi"]mer30 agha mahan.
pco in soalaro kheili bahal hal mikone.#
hamin soalo bebin:
[url]http://www.mathlinks.ro/viewtopic.php?t=323393[/url][/quote]
Khahesh mikonam.
[b]pco[/b] ghool tarin adamie  :o  ke man too Algebra (be gheire Inequality) mishnasam.
man harmoghe bebinam pco online hastesh sai mikonam ye soale dorost hesabi azash beporsam.(chon midoonam avallin nafar javab mide!!) :wink:

[quote="irantst"]همیشه سوالای من رو جواب بده و دقیقا درست
یادم میاد یه بار از خودم یه معادله تابع دادم واقعا وحشتناک بود
حدودا 3 صفحه می شد جوابش و کاملا درست
از پرسیدم
55 ساله شه[/quote]

Eival,55 saleshe bad hanoozam miad inja,kheili bahale pas,
man fekr mikardam azam koochiktare!!!

سلام
من خودم کنجکاو شدم بدونم که چی کاره است و چند سالشه
مشاور امنیت کامپیوتره و 55 ساله شه
دلم می خاد که با پیگیری مباحثش من هم بتونم جبر خودم رو تقویت کنم


Jo adds up all the positive integers from 1 to 50. Kate does a similar thing with the first 50 positive integers; however, she first rounds every integer to its nearest multiple of 10 (rounding 5s up) and then adds the 50 values. What is the positive difference between Jo's sum and Kate's sum?

sum of jo= n(n+1)/2 = 50*51/2=1275
sum of kate = for no from 0to4 sum=0
                    for no. from 0to14(10 numbers) sum=10*10 (rounded to 10)
                    for no   from 15to34  sum=20*10
so sum till 44=10(10+20+30+40)=1000
              sum frm 45 to50 =6*50=300
            so total sum=1300
 and thus positive difference is 25

There is actually an easier way to solve this problem. When Kate rounds off the first number, her final answer is 1 less than Joe's answer. However, when she rounds off the 9th number, this inequality will be balanced again, so (as I will say) they [i]canceled out[/i]. When she continues, the 2nd and 8th will also cancel(do not worry about the 10's digit), and the 3rd and 7th, and the 4th and 6th. However, each time through each units digit, the five will be rounded up, adding five to the correct total. Since there are 5 cycles through each unit, (1's, 10's, 20's, 30's, and 40's) you will be adding an extra five [b]five times[/b], meaning that in total, Kate added [u][i][b]25[/b][/i][/u] to Joe's answer.


Given some mxn table, and some numbers in its fields.  You are allowed
to change the sign in one row or one column simultaneously.
Prove that You can obtain a table, with nonnegative sums over each row
and over each column.

If you have columns with negative sums, change the signs of those columns. The overall sum will increase. Do the same for rows with negative sum. Since there are finitely many sums (there are at most $2^{mn}$ ways of filling in the signs of the entries), the operation can't go on indefinitely, so we must stop at some point.

[quote="grobber"]Since there are finitely many sums (there are at most $2^{mn}$ ways of filling in the signs of the entries), the operation can't go on indefinitely, so we must stop at some point.[/quote]

How do you prove you can't get a loophole?

You seem correct but I don't get why it is that rigorous...

What sort of loophole might that be? What's unclear? If the operation is performed, then the sum increases strictly, and it cannot do so indefinitely. It is rigorous..