Denote $ABCD$ the tangential quadrilateral and $r_A, r_B, r_C, r_D$ radii of the circles $a, b, c, d$ inscribed in the corners $A, B, C, D$. Let $t_A, t_B, t_C, t_D$ be their tangent lengths from the vertices $A, B, C, D$ and $t_{AB}, t_{BC}, t_{CD}, t_{DA}$ the segment lengths of the common external tangents of the circle pairs $(a, b), (b, c), (c, d), (d, a)$: $AB = t_A + t_{AB} + t_B$ $BC = t_B + t_{BC} + t_C$ $CD = t_C + t_{CD} + t_D$ $DA = t_D + t_{DA} + t_A$ Since the quadrilateral is tangential, $AB + CD = BC + DA$ and $t_A + t_{AB} + t_B + t_C + t_{CD} + t_D = t_B + t_{BC} + t_C + t_D + t_{DA} + t_A$ $t_{AB} + t_{CD} = t_{BC} + t_{DA}$ Since the circle pairs $(a, b), (b, c), (c, d), (d, a)$ pairwise touch, $t_{AB} = 2 \sqrt{r_Ar_B}$ $t_{BC} = 2 \sqrt{r_Br_C}$ $t_{CD} = 2 \sqrt{r_Cr_D}$ $t_{DA} = 2 \sqrt{r_Dr_A}$ $\sqrt{r_Ar_B} + \sqrt{r_Cr_D} = \sqrt{r_Br_C} + \sqrt{r_Dr_A}$ Squaring this equation: $r_Ar_B + 2 \sqrt{r_Ar_Br_Cr_D} + r_Cr_D = r_Br_C + 2 \sqrt{r_Br_Cr_Dr_A} + r_Dr_A$ $r_Ar_B + r_Cr_D = r_Br_C + r_Dr_A$ $r_Ar_B - r_Br_C = r_Dr_A - r_Cr_D$ $r_B(r_A - r_C) = r_D(r_A - r_C)$ Either $r_A - r_C = 0$, i.e., $r_A = r_C$, or $r_A - r_C \neq 0$, but then $r_B = r_D$. It does not seem possible that such circles exist, unless the diagonals $AC, BD$ of the tangential quadrilateral $ABCD$ are perpendicular to each other, i.e., unless $ABCD$ is at least a kite, or unless at least two of the circles ($a, c$ or $b, d$) coincide with the quadrilateral incircle.

Yetti wrote : ........ It does not seem possible that such circles exist, unless the diagonals AC, BD of the tangential quadrilateral ABCD are perpendicular to each other, i.e., unless ABCD is at least a kite ... ...................... Why not start the solution here as a physicist (what you claim you are) and not as a mathematician (what you always wanted to be), without being carried away by algebra? Maj. Pestich

pestich wrote: Why not start the solution here as a physicist (what you claim you are) and not as a mathematician (what you always wanted to be), without being carried away by algebra? Maj. Pestich You call this being carried away with algebra, major?

The following is going to be a combination of Yetti's solution of the original problem (slightly simplified) with some combinatorial-geometrical ideas to answer positively Yetti's question. Theorem 1. A convex quadrilateral ABCD has an incircle. Four circles x, y, z, w are given such that the circle x touches the segments AD and AB, and so on cyclically, and such that the circles x and y are externally tangent to each other, and so on cyclically. Then, either the quadrilateral ABCD is a kite, or the circles x and z coincide, or the circles y and w coincide. Proof. We will require two lemmata: Lemma 2. If two circles with radii u and v touch each other externally, and one of the two external common tangents of these two circles touches them at the points P and Q, respectively, then $PQ=2\sqrt{uv}$. Lemma 2 was proven at http://www.mathlinks.ro/Forum/viewtopic.php?t=104489 . Lemma 3. Given a line g through a point A, some positive number r, and a circle k touching the line g at a point L. Then, there exists at most one circle t which is externally tangent to the circle k, has radius r, lies in the same halfplane with respect to the line g as the circle k and touches the line g at a point lying on the extension of the segment AL beyound L. We will call this circle t (when it exists) the (A, g, r, k)-circle. The main purpose of Lemma 3 is to show that this notation is well-defined, i. e. the (A, g, r, k)-circle is (if it exists) uniquely determined from A, g, r, k. (By the way, it always exists, but we don't need to know that.) Proof of Lemma 3. We prove Lemma 3 by contradiction: Assume that there are two different circles $t_{1}$ and $t_{2}$ which are both externally tangent to the circle k, have radius r, lie in the same halfplane with respect to the line g as the circle k and touch the line g at two points both lying on the extension of the segment AL beyound L. Let $T_{1}$ and $T_{2}$ be the centers of these two circles $t_{1}$ and $t_{2}$, and let $M_{1}$ and $M_{2}$ be the points where these two circles touch the line g. Both of these points $M_{1}$ and $M_{2}$ lie on the extension of the segment AL beyound L. The points $T_{1}$ and $T_{2}$ must be distinct (since, if they were equal, the circles $t_{1}$ and $t_{2}$ would be equal, having the same center $T_{1}=T_{2}$ and the same radius r). Let K be the center and $r_{k}$ the radius of the circle k. Since the circle $t_{1}$ with center $T_{1}$ and radius r is externally tangent to the circle k with center K and radius $r_{k}$, we have $KT_{1}=r_{k}+r$. Similarly, $KT_{2}=r_{k}+r$. Thus, $KT_{1}=KT_{2}$. Since the circles $t_{1}$, $t_{2}$, k have centers $T_{1}$, $T_{2}$, K and touch the line g at the points $M_{1}$, $M_{2}$, L, respectively, the lines $T_{1}M_{1}$, $T_{2}M_{2}$ and KL are all perpendicular to g, and since r is the radius of the circles $t_{1}$ and $t_{2}$ which pass through the points $M_{1}$ and $M_{2}$, respectively, we have $T_{1}M_{1}=r$ and $T_{2}M_{2}=r$. Since the circle $t_{1}$ with center $T_{1}$ and radius r touches the line g, the distance from the point $T_{1}$ to the line g equals r. Similarly, the distance from the point $T_{2}$ to the line g equals r. Thus, the points $T_{1}$ and $T_{2}$ are equidistant from the line g. Also, the points $T_{1}$ and $T_{2}$ lie in the same halfplane with respect to the line g (in fact, the circles $t_{1}$ and $t_{2}$ must lie in the same halfplane with respect to the line g as the circle k; hence, the centers $T_{1}$ and $T_{2}$ of these two circles must also lie in this halfplane). Thus, $T_{1}T_{2}\parallel g$. Hence, the perpendicular bisector of the segment $T_{1}T_{2}$ must be perpendicular to the line g. On the other hand, this perpendicular bisector of the segment $T_{1}T_{2}$ passes through the point K (since $KT_{1}=KT_{2}$). Hence, this perpendicular bisector is the perpendicular to the line g through the point K. Since $KL\perp g$, this perpendicular also passes through the point L. Thus, we have shown that the perpendicular bisector of the segment $T_{1}T_{2}$ passes through the point L, and this entails $LT_{1}=LT_{2}$. Together with $T_{1}M_{1}=T_{2}M_{2}$ (because $T_{1}M_{1}=r$ and $T_{2}M_{2}=r$), this yields the congruence of the right-angled triangles $LT_{1}M_{1}$ and $LT_{2}M_{2}$; thus, $LM_{1}=LM_{2}$. Consequently, either the points $M_{1}$ and $M_{2}$ coincide, or the point L is the midpoint of the segment $M_{1}M_{2}$. But the points $M_{1}$ and $M_{2}$ cannot coincide because this would yield $T_{1}T_{2}\perp g$ (since both points $T_{1}$ and $T_{2}$ would lie on the perpendicular to the line g at the point $M_{1}=M_{2}$), contradicting $T_{1}T_{2}\parallel g$; and the point L cannot be the midpoint of the segment $M_{1}M_{2}$, since both points $M_{1}$ and $M_{2}$ lie on the extension of the segment AL beyound L. Hence, we get a contradiction, and our assumption that there exist two different circles $t_{1}$ and $t_{2}$ both satisfying the condition of Lemma 3 was wrong. Thus, we have shown that there exists at most one such circle, and Lemma 3 is proven. Let's solve the problem now. Let the circle x have center X and radius $r_{x}$ and touch the sides AD and AB at the points $X_{d}$ and $X_{b}$, respectively. Similarly define the objects Y, $r_{y}$, $Y_{a}$, $Y_{c}$ for the circle y, the objects Z, $r_{z}$, $Z_{b}$, $Z_{d}$ for the circle z, and the objects W, $r_{w}$, $W_{c}$, $W_{a}$ for the circle w. Since the points $X_{b}$ and $Y_{a}$ both lie on the segment AB, two cases are possible: Either the points A, $X_{b}$, $Y_{a}$, B lie on the line AB in this order, or the points A, $Y_{b}$, $X_{b}$, B lie on the line AB in this order. But the second of these two cases, the case when the points A, $Y_{b}$, $X_{b}$, B lie on the line AB in this order, is actually impossible: if it was the case, the segments $X_{b}X_{d}$ and $Y_{a}Y_{c}$ would intersect at a common interior point; this point would then be an interior point of both circles x and y, but these two circles x and y are externally tangent to each other and therefore cannot have common interior points. Hence, the points A, $X_{b}$, $Y_{a}$, B lie on the line AB in this order. Similarly, the points B, $Y_{c}$, $Z_{b}$, C lie on the line BC in this order; the points C, $Z_{d}$, $W_{c}$, D lie on the line CD in this order; the points D, $W_{a}$, $X_{d}$, A lie on the line DA in this order. Since the quadrilateral ABCD has an incircle, AB + CD = BC + DA. In other words, $\left(AX_{b}+X_{b}Y_{a}+Y_{a}B\right)+\left(CZ_{d}+Z_{d}W_{c}+W_{c}D\right)=\left(BY_{c}+Y_{c}Z_{b}+Z_{b}C\right)+\left(DW_{a}+W_{a}X_{d}+X_{d}A\right)$. But since the two tangent segments from a point to a circle are equal, we have $AX_{b}=X_{d}A$, $BY_{c}=Y_{a}B$, $CZ_{d}=Z_{b}C$, $DW_{a}=W_{c}D$, so this equation simplifies to $X_{b}Y_{a}+Z_{d}W_{c}=Y_{c}Z_{b}+W_{a}X_{d}$. Now, since the circles x and y with radii $r_{x}$ and $r_{y}$ touch each other externally, and the line AB, which is one of their external common tangents, touches them at the points $X_{b}$ and $Y_{a}$, by Lemma 2 we have $X_{b}Y_{a}=2\sqrt{r_{x}r_{y}}$. Similarly, $Y_{c}Z_{b}=2\sqrt{r_{y}r_{z}}$, $Z_{d}W_{c}=2\sqrt{r_{z}r_{w}}$ and $W_{a}X_{d}=2\sqrt{r_{w}r_{x}}$. Hence, the equation $X_{b}Y_{a}+Z_{d}W_{c}=Y_{c}Z_{b}+W_{a}X_{d}$ becomes $2\sqrt{r_{x}r_{y}}+2\sqrt{r_{z}r_{w}}=2\sqrt{r_{y}r_{z}}+2\sqrt{r_{w}r_{x}}$ $\Longleftrightarrow\ \ \ \ \ \sqrt{r_{x}r_{y}}+\sqrt{r_{z}r_{w}}=\sqrt{r_{y}r_{z}}+\sqrt{r_{w}r_{x}}$ $\Longleftrightarrow\ \ \ \ \ \sqrt{r_{x}r_{y}}+\sqrt{r_{z}r_{w}}-\sqrt{r_{y}r_{z}}-\sqrt{r_{w}r_{x}}=0$ $\Longleftrightarrow\ \ \ \ \ \left(\sqrt{r_{x}}-\sqrt{r_{z}}\right)\left(\sqrt{r_{y}}-\sqrt{r_{w}}\right)=0$ $\Longleftrightarrow\ \ \ \ \ \sqrt{r_{x}}-\sqrt{r_{z}}=0\text{ or }\sqrt{r_{y}}-\sqrt{r_{w}}=0$ $\Longleftrightarrow\ \ \ \ \ r_{x}=r_{z}\text{ or }r_{y}=r_{w}$. We can WLOG assume that $r_{x}=r_{z}$. Denote $R=r_{x}=r_{z}$. Now, consider the line AB through the point B, the positive number R, and the circle y touching the line AB at the point $Y_{a}$. The circle x is externally tangent to the circle y, has radius $r_{x}=R$, lies in the same halfplane with respect to the line AB as the circle y (in fact, both circles x and y lie in the halfplane with respect to the line AB which contains the quadrilateral ABCD, because the circle x touches the segment AD and the circle y touches the segment BC) and touches the line AB at the point $X_{b}$ lying on the extension of the segment $BY_{a}$ beyound $Y_{a}$. Thus, the circle x is the (B, AB, R, y)-circle. Similarly, the circle z is the (B, BC, R, y)-circle. Now, the reflection with respect to the angle bisector of the angle ABC maps the line AB to the line BC, the point B to itself, and the circle y to itself (since the center of the circle y lies on the angle bisector of the angle ABC, because the circle y touches the sides BA and BC of this angle). Hence, this reflection maps the (B, AB, R, y)-circle to the (B, BC, R, y)-circle. In other words, it maps the circle x to the circle z. Hence, this reflection maps the center X of the circle x to the center Z of the circle z. Now, if X = Z, then the circles x and z coincide (because they have the same center X = Z and the same radius $r_{x}=r_{z}$) and Theorem 1 is proven. So it remains only to consider the case when $X\neq Z$. In this case, the fact that the reflection with respect to the angle bisector of the angle ABC maps the point X to the point Z can be rewritten as follows: The angle bisector of the angle ABC is the perpendicular bisector of the segment XZ. Similarly, the angle bisector of the angle ADC is the perpendicular bisector of the segment XZ. Hence, the angle bisectors of the angles ABC and ADC are one and the same line; this line must be the line BD. Hence, $\measuredangle ABD=\measuredangle CBD$ and $\measuredangle ADB=\measuredangle CDB$. Together with BD = BD, this yields that the triangles ABD and CBD are congruent, so the quadrilateral ABCD is a kite. Theorem 1 is thus proven. Darij