Fuss’ problem To find the relation between the radii and the line joining the centers of the circles of circumscription and inscription of a bicentric quadrilateral. It’s useful to mention certain properties of chordal and tangential quadrilaterals before investigating bicentric quadrilateral. Chordal or cyclic quadrilateral is a quadrilateral admitting circumcircle. The sum of its internal angles is the straight angle. Conversely, every convex quadrilateral with internal angles giving together the straight angle is chordal. Tangential quadrilateral is a quadrilateral admitting incircle. Let a, b, c, d be the lengths of its sides clockwise or counterclockwise, then a + c = b + d and conversely.

Bicentric quadrilateral The tangency chords of the two pairs of opposite sides of a bicentric quadrilateral are perpendicular to each other (see figure 1).

C

M

M

D

.

D

0

L O

γ ω

/

B K

N

C

N

/

1

O 0 .

1

ω

L B

K

α

A

Figure 1: Bicentric quadrilateral

A

Figure 2: Designating the angles

Proof: Let ABCD be a bicentric quadrilateral; α and γ be the angles by the vertices A and C; ¯ L, ¯ M ¯ and N ¯ the angles by the tangent points K, L, M and N . The investigated angle K, formed by tangency chords is designated ω (see figure 2). The lines AB and CD touch the incircle in points M and K, thus MK is the tangency chord and the angles by points M and K are congruent. The same stands for L and N . For quadrilaterals AKON and CMOL: ¯ +L ¯ + ω = 360◦ γ+M ¯ +N ¯ + ω = 360◦ α+K ¯ +K ¯ +L ¯+N ¯ +2ω = 720◦ α + γ +M | {z } | {z } | {z }

180◦

180◦

⇒

ω = 90◦

180◦

2

Conversely, every cyclic quadrilateral where the tangency chords of the two pairs of opposite sides are perpendicular to each other is bicentric. Consequently starting with two perpendicular chords of a circle, constructing tangents in their extremities, the bicentric quadrilateral is created.

The locus Lines KM and LN divide the bicentric quadrilateral ABCD in four quadrilaterals AKON , BLOK , CMOL and DNOM with some common properties. Examining this sort of quadrilateral OXPY by investigating locus of points P is conductive: P

Y

• Let k be a circle, O be the point inside, • X, Y be the points on k giving right angle XOY , • P be the point of intersection of tangents touching k in X and Y .

X M O k

Visualizing the locus (see figure 3) the locus seems to be a circle.

P

Y

P

Y N

F

X M O

Figure 3: Visualization of the locus

ϕ

X

ρ M e O

Figure 4: Designating of situation

Proof: Designating points see figure 4. Let e be the length of MO, ϕ the size of the angle OMP , ρ the radius of k and p the length of MP . In the right-angled triangle OXY |OF |2 = |FX | · |FY |. The line MP is the bisector of XY , thus |NX | = |NY | and the angle MNY is right. Furthermore |NF| = e·sinϕ. Therefore |OF| = |MN | − e · cos ϕ |FX | = |NX | − e · sin ϕ |FY | = |NX | + e · sin ϕ Substituting foregoing into |OF|2 = |FX | · |FY | (|MN | − e · cos ϕ)2 |MN |2 − 2 |MN | e · cos ϕ + e2 cos2 ϕ |MN |2 − 2 |MN | e · cos ϕ + e2

= (|NX | − e · sin ϕ)(|NX | + e · sin ϕ) = |NX |2 − e2 · sin2 ϕ = |NX |2

As |MX | = ρ, then in the right-angled triangle MXN |NX |2 = ρ2 − |MN |2 . Then 2 |MN |2 − 2 |MN | e · cos ϕ + e2 = ρ2 . In right-angled triangle M XP is |MX |2 = |MP| · |MN | or ρ2 = p |MN |. Substituting for |MN | and arranging: ρ2 e 2ρ4 = 2 p cos ϕ + p2 . ρ2 − e2 ρ2 − e2 p and ϕ are variables dependent on the position of point P ; ρ and e are invariable for given circle k and point O. Assuming P lying on a circle with the radius r = |SP | and the centre S lying on M O, let x = |SM | be the distance of centers. Then in the triangle SMP there is r2 = x2 + p2 + 2xp cos ϕ. 2 Comparing last two equations, r is constant for x = ρ2ρ−ee 2 . Hence the desired locus is a circle with the centre S and radius r. Substituting for x and eliminating e we get the relation between the radii of given circle ρ, found circle r and the distance of their centers x: 2ρ2 (r2 + x2 ) = (r2 − x2 )2 . 2

Conclusion

Given two circles, one inside another. Starting in P , drawing a tangent to the inner circle, from the point of intersection with the outer circle a tangent to inner one again and so on, we return to P and get the bicentric quadrilateral. The position of P on the outer circle is arbitrary, moving P we obtain all bicentric quadrilaterals for given incircle and circumcircle. Foregoing procedure leads not only to proving the locus is a circle but above all to the solution of the Fuss’ problem: Given ρ and r the radii of incircle and circumcircle, then the distance x of their centers satisfies the equation 2ρ2 (r2 + x2 ) = (r2 − x2 )2 . √ For x ∈ (0, r − ρ), r ≥ ρ 2 this equation has one solution r

x=

r2

+

ρ2

q

− ρ 4r2 + ρ2 .

√ For r < ρ 2 it has no solution and no bicentric quadrilateral can be constructed. Derived equation can be arranged: 1 1 1 + = 2. 2 2 (r − x) (r + x) ρ

References [1] D¨orrie, H. 100 Great Problems of Elementary Mathematics, Their History and Solution. Dover Publications Inc., New York. 1965 [2] Weisstein, E. W. Bicentric Quadrilateral. MathWorld – A Wolfram Web Resource. URL