CIRCLES HOLDING CONVEX

CIRCLES HOLDING CONVEX

POLYHEDRA

MARCH 2011

Yuichi Tanoue Graduate School of Science and Technology

KUMAMOTO UNIVERSITY

Yuichi TANOUE March 3, 2011

Contents

Introduction

1 Circles holding regular triangular pyramids 1.1 Holding circles . . . , ' . . . 1.2 Triangular pyramids with a regular triangle 1.3 Triangular pyramids with an isosceles triangle . 2 Circles holding a regular triangular prism

2.1 Circles circumscribed to the symmetric pentagon 2.2 Regular triangular prism held by a circle.

3 Circles holding right prisms

3.1 Circles circumscribed to the pentagon 3.2 Rotations and translations . . . . 3.3 Holding line of the general polygonal prism 4 Circles fixing the Platonic solids

4.1 Fixing circles . . . . 4.2 Circles holding or fixing the regular tetrahedron. .

1

3 8 8 9 14 16 16 20 23

23

26 34

40 40 40

4.3 Circles holding or fixing the cube . . . . 4.4 Circles holding or fixing the regular octahedron 4.5 Circles fixing the regular dodecahedron

4.6 Circles fixing the regular icosahedron 4.7 Sufficient conditions of fixing by circles .

41 42 42 46 47 5 Tetrahedra passing through a circular or square hole 52

5.1 Passing through a hole. 52

5.2 Circular hole 52

5.3 Square hole . 55

Introduction

In this thesis we study the circles holding some convex polyhedra in 3- dimensional Euclidian space IR3. First, we define the holding circle and the fixing circle as follows. [1] [9] [12]

Definition 1.2 For a convex body B, and a circle C, a circular frame C is the boundary of a convex circular disk in a plane in IR3. A circular frame C is said to be attached to a convex body B c IRa if C n intB = 125 and convC n B =1= 125, where convC denotes the convex hull of C and intB denotes the interior B. A circular frame C attached to B can slip out of B ~f it is possible to make convC n B = 125 by a continuous rigid motion of C ( or B) in IR3 with keeping C n intB = 125. If a circular frame C can be attached to B so that it cannot slip out of B, then we say C can hold Band C is the holding circle.

Definition 4.1 If the holding circle C cannot move at all, we say that the convex body B is held by the fixing circle C.

In 1957, A. S. Besicovitch [2] [3] considered the problems on a net holding a unit ball and a convex polyhedron containing a unit ball. In 1959, H. S.

M. Coxeter [5] proposed a problem on a cage (a frame formed by edges of a convex polyhedron) holding a unit ball : The problem is : Find a cage of minimum sum, of edge lengths, holding a unit ball (not permitting it to slide out). He conjectured that it is a right triangular prism all of whose edges are equal to }3, so that the total length of all edges is 9}3 = 15.59 ....

But this conjecture was false. In 1963, A. S. Besicovitch [4] proved that the greatest lower bound of the sum of edges of a cage to hold a unit ball is at most

,=

Can a very simple instrument such as a (rigid) circle be used to hold a convex body? Certainly not in the case of a ball. More generally, it cannot be

used to hold any ellipsoids, circular cylinders, or other usual convex bodies.

In 1995, T. Zamfirescu [12J showed not only that these convex bodies do exist, but also that they forIll a large majority. More precisely, he proved that the convex bodies in lRa which cannot be held by a circle form a nowhere dense subset of the space of all convex bodies in lR.³ with the Hausdorff metric.

But the concrete examples which is held by a circle are not well known. In the main part of this thesis, we treat several concrete polyhedra containing Platonic solids and investigate whether they can be held by a. circle. More- over, in the case of Platonic solids, we study the diameter of fixing circles.

At the end of this thesis, we study a relational topic.

In chapter 1, we consider the regular triangular pyramids. Let P be a regular triangular pyramid which has the apex a above the barycenter of the equilateral triangle bcd of side length 1, let h be the height of P. The lateral sides are congruent isosceles triangles. We show the following theorem.

Theorem 1.3 A regular triangular pyramid can be held by a circle if its height is greater than ho, where

ho =

~J -~

V;

Theorem 2.7 A regular triangular prism with all edges of length 1 can be held by a circle.

In chapter 3, as an extension of chapter 2, we consider the more general case of the right prism which has bases of scalene triangles and has arbitrary height. We show the following theorems.

Theorem 3.6 Any right triangular prism, can be held by a circle.

Theorem 3.8 The regular right prism, whose base is a regular (2n-l)-gon can be held by a circle.

Theorem 3.9 Let P be a convex n-polygon. ala2'" ai'" aj ... an. We assume that there is at least one vertex ai and the edge ajaj+l such that the foot ri of the orthogonal projection from ai on the line ajaj+l is contained in the edge ajaj+l except both ends. If the right polygonal prism, has the basis P and its height is very small, then the prism can be held by a circle.

In the first half of chapter 4, we present the next theorems for holding circles.

Theorem 4.2 The tetrahedron T of the edge length 1 can be. held precisely by the circles of diameter d satisfying

V2 ^{t6 - to} + 1

2 ^{~ d} < = 0.8957 ... ,

jit5 - to + 1

(0.2)

where to is the solution of the equation 3t³ - 6t2 + 7t - 2 = O.

Theorem 4.3 The cube C of the edge length 1 can be held by the circles of diameter d satisfying

V2 ~ ^d < V2(t6 - 2to + ³⁾ = 1.5347 ... , .Jtg - 4to + 6

where to is the solution of the equation t³ - 6t² + 13t - 6 = O.

(0.3)

Theorem 4.4 The octahedr'On 0 of the edge length 1 can be held by the

circles of diameter d satisfying

1 ~ ^d < 2(1 + t6) = 1.1066 ... ,

J3tg + 2to + 3 (0.4)

where to is the solution of the equation 3t³ + 3t² + 3t - 1 = O.

For the dodecahedron and icosahedron, we cannot estimate the upper bound of the diameter of the holding circles. It is more complicated than the tetrahedron, the cube and the oc.tahedron.

Also, in the latter half of chapter 4, we present other examples for fixing circles. We consider the Platonic solids can be held by the circle rigidly. We show the following theorerns.

Theorem 4.5 Platonic solids have fixing circles passing through 0 as follows. Tetrahedron, cube, octahedron have at least one fixing circle C T, CH,

Co respectively. Dodecahedron has at least three fixing circles Coo, COl, Co_{2 ,} and icosahedron has at least two fixing circles CIO, Cil .

In chapter 5, we study the other related subjects. Suppose a plane in lR³ has a hole. We determine the smallest circular and the smallest square hole through which a regular tetrahedron of given size can pass.

In 2005, J. !toh and T. Zamfirescu [7] studied the shape of convex hole' H of diameter and width as small as possible such that the regular tetrahedron T of edge length 1 can pass through it. They found such a hole with diameter

4, the width of a face of T, and with width

1',

We investigate the smallest diameter for the case of a circular hole which a regular tetrahedron can pass in Theorem 5.1. In Theorem 4.2, we estimate the diameter of circles holding the regular tetrahedron. Theorem 5.1 states the same result of Theorem 4.2 in other words. Moreover, we study the smallest diameter for the case of a square hole in TheorerIl 5.2.

Theorem 5.1 Assume that the hole H C P is a disk. The smallest diameter of H such that T can pass through H is

where

t5 - to + 1 = 0.8957 ... ,

J~t5. - ^{to + 1}

2+ {/)43-4- {/)43+4

to = .

. 3

(0.5)

(0.6)

Theorem 5.2 Assume that the hole H C P is a square. The smallest such hole allowing T to pass through it has diagonal length 1.

Acknowledgements

The author would like to thank Professors O. Kobayashi and J. Itoh for their encouragement and discussions. He also wishes to express gratitude to Prof.

T. Zamfirescu for his constant encouragement and valuable suggestions, and Dr. C. Vllcu for his helpful comments.

1 Circles holding regular triangular pyramids

1.1 Holding circles

Problems about the shortest total length of a net or a cage to hold a unit ball so that the ball cannot slip out of it, have been considered by A. S.

Besicovitch [2], [3], [4] and H. S. M. Coxeter [5]. More recently, T. Zamfirescu considered the problem of holding a convex body by using a very simple instrument, namely a circle instead of a cage or a net [12]. Do there exist any convex bodies which can be held using a circle? Certainly a ball cannot be held. A circular cylinder cannot be held by a circle either. On the other hand, tetrahedra can.

T. Zamfirescu showed not only that such convex bodies exist, but also that they form a large majority. More precisely, he proved that the convex bodies which cannot be held by a circle fonn a nowhere dense subset, in the space of all convex bodies with the Hausdorff metric.

Let B be the all convex bodies, and C be the space of all circles in IR3 endowed with the Hausdorff metric 8. We define the Hausdorff metric 8 on B ( or C ) as follows.

Definition 1.1 For any K1,K2 E B (orC), the Hausdorjjmetric8(Kl,K2) is defined by

We define the holding circle of compact convex bodies as follows [1] [9].

Definition 1.2 For a convex body B c B, and a circle C c C, a circular

!ram,e C is the boundary of a convex circular disk in a plane in ill.^3• A circular frame C is said to be attached to a convex body B c IR3 if C n

intB = 0 and convC n B i= ^0, where convC denotes the c~nvex hull of C and intB denotes the interior B. A circular frame C attached to B can slip out

of B if it is possible to m,ake convC n B = 0 by a continuous rigid motion of C (or B) in m.³ with keeping C n intB = ^0. If a circular frame , C can be attached to B so that it cannot slip out of B, then we say C can hold B and C is the holding circle.

T. Zamfirescu defined the holding cir"Cle as follows [12]. His definition states the same content of Definition 1.2 in other words.

Let C be the space of all circles in 1R ³ endowed with the Hausdorff metric 8. For a convex body Be m⁸ to be held by the circle C means that intB n

C = ⁰ and, for some natural number m, there is no continuous mapping, f : [0, 1] ^-1- C such that f(O) = c, 8(f(0), f(l)) > m and, for all A E [0, 1], f(A) is congruent with C and f(A) (1 intB = ^0.

Related problems were treated by J. Itoh and T. Zamfirescu [7]. They looked for the shape of a convex hole of diameter and wi~th as small as possible such that the regular tetrahedron of edge length 1 can pass through it. In 1920, K. Zindler [14] already cqnsidered convex bodies moving through a circular hole.

1.2 Triangular pyramids with a regular triangle

Let P be a regular triangular pyramid which has the apex a. above the barycenter of the equilateral triangle bcd of side length 1, let h be the height of P. The lateral sides are congruent isosceles triangles.

Theorem 1.3 A regular triangular pyramid can be held by a circle if its height is greater than ho, where

ho =

~J-~

First, we will take the coordinates as follows. Let 0 be the midpoint of the edge bd as the origin, x-axis be the line through Oc, y-axis be the line through Od, and z-axis be the line orthogonal to the xy-plane through O.

The triangle Oca is included in the zx-plane.

Let Bo be the angle between the z-axis and Oa. We rotate the triangu- lar pyramid P around the y-axis fixing the edge bd until a comes to the z-axis. Continuing the rotation, Oa goes through the z-axis and we stop the rotation when Oc comes to the z-axis. Let P( B) be the above rotating triangular pyramid such that the angle between the z-axis and Oa is B (0 ::::;; B ::::;; ~ - Bo). Thus, the initial position P(O) has a(O) on the z-axis.

We denote vertices of P(B) by a(B), b, c(B), d. Let O(B) be the union of the four edges a(B)b, bc(B),' c(B)d, da(B).

z

a

Figure 1=

We denote the orthogonal projection from lll3 to the xy-plane by II. When

o < B < ~ - _Bo, II (f2( B)) becornes a kite symmetrical about the x-axis. Its initial and final positions II (0(0)) and II (O(~ - Bo)) are isosceles triangles.

As IT (O( e)) is a kite or a triangle, there is an inscribed circle f( e) to it.

Let p' (e), q' (e), r' (e), s' (e) be the points of contact of f( e) with the sides II (a( e)) b, bIT (c( e)), II (c( e)) d, dII (a( e)) respectively. We define p( e) = II-I (p'(e)) n O(e) and, analogously, q(e), r(e), 8(e). These four points are determined only bye. As p(e), q(e) are symmetrical to s(e), r(e) with respect to the zx-plane, there exists a plane containing these four points.

We denote this plane by ~(e). The boundary p(e) n ^~(e) is the quadrilateral p(e)q(e)r(e)s(e). Before giving a proof to our theorem, we establish the following preparatory lemma.

y y

c(O)

a(O) _a

x ~---~~ x

q' ⁰ ~(c(O)) q' 0 7r(c(O))

Figure 2:

Lemma 1.4 When h > ho, there is a. unique

e

e

Proof. When e = 0, IT (0(0)) is the triangle bIT (c(O)) d. The circle r(O) inscribed to bIT (c(O))d has the points of contact as 0, q'(O), r'(O) on the sides bd, bIT (c(O)), IT (c(O)) d, and the intersection P(O) n ~(O) is the triangle a(O) (= p(O) = s(O) )q(O)r(O).

Similarly, When B = ~ - Bo, II (n (~ - Bo)) is the triangle II (o,( ~ - Bo)) bd.

The circle r(~-Bo) inscribed to II (o,(~ - Bo)) bd has the points of contact as O,p'(~-Bo), 8'(~-Bo) on the sides bd, II (o,(~ - Bo)) b, II (a(~ - Bo)) d, and the intersection P (~ - Bo) n e ^(~ - ^Bo) is the triangle c( ~ - Bo) (= q( ~ - Bo)

= r(~ - Bo)) p (~-Bo) 8 (~-Bo). Let a be the plane passing through 0,(0)

p~allel to the xy-plane, and f3 be the plane of the triangle bc(O)d. The intersection a n f3 is a line l. We denote the distance from a(O) to 1 by d.

From elementary calculations, d = 2v'3hV h² + 11

2, On the other hand, the intersection P(O) n e(O) becomes a triangle o,(O)q(O)r(O). The height of the triangle o,(O)q(O)r(O) from a(O) is a(O)h(O) = V3h , where h(O) is the

4Jh

mid-point of the side q(O)r(O). If d > a(O)h(O), the triangle a(O)q(O)r(O) is situated under Q. If d < a(O)h(O), the triangle a(O)q(O)r(O) is situated over Q. If d = a(O)h(O), the triangle a(O)q(O)r(O) is contained in Q, and the height h of P(O) becomes

ho = -15+ v'1377 ₂₈₈ =

~J -~

(1.4)

Note that the plane e(B) is determined by the angle B only, and changes continuously from e(O) to €(~-Bo) as B increases. We denote the z-coordinate of p(B) and 8(B) by z(p(B)), the z-coordinate of q(B) and r(B) by z(q(B)).

Let f (B) be the function defined by

f(B) = z(p(B)) - z(q(B)). (1.5) Obviously, the function f is continuous and monotone. If B = 0 and h > ho, then f(O) > 0, for the triangle a(O)q(O)r(O) is situated under the plane Q.

For 8 = ~ - _80, f(~ - 80) < 0 for all h. By the intermediate value theorem, there is a unique 8 such that f(8) = ^0, i.e., the plane ~(8) is parallel to the xy-plane and the lemma is proved.

We will give a proof of Theorem 1.3.

Proof. The intersection II-I (r(8)) n ^~(8) is an ellipse A(8). From our Lemma, there is a unique 8 such that the plane ~ (8) is parallel to the xy- plane. Then the ellipse A(O) becomes a circle r congruent with r(O). We will prove that the triangular pyramid P(8) can be held by the circle r. We claim that each circle rt satisfying 0 < ~

cr,

Remark that 0(8) rneets the plane k(~(O)) inside of rt, if rt n intP(O) = ¢.

The orthogonal projection II(rt) onto the xy-plane is (i) a circle congruent with r ^{or (ii)} an ellipse whose long axis equals to the diameter of :f. The orthogonal projection II(O(O))=II(a(O))b U bII(c(O)) U II(c(O))d U dII(a(O)) is the kite circumscribed to r( 8).

y

~8) d

q'(8) x

n-(a(8)) n-( c( 8)) x

Figure 3:

y

In case (i), the orthogonal projections of r, ^rt on to the xy-plane are the circles r(B), lI(rt) respectively. The circle lI(rt) is congruent with r(B). If r(B) = lI(rt), lI(rt) is the circle inscribed the kite lI(n(B)). k(e(B)) n P(B) is the quadrilateral t'l.£VW, where the points t, 'l.£, V, W exist on the edges a(8)b, bc(e), c(e)d, dace) respectively. The orthogonal projections lI(t), lI('l.£) , lI(v), lI(w) exist on the edges lI(a(e))b, blI(c(B)), lI(c(B))d, dII(a(B)). As these four points should be different from p' (e), q' (e), r' (B), s' (e) respectively, the circle lI(rt) intersects the interior of the quadrilateral lI(t)lI(u)lI(v)lI(w).

Namely, rt intersects the interior of the quadrilateral tuvw. This implies rt n intp(e) =F 4>. If r(B) =F lI(rt), II(rt) does not meet at least one of the edges in lI(n(B)). This implies that the convex hull of rt does not meet at least one of the edges in n ( e), therefore rt n int P ( B) =F 4>.

In case (ii), the orthogonal projection of rt onto the xy-plane is the ellipse lI(rt) whose long axis equals to the diameter of r. As a short axis of the ellipse lI(rt) is smaller than the radius of r(B), rt does not meet at least one of the edges in nco). This implies that the convex hull of rt does not meet at least one of the edges in ncO) and, therefore rt n intP(O) =F cP. This is a contradiction, which ends the proof.

1.3 Triangular pyramids with an isosceles triangle We can strengthen Theorern 1.3 as follows.

Theorem 1.5 Let the triangular pyramid Pt have the apex a above the barycenter of the base, which is an isosceles triangle bcd with Ibdl = 1, Ibcl =

Icdl ⁼ t ~ 1. Then P_t can be held by a circle if its height is greater than hI,' where'

hI = ^{-( 4t}

2 + 1) + J208t⁴ - 56t² + 1

96t² (1.6)

Proof. As in the proof of Theorem 1.3, let P_{t (())} be the rotating triangular pyramid such that the angle between the z-axis and Oa is () (0 ~ () ~ ~-eo).

If () = 0, then the intersection ~t(O) n Pt(O) is a triangle a(O)qt(O)rt(O). The height .of a triangle a(O)qt(O)Tt(O) from a(O) is

(1.7)

where ht(O) is the mid-point of the side qt(O)Tt(O). On the other-hand, the intersection at n f3t is a line It, where at is the plane passing through a(O) parallel to the xy-plane, and f3t is the plane of the triangle bCt(O)d. We denote the distance from a(O) to It by dt:. An elernentary calculation,

(1.8) The distance dt is decided by h only. From the equation a(O)ht(O) = dt, namely

(1.9) we get the value of hI. If h > hI, then the triangle a(O)qt(O)Tt(O) is situated under at. As in the proof of the Lemma 1.4, we deduce the existence of a unique Ot such that the plane ~t(Ot) is parallel to the xy-plane. This theorem is proved.

2 Circles holding a regular triangular prism

2.1 Circles circumscribed to the symmetric pentagon

Figure 4:

In this chapter, we will investigate the non-obvious question whether a prism B can be held by a circle. We consider the regular triangular prism abcdef with all edges of length 1 (see Fig. 4). Let l be the orthogonal projection of a on be and rn be the orthogonal projection of d on ef. Take points t on the side le, and p on the side ac with pt II al. Let x be the distance Iltl from l to t (0 ~ x < ~). Denote by q,o the mid points of adl lm respectively. Let

7r be the plane through the points p, q, t, as in Figure 4. The intersection

7r n B is a pentagon pqrst symmetric with respect to qo. The quadrilateral prst is a rectangle. We denote the radius of the circle circumscribed to the rectangle pr st by R( x). From elementary calculation, it follows that

(2.1) R(x) has a (single) minimurn at x = ^T = ¹³₄ = ^{0.24 ...}

We denote the radius of the circle circumscribed to the triangle tqs by T(x)~

and find out that

(2.2) Since x ~ 0, T(x) is an increasing function. The triangle tqs is always acute. We denote the value of x which realizes R(x)=T(x) by 0". We define the function

M(x) = max {R(x), T(x)}. (2.3)

Lemma 2.1 M(x) has a unique minimum at X=O".

Proof . . The value of 0" is the solution of the equation

1.1 ^{7 1}

2 V ^{7 x2 - 3x} + 4 ⁼ V3 (x² + 1) , (2.4) i.e.

16x^{4 -} 52x² + 36x - 5 = 0. (2.5)

a =

~ (-.,f{,

(4;~13)v1)

~ = Vl09 cos ~ + 13 and 6 3 6

c o s a = - - - . 163

VI09³ (2.7)

Hence,O" < ^T. Since R(x) is decreasing and T(x) is increasing in [0,0"], M(x) = R(x) is decreasing there. Since both R(x) and T(x) are increasing in [0", ~], M(x) is also increasing there. Therefore, M(x) has a unique minimum at ^(J".

Next, we consider the case of moving the plane 7r in a neighb,orhood of x = 0", in various directions. We prepare now the proof of Theorem 2.7.

y

-, , _, , ,

0 ^(J or 1

2

Figure 5:

Lemma 2.2 Let rr' be the plane obtained from rr after a rotation of angle 8 around ts. We denote by p', r' the intersections of rr' with the edges ac, de respectively. The intersection rr' n ^B becomes the trapezoid p'r'st, and denote by p, p' the radii of the circles circumscribed to the rectangle prst and the trapezoid p' r' st. Then p ~ pI if 0 ~ x ~ ^130,'

Proof. Assume 8 > 0, i.e. a counterclockwise rotation. Then

p,2= x2+i (3(~_X)2 4(x2 ~)_2J3X(~-x)sin8)

4x² cos² 8 + 1 4 sin ² (* - 8) + + 4 sin (* - 8)

(2.8) We have

(2.9) Therefore J~

( d;;2) ;;,

because 0 < a < 13 0'

By symmetry, p'(8) = p'(-8), hence p is a minimum of p'.

Lemma 2.3 Consider the point.e; p" on the line pt and r" on the line rs. If Iptl > Iplltl and Irsl < Ir" sl, then the radius p of the circle circumscribed to

the rectangle prst is smaller than the radius p" of the trapezoid p"r" st. If Iptl < Ip"tl and Irsl > ^Ir" sl, then we get the same conclusion.

Proof. Clearly, p = gt. Now, in case Iptl > Ip"tl and Irsl < Ir" sl, the radius Ir"tl Ir"tl Irtl of the circle circumscribed to p"r" st is -2- and so p" = ^-2- ~ "2 = p.

The case Iptl < Ip"tl and Irsl ^{> Ir"} sl is analogous.

Lemma 2.4 Let ho > O. Among all triangles having fixed side and the corresponding height h ~ ho, the isosceles one of height ho has smallest circumscribed circle.

Proof. It is obvious.

Lemma 2.5 Any plane in some neighborhood of the plane 7r can be obtained /rom,7r by a rotation around qo, a rotation around st, and a translation to lc.

Proo f. This is common knowledge.

Lemma 2.6 Let abcd be a trapezoid with ad II ^{bc, labl = Idcl = w, ladl =.}

u, Ibcl = v (u > v), and abcd be a rectangle with ladl = Ibcl = u,t'lJ, labl = Icdl = w. We define the radii of the circles circumscribed to abcd, abcd by

p,p. Ifw > 'lJ'2"v, then p > p.

Proof. From direct calculation,

P^{2 = w}

2(uv + w²) 1

P-2 = 16 {(u + v)2 + 4w2} , (2 10)

{2w+(u-v)}{2w-(u-v)}' .

and

2 -2 _ ( U + v) ² (u - v) ² 0

P -p - > .

16 {2w + (u - v)} {2w - (u - v)} ^(2.11)

2.2 Regular triangular prism held by a circle We now proceed to the proof of Theorem 2.7.

Theorem 2.7 A regular triangular prism with all edges of length 1 can be held by a circle.

Proof of Theorern 2.7. By Lemma 2.5, we have to consider the following three transformations of the plane cutting B. First, a rotation around the axis qo, next, a rotation about the axis st, next, a translation along bc.

Suppose 7ro is the position of the plane for x = CJ, 7r is the position of the plane after the first rotation, 7r' the. position after the second rotation, and

7r" the position after the translation. The intersection ^7r' n B is a pentagon

p' q' r' st. the triangle tq's becomes non-isosceles and the quadrilateral p' r' st becomes a trapezoid.

In case a < x < ~, we will prove that T(a) is less than the radius of the circle circumscribed to all intersections with B of planes in a neighborhood of ^7r. Indeed, this is clear by Lemma 2.4.

In case 0 < x < a, we will prove that R( CJ) is the smallest radius of the circle circumscribed to the quadrilateral p"r" s"t" determined by 7r". We denote the radii of the circles circumscribed to the trapezoids p'r'st, p"r" s"t" by R'(x, 8), R"(x, 8, Llx) respectively.

For 0 < x < 13

0' by Lemma 2.2, the radius R'(x,8) increases with 8. But, by the translation parallel to 7r', R" (x, 8, .6. x ) decreases in the following two cases. We must consider these two cases carefully.

(i) The case of a counterclockwise rotation (8 > 0) and of a translation Llx > 0 (in direction bc).

(ii) The case of a clockwise rotation (8 < 0) and of a translation Llx < 0 (in direction cb).

The two cases are symmetric, so we shall discuss only case (i).

In case (i), the diagonal p's of the trapezoid p'r'st is longer than the diagonal r't. But the following translation .6.x yields

I " "I p s < I ' I p s an ' d Ir"t"l > Ir'tl·

Thus for some.6.; = ^II. the position p" r" s" t" of the trapezoid p"r" s"t"

, f""" (f-L) (f-L) (f-L) (f-L)

is such that Ip(~) s(~) 1 = Ir(~) ^t'(f-L) I. Then Ip(~) r(~) I = 1 s(~) ^t'(f-L) I, the trapezoid is isosceles. We show now that R"(x, 0, J.L) is the smallest radius of the circle circumscri bed to all the quadrilaterals p" r" s" til, for D.x in a neighbor hood of J.L.

From direct calculation,

We have

and

3v'3 sinO 213 . 0

- 4"" sin( % +0) sine i ^{-0) - V .:> SIn} v'3

J.L = ^{.Q cosO} x + 2: tanO.

4 sine i+O) sine i-Oj

(2.12)

(2.13) We will prove that the radius jj(J.L) of the circle circumscribed to the rectangle P(f-L)T(f-L)s(f-L)l(f-L) is larger than the radius R(x) (see the notation of

Figure 6:

Lemma 2.2). As Iprl = Istl ⁼ Ip(JL)T(JL) I ⁼ IS(JL/(JL) I, we will compare Ip(JL)i(JL) I

with Iptl. Let f be the function defined by

f(x,

G - ^x)

G - ^x)

(2.14) From direct calculation,

f( 0) ⁼ 0 l' df(x,O) = 0 l' d² f(x, B) = v'3(6 1) 0

x, ,1m _(J~O dB ,1m _(J~O d0₂ 2 x + >. ^(2.15)

Therefore, for fixed x, f(x, 0) has a local minimum at B = 0, and f(x, 0) > 0 in a whole neighborhood of 0 except {a}.

From Lemma 2.2, we have R"(x, 0, J-L) > p(J-L). Clearly, P(I1J) > R(x), and R(x) > R(cr), so R(cr) < R"(x,O,J-L). The theorem is proven.

3 Circles holding right prisms

3.1 Circles circumscribed to the pentagon

e

Figure 7:

In this chapter, we will investigate the question whether an arbitrary right prism can be held by a circle. First, we consider a right triangular prism B which has bases of scalene triangles with its sides v, U (v:::;; u) and has height h, let 7r be the plane including the circle C. \\Then the intersection

7r n B is a triangle or a quadrilateral, it is clear that B cannot be held by C. The circle C can slip out through to the certain vertex. When the intersection 7r n B is a pentagon, there is a circle C such that B can be held.

Let triangle abc be a vertex a with acute angles at band c. We consider the right triangular prism abcdeJ with conjugate faces abc and deJ, labl = v, lacl = u (v :::;; u) and height ladl =h. Let I be the orthogonal projection of a on bc and m, be the orthogonal projection of d on ef. Put L.cal=a. and L.bal={3, and assume {3 :::;; a.. Take points t on the side lc~ and p on the side

ac with pt " al. Let x be the distance Iltl from l to t (0 ~ x ~ usina). We denote by q,o with

aq lo usina qd = om, = v sin f3 .

Let ⁷⁽ be the plane through the points p, q, t as in Figure 7. The intersection

7( n B is a pentagon pqrst. The quadrilateral prst is a rectangle, and to usina

os = v sin f3'

Take points tf on the side lb, and p' on the side ab with p't' II al. Let y be the distance Ilt'l from l to t' (O~ Y ~ v sin f3). Denote by q', 0' with

aq' lo' v sin f3

q'd = o'm = usina' '

Let ^7(' be the plane through points p' ,q', t'. The intersection ^7(' n B is a pentagon p' q' r' s' t'. The quadrilateral p' r' s' ^tt' is a rectangle. In this case, turn this right prism upside down, so we find the same case of ^7(. Therefore, we only consider the case of the plane 7(.

We denote the radius of the circle circumscribed to the rectangle prst by R( x). From the elementary calculation, it follows that

1 [{ 2} { ^{u sin a }} 2

R (x) = 2 tan a 1 + (tan a + tan {3) x - 2

1 + (tan a + tan{3)

1

+ ^{I + ^{(tan a} + ^{tan{3)2} h2 tan2 a} + ^u2 sin2 a (tan a + ^{tan {3)2] 2}

(3.1) 1 + (tan a + tan {3)2

We denote the radius of the circle circumscribed to the triangle tqs by T(x).

Similarly, it follows that

T(x) =

2 2 U v

{ (-2

-2)2}]

+U cos a 1 - 4U₂V₂ '

h - . - . ^I-l 1 + tan (3 M

were ^U=U SIn ^0:, v = v SIn ^{fJ ~} C = tan ^{0: •} oreover,

st 1 L(x) = - = -

2 2 ( tan(3) ²

1 + - - ^x2+h2•

tan ^0:

(3.3)

(3.4) We denote the value of x which realizes the minimum of R(x) (0 ~ x ~ U sin ^0:) by T, namely

usina T = - - - : : : -

1 + (tan ^0: + tan (3)2' and the value of x which realizes R(x)=T(x) bya.

Let be defined by

and,

TW

L(x)

if Ltqs is non obtuse if Ltqs is obtuse

lvl(x) = max {R(x),

TW} .

(3.5)

(3.6)

(3.7)