Minimum area isosceles containers
aa r X i v : . [ m a t h . M G ] S e p MINIMUM AREA ISOSCELES CONTAINERS
GERGELY KISS, J ´ANOS PACH, AND G ´ABOR SOMLAI
Abstract.
We show that every minimum area isosceles triangle containing a given triangle T sharesa side and an angle with T . This proves a conjecture of Nandakumar motivated by a computationalproblem. We use our result to deduce that for every triangle T , (1) there are at most 3 minimumarea isosceles triangles that contain T , and (2) there exists an isosceles triangle containing T whosearea is smaller than √ T . Both bounds are best possible. Introduction
Given two convex bodies, T ′ and T , in the plane, it is not easy to decide whether there is a rigidmotion that takes T ′ into a position where it covers T . Suppose, for instance, that we place a 2-dimensional convex body T ′ in the 3-dimensional space, and let T denote the orthogonal projection of T ′ onto the x - y plane. The area of T ′ is at least as large as the area of T , and it looks plausible that T ′ can be moved to cover T . However, the proof of this fact is far from straightforward; see [3, 12].As Steinhaus [21] pointed out, it is not even clear how to decide, whether a given triangle T ′ can bebrought into a position where it covers a fixed triangle T . The first such algorithm was found by Post[17] in 1993, and it was based on the following lemma. Lemma 1.1 (Post) . If a triangle T ′ can be moved to a position where it covers another triangle T ,then one can also find a covering position of T ′ with a side that contains one side of T . In many problems, the body T ′ is not fixed, but can be chosen from a family of possible “containers,”and we want to find a container which is in some sense optimal. To find a minimum area or minimumperimeter triangle, rectangle, convex k -gon, or ellipse (L¨owner-John ellipse) enclosing a given set ofpoints are classical problems in geometry with interesting applications in packing and covering, approx-imation, convexity, computational geometry, robotics, and elsewhere [1, 2, 4, 5, 6, 8, 9, 10, 16, 18, 19].Finding optimal circumscribing and inscribed simplices, ellipsoids, polytopes with a fixed number ofsides or vertices, etc., are fundamental questions in optimization, functional analysis, and numbertheory; see e.g. [7, 13, 11, 20, 22].Motivated by a computational problem, R. Nandakumar [15] raised the following interesting specialinstance of the above question: Determine the minimum area of an isosceles triangle containing a giventriangle T . The aim of the present note is to solve this problem and to find all triangles for which theminimum is attained. We call these triangles minimum area isosceles containers for T . It is easy toverify that every triangle has at least one minimum area isosceles container (see Lemma 3.1). However,we will see that in some cases the minimum area isosceles container is not unique.Our main objective is to prove the following statement conjectured by Nandakumar [15]. Theorem 1.2.
Let T be a triangle and let T ′ ⊇ T be one of its minimum area isosceles containers.Then, T ′ and T have a side in common, and their angles at one of the endpoints of this side are equal. For any two points, A and B , let AB denote the closed segment connecting them, and let | AB | stand for the length of AB . To unify the presentation, in the sequel we fix a triangle T with vertices A, B, C, and side lengths a = | BC | , b = | AC | , c = | AB | . If two sides are of the same length, then T is the unique minimum area isosceles container of itself, so there is nothing to prove. Therefore, fromnow on we assume without loss of generality that a < b < c .To establish Theorem 1.2 and to formulate our further results, we need to introduce some specialisosceles triangles associated with the triangle ABC , each of which shares a side and an angle with
ABC . Special containers of the first kind.
Let B ′ denote the point on the ray ~CB , for which | B ′ C | = | AC | = b (see Fig. 1). Analogously, let C ′ (and C ′′ ) denote the points on ~AC (resp., ~BC ) such that | AC ′ | = c (resp., | BC ′′ | = c ). Obviously, the triangles AB ′ C , ABC ′ , and ABC ′′ are isosceles. We callthem special containers of the first kind associated with ABC . Special containers of the second kind.
Let B denote the point on the ray ~AB , different from A , for which | B C | = | AC | = b (see Figure 2). Analogously, let C (resp., C ) denote the point on ~AC B Aa bc C C ′′ C ′ B ′ Figure 1.
Special containers of the first kind AB ′ C, ABC ′ , and ABC ′′ .(resp., ~BC ) for which | BC | = | AB | = c and C = A (resp., | AC | = | AB | = c and C = B ). Thetriangles AB C , ABC , and ABC are called the special containers of the second kind associated with ABC . B Aa bc cb cC C C B Figure 2.
Special containers of the second kind AB C , ABC , and ABC . Special containers of the third kind.
Let A be the intersection of the perpendicular bisectorof BC and the line AC . Since we have b = | AC | < | AB | = c , the point A lies outside of ABC .Analogously, denote by B (resp., C ) the intersection of the perpendicular bisector of AC (resp. AB )and the line BC . Note that ABC and
ABC do not contain
ABC if ∢ BCA ≥ ◦ . The triangles ABC , ABC , and
ABC are called special containers of the third kind associated with
ABC , providedthat they contain
ABC . Thus, if
ABC is acute, then it has three special containers of the third kind.Otherwise, it has only one (see Figure 3).
B CA CAB B AC C
Figure 3.
Special containers of the third kind in the acute and in the non-acute cases.
INIMUM AREA ISOSCELES CONTAINERS 3
All special containers share a common angle and a common side with the original triangle
ABC .Obviously, there is no other isosceles container having the same property. Indeed, for each vertex of
ABC , there are at most 3 isosceles triangles that share this vertex and the angle at this vertex with
ABC , and also have a common side with
ABC .Therefore, Theorem 1.2 is an immediate corollary of the following statement.
Theorem 1.3.
All minimum area isosceles containers for a triangle are special containers of the firstkind, or of the second kind, or of the third kind.
Whenever a minimum area isosceles container of a triangle is acute, we can be more specific.
Theorem 1.4.
If a minimum area isosceles container of a triangle is acute, then it is a special containerof the first kind.
One is tempted to believe that if a triangle is acute, then all of its minimum area isosceles containersare acute and, hence, all of them are special containers of the first kind. However, this is not thecase: Example 5.1 demonstrates that there are acute triangles with obtuse minimum area isoscelescontainers. As all special containers of the first kind and the third kind of an acute triangle are acute,Theorem 1.4 implies the following statement.
Corollary 1.5.
A minimum area isosceles container for an acute triangle is obtuse if and only if it isa special container of the second kind.
It follows from Theorem 1.3 that every triangle has at most 9 minimum area isosceles containers:at most 3 special containers of each kind. In the next section, we prove that there are no minimumarea isosceles triangles of the third kind (see Lemma 2.2). Thus, every triangle can have at most 6minimum area isosceles triangles. In fact, this bound can be further reduced to 3.
Theorem 1.6.
Every non-isosceles triangle
ABC has at most minimum area isosceles containers, AB ′ C , ABC ′ , and AB C . In particular, every minimum area isosceles container is a special containerof the first or the second kind.There is a unique triangle T ∗ , up to similarity, which has precisely different minimum area isoscelescontainers. Its angles are α ∗ ≈ . ◦ , α ∗ , and ◦ − α ∗ , where α ∗ is the unique solution of sin( α ) sin(2 α ) − sin (3 α ) = 0 in the interval [36 ◦ , ◦ ] . Finally, we discuss how large the area of a minimum area isosceles container for a triangle T can berelative to the area of T . We also consider the same question for special containers of the first kind. Theorem 1.7. (a) Every triangle of area has an isosceles container whose area is smaller than √ .(b) Every triangle of area has a special container of the first kind, whose area is smaller than √ .Both bounds are best possible. As (b) is best possible, there exists a triangle of area 1 for which every special container of the firstkind has area larger than √ . Therefore, by (a), none of its special containers of the first kind canbe a minimum area isosceles container. This disproves an earlier conjecture of Nandakumar, accordingto which every triangle T admits a minimum area isosceles container which is a special container ofthe first kind.Our paper is organized as follows. In Section 2, we prove some useful inequalities for the areas ofspecial containers. In particular, we prove Theorem 1.7 (b). In Section 3, we establish some elementaryproperties of minimum area isosceles containers. Section 4 contains the proofs of Theorems 1.3, and1.4. Finally, Theorem 1.6, and 1.7 (a) are proved in Section 5.2. Preliminaries—Proof of Theorem 1.7 ( b)In this section, we collect some basic facts about special containers and establish Theorem 1.7 (b).First, we consider special containers of the first kind, because their areas can be easily comparedto the area of the triangle
ABC . As everywhere else, we assume that the side lengths of
ABC satisfy a < b < c . The area of
ABC is denoted by t ( ABC ). Lemma 2.1.
For any non-isosceles triangle
ABC , we have(a) t ( ABC ′′ ) > t ( ABC ′ ) ,(b) t ( AB ′ C ) > t ( ABC ′ ) (resp., t ( AB ′ C ) ≥ t ( ABC ′ ) )if and only if b > ac (resp., b ≥ ac ). GERGELY KISS, J´ANOS PACH, AND G´ABOR SOMLAI
Proof.
Let m be the length of the altitude of ABC perpendicular to the side BC . We have t ( ABC ) = a · m and t ( AB ′ C ) = b · m . Thus, the ratio t ( AB ′ C ) /t ( ABC ) = b/a .Similar arguments show that t ( ABC ′ ) t ( ABC ) = cb and t ( ABC ′′ ) t ( ABC ) = ca . (a)
Since b > a , we have t ( ABC ′′ ) > t ( ABC ′ ). (b) Straightforward. (cid:3)
Next, using Lemma 2.1, we determine the supremum of the ratio of the area of the smallest isoscelescontainers of the first kind to the area of the original triangle
ABC . This will also provide an upperbound for the ratio of the area of a smallest area isosceles containers to the area of the original triangle.This fact will be used in the sequel.
Proof of Theorem 1.7 (b).
Let r ∗ denote the supremum of the ratio of the area of a smallestcontainer of the first kind associated with ABC and the area of
ABC , over all triangles
ABC withthe above property. We show that r ∗ = √ . Suppose without loss of generality that a = 1. By ourassumptions and the triangle inequality, we have 1 < b < c < b + 1. Let r ( b, c ) := ( b if b ≤ c,c/b if b > c. By Lemma 2.1, r ( b, c ) = min (cid:16) t ( AB ′ C ) t ( ABC ) , t ( ABC ′ ) t ( ABC ) (cid:17) , and r ∗ = sup c and b < √ , then r ( b, c ) = cb < b < √ . Otherwise, if b ≥ √ (and, hence, b > c ),then r ( b, c ) = cb < b +1 b = 1 + b ≤ √ = √ . Thus, we obtain that r ( b, c ) < r ∗ ≤ √ , for all1 < b < c < b + 1, i.e, the supremum r ∗ is not attained for any triangle ABC .The supremum of r ( b, c ), restricted to the parabola arc c = b < b + 1 in the ( b, c ) plane, is √ .Since every point ( b, c ) of this arc corresponds to a triangle with side lengths 1 , b, c , we obtain that r ∗ = √ , as required. (cid:3) We show that for any special container of the third kind, there exists a special container of thesecond kind whose area is smaller.
Lemma 2.2.
For any triangle
ABC we have t ( AB C ) < t ( ABC ) . If ABC is acute, then we also have t ( ABC ) < t ( ABC ) and t ( ABC ) < t ( ABC ) . Thus, a minimum area isosceles container can never bea special container of the third kind. Proof.
We verify only the first inequality; the other two statements can be shown analogously.Assign planar coordinates to the points. We can assume without loss of generality that A = (0 , B = (2 , C = ( p, q ), and C = (1 , d ). Then t ( ABC ) = d . The equation of the line passing through B , C , and C is dx + y = 2 d . Taking p = 1 + s for some 0 < s <
1, we have q = d (1 − s ) and B = (2(1 + s ) , t ( AB C ) = d (1 − s ) < d = t ( ABC ). (cid:3) A = (0 , B = (2 , s, C = (1 , d ) B C = (1 + s, d (1 − s )) Figure 4.
Proof of Lemma 2.2.
INIMUM AREA ISOSCELES CONTAINERS 5 Four useful lemmas
The aim of this section is to prepare the ground for the proofs of the main results that will be givenin the next two sections.First, we show that the problem is well-defined, that is, for every triangle
ABC , there is at leastone isosceles triangle containing
ABC , whose area is smaller than or equal to the area of any otherisosceles container.
Lemma 3.1.
Every triangle
ABC has at least one minimum area isosceles container.
Proof.
It follows from Theorem 1.7 (b) that the area of a minimum area isosceles container is at most √ times larger than the area of the original triangle ABC . Therefore, the vertices of any minimumarea isosceles container must lie within a bounded distance from
ABC , and the statement follows bya standard compactness argument. (cid:3)
Lemma 3.2.
Let
ABC be a triangle and
SP R a minimum area isosceles container for
ABC . Then
A, B, C must lie on the boundary of
SP R , and each side of
SP R contains at least one of them.
Proof.
First, we show that each side of
SP R contains a vertex of
ABC . Indeed, if one of the sides didnot contain any vertex of
ABC , then we could slightly move it, parallel to itself, to obtain a smallerisosceles container.Assume next, for contradiction, that A , say, does not lie on the boundary of SP R . Then we couldslightly rotate
ABC about B or C , to bring it into a position where only one of its vertices lies on theboundary of SP R . However, in that case, at least one of the sides of
SP R would contain no vertex of
ABC . (cid:3) Let
SP R be an isosceles triangle and let m S , m P , m R denote the midpoints of the sides P R , SR ,and SP , respectively. The boundary of SP R splits into three polygonal pieces, \ m P m R , \ m R m S and \ m S m P , each of which consists of two closed line segments. Namely, \ m P m R = m P S ∪ Sm R , \ m R m S = m R P ∪ P m S , \ m S m P = m S R ∪ Rm P . See Figure 5.
Lemma 3.3.
Let
ABC be a triangle and
SP R a minimum area isosceles container for
ABC . Then,each of the closed polygonal pieces \ m P m R , \ m R m S , and \ m S m P contains precisely one vertex of ABC . Proof.
By Lemma 3.2, the vertices
A, B, C lie on the boundary of
SP R . Suppose for contradictionthat the closed polygonal piece \ m P m S contains two vertices of ABC . We may and do assume withoutloss of generality that these vertices are A and C .Let T and T denote the intersection points of the segment m P m S with AB and CB , respectively.The quadrilateral CAT T ⊆ Rm P m S , so that t ( CAT T ) ≤ t ( Rm P m S ). Since | T T | ≤ | m P m S | , wehave t ( T T B ) ≤ t ( m S m P m R ). Consequently, we get t ( ABC ) ≤ t ( Rm P m S ) + t ( m S m P m R ) = 12 t ( SP R ) . Equality holds if and only if A = R, B = S, C = m S or A = m P , B = P, C = R .On the other hand, by Theorem 1.7 (b), we obtain t ( SP R ) < √ t ( ABC ), the desired contradic-tion. (cid:3)
GERGELY KISS, J´ANOS PACH, AND G´ABOR SOMLAI
S PRT T m R m S m P BA C
Figure 5.
Illustration for the proof of Lemma 3.3.
Lemma 3.4.
Let
ABC be a triangle and
SP R a minimal area isosceles container for
ABC . Then
ABC and
SP R have a common vertex.
Proof.
By Lemma 3.2, the points
A, B, C must lie on the boundary
SP R . Suppose for contradictionthat none of them is a vertex of
SP R . In view of Lemmas 3.2 and 3.3, there are two possibilities:each of the segments m P R, m R S, m S P contains precisely one vertex of ABC , or each of the segments m P S, m R P, m S R contains precisely one vertex of ABC . Suppose without loss of generality that A ∈ m P R, B ∈ m R S, C ∈ m S P , as in Figure 6.Let Q denote the center of the circle circumscribed around SP R . It is easy to see that a smallclockwise rotation about Q will take ABC into a position such that all of its vertices lie in the interiorof the triangle
SP R . This contradicts the minimality of
SP R and Lemma 3.2. (cid:3)
S PRm R m P m S AB CQ
Figure 6.
Illustration for the proof of Lemma 3.4.4.
Nandakumar’s conjecture—Proofs of Theorems 1.3 and 1.4
We start with the proof of Theorem 1.3, which immediately implies Nandakumar’s conjecture (The-orem 1.2).
Proof of Theorem 1.3.
Let
SP R be a minimum area container for
ABC with apex R . By Lemma3.2, A, B , and C are on the boundary of SP R , and, by Lemma 3.4, the triangles
SP R and
ABC sharea vertex. Using Lemma 3.3 under the assumption that
ABC = SP R , we can distinguish 8 cases, upto symmetry (see Figure 7). Cases (1)–(3) represent those instances when
ABC and
SP R have twocommon vertices. In these cases,
SP R is a special container of the first, the second, and the third kind,respectively, so we are done.In the remaining cases,
ABC and
SP R have only one vertex in common. In cases (4)–(6), thisvertex is a base vertex (say, S ) of SP R . Finally, in cases (7)–(8), R is the unique common vertex of ABC and
SP R . It is sufficient to show that in cases (4)–(8), the area of
SP R is not minimal.First, we discuss cases (5)–(8). Case (4) is more delicate and is left to the end of the proof.Cases (5) and (6) are analogous. Let D denote the vertex of ABC lying on RP . In both cases, wehave ABC ⊆ SP D . Clearly,
SP R is a special container of the third kind associated with
SP D , and
INIMUM AREA ISOSCELES CONTAINERS 7
S PR (1) m R m S m P S PR (2) m R m S m P S PR (3) m R m S m P S PR (4) m R m S m P S PR (5) m R m S m P D D S PR (6) m R m S m P DS PR (7) m R m S m P S PR (8) m R m S m P D Figure 7.
The 8 cases up to symmetry. Triangle
ABC is shaded.by Lemma 2.2, it cannot be minimal. Since every container for
SP D is also a container for
ABC , weconclude that
SP R is not a minimum area container for
ABC .In case (7), we can find an isosceles triangle with apex R which contains ABC and whose base isproperly contained in
P S . Thus,
SP R was not minimal.In case (8), one vertex of
ABC is R , another (denoted by D ) belongs to Sm R , and the third lies on P m S . Since SP R is an isosceles triangle, we have ∢ RDS ≥ ◦ . Hence, ABC can be slightly rotatedabout R so that it remains within SP R , which leads to a contradiction.It remains to handle case (4). We distinguish two subcases. Denote the apex angle ∢ SRP by δ . If δ ≥ ◦ , then we can rotate ABC about S . Indeed, vertex D of ABC belongs to m S P , while the baseof the altitude belonging to P R lies on m S R . Hence, we have ∢ SDP ≥ ◦ , and the image of ABC through a small clockwise rotation about S is still contained in SP R . Therefore, in this case,
SP R cannot be minimal either.Therefore, from now on we assume δ < ◦ . Choose a suitable coordinate system, in which thevertices of ABC are (0 , s, D = ( p, q ). We also have S = (0 ,
0) and R = ( s + x,
0) for some x >
0. Since δ < ◦ < ◦ , vertex D is to the left of R , that is, p < s + x . GERGELY KISS, J´ANOS PACH, AND G´ABOR SOMLAI (4/a) (4/b) S = (0 , R = ( s + x, D = ( p, q )( s, m S m R m P P δm S = (0 , R = ( s + x, s, D = ( p, q ) m S m R m P P δm
Figure 8.
Possible realizations of Case (4) when δ < ◦ .By simple calculation,(1) P = ( s + x,
0) + ( s + x ) p ( p − ( s + x )) + q ( p − ( s + x ) , q ) . Denote by m the length of the altitude of SP R belonging to the side SR . Then, m is equal to thesecond coordinate of P (see Figure 8). We have m = q ( s + x ) p ( p − ( s + x )) + q . Let us compute the derivative of the function f ( x ) = 2 t ( SP R ) = q ( s + x ) · (cid:0) ( p − ( s + x )) + q (cid:1) − . We obtain f ′ ( x ) = 2 q ( s + x ) (cid:0) ( p − ( s + x )) + q (cid:1) − + q ( s + x ) ( p − ( s + x )) (cid:0) ( p − ( s + x )) + q (cid:1) − = q ( s + x ) (cid:0) ( p − ( s + x )) + q (cid:1) − (cid:2) ( p − ( s + x ))(2 p − ( s + x )) + 2 q (cid:3) = q ( s + x ) (cid:0) ( p − ( s + x )) + q (cid:1) − | {z } > "(cid:18) p − ( s + x ) (cid:19) − p + 2 q . Case (4/a1) : q ≥ p . Then, − p + 2 q >
0, and hence, f ′ ( x ) > x ≥
0. Thus, f is strictlyincreasing and since x cannot be negative, f takes its minimum at x = 0. This means that the area ofa special container of the first kind where x = 0 (see the triangle with dashed sides on Figure 8 (4/a))is smaller than the area of SP R for x > Case (4/a2) : 2 p < s + x . Then p < ( s + x ) − p , so that ( p − ( s + x )) − ( p ) >
0. Again, wehave f ′ ( x ) > x ≥ SP R (see Figure 8 (4/a)).
Case (4/b) : q < p and 2 p ≥ s + x . Let ∆ denote the triangle with vertices (0 , p, q ), and (2 p, p ≥ s + x that ∆ is an isosceles container of the second kind associatedwith ABC (see Figure 8 (4/b)). We show that ∆ has smaller area than
SP R . To prove this, we haveto verify that t (∆) = pq < q ( s + x ) p ( p − ( s + x )) + q = t ( SP R ) . Using our assumption that ( p, q ) ∈ m S P , we obtain p ≤ s + x + ( s + x )( p − ( s + x ))2 p ( p − ( s + x )) + q . The right-hand side of the last inequality is the first coordinate of the midpoint m S of P R , where P isgiven by formula (1) and R = ( s + x, s + x ) + ( s + x )( p − ( s + x ))2 p ( p − ( s + x )) + q < ( s + x ) p ( p − ( s + x )) + q , which reduces to 3 p + 4 q < s + x ) p . The last inequality holds, because p < s + x and, by ourassumption, 2 q ≤ p . This completes the proof of Theorem 1.3. (cid:3) INIMUM AREA ISOSCELES CONTAINERS 9
Proof of Theorem 1.4.
By Theorem 1.3, every minimum area isosceles container for a triangle
ABC is one of its special containers. By Lemma 2.2, it must be a special container of the first or the secondkind.Suppose for a contradiction that a minimum area special container
SP R associated with the triangle
ABC is acute, but it is of the second kind. Assume without loss of generality that S = B and R = A .(The other cases can be treated in a similar manner.) By our notation, P = C . We prove that t ( ABC ′ ) < t ( ABC ) = t ( SP R ) (see Figure 9). B = S P = C A = Rm R m S m P CC ′ Figure 9.
An acute minimum area isosceles container cannot be a special containerof the second kind.Indeed, we have | AB | = | AC | = | AC ′ | and ∢ C ′ AB < ∢ C AB . Since SP R = ABC is acute, itfollows that t ( ABC ′ ) < t ( ABC ). (cid:3) Corollary 4.1.
If a minimum area isosceles container for
ABC is a special container of the secondkind, then it must be AB C . Proof.
By Theorem 1.4, if a special container of the second kind has minimum area, then it hasto be non-acute. If
ABC is non-acute, then ABC is obtuse and t ( ABC ) > t ( ABC ), because | AB | = | AC | = | AC | and ∢ ABC > ∢ BAC ≥ ◦ . On the other hand, as AB C and ABC sharea base angle at A and b < c , it follows that t ( ABC ) > t ( AB C ). (cid:3) Quantitative results—Proofs of Theorems 1.6 and 1.7 ( a) Proof of Theorem 1.6.
By Theorem 1.3, a minimal area isosceles container for
ABC is a specialcontainer associated with
ABC . In view of Lemma 2.2, it must be a special container of the first orsecond kind. By Lemma 2.1 (a) and Corollary 4.1, among special containers of the first kind, it isenough to consider
ABC ′ and AB ′ C , and among special containers of the second kind, only AB C .These immediately show that every triangle ABC admits at most 3 minimum area isosceles containers.If
ABC is an obtuse or right triangle, then t ( AB ′ C ) > t ( AB C ). Indeed, in this case | AC | = | CB ′ | = | CB | , both AB ′ C and AB C are obtuse or right triangles, and their apex angles satisfy ∢ ACB ′ < ∢ ACB . Thus, there are only two candidates for a minimum area isosceles container: ABC ′ and AB C .If ABC is an acute triangle and it has 3 minimum area isosceles containers, then t ( AB ′ C ) = t ( ABC ′ ) = t ( AB C ). Since t ( BCB ) = t ( BCC ′ ), we obtain(2) ( c − b ) sin( α + β ) = b sin( β − α ) . Note that this equation also holds when
ABC is obtuse.It follows from equation (2) that cb = β ) cos( α )sin( α + β ) . By Lemma 2.1 (b), the equation t ( AB ′ C ) = t ( ABC ′ ) reduces to cb = ba . Thus, sin( β )sin( α ) = β ) cos( α )sin( α + β ) , so that sin( α + β ) = sin(2 α ). Therefore, either α = β , which is impossible, or 180 ◦ − ( α + β ) = γ = 2 α .It follows from cb = ba that(3) sin(2 α )sin(3 α ) = sin(3 α )sin( α ) . Since
ABC is acute, its smallest angle is α , and 180 ◦ − α = β < γ = 2 α , we have that 36 ◦ < α < ◦ .Simple analysis shows that equation (3) has exactly one solution α ∗ in the interval [36 ◦ , ◦ ]. It can be approximated by computer. The other two angles of the corresponding triangle are β ∗ = 180 ◦ − α ∗ and γ ∗ = 2 α ∗ . (cid:3) Example 5.1.
By the proof of Theorem 1.6, any minimal area isosceles container for
ABC is either AB ′ C , or ABC ′ , or AB C . Here, we construct a family of acute triangles ABC whose only minimalarea isosceles containers are special containers of the second kind, i.e., AB C . Moreover, AB C isobtuse.Let α > α ∗ and 90 ◦ > γ > α > γ ∗ . Then, denoting by β = 180 ◦ − α − γ we obtain thatsin( γ )sin( β ) > sin( γ ∗ )sin( β ∗ ) = sin( β ∗ )sin( α ∗ ) > sin( β )sin( α ) , which implies that t ( AB ′ C ) < t ( ABC ′ ). The triangles AB C and AB ′ C are isosceles with legs oflength b , so it is enough to show that sin( ∢ ACB ) = sin(180 ◦ − α ) < sin( γ ) = sin( ∢ ACB ′ ). However,this follows from the inequalities 90 ◦ > γ > α . The base angle of AB C satisfies α < ◦ , so that AB C is obtuse. Proof of Theorem 1.7 (a):
Let r ∗ denote the supremum of the ratio of the area of a minimum areaisosceles container of a triangle to the area of the triangle itself. In view of Theorem 1.6, we have r ∗ = sup triangle ABC min (cid:16) t ( AB ′ C ) t ( ABC ) , t ( ABC ′ ) t ( ABC ) , t ( AB C ) t ( ABC ) (cid:17) . If β ≥ ◦ , then sin( β ) ≥ √ . Using the proof of Lemma 2.1(b) and the law of sines, we obtain t ( ABC ′ ) t ( ABC ) = cb = sin( γ )sin( β ) ≤ √ = √ . Equality holds here if and only if β = 45 ◦ and γ = 90 ◦ , in which case ABC is an isosceles triangle andthe ratio of the area of the minimum isosceles container to the area of
ABC is 1.If β < ◦ , then γ > ◦ . Hence, ABC is obtuse and, by the proof of Theorem 1.6, the minimumarea isosceles container is
ABC ′ or AB C . For fixed β and c , we can express the ratios of the areas asfunctions of α . Let f ( α ) = t ( ABC ′ ) t ( ABC ) = cb and g ( α ) = t ( AB C ) t ( ABC ) = 2 b cos( α ) b cos( α ) + a cos( β ) = 1 + tan( α )2 tan( β ) , where 0 < α < β . Obviously, f ( α ) is strictly increasing and g ( α ) is strictly decreasing on the openinterval (0 , β ), and both functions are continuous. We havelim α → f ( α ) = 1 , < lim α → β − f ( α ) < , lim α → g ( α ) = 2 , lim α → β − g ( α ) = 1 . Therefore, the graphs of f and g intersect at a unique point z . Thus, max ◦ <α<β (min ( f ( α ) , g ( α ))) = f ( z ) = g ( z ), which implies t ( ABC ′ ) = t ( AB C ). This means that t ( BCB ) = t ( BCC ′ ), so thatequation (2) above holds. Using the law of sines, we obtain ( cb ) = 2 cos( z ) < cb < √ β →
0, then z → c/b → √
2. This implies that r ∗ = √
2, but the supremum is not realized byany triangle
ABC . (cid:3) Acknowledgement.
The authors would like to express their gratitude to the anonymous referee for heruseful suggestions. Research by the first author was supported by a Premium Postdoctoral Fellowshipof the Hungarian Academy of Science, and by NKFIH (Hungarian National Research, Development andInnovation Office) grant K-124749. Research by the second author was partially supported by NKFIHgrants K-176529 and KKP-133864, Austrian Science Fund grant Z 342-N31, and by the Ministry ofEducation and Science of the Russian Federation in the framework of MegaGrant No. 075-15-2019-1926.Research by the third author was partially supported by NKFIH grant K-115799.
INIMUM AREA ISOSCELES CONTAINERS 11
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Gergely Kiss, Alfr´ed R´enyi Institute of Mathematics, POB 127, Budapest H-1364, Hungary
E-mail address: [email protected]´anos Pach, Alfr´ed R´enyi Institute of Mathematics, POB 127, Budapest H-1364, Hungary andMIPT, Dolgoprudny, Moscow Oblast 141701, Russia
E-mail address: [email protected]´abor Somlai, L´or´and E¨otv¨os University, P´azm´any P´eter s´et´any 1/C, Budapest H-1117, Hungary