Targil 8.
1. What is the radius of the largest planar disc inside the unit cube?
3
.
2
Solution. Unless the plane is parallel to one of the faces, the intersection might be
defined as follows. Each pair planes of parallel faces define a strip on that plane – a
part of plane between two parallel lines. The planar section is the intersection of 3
such strips. Each strip has its width, and the diameter of the disc is bounded by that
width. If normal vector to the plane is a unit vector (a, b, c), then the width of
1 1 1
, ,
parallel strips are
(easy exercise to the reader). The largest among
a b c
Answer.
a , b , c must be at least
1
, since sum of their squares is 1. Therefore the
3
thinnest of the three strips is of width
3 at most, and the radius is bounded by
3
anyway.
2
The equality can be achieved: when the plane is the perpendicular bisector of
cube’s diagonal, the section is a regular hexagon, and the disc is tangent to all its
sides, hence its diameter equals the width of all three strips. But in this case
1
, therefore the diameter is 3 .
abc
3
2. A box {0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , 0 ≤ z ≤ A }, where A is a positive real, is
intersected by a family planes {x + y + z = n + α}, where α is a real number, and n
are all possible integer numbers. The intersections of this family of planes and the
box gives a family of planar polygons.
a. Prove that sum of areas of these polygons does not depend on α.
b. Prove that mass center of all these polygons is the center of the box.
Solution. By shifting the polygons in x direction by integer numbers let us move
all of them to the plane x + y + z = α.
We shall get the intersection of plane x + y + z = α and infinite box
{0 ≤ y ≤ 1 , 0 ≤ z ≤ A }. The intersection is a parallelogram.
Its area obviously doesn’t depend on a, and its mass center is the same as its center
of symmetry. Therefore, its mass center has y = ½ , z = A/2.
The same can be said about the original system of polygons, because we moved
everything in x direction, so y and z of the mass center are unchanged.
But the original system is symmetric w. r. t. the plane x = y, so the center of mass
also has x = ½ . So it is in the center of the box, regardless of α.
3. What is the greatest possible area of the orthogonal projection of the unit cube?
(among all possible directions)
Solution. We shall prove a nice lemma.
Lemma. Area of the orthogonal projection of the unit cube to a plane equals length
of orthogonal projection of that cube to the line, which is orthogonal to the plane.
From this lemma it directly follows that the greatest projection is to the plane,
which is orthogonal to cubes diagonal, and its area is equal to the length of the
diagonal, which is
3.
Proof of lemma. Projection of each face is a parallelogram. These parallelograms
are congruent. One of projected parllelograms is ABCD (we may assume it is nondegenerate), another is KLMN (shifting by vector AK moves A to K, B to L, C to
M, D to N).
We may assume WLOG that when vector AK is expressed as a
linear combination of vectors AB and AD, the coefficients are
nonnegative. In this case, the projection is a hexagon ABLMND
(this hexagon is convex though in some cases, it degenerates into
rectangle).
The intervals AL, LN, NA cut parallelograms BAKL, KLMN, DAKN into pairs of
equal triangles; hence the area of triangle LAN is half the area of the entire
hexagon.
If we have a planar polygon P in space, and an interval I which is orthogonal to the
plane of that polygon, then the area of projection from P to the plane is orthogonal
to the length of projection of I to the line orthogonal to that plane, because both the
length and the area are multiplied by cosine of the same angle.
Hence in our case, we see that the area of projection of the cube is twice the area of
projection of a triangle, which is proportional to the length of projection of the
diagonal of the cube to the orthogonal line. This diagonal is formed by two
opposite vertexes, both of which are projected inside the hexagon, so one of them
is the upmost, and another downmost with respect to the plane of projection.
Therefore, projection of diagonal to the orthogonal line coincides with the
projection of the cube to the same line.
So, the length of the projection of the cube to the line is proportional to the area of
its projection to the orthogonal plane. It only remains to compute the
proportionality coefficient. This is easy – just consider a plane which is parallel to
a face of the cube.
4. a. What is the greatest triangular planar section of a tetrahedron (not necessarily
regular)?
b*. What is the greatest possible planar section of a tetrahedron?
Answer. The greatest face.
Solution. a. Take some triangular section: its vertexes are 3 points A, B, C on three
different edges. While keeping A and B stable, move C along its edge.
The basis AB of the triangle ABC is stable; so the only thing that influences the
area is the distance from line AB to C. Since distance from a line to a point is a
convex function of the point, the maximum will be achieved in one of the
endpoints, which is a vertex of the tetrahedron.
Therefore, in the greatest triangular section C will be a vertex; similarly, A and B
will be vertexes. So it will be a face. QED.
b. In the previous section, we saw that the greatest triangular section is a face.
However, there are also quadrilateral sections. So, now we shall consider
quadrilateral section KLMN of a tetrahedron ABCD. The cutting plane splits
between two non-adjacent edges: WLOG, those are AB and CD, so we shall
assume that K is on AC, L is on CB, M is on BD, N is on DA.
Lemma. (three-dimensional version of Menelaus theorem)
AK CL BM DN



 1.
KC LB MD NA
Proof of lemma. Let l be a line, orthogonal to the plane KLMN.
We shall take the orthogonal projection of the whole picture to the line l.
The points A, B, C, D will be projected to A ', B ', C ', D ' , the point K, L, M, N will
all be projected to the same point O.
AK A ' O
CL C ' O
BM B ' O
DN D ' O

,

,

,

KC OC '
LB OB '
MD OD '
NA OA '
Multiplication of these 4 fractions gives the result.
Now, we shall prove that the quadrilateral section is smaller than one of the faces.
Project orthogonally the tetrahedron to the plane KLMN: the vertexes A, B, C, D of
will go to the points A1, B1, C1, D1 on the plane. We shall prove that the area of
KLMN is not greater than projection of one of the faces, so before the projection it
was greater that face was greater.
Assume that we know the ratios in which K, L, M N divide the sides:
A1K AK

CL C1L

BM B1M

D N DN



,


,


, 1 

KC1 KC  
LB LB1  
MD MD1 1  
NA1 NA 1  
Denote S  S B1C1A1D1 , S A  SC1A1D1 , SC  SB1C1A1 , SB  SC1B1D1 , SD  SB1D1A1 .
The maximum between SA, SB, SC, SD will be denoted Smin, the maximum Smax.
Of course, S  S A  S B  SC  S D  S min  S max .
We want to prove SKLMN ≤ Smax. To compute SKLMN, we shall compute the rest of the
area inside S:
S NKA1      S A , S KLC1      SC , S LMB1      S B , S MND1      S D
Hence
S  S KLMN      S A      SC      S B      S D 
 Smin                    
By the lemma, we have the condition:   1   1   1   1    .
Therefore,
0  1                             .
                    1            
We shall prove that the last expression is at least 1. From this we get directly:
S  SKLMN  Smin                      Smin
S max  S KLMN , QED.
So it remains to prove: 1              1 .
In other words,            .
Recall that 0   ,  ,  ,  1 .
1
1 1 1 1 1
So the required result is:

    .






 1  1   1  1 
In other words,   1   1    1   1  2
   

 
 
1
 1 
 1  1 
Denote U    1   1 , V    1   1
   



 
We already know   1   1   1   1    , if we divide both sides of
this condition by  we get: UV = 1.
U V
Hence by AMGM (Cauchy inequality) we get
 UV  1.
2
That means U  V  2 , QED.
5. a**. What is the greatest possible area of a planar section of a unit cube?
b**. Same question for the box a×b×c.
Answer. a.
2
b. If a < b,c it is b 2  c 2
Anyway, the greatest section is “diagonal” rectangle.
Solution. Let us start with easy steps:
Lemma 1. We may assume that the greatest section passes through the center.
Lemma 2. The section through center is a central-symmetric polygon:
quadrilateral or hexagon.
Lemma 3. The greatest quadrilateral section is the diagonal.
Proof of lemma 1. Let P be a section which doesn’t go through the center. Let Q
be another section, symmetric to P w.r.t. the center of the box (Q is symmetric, and
hence congruent to P). Consider a family of straight parallel lines in the plane of P,
that intersect P. The extreme two lines in this family will be denoted l1, l2.
Line k2, k1 are symmetric to l1, l2 respectively with respect to the center of the box.
Consider the family F of parallel planes, among which are the plane through l1 and
k2, the plane through l2 and k1, and all the planes between them.
Let M be the planar section of the same box, is parallel to both P and Q which and
passes through the center of the box.
Consider intersection of any plane from F with P, Q, and M. All three intersections
are intervals; intervals of intersection with P and Q form a trapezoid, which is
entirely contained in the box (since the box is convex) so the mid-segment of the
trapezoid is entirely contained in M. So, length intersection of the plane with M is
greater or equal to the average of intersections with P and Q. Since area of each
section can be computed as integral of intersection lengths with all planes, parallel
to a certain direction, we see that are of section M is greater or equal than the area
average area of P and Q which is equal to the area of P. So, if instead of P we
consider parallel section through the center, it will have greater or equal area.
Proof of lemma 2. Each face of a box defines a half-space, and the box itself is
intersection of these half-spaces. In every plane, this half-spaces define half-planes
(though sometimes they give an empty set or the entire plane). So, we have a
polygon with no more than 6 sides (because we start with no more than 6 halfspaces). The plane and the box are symmetric w. r. t. the center of the box, hence
the polygon is also symmetric w. r. t. the same center. Hence number of sides is
even (there are pairs of opposite sides). Therefore number of sides is 4 or 6 (2 is
impossible).
Proof of lemma 3. If we have quadrilateral, it doesn’t intersect a pair of opposite
faces. WLOG these are horizontal faces (if not, rotate the box). Therefore it can be
projected orthogonally onto each of those two faces. Hence the area of the section
is the area of the face divided by the cosine of the slope. We want that cosine to be
minimal.
Unless the cutting plane is horizontal (but then the cosine is maximal), let us walk
along the steepest descent direction along the cutting plane, until we reach the
plane of the horizontal face. We shall cover y = half the height of the box
vertically, and distance x horizontally. The tangent of the slope angle will be y/x:
we want it to be maximal, then the slope would be maximal, the cosine minimal,
and the area maximal. So, we want x to be minimal. It is the distance in horizontal
projection to a point which is in the plane of the horizontal face but not inside the
face itself. The closest point in that set is obviously the center of the longest edge.
But in this case we have the “diagonal” section, QED.
Now it remains to study the hexagonal case, and to prove it cannot be of greater
area than the “diagonal” rectangle.
Lemma 4. Consider a polytope. For each face, consider an outside normal vector
of this face which is of length equal to the area of that face. Then sum of those
vectors is a zero vector.
Lemma 5. Let consider the closed broken line A1A2…A2n, each interval of which
intersects certain plane: A2nA1 intersects that plane at B1, A1A2 at B2, A2A3 at B3 etc.
A B AB A B
A B
Then 2 n 1  1 2  2 3  ...  2 n1 2 n  1 .
B1 A1 B2 A2 B3 A3
B2 n A2 n
Lemma 5 is very similar to the lemma from the solution of problem 4, so we shall
leave the proof as an exercise to the reader.
Lemma 4 has and extremely beautiful physical proof, though some of you might
not accept it as a proof, hence we shall give two proofs.
First proof of lemma 4. Consider a large part of space of air (or water) standing
still with no wind (no current). Consider inside a smaller part of space in the form
of that polytope. It doesn’t move. Hence the sum of forces applied to it is zero.
Forces applied to it are forces of air pressure (‫ – )לחץ אוויר‬they are proportional to
the areas of the faces and directed into the faces. Reverse the signs, QED.
Second proof of lemma 4 (which is just a translation of the first proof to a
rigorous language). We shall prove that z coordinate of the sum of normal vectors
is zero, it is the same for x and y coordinates. The z coordinate of normal vector
equals to the oriented area of projection of that face to xy plane: it is considered
positive, it the original face was facing up, negative if down. So, the z coordinate is
sum of areas of xy-projections facing up minus sum of xy-projections facing down.
Since above each point there’s the same number of both kinds, they cancel out.
So, back to our problem. We cut the box through the
center in two halves by a hexagon. This hexagon cuts
6 edges into 12 sub-intervals of lengths:
u, a – u, v, b – v, w, c – w, u, a – u, v, b – v, w, c – w
(recall that lengths of edges are a, b, c, and last six
repeat first six because of the symmetry).
u
v
w
u
v
w
By lemma 5 we get





 1.
a u bv c  w a u bv c  w
u
v
w
Or simply


 1.
a u bv cw
We shall find the expression for the area using lemma 4. Consider one of the
halves of the box as a polytope with 7 faces. Sum of normal vectors is zero.
Take the two vertical vectors. One corresponds to a×b rectangle with right-angled
triangle (with legs a – u and v) cut out, another is a triange with opposite
orientation with legs a – u and v.
The contribution of these two sides together is the same as we would get from a×b
rectangle if rectangle (a – u)×v would be cut out. So, total of these two vectors is
of length and looking down ab – (a – u)v.
Similar things can be said about other pairs of parallel faces. So the total of six
   ab   a  u  v  


vectors (all of them except the hexagon) is    bc   b  v  w   .


   ca   c  w  u  


The total of all seven vectors is zero, so the vector that corresponds to the hexagon
can be described by the same expression with opposite sign in each coordinate.
For the case of cube, finding the hexagonal section of greatest are is reformulated
u
v
w


 1 , find the maximum of
as follows: given the condition
1 u 1 v 1 w
1  1  u  v   1  1  v  w  1  1  w u  .



For the case of generic box, if we denote C  ab, A  bc, B  ca (areas of the faces)
and change u, v, w by au, bv, cw, the condition will be the same as for the cube, but
the function to minimize will be
C 2 1  1  u  v   A2 1  1  v  w  B2 1  1  w u  .



This inequality is not easy, because we have to take the condition into account.
The condition reminds of
Ceva theorem. Consider triangle KLM, point P on the side LM, point Q on KL,
KR MP LQ
point R on LK. Assume KP, LQ, MR meet in one point. Then


 1.
RM PL QK
We shall give the proof (though it is a standard fact from
elementary geometry, but that is one of those things that are too
elementary for the university and too hard for the high-school, so
many people might not have heard of it).
Proof of Ceva theorem. Put mass k in point K and mass l in point L so that the
mass center of these two points will be at R. Put mass m in point M such that the
mass center of K and M will be at R. Then the mass center G of all three points
would be both on KP and on MR. Also, G will also be on MQ' , where Q' is the
KR MP LQ' m k l
mass center of K and M. It is easy to see that


    1.
RM PL Q'K k l m
But since G is intersection of KP and MR, also LQ goes through it, hence Q and
Q' coincide. QED.
u
v
w


 1 . We can say that what we actually
1 u 1 v 1 w
have are Ceva picture with equilateral triangle KLM, all sides of which are 1, and
(in the notations of Ceva theorem as we formulated it):
KR = w, RM = 1 – w, MP = u, PL = 1 – u, LQ = v, QK = 1 – v.
Our problem was that this parametrization was hard to handle: it comes with an
ugly condition. Let us consider reformulating it in terms of masses (like in the
proof of Ceva theorem) the intersection point is a mass center of mass x at point K,
mass y at point L, and mass z at point R. Then, since points P, Q, R are centers of
mass of couples of points we get
So, we had the condition
y
x
z
y
x
z
, 1 w 
, u
, 1 u 
, v
, 1V 
x y
x y
yz
yz
xz
yz
Where x, y, z, are any real numbers. Inequalities about arbitrary real numbers are
usually easier than inequalities about numbers satisfying some fancy condition.
x
z
 x  y  y  z   xz  y  x  y  z 
1  1  w u  1 


x y yz
 x  y  y  z 
 x  y  y  z 
w
Similar for the other two expressions, hence:
C 2 1  1  u  v   A2 1  1  v  w   B 2 1  1  w  u  



 x x  y  z   y x  y  z   z  x  y  z 
A
 B
  C

x

y
x

z
x

y
y

z
x

y
y

z










 
 

And we have to find, for which positive x, y, z will that achieve its maximal value.
2
2
2
Consider yet another geometric picture. Consider triangle ABC, its sides
AB = x + y , BC = y + z , CA = z + x (yes, this creature really exists).
In this context, x, y, z have a geometric meaning: the tangency
points of the incircle and triangles sides divide the sides of
triangle into the intervals of lengths x, y, z.
Denote angles of the triangle BAC = 2α, ABV = 2β, ACB = 2γ.
x x  y  z
2
Easy exercise.
  cos   .
 x  y  x  z 
Hints. (1). Denote p = x + y + z. (2) First prove that S = pr, where r is the radius of
incircle.
So, finding for non-negative x, y, z of the maximum of the function,
 x x  y  z   y x  y  z   z  x  y  z 
A
 B
  C

x

y
x

z
x

y
y

z
x

y
y

z










 
 

Is the same as for nonnegative  ,  ,  such that       1 finding the
2
2
2
maximum of the function A2 cos4   B2 cos4   C 2 cos4  .
True, we now have the condition again, but not that ugly and also the function is
much nicer.
The last lemma. The maximum of that function will be in one of the vertexes of
the domain (which is triangle). That is  ,  ,  = 0, 0, 1 in some order.
This case corresponds in the original problem to degenerate case, when the
hexagon becomes the rectangle. In this case the value of the function (which was
the squared are of the hexagon) is something like A2 + B2, which is squared area of
the diagonal rectangle.
So, it remains to prove the last lemma.
Proof of the last lemma. (Alexey Gladkich) Our domain is triangle:  ,  ,   0 ,
      1 . The maximum can be either in the vertex, or on the side, on in the
interior.
Assume that the maximum is on the side   0 ,     1. The function is
A2 cos4   B 2 cos4   C 2  A2t 2  B 2 1  t   C 2 , where t  cos 2    0,1 .
2
As a function of t, it is convex; so maximum is in one of the endpoints of the
domain, hence at the vertex of the triangle.
It remains to exclude the possibility of the maximum inside the triangle.
Then we can differentiate:
df    ,    ,  
0
 A2 cos3  sin   B 2 cos3  sin 
d
 0
Hence, A2 cos3  sin   B2 cos3  sin  . Doing the same for another couple of
coordinates, we get A2 cos3  sin   B2 cos3  sin   C 2 cos3  sin  .
Therefore, we can apply some scaling to the coefficients (that won’t shift the
position of the maximum) and assume:
1
A2 
cos  sin 
1
B2 
cos3  sin 
1
C2 
cos3  sin 
3
Hence at the maximal point f  ,  ,    cot    cot     cot   .
1
At vertexes we have things like f  0,0,1 
cos3  sin 

1
cos3  sin 
.
We shall assume that      , and prove that
1
1
,
cot    cot     cot   

cos3  sin  cos3  sin 
and that will complete the proof.
First, notice that
1
 cos   sin  

2
2
cos3  sin 
cos3  sin 
0     / 2 . Hence it is enough to prove that
2
 cot   2 tan  , at least for
cot    cot     cot     cot    2 tan    cot     2 tan   
cot     2 tan    2 tan   
 
. Then cot    tan      tan  2  .
2
In the relevant domain tan is a convex function (first derivative is 1/cos 2 and it
tan    tan   
   
 tan 
grows monotonically). Hence
  tan   .
2
 2 
We are supposed to prove cot    2tan    2tan    , but it is enough to prove
Let’s denote  
that tan  2   4tan   , because 4tan    2  tan    tan     .
By assumption      , hence  
Hence the claim tan  2  
 
2

   
3


6
. So tan   
2 tan  
 4 tan   is obvious. QED.
1  tan 2  
1
.
3