# sol round2 9 12 by HC120918125621

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```									                                               2002-2003 Problems

Maine Mathematics Science and
Engineering Talent Search

1. Connect the midpoints of two opposite sides of a convex quadrilateral to the corners as
shown. Showe that the area of the blue region is equal to the area of the red.

Solution . Let ABCD the convex quadrilateral, M and m the midpoints , Dh and BH the
distance between sides AB and CD as shown. The diagonal BD (as shown in blue color
separates the area of the quadrilateral for the area of triangles ABD and BCD.

H
C
M

D

A
B
h                    m
1             1
Area of (ABCD) = Area of (ABD) + Area of (BCD) =               ( AB  hD)  (CD  HB) 
2             2

1             1              1            1
 ( Am  hD)  (mB  hD)  ( MC  HB)  ( MD HB) =
2             2              2            2
1             1                            1              1
Blue region{ (mB hD) + ( MD HB) } +Red region { ( Am hD) + ( MC HB)                        }
2            2                            2              2
As m and M are the midpoints of the sides AB and BD respectively, we know that
Am = mB and CM = MD : substituting these terms in the equation,

1             1                               1               1
Blue region{ (mB hD) + ( MD HB) }+Red region { (mB hD) + ( MD HB) }
2             2                               2               2
The terms multiplied together are identical, wherefore the areas of blue region and red
region are equal and ½- ½ of the area of the convex quadrilateral.

2.Given three spheres that pairwise touch each other and also touch the plane P,
determine the radius of the sphere that has its center on the plane P and that touches all
the three spheres.

Solution

First, we have to impose some conditions on the spheres. Draw the three spheres and
connect the respective centers and also connect the projections of the centers on the
plane.

The centers are C1 , C 2 , C 3 and projections P1 , P2 , P3 . Since C1C2 connects the spheres of
two tangent spheres,

C1C2  r1  r2 , likewise C2C3  r2  r3 . and C1C3  r1  r3            (1)
If the spheres are tangent to the plane, the triangle P P2 P3 must exist.
1

Using the Pythagorean theorem

(C1C3 )2  (r1  r3 )2  P P         P P3  (r1  r3 ) 2  (r1  r3 ) 2 =    2r1r3  (2r1r3 )  2 r1r3
2
1 3          1

Equivalently,            P P2  2 r1r2
1                 and      P2 P3  2 r2 r3                        (2)

So, since the triangle must exist, its sides must obey the triangle inequality.

2 r1r2  2 r2 r3  2 r1r3                r1r2  r2 r3  r1r3

Likewise      r3r1  r2 r3  r1r2 and r1r3  r1r2  r2 r3
These conditions are necessary for the spheres to be tangent to each other and the plane.
There are two half-spheres with centers on the plane and tangent to the three given
spheres.

The First sphere
The small half-sphere shaded with a darker tone of gray is the first sphere. Let P the
center of the small half-sphere sphere and x be the radius of the sphere.

PC1  x  r1             PP1  ( x  r1 ) 2  r1  x  2r1 x
2      2

PC 2  x  r2           PP2  x  2r2 x
2

PC3  x  r3             PP3  x  2r3 x
2

Drawing the triangle P1P2P3, the area of the triangle equals the sum of the areas of P1PP2
and P1PP3 and P2PP3 together. Using the Heron formula for the areas the radios x can be
expressed in terms of r1, r2 and r3.

P2

P1
P

P3
Second sphere: the second sphere is covering the three spheres

Let P be the center of the half-sphere and x be the radius.
PC1  x  r1

PC2  x  r2

PC3  x  r3
 PP  ( x  r1 ) 2  r1  x 2  2r1 x
2
1

PP  x 2  2r2 x
2

PP  x 2  2r3 x
3

By a similar method, using the area of the respective triangles that are formed, we can
determine x as a function of r1, r2 and r3.

.

3. Prove that the following numbers are all composite.

     1000...001   (with 1991 0's)
     111...1111    (with 1989 1's)
     111...1111     (with 1990 1's)

Solution
First, we prove that the number: 1000...001                  (with 1991 0's) is a composite number.

We’ll use the fact that

a n  1  (a  1)( a n 1  a n  2  a n 3  .....  a 2  a  1) if n odd

1000 ...... 01  101990  1  (1010 )199  1  (1010  1)(...) the last expression is the
product of two numbers, therefore the original number is composite.

Next, show that 111...1111 (with 1989 1's) is composite number. The sum of the digits
(that is, the sum of the 1989 1’s) equal 27. As 27 is divisible by 3, therefore the number
written as 1989 1’s is divisible by 3. Therefore it is composite number.

Finally to show that 11……..1               (with 1990 1's) is composite number.

As 1990 is an even number, the number 11…….1 is divisible by 11.

More precisely, 11……..1 = 111010..... therefore, divisible by 11 and so, it is a
01
composite number.

4. Connect any point in the interior of a parallelogram with the four corners. Show that
the area of the blue region is equal to the area of the red.

Solution

Let P the point as described, inside the parallelogram. The perpendicular distance of AB
and CD is DH2. The parallel sides AB = CD and AD = BC.

Draw perpendiculars from P to all four sides. The area of the blue triangles:
1            1
 AD  PH3 ' BC  PH3 the ½ of the area of the parallelogram
2            2
1           1
The area of the red triangles      AB  PH1  DC  PH1 ' the ½ of the area of the
2           2
parallelogram

5. We have twenty coins of the same size and external appearance some are real gold,
while others are counterfeit. The counterfeit coins are lighter. Using at most eleven
weighings on a pan balance, how can we determine how many coins are counterfeit?

Solution

First we make a “weight trial” by placing one coin on each pan. There are three possible
outcomes:
1. One of the pans is heavier. Then one of the coins is gold, the other is the
lighter counterfeit coin. Place both coins in one pan and weigh them against
pairs of remaining blocks (those being divided into nine pairs arbitrarily).
Any pair of coins which outweigh the first pair must consist of two gold
coins. If the first pair is the heavier then both coins in the second pair are
counterfeit. If both pair balance, the second pair has one gold and one
counterfeit.
2. On the first trial weigh both pans balance. In this case both coins are gold or
both coins are counterfeit. Just as before, place both coins in one pan and
weigh them against pairs of the remaining coins. Assume that the first k pairs
of blocks from the nine blocks have the same weight as the test blocks and
that the (k+1)st pair tested have a different weight. If k=9then all the coins
weigh the same and so there are no counterfeit coins. Assume, without loss
of generality that the ( k+1 )st pair tested is lighter. Then the original two
blocks as well as all those of the first k pairs tested must be gold coins.
Therefore in the 1+(k+1) = k+2 weight trials already made we have found
(k+1) gold coins. Now we compare the two blocks of the (K+1)st (heavier)
pair.
(This is the (k=3)rd trial). If both blocks are the same weight, they both must
be gold. If they are not of the same weight, one is gold and one is counterfeit.
In either event we are able after k+3 weight trial to display a pair of blocks of
which one is counterfeit and the other gold.
3. By using this pair we can determine in, at most, 8-k weight trials how many
gold coins remain among the 20-2(k+2)=16-2k unweighed ones. Using the
techniques employed in the first event. The number of weight trials used in
the second event is then equal to k+3+(8-k) = 11

6. We have a faulty clock, that is uniformly fast, and is ahead every day less than an
hour. On January 1st 2002 it showed the correct time, and the same day at 1:05 p.m. the
two hands of the clock were in perfect overlap. When will this faulty clock show the
correct time again?

This problem will be on Round 4 again.

7. A volleyball net has the form of a rectangular lattice with dimensions of 50 x 600.
What is the maximum number of unit strings you can cut before the net falls apart into
more than one piece?

Solution
We consider the volleyball net as a graph, its nodes as vertices, and the strings as
edges.
A cycle is any closed path in a graph which does not pass through the same vertex of the
graph twice . A connected graph without cycles is a tree.

Our objective is to erase as many edges as possible, while keeping the graph
connected. We delete the edges one by one as long as we can.. Notice, that if the graph
has a cycle , then we can delete any of the edges in this cycle. But a connected graph
without a cycle is a tree, thus, when we have obtained a tree, we cannot delete any more
of the graph’s edges.
Let us calculate the number of edges in our graph at this final moment. The number of
vertices is the same as originally. , that is, equals 51 601  30651. On the other hand a
tree with this many vertices must have 30651-1 = 30650 edges. At the very beginning we
had 601 50  600 51  60650 edges. Thus we can delete no more than 30000 edges -
and it is easy to see that we actually can do this.

8.

Two knights of each color sit on an irregular fragment of a
chessboard as shown. They move as in chess (without the
requirement that alternate moves be by alternate colors) and must
remain on the fragment. Can they be moved so that the white and
black knights exchange positions?(In my idea folder)
Solution.

Number the squares as shown in the figure and build a graph by connecting
two nodes if they are associated with squares that a knight can traverse in a
single move. The graph, annotated to show the initial positions of the knights,
can be seen here. Node 3 provides convenient "off-street parking" for any
knight the player wishes to place in the next available upper position; that is,
the knights can be shuttled into the lower nodes while dropping off any
desired knight at node 3. That knight can then be moved upward and the
process repeated. The puzzle is so flexible that more complicated variations
can also be solved. If the knights were individually distinguished, say by the
letters A through D, you could create any of the 4!=24 possible permutations
of their initial positions -- even if squares 4 and 5 were removed from the
board fragment!

9. There are seven lakes in Lakeland. They are connected by 10 canals so that one can
swim through the canals from any lake to any other. How many islands are there in
Lakeland ? Mathematical Circles Page 145
(Note: to solve this problem, you need to know Euler’s formula for planar graphs. You
may find in our links, or by using any search engine. If you need help to find it, please
e-mail us)
If a planar graph is depicted properly, then it divides the plane into several regions called
faces. Let us denote the number of faces by F, the number of vertices by V and the
number of edges of the graph by E.
Euler’s formula for planar graphs:

for a properly depicted connected planar graph the equality

V-E +F = 2

Solution. We have a planar graph consisting of 7 lakes (these are the vertices), 10 canals
(these are the edges) and we want to determine how many islands (faces) we have.
If the graph is connected, we have

V-E + F = 2                              7 – 10 + F = 2               f=5
that is 4 islands and the land outside

If any edge is disconnected, the total number of the connected portion is V-1 and the total
number of edges is E-1 . The number of faces doesn’t change it is always 4, as long as the
canals don’t cross.