North Dakota Mathematics Talent Search 2004-2005
SOLUTION SET 1
(Posted on 12/01/2004)
- To start let's examine all of the information supplied by
the conversation carefully. If the product of the ages
of the three sons in 36, there are only eight possibilities
to consider:
| Son 1 | Son 2 | Son 3 |
| 36 | 1 | 1 |
| 18 | 2 | 1 |
| 12 | 3 | 1 |
| 9 | 4 | 1 |
| 9 | 2 | 2 |
| 6 | 6 | 1 |
| 6 | 3 | 2 |
| 4 | 3 | 3 |
The second piece of information was that the sum of the sons'
ages is the same as the number of windows in the building. We
have to assume that the mathematician knew the number of windows,
so he knew the total. What are the possibilities? Adding the
numbers for all eight cases yields the following totals:
Suddenly, everything is clear. If the number of windows in the
building had been 21 (or 38, 16, 14, 11, or 10), the mathematician
would have given the answer immediately. Instead, he said that
he still needed an additional piece of information. This
indicates that the number of windows was 13, thereby leaving only
two options: (9,2,2) or (6,6,1). As the second option does not
have an oldest son, the ages of the three sons must be
(9,2,2).
- The reason that this problem is a challenge is that there is no
obvious starting point, but it is really quite simple to develop
an appropriate graphical model for the problem. The only
information provided in this problem is that Mr.Smith asked
every person in the room how many times they shook someone's
hand, and that all of the answers he received were different.
What answers did he get then? Well, the minimum number of
handshakes is 0: it is possible that some antisocial person
didn't shake anyone's hand at all. But what is the maximum
number of handshakes for a single person? It is more difficult
to think of asking this question than it is to answer it.
The maximum number of handshakes for a person is eight, as there
are ten people in the room, and a person can't shake hands with
himself or herself nor with his or her spouse.
Let's collect the facts we have now. Mr.Smith asked the
question to nine people in the room, and all of the answers were
different. Furthermore, each answer was a number between 0 and 8.
Therefore, the answers he received were 0, 1, 2, 3, 4, 5, 6, 7, 8.
Let's try to draw all of the handshakes that were exchanged. The
person 8 (we'll name everyone except Mr.Smith by the number of
handshakes they exchanged) shook hands 8 times, i.e. with
everyone else in the room except himself or herself and his or
her spouse. The spouse of person 8 must be person 0. Did the
proverbial light bulb just turn on? Let's turn our attention
to person 7. We can conclude that the spouse of person 7 is
person 1. We can repeat this reasoning for persons 6 and 5.
We see that their spouses are person 2 and person 3 respectively.
Using these conclusions, we know what the spouse of Mr.Smith is
person 4. Therefore, Mrs.Smith exchanged 4 handshakes.
- The average speed is equal to the ratio of the total distance
traveled over the total time of the trip. Let
X
be the distance between the two cities in question. Then the
time of the trip from Washington D.C. to New York City was
X/40 and the time of the return trip was
X/60. Therefore the averaged speed was equal to
2*X / (X/40 + X/60) = (240*X) / (5*X) = 48 miles/hour.
- Let's introduce a few variables:
X - amount of water
filled with the large pipe in an hour, Y - amount
of water filled with the small pipe in an hour, Z
- amount of water in the pool when it is completely filled.
Then
4*X = Z and 6*Y = Z,
in particular 4*X = 6*Y.
So, the time required to fill the pool when using both pipes is
equal to
Z / (X+Y) = 4*X / (X+ (4/6)*X) = 12/5 =
2 hours 24 minutes, which is less than 2 hours 25 minutes.
- The initial step is the hardest part of the problem. Indeed,
the problem does not provide us with any convenient starting
point. In situation like this, it is often advantageous to
introduce some parameter ourselves: Let's consider a polyhedron
with
F faces. Now we have something, a variable,
to work with.
We have to prove something about the edges of faces of this
polyhedron. Each face has a number of edges. To say
something about these numbers, it would be nice at least to
know the range of values that we might expect. So, let's
ask the question: what is the minimum number of edges that a
face may have? The answer is 3, and in that case the face is
a triangle. This minimum number is independent of the total
number F of faces of the polyhedron.
And what is the maximum number of edges a face may have? Well,
each edge belongs to precisely two faces, so if we have a face
with six edges, we know that the face is part of a polyhedron
with at least seven faces (the current face plus six new faces,
one for each edge). Thus, any face of a polyhedron that has
F faces in total can't have more than
F-1 edges, since a new face originates from each
edge. This concludes the proof.
If this fact surprises you then take note that you have lost
sight of what you are trying to prove. Go back and remind
yourself of the problem. The reason the proof is complete
is because if there are F faces in total, and
if each face must have a number of edges between 3 and
F-1, then some repetitions in the number of edges
must occur across the F faces. There are more
faces than possibilities for the number of edges, so you'll
have to use some of these numbers more than once. Therefore,
at least two faces will have the same number of edges.
The keys to solving this problem were to invent a starting
point to pursue and not lose sight of the goal.
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