Download 1. A tetrahedron has four vertices. Let us name these vertices a, b, c

Wisconsin Mathematics, Science and Engineering Talent Search
Problem Set II
Sophia Sun
November 2013
1. A tetrahedron has four vertices. Let us name these vertices a, b, c and d. We are given that
the sum of the numbers on each face of the tetrahedron is divisible by five. This means that
(a+b+c), (a+b+d), (a+c+d), and (b+c+d) are all divisible by five. If we add these four expressions,
we get (3a+3b+3c+3d). This sum must also be divisible by five. We can factor out 3 to get
3(a+b+c+d). This shows that (a+b+c+d) must be divisible by five, as 3 is not divisible by five. For
(a+b+c+d) to be divisible by five, given that (a+b+c) is divisible by five, we know that d must also
be divisible by five. Similarly, given that (a+b+d) is divisible by five, we can conclude that c is
also a multiple of five. Continuing on, we can see that if (a+c+d) and (b+c+d) are divisible by five,
b and d, respectively, must also be divisible by five. Therefore, if the sum of the numbers on
each face of our tetrahedron is divisible by five, then all the numbers on each vertex must also
be divisible by five.
Nicholas Janovetz
Mukwonago
Talent Search
Problem Set 2
Problem(II(
Since DAB+ABC = 120 degrees, if we extend AD and BC they intersect at a 60 degree
angle at point M, making the triangles ABM, DMB, AMC, and DMC. As equilateral
triangles are also equiangular, each angle of ACE, DCG, and DBF measures 60 degrees.
From this, we can prove M, C, A, and E are four points on a circle because angles AEC
and AMC are equal to 60 degrees and inscribe the segment AC into that circle. We can
also prove M, G, C, and D are four points on a circle because angles DGC and DMC are
equal to 60 degrees and inscribe the segment DC into that circle. We can also prove M, F,
B, and D are four points on a circle because angles DFB and DMB are equal to 60
degrees and inscribe the segment DB into that circle. In summary, M, C, A, and E are
four points on a circle, M, G, C, and D are four points on a circle, and M, F, B, and D are
four points on a circle. Angles EMA and ECA are both equal to 60 degrees because they
are inscribed angles inscribing the same segment EA. Angle AMC, the same angle as
AMB, is 60 degrees as proven earlier. Angles GMC and GDC are both equal to 60
degrees because they are inscribed angles inscribing the same segment GC. Thus angles
EMA, AMC, and GMC all sum to 180 degrees, making the points E, M, and G collinear.
Angles FMB and FDB are both equal to 60 degrees because they are inscribed angles
inscribing the same segment FB. Thus angles EMA, AMB, and FMB all sum to 180
degrees, making points E, M, and F collinear. Combing our results, E, M, G, and F are
collinear, meaning E, F, and G are all collinear. Graphic provided below.
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2
WENTONG ZHANG
Problem 2
Sorry. No time to bash.
Problem 3
We can use Newton Divided Di↵erences to solve for the general formula of the polynomial.
Let x1 = 1, x2 = 2, . . . , xn+1 = n + 1. Then, we see that we have for the first Divided
Di↵erences, we have that f [xi , xi+1 ] = 2 for i = 1, 2, . . . , n 1. However, f [xn , xn+1 ] = 4. As
a result, we know that f [xi , xi+1 , . . . , xi+k ] = 0 for all k = 1, 2, . . . , n 1 and i = 1, 2, . . . , n 2
such that i + k < n + 1. Basically, if we draw out the Newton Divided Di↵erences, we see
that the whole top triangle is filled with zeros, and the only non-zero values past the First
Divided Di↵erences are on the bottom. Through simple calculation, it is easily shown that
2
f [x(n+1) i , . . . , xn+1 ] =
i!
for i > 1. Consequently, we know that f [x1 , . . . , xn+1 ] = n!2 . However, now we know all the
values of the top row: values of f [x1 , . . . , xk ] for k = 2, 3, . . . , n + 1. In fact, we know that
when k = 2, f [x1 , . . . , xk ] = 2. Then, when k = 3, 4, . . . , n, f [x1 , . . . , xk ] = 0. Finally, when
k = n + 1, f [x1 , . . . , xk ] = n!2 . Thus, we can write our interpolating polynomial as:
!
n+1
i 1
X
Y
P (x) =
f [x1 , . . . , xi ] ·
(x xj )
i=1
j=1
Since we have so many terms as zero, we end up with:
P (x) = f [x1 ] + f [x1 , x2 ] · (x
= 3 + 2(x
x1 ) + f [x1 , . . . , xn+1 ] ·
n
2 Y
1) +
·
(x
n! j=1
n
Y
(x
xn )
j=1
j)
Note that this polynomial has degree n, so it is the one we want for the interpolant. Now,
evaluate this interpolant at x = n + 3. We obtain, with a bit of intuitive reindexing:
n
2 Y
P (n + 3) = 3 + 2(n + 3 1) +
·
((n + 3) j)
n! j=1
Q
2 · nj=1 (2 + j)
= 3 + 2n + 4 +
n!
(n + 2)!
= 3 + 2n + 4 +
n!
= 3 + 2n + 4 + (n + 1)(n + 2)
= 3 + 2n + 4 + n2 + 3n + 2
= n2 + 5n + 9
Thus, we obtain the solution that
P (n + 3) = n2 + 5n + 9
⇤
Problem 5
5)
Claim: If 2013 students each roll 9 six-sided dice and record how many times each of the
numbers 1, 2, 3, 4, 5, 6 appear, there are at least two students who recorded the same result.
Proof:
If the number of arrangements possible when rolling 9 six-sided dice is smaller than or equal to
2012, and if every one of 2012 students records a different possibility, the 2013th student has to
end up recording the same result as someone else because there isn’t another possible result.
So we must show the number of arrangements possible is smaller than or equal to 2012.
This problem is essentially the same as the number of combinations of size 9 from a set with 6
elements when the repetition of elements is allowed (because each of the 9 dice has to have a
value and the number on each dice can be repeated).
We can imagine this problem as putting 9 indistinguishable balls (the 9 dice, we don’t care which dice has which number so they’re indistinguishable) into 6 distinguishable boxes (the
choices for the 6 faces of the dice).
To visualize this, we can put 6-1=5 dividers between the 6 boxes.
And we can put any number of balls into any of the boxes provided that every single ball is used,
so a combination might look something like this.
If we look at just the balls (represented by “x”) and the dividers (represented by “|”), we see that
in order to have a place for each of the 5 dividers and 9 balls, we must need 14 places.
Problem 5
x
x
|
x
|
X
x
x
|
X
x
|
x
|
The 9 balls can be in any of the 14 places and the 5 dividers must take up 5 of the places
(because it’s one ball or one divider per place). So we’re choosing 9 places for the balls to be in out of a total of 14 possibilities.
𝐶(14, 9) = 14
= 2002
9
(The general formula for the number of k-combinations from set of n elements is 𝐶(𝑛 + 𝑘 −
1, 𝑘)
So there are 2002 combinations, which is less than 2013. So two or more students must have
recorded the same result because there are not enough combinations for everyone to have a
different combination.