Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution: As $f(x)=ax^2 + bx + c$ , we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$
To compute $f(1)$ , set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$

Answer:  2
Problem: Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution: Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$
We can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$
Comparing coefficients we have $a=3$ $6a+b=7$ , and $9a+3b+c = 4$ . From the second equation we get $b=-11$ , and then from the third we get $c=10$ . Hence $a+b+c = 3-11+10 = \boxed{2}$

Answer:  2
Problem: How many positive integers less than $1000$ are $6$ times the sum of their digits?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$

Solution: The sum of the digits is at most $9+9+9=27$ . Therefore the number is at most $6\cdot 27 = 162$ . Out of the numbers $1$ to $162$ the one with the largest sum of digits is $99$ , and the sum is $9+9=18$ . Hence the sum of digits will be at most $18$
Also, each number with this property is divisible by $6$ , therefore it is divisible by $3$ , and thus also its sum of digits is divisible by $3$ . Thus, the number is divisible by $18$
We only have six possibilities left for the sum of the digits: $3$ $6$ $9$ $12$ $15$ , and $18$ , but since the number is divisible by $18$ , the digits can only add to $9$ or $18$ . This leads to the integers $18$ $36$ $54$ $72$ $90$ , and $108$ being possibilities. We can check to see that $\boxed{1}$ solution: the number $54$ is the only solution that satisfies the conditions in the problem.

Answer:  1
Problem: How many positive integers less than $1000$ are $6$ times the sum of their digits?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$

Solution: The sum of the digits is at most $9+9+9=27$ . Therefore the number is at most $6\cdot 27 = 162$ . Since the number is $6$ times the sum of its digits, it must be divisible by $6$ , therefore also by $3$ , therefore the sum of its digits must be divisible by $3$ . With this in mind we can conclude that the number must be divisible by $18$ , not just by $6$ . Since the number is divisible by $18$ , it is also divisible by $9$ , therefore the sum of its digits is divisible by $9$ , therefore the number is divisible by $54$ , which leaves us with $54$ $108$ and $162$ . Only $54$ is $6$ times its digits, hence the answer is $\boxed{1}$

Answer:  1
Problem: For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$

Solution: We know that $i^x$ cycles every $4$ powers so we group the sum in $4$ s. \[i+2i^2+3i^3+4i^4=2-2i\] \[5i^5+6i^6+7i^7+8i^8=2-2i\]
We can postulate that every group of $4$ is equal to $2-2i$ .
For 24 groups we thus, get $48-48i$ as our sum. 
We know the solution must lie near
The next term is the $24*4+1=97$ th term. This term is equal to $97i$ (first in a group of $4$ so $i^{97}=i$ ) and our sum is now $48+49i$ so $n=97\Rightarrow\boxed{97}$ is our answer

Answer: D 97
Problem: For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$

Solution: Obviously, even powers of $i$ are real and odd powers of $i$ are imaginary. 
Hence the real part of the sum is $2i^2 + 4i^4 + 6i^6 + \ldots$ , and 
the imaginary part is $i + 3i^3 + 5i^5 + \cdots$
Let's take a look at the real part first. We have $i^2=-1$ , hence the real part simplifies to $-2+4-6+8-10+\cdots$ .
If there were an odd number of terms, we could pair them as follows: $-2 + (4-6) + (8-10) + \cdots$ , hence the result would be negative. As we need the real part to be $48$ , we must have an even number of terms. If we have an even number of terms, we can pair them as $(-2+4) + (-6+8) + \cdots$ . Each parenthesis is equal to $2$ , thus there are $24$ of them, and the last value used is $96$ . This happens for $n=96$ and $n=97$ . As $n=96$ is not present as an option, we may conclude that the answer is $\boxed{97}$

Answer:  97
Problem: For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$

Solution: Some may know the equation:
\[\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\]
(For those curious, this comes from differentiating the equation for finite geometric sums)
Using this equation, we have
\[48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}\] \[=\frac{1-(n+1)i^n+ni^{n+1}}{-2}\] \[=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}\]
Since the imaginary and the real part must be positive, we know that $i^{n+1}=-1$ or $i^{n+1}=-i$ . By the same line of reason, $i^{n}=1,i$ . This only works for $n\equiv 1 \mod 4$ . Therefore, we have:
\[\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i\]
Solving either the real or imaginary part gives $\boxed{97}$

Answer: D 97
Problem: A circle with center $C$ is tangent to the positive $x$ and $y$ -axes and externally tangent to the circle centered at $(3,0)$ with radius $1$ . What is the sum of all possible radii of the circle with center $C$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$ . For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$ , the distance between $C$ and $(3,0)$ must be exactly $r+1$
By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$ , hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$
Simplifying, we obtain $r^2 - 8r + 8 = 0$ . By Vieta's formulas the sum of the two roots of this equation is $\boxed{8}$

Answer:  8
Problem: Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term.  The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ .  What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$

Solution: Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$
Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$

Answer:  1
Problem: Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term.  The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ .  What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$

Solution: Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
Which can be further rewritten as $r_1-r_1^2 = r_2-r_2^2$ .
Rearranging the equation we get $r_1-r_2 = r_1^2-r_2^2$ .
Expressing this as a difference of squares we get $r_1-r_2 = (r_1-r_2)(r_1+r_2)$
Dividing by like terms we finally get $r_1+r_2 = \boxed{1}$ as desired.

Answer:  1
Problem: Let $p(x) = x^3 + ax^2 + bx + c$ , where $a$ $b$ , and $c$ are complex numbers.  Suppose that
What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$

Solution: Consider the graph of $x^4$ . It is similar to a parabola, but with a wider "base". First examine $x^4-2009$ and $x^4-9002$ . Clearly they are just being translated down some large amount. This will result in the $x$ -axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra we know that a polynomial must have exactly as many roots as its highest degree, so we are left with $4-2$ or $2$ nonreal roots for each of the graphs. For the graph of $x^4-(2009+9002\pi i)$ , it's not even possible to graph it on the Cartesian plane, so all $4$ roots will be nonreal. This is $2+2+4 = 8$ total nonreal roots $\Rightarrow \boxed{8}$

Answer: C 8
Problem: The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$
What is $|a_{2009}|$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3$

Solution: Consider another sequence $\{\theta_1, \theta_2, \theta_3...\}$ such that $a_n = \tan{\theta_n}$ , and $0 \leq \theta_n < 180$
The given recurrence becomes
It follows that $\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$ . Since $\theta_1 = 45, \theta_2 = 30$ , all terms in the sequence $\{\theta_1, \theta_2, \theta_3...\}$ will be a multiple of $15$
Now consider another sequence $\{b_1, b_2, b_3...\}$ such that $b_n = \theta_n/15$ , and $0 \leq b_n < 12$ . The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}$
As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the first few terms of the sequence until it starts to repeat.
$\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}$
Note that $b_{25} = b_1 = 3$ and $b_{26} = b_2 = 2$ . Thus $\{b_n\}$ has a period of $24$ $b_{n + 24} = b_n$
It follows that $b_{2009} = b_{17} = 0$ and $\theta_{2009} = 15 b_{2009} = 0$ . Thus $a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.$
Our answer is $|a_{2009}| = \boxed{0}$

Answer: A 0
Problem: The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$
What is $|a_{2009}|$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3$

Solution: First, some intuition. The given recurrence relation: $a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}$ looks much like the tangent addition formula. So, we let $a_n=\tan{\theta_n}$ . Then, we get: \[\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}\] This gives us: \[\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}\] Now, observe that \[\theta_1=45\] \[\theta_2=30\] We know that this sequence of values for $\theta_n$ will repeat eventually because it is mod $180$ .
It is just a matter of when, so we start bashing:
\[\theta_3=75\] \[\theta_4=105\] \[\theta_5=0\] \[\theta_6=105\] \[\theta_7=105\] \[\theta_8=30\] \[\theta_9=135\] \[\theta_{10}=165\] \[\theta_{11}=120\] \[\theta_{12}=105\] \[\theta_{13}=45\] \[\theta_{14}=150\] \[\theta_{15}=15\] \[\theta_{16}=165\] \[\theta_{17}=0\] \[\theta_{18}=165\] \[\theta_{19}=165\] \[\theta_{20}=150\] \[\theta_{21}=135\] \[\theta_{22}=105\] \[\theta_{23}=60\] \[\theta_{24}=165\] \[\theta_{25}=45\] \[\theta_{26}=30\] And there is the repetition. So, this series has a period of 24. $2009 \equiv 17 \pmod{24}$ , so $|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{0}$ ~Firebolt360

Answer: A 0
Problem: Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ .  What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution: Because the line $x + y = 7$ is parallel to $\overline {AB}$ , the area of $\triangle ABC$ is independent of the location of $C$ on that line.  Therefore it may be assumed that $C$ is $(7,0)$ .  In that case the triangle has base $AC = 4$ and altitude $3$ , so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed{6}$

Answer:  6
Problem: Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ .  What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution: The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$ .  Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$ , which is $\frac 4{\sqrt 2} = 2\sqrt 2$ . Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$

Answer: A 6
Problem: The fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively.  What is the first term?
$\mathrm{(A)}\ 60\qquad \mathrm{(B)}\ 75\qquad \mathrm{(C)}\ 120\qquad \mathrm{(D)}\ 225\qquad \mathrm{(E)}\ 315$

Solution: Let the $n$ th term of the series be $ar^{n-1}$ .  Because \[\frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8,\] it follows that $r = 2$ and the first term is $a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}$

Answer: E 315
Problem: Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$
$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$

Solution: Let $D$ be the foot of the altitude to $\overline BC$ .  Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$ .  Thus $BC = BD + BC = 5 + 9 = 14$ or  assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ .  The sum of the two possible values is $14 + 4 = \boxed{18}$

Answer: D 18
Problem: Trapezoid $ABCD$ has $AD||BC$ $BD = 1$ $\angle DBA = 23^{\circ}$ , and $\angle BDC = 46^{\circ}$ .  The ratio $BC: AD$ is $9: 5$ .  What is $CD$
$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$

Solution: Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$ .  Then
\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}
Thus $\triangle BDE$ is isosceles with $DE = BD$ .  Because $\overline {AD} \parallel \overline {BC}$ , it follows that the triangles $BCE$ and $ADE$ are similar.  Therefore \[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\] so $CD = \boxed{45}.$

Answer:  45
Problem: For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$

Solution: To find the answer it was enough to play around with $f$ . One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$ , and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$

Answer:  802
Problem: For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$

Solution: We will now show a complete solution, with a proof that no other values are prime.
Consider the function $g(x) = x^2 - 360x + 400$ , then obviously $f(x) = g(x^2)$
The roots of $g$ are: \[x_{1,2}  = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2  = 180 \pm 80 \sqrt 5\]
We can then write $g(x) = (x - 180 - 80\sqrt 5)(x - 180 + 80\sqrt 5)$ , and thus $f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 + 80\sqrt 5)$
We would now like to factor the right hand side further, using the formula $(x^2 - y^2) = (x-y)(x+y)$ . To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.
We are looking for rational $a$ and $b$ such that $(a+b\sqrt 5)^2 = 180 + 80\sqrt 5$ . Expanding the left hand side and comparing coefficients, we get $ab=40$ and $a^2 + 5b^2 = 180$ . We can easily guess (or compute) the solution $a=10$ $b=4$
Hence $180 + 80\sqrt 5 = (10 + 4\sqrt 5)^2$ , and we can easily verify that also $180 - 80\sqrt 5 = (10 - 4\sqrt 5)^2$
We now know the complete factorization of $f(x)$
\[f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)\]
As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.
We have $(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20$ ,
and $(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20$
Hence we obtain the factorization $f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)$
For $x\geq 20$ both terms are positive and larger than one, hence $f(x)$ is not prime. For $1<x<19$ the second factor is positive and the first one is negative, hence $f(x)$ is not a prime. The remaining cases are $x=1$ and $x=19$ . In both cases, $f(x)$ is indeed a prime, and their sum is $f(1) + f(19) = 41 + 761 = \boxed{802}$

Answer:  802
Problem: A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution: Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has $\frac 12 \cdot 3 \cdot 200 = \boxed{300}$ edges.

Answer:  300
Problem: A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution: At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex.  Thus the total number of new edges is the total number of endpoints of the original edges, which is $200$ .  A middle portion of each original edge is also present in $R$ , so $R$ has $100 + 200 = \boxed{300}$ edges.

Answer:  300
Problem: A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution: Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$ , where F is the number of faces of $Q$ .  Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices.  Each cut by a plane $P_k$ creates an additional face on $R$ , so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$ , where $E$ is the number of edges of $R$ .  Subtracting the first equation from the second gives $300 - E = 0$ , whence $E = \boxed{300}$

Answer: C 300
Problem: A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution: Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.
[asy] pair A,B,C,D,E,F; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); draw(A--C--B); draw(C--D); draw(E--D--F); [/asy]
But because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted $\times 2$
[asy] pair A,B,C,D,E,F,G,H,I,J,K; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7);  G=(-107.5,236.2); H=(-107.5,-236.2); I=(370.7,285); J=(370.7,-285); K=(441.4,0);  draw(A--C--B); draw(C--D); draw(E--D--F);  draw(G--A, red); draw(H--B, red); draw(I--E, red); draw(J--F, red);  draw(A--B, dashed); draw(E--K, dashed); draw(F--K, dashed);  [/asy]
Since there are $100$ edges to start with, $400/2=200$ new edges result. So there are $100+200=\boxed{300}$ edges in the figure.

Answer: C 300
Problem: A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution: The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and $2 \cdot 100 = 200$
$\frac{200}{4} = 50$ , so there are $50$ vertices.
The base of a pyramid has 4 edges, so each sliced vertex would add four edges to $R$
$100 + 4 \cdot 50$ $\boxed{300}$

Answer: C 300
Problem: Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$

Solution: Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is \[\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.\]

Answer:  89
Problem: Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$

Solution: Recall that the number of ways to arrange 1x1 and 2x1 blocks to form a 1xn rectangle results in Fibonacci numbers . Clearly, $\boxed{89}$ is the only fibonacci number, so no calculation is needed for this problem.

Answer:  89