Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$

Solution: Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$
There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$ . Connect each of these points to $G$
[asy] size(8cm); pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2; B=(0,0); A=(5,12); C=(14,0); E=(12.6667,8); D=(7.6667,-4); F=(-1.3333,8); G=(6.3333,4); B_1=(4.6667,0); B_2=(1.6667,4); A_1=(3.3333,8); A_2=(8,8); C_1=(11,4); C_2=(9.3333,0); dot(A); dot(B); dot(C); dot(G); dot(D); dot(E); dot(F); dot(A_1); dot(B_1); dot(C_1); dot(A_2); dot(B_2); dot(C_2); draw(B--A--C--cycle); draw(E--D--F--cycle); draw(B_1--A_2); draw(A_1--C_2); draw(C_1--B_2); label("$B$",B,WSW); label("$A$",A,N); label("$C$",C,ESE); label("$G$",G,S); label("$B'$",E,ENE); label("$A'$",D,S); label("$C'$",F,WNW); [/asy]
There are $12$ smaller congruent triangles which make up the desired area. Also, $\Delta ABC$ is made up of $9$ of such triangles. 
Therefore, $\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}$

Answer:  112
Problem: In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$

Solution: First, find the area of $\Delta ABC$ either like the first solution or by using Heron’s Formula. Then, draw the medians from $G$ to each of $A, B, C, A’, B’,$ and $C’$ . Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians $GA$ and $GB’$ , and let’s call the points that $GA$ intersects $C’B’$ $H$ ” and the point $GB’$ intersects $AC$ $I$ ”. From the previous property and the fact that both $\Delta ABC$ and $\Delta A’B’C’$ are congruent, $\Delta GHB’$ has the same area as $\Delta GIA$ . Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles).
Also, since the centroid of a triangle divides each median with the ratio $2:1$ , along with the previous fact, each outer triangle has $1/9$ the area of $\Delta ABC$ and $\Delta A’B’C’$ . Thus, the area of the region required is $\frac{4}{3}$ times the area of $\Delta ABC$ which is $\boxed{112}$

Answer:  112
Problem: In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$

Solution: $[ABC]$ can be calculated as 84 using Heron's formula or other methods. Since a $180^{\circ}$ rotation is equivalent to reflection through a point, we have a homothety with scale factor $-1$ from $ABC$ to $A'B'C'$ through the centroid $G$ . Let $M$ be the midpoint of $BC$ which maps to $M'$ and note that $A'G=AG=2GM,$ implying that $GM=MA'.$ Similarly, we have $AM'=M'G=GM=MA'.$ Also let $D$ and $E$ be the intersections of $BC$ with $A'B'$ and $A'C',$ respectively. The homothety implies that we must have $DE || B'C',$ so there is in fact another homothety centered at $A'$ taking $A'DE$ to $A'B'C'$ . Since $A'M'=3A'M,$ the scale factor of this homothety is 3 and thus $[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].$ We can apply similar reasoning to the other small triangles in $A'B'C'$ not contained within $ABC$ , so our final answer is $[ABC]+3\cdot\frac{1}{9}[ABC]=\boxed{112.}$

Answer:  112.
Problem: Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively.

Solution: The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$ . The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$
The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$ , where $a$ $b$ , and $c$ are the sides and $R$ is the circumradius. Thus, the area of $\triangle ABD$ is $ab=2a(a^2+b^2)/(4\cdot12.5)$ . Also, the area of $\triangle ABC$ is $ab=2b(a^2+b^2)/(4\cdot25)$ . Setting these two expressions equal to each other and simplifying gives $b=2a$ . Substitution yields $a=10$ and $b=20$ , so the area of the rhombus is $20\cdot40/2=\boxed{400}$

Answer:  400
Problem: Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively.

Solution: Let $\theta=\angle BDA$ . Let $AB=BC=CD=x$ . By the extended law of sines, \[\frac{x}{\sin\theta}=25\] Since $AC\perp BD$ $\angle CAD=90-\theta$ , so \[\frac{x}{\sin(90-\theta)=\cos\theta}=50\] Hence $x=25\sin\theta=50\cos\theta$ . Solving $\tan\theta=2$ $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}$ . Thus \[x=25\frac{2}{\sqrt{5}}\implies x^2=500\] The height of the rhombus is $x\sin(2\theta)=2x\sin\theta\cos\theta$ , so we want \[2x^2\sin\theta\cos\theta=\boxed{400}\]

Answer:  400
Problem: Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Solution: If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$ $f(2)=1716$ , and $f(3)=1848$ . Plugging in the values for x gives us a system of three equations:
$a+b+c=1440$
$4a+2b+c=1716$
$9a+3b+c=1848$
Solving gives $a=-72, b=492,$ and $c=1020$ . Thus, the answer is $-72(8)^2+492\cdot8+1020= \boxed{348}.$

Answer:  348
Problem: Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Solution: Use the same rationale as in solution 1; instead of using terms $1,2,3$ , we use $-1,0,1$ and solve the $6$ th term.
$a-b+c=1440$
$c=1716$
$a+b+c=1848$
Accordingly we will solve
$a=-72, b=204, c=1716$
$36a+6b+c= \boxed{348}.$

Answer:  348
Problem: Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Solution: Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$ , we can set up the following equalities
$ab = 1440$
$(a+r_1)(b+r_2)=1716$
$(a+2r_1)(b+2r_2)=1848$
We want to find $(a+7r_1)(b+7r_2)$
Foiling out the two above, we have
$ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$
Plugging in $ab=1440$ and bringing the constant over yields
$ar_2 + br_1 + r_1r_2 = 276$
$ar_2 + br_1 + 2r_1r_2 = 204$
Subtracting the two yields $r_1r_2 = -72$ and plugging that back in yields $ar_2 + br_1 = 348$
Now we find
$(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}$

Answer:  348
Problem: Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$

Solution: When we use long division to divide $P(x)$ by $Q(x)$ , the remainder is $x^2-x+1$
So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$
Now this also follows for all roots of $Q(x)$ Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]
Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$ ,
so by Newton's Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$
$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$
$(1)(s_2)+(-1)(1)+2(-1)=0$
$s_2-1-2=0$
$s_2=3$
So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$

Answer:  006
Problem: Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$

Solution: Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]
Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$

Answer:  6
Problem: Two positive integers differ by $60$ . The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

Solution: Call the two integers $b$ and $b+60$ , so we have $\sqrt{b}+\sqrt{b+60}=\sqrt{c}$ . Square both sides to get $2b+60+2\sqrt{b^2+60b}=c$ . Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$ , and $(b+n+30)(b-n+30)=900$ . The sum of these two factors is $2b+60$ , so they must both be even. To maximize $b$ , we want to maximixe $b+n+30$ , so we let it equal $450$ and the other factor $2$ , but solving gives $b=196$ , which is already a perfect square, so we have to keep going. In order to keep both factors even, we let the larger one equal $150$ and the other $6$ , which gives $b=48$ . This checks, so the solution is $48+108=\boxed{156}$

Answer:  156
Problem: Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

Solution: We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$ $MN=BN-BM$ , and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get $CN=\frac{168} {25}.$
By the Pythagorean Theorem in $\Delta BCN,$ we have
$BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$
Thus, $MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$
In $\Delta ADM$ , we use the Pythagorean Theorem to get $DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.$
Thus, $[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$
Hence, the answer is $527+11+40=\boxed{578}.$

Answer:  578
Problem: Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

Solution: Let $E$ be the intersection of lines $BC$ and $DM$ .  Since triangles $\Delta CME$ and $\Delta CMD$ share a side and height, the area of $\Delta CDM$ is equal to $\frac{DM}{EM}$ times the area of $\Delta CME$ .
By AA similarity, $\Delta EMB$ is similar to $\Delta ACB$ $\frac{EM}{AC}=\frac{MB}{CB}$ . Solving yields $EM=\frac{175}{48}$ . Using the same method but for $EB$ yields $EB=\frac{625}{48}$ . As in previous solutions, by the Pythagorean Theorem $DM=\frac{5\sqrt{11}}{2}$ . So, $\frac{DM}{EM}=\frac{24\sqrt{11}}{35}$ .
Now, since we know both $CB$ and $EB$ , we can find that $CE=\frac{527}{48}$ . The height of $\Delta CME$ is the length from point $M$ to $CB$ . Since $M$ is the midpoint of $AB$ , the height is just $\frac{1}{2}\cdot7=\frac{7}{2}$ . Using this, we can find that the area of $\Delta CMD$ is $\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$ , giving our answer of $\boxed{578}$

Answer:  578
Problem: The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates.  For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate.  What was the smallest possible number of members of the committee?

Solution: Let $v_i$ be the number of votes candidate $i$ received, and let $s=v_1+\cdots+v_{27}$ be the total number of votes cast. Our goal is to determine the smallest possible $s$
Candidate $i$ got $\frac{v_i}s$ of the votes, hence the percentage of votes they received is $\frac{100v_i}s$ . The condition in the problem statement says that $\forall i: \frac{100v_i}s + 1 \leq v_i$ .  ( $\forall$ means "for all", so this means "For all $i$ $\frac{100v_i}s + 1 \leq v_i$ is true")
Obviously, if some $v_i$ would be $0$ or $1$ , the condition would be false. Thus $\forall i: v_i\geq 2$ . We can then rewrite the above inequality as $\forall i: s\geq\frac{100v_i}{v_i-1}$
If for some $i$ we have $v_i=2$ , then from the inequality we just derived we would have $s\geq 200$
If for some $i$ we have $v_i=3$ , then $s\geq 150$
And if for some $i$ we have $v_i=4$ , then $s\geq \frac{400}3 = 133\frac13$ , and hence $s\geq 134$
Is it possible to have $s<134$ ? We just proved that to have such $s$ , all $v_i$ have to be at least $5$ . But then $s=v_1+\cdots+v_{27}\geq 27\cdot 5 = 135$ , which is a contradiction. Hence the smallest possible $s$ is at least $134$
Now consider a situation where $26$ candidates got $5$ votes each, and one candidate got $4$ votes. In this situation, the total number of votes is exactly $134$ , and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is $s=\boxed{134}$

Answer:  134
Problem: The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates.  For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate.  What was the smallest possible number of members of the committee?

Solution: Let there be $N$ members of the committee.
Suppose candidate $n$ gets $a_n$ votes.
Then $a_n$ as a percentage out of $N$ is $100\frac{a_n}{N}$ . Setting up the inequality $a_n \geq 1 + 100\frac{a_n}{N}$ and simplifying, $a_n \geq \lceil(\frac{N}{N - 100})\rceil$ (the ceiling function is there because $a_n$ is an integer.
Note that if we set all $a_i$ equal to $\lceil(\frac{N}{100 - N})\rceil$ we have $N \geq 27\lceil(\frac{N}{100 - N})\rceil$ . Clearly $N = 134$ is the least such number that satisfies this inequality. Now we must show that we can find suitable $a_i$ . We can let 26 of them equal to $5$ and one of them equal to $4$ . Therefore, $N = \boxed{134}$ is the answer.
- whatRthose

Answer:  134
Problem: The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates.  For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate.  What was the smallest possible number of members of the committee?

Solution: Let $n$ be the total number of people in the committee, and $a_i$ be the number of votes candidate $i$ gets where $1 \leq i \leq 27$ .  The problem tells us that \[\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}.\] Therefore, \[\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100},\] and so $n(n-127) \geq 0 \implies n \geq 127$ .  Trying $n = 127$ , we get that \[a_i \geq \frac{127}{27} \approx 4.7 \implies a_i \geq 5 \implies \sum^{27}_{a_i} a_i \geq 5 \cdot 27 = 135 \geq 127,\] a contradiction.  Bashing out a few more, we find that $\boxed{134}$ works perfectly fine.

Answer:  134
Problem: A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution: Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\frac{1}{2}$ chance of reaching the starting vertex in the next move. Thus $P_n=\frac{1}{2}(1-P_{n-1})$ $P_0=1$ , so now we can build it up:
$P_1=0$ $P_2=\frac{1}{2}$ $P_3=\frac{1}{4}$ $P_4=\frac{3}{8}$ $P_5=\frac{5}{16}$ $P_6=\frac{11}{32}$ $P_7=\frac{21}{64}$ $P_8=\frac{43}{128}$ $P_9=\frac{85}{256}$ $P_{10}=\frac{171}{512}$
Thus the answer is $171+512=$ $\boxed{683}$

Answer:  683
Problem: A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution: Consider there to be a clockwise and a counterclockwise direction around the triangle.  Then, in order for the ant to return to the original vertex, the net number of clockwise steps must be a multiple of 3, i.e., $\#CW - \#CCW \equiv 0 \pmod{3}$ .  Since $\#CW + \#CCW = 10$ , it is only possible that $(\#CW,\, \#CCW) = (5,5), (8,2), (2,8)$
In the first case, we pick $5$ out of the ant's $10$ steps to be clockwise, for a total of ${10 \choose 5}$ paths.  In the second case, we choose $8$ of the steps to be clockwise, and in the third case we choose $2$ to be clockwise.  Hence the total number of ways to return to the original vertex is ${10 \choose 5} + {10 \choose 8} + {10 \choose 2} = 252 + 45 + 45 = 342$ .  Since the ant has two possible steps at each point, there are $2^{10}$ routes the ant can take, and the probability we seek is $\frac{342}{2^{10}} = \frac{171}{512}$ , and the answer is $\boxed{683}$

Answer:  683
Problem: A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution: Label the vertices of the triangle $A,B,C$ with the ant starting at $A$ . We will make a table of the number of ways to get to $A,B,C$ in $n$ moves $n\leq10$ . The values of the table are calculated from the fact that the number of ways from a vertex say $A$ in $n$ steps equals the number of ways to get to $B$ in $n-1$ steps plus the number of ways to get to $C$ in $n-1$ steps.
\[\begin{array}{|l|ccc|} \multicolumn{4}{c}{\text{Table}}\\\hline \text{Step}&A&B&C \\\hline 1 &0 &1 &1 \\ 2 &2 &1 &1 \\ 3 &2 &3 &3\\ \vdots &\vdots&\vdots&\vdots \\ 10 &342 &341 &341 \\\hline \end{array}\] Therefore, our answer is $512+171=\boxed{683}.$

Answer:  683
Problem: A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution: This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to $341.33/1024$ . We can either round up or down. If we round down, we see $341/1024$ cannot be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up, $342/1024$ can be reduced to $171/512$ where the sum 171+512= $\boxed{683}$ is an accepted answer.

Answer:  683
Problem: A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution: Start of with any vertex, say $A$ . Denote $a_n$ the number of paths ending where it started. Then notice that for a path to end in the vertex you started on, you must have for the $(n-1)$ case that of the $2^{n-1}$ total paths, we must take away the paths that end in the $(n-1)$ -st term where it started off. (Because imagine on the $(n-1)$ move that it landed on vertex $A$ . Then if we wanted to find the number of length $n$ paths that end at $A$ , we would be able to find that path, because on the $(n-1)$ -st move it landed at $A$ . You can't go from $A$ to $A$ ). Hence we have the recursion $a_n=2^{n-1}-a_{n-1}$ , with $a_3 = 2$ . Now reiterating gives us $a_{10} = 342$ , so that the probability is $\frac{342}{2^{10}} = \frac{171}{512}$ . So we have $171 + 512 = \boxed{683}$ .
~th1nq3r

Answer:  683
Problem: Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$

Solution: The y-coordinate of $F$ must be $4$ .  All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting $F = (f,4)$ , and knowing that $\angle FAB = 120^\circ$ , we can use rewrite $F$ using complex numbers: $f + 4 i = (b + 2 i)\left(e^{i(2 \pi / 3)}\right) = (b + 2 i)\left(-1/2 + \frac{\sqrt{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i$ .  We solve for $b$ and $f$ and find that $F = \left(-\frac{8}{\sqrt{3}}, 4\right)$ and that $B = \left(\frac{10}{\sqrt{3}}, 2\right)$
The area of the hexagon can then be found as the sum of the areas of two congruent triangles ( $EFA$ and $BCD$ , with height $8$ and base $\frac{8}{\sqrt{3}}$ ) and a parallelogram ( $ABDE$ , with height $8$ and base $\frac{10}{\sqrt{3}}$ ).
$A = 2 \times \frac{1}{2} \times 8 \times \frac{8}{\sqrt{3}} + 8 \times \frac{10}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3}$
Thus, $m+n = \boxed{051}$

Answer:  051