Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution: Take point $N$ inside $\triangle ABC$ such that $\angle CBN = 7^\circ$ and $\angle BCN = 23^\circ$
$\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ$ . Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM = CN$ . Hence $\triangle CMN$ is an equilateral triangle , so $\angle CNM = 60^\circ$
Then $\angle MNB = 360^\circ - \angle CNM - \angle CNB = 360^\circ - 60^\circ - 150^\circ = 150^\circ$ . We now see that $\triangle MNB$ and $\triangle CNB$ are congruent. Therefore, $CB = MB$ , so $\angle CMB = \angle MCB = \boxed{83}$

Answer:  83
Problem: Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution: From the givens, we have the following angle measures $m\angle AMC = 150^\circ$ $m\angle MCB = 83^\circ$ .  If we define $m\angle CMB = \theta$ then we also have $m\angle CBM = 97^\circ - \theta$ .  Then apply the Law of Sines to triangles $\triangle AMC$ and $\triangle BMC$ to get
\[\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin (97^\circ - \theta)}\]
Clearing denominators , evaluating $\sin 150^\circ = \frac 12$ and applying one of our trigonometric identities to the result gives
\[\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta\]
and multiplying through by 2 and applying the double angle formulas gives
\[\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta\]
and so $\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta \Longleftrightarrow \tan 7^{\circ} = \cot \theta$ ; since $0^\circ < \theta < 180^\circ$ , we must have $\theta = 83^\circ$ , so the answer is $\boxed{83}$

Answer:  83
Problem: Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution: Without loss of generality , let $AC = BC = 1$ .  Then, using the Law of Sines in triangle $AMC$ , we get $\frac {1}{\sin 150} = \frac {MC}{\sin 7}$ , and using the sine addition formula to evaluate $\sin 150 = \sin (90 + 60)$ , we get $MC = 2 \sin 7$
Then, using the Law of Cosines in triangle $MCB$ , we get $MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1$ , since $\cos 83 = \sin 7$ . So triangle $MCB$ is isosceles, and $\angle CMB = \boxed{83}$

Answer:  83
Problem: Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution: Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros
First, take point $E$ outside of $\triangle{ABC}$ so that $\triangle{CEB}$ is equilateral. Then, connect $A$ $C$ , and $M$ to $E$ . Also, let $ME$ intersect $AB$ at $F$ $\angle{MCE} = 83^\circ - 60^\circ = 23^\circ$ $CE = AB$ , and (trivially) $CM = CM$ , so $\triangle{MCE} \cong  \triangle{MCA}$ by SAS congruence. Also, $\angle{CMA} = \angle{CME} = 150^\circ$ , so $\angle{AME} = 60^\circ$ , and $AM = ME$ ,
making $\triangle{AME}$ also equilateral. (it is isosceles with a $60^\circ$ angle) $\triangle{MAF} \cong \triangle{EAF}$ by SAS ( $MA = AE$ $AF = AF$ , and $m\angle{MAF} = m\angle{EAF} = 30^\circ$ ), and $\triangle{MAB} \cong \triangle{EAB}$ by SAS ( $MA = AE$ $AB = AB$ , and $m\angle{MAB} = m\angle{EAB} = 30^\circ$ ). Thus, $\triangle{BME}$ is isosceles, with $m\angle{BME} = m\angle{BEM} = 60^\circ + 7^\circ = 67^\circ$ . Also, $\angle{EMB} + \angle {CMB} = \angle{CME} = 150^\circ$ , so $\angle{CME} = 150^\circ - 67^\circ  = \boxed{83}$

Answer:  83
Problem: Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution: Noticing that we have three concurrent cevians, we apply Ceva's theorem:
\[(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\sin \angle BCM)\] \[(\sin 23)(\sin 30)(\sin x) = (\sin 7)(\sin 37-x)(\sin 83)\]
using the fact that $\sin 83 = \cos 7$ and $(\sin 7)(\cos 7) = 1/2 (\sin 14)$ we have:
\[(\sin 23)(\sin x) = (\sin 14)(\sin 37-x)\]
By inspection, $x=14^\circ$ works, so the answer is $180-83-14= \boxed{083}$

Answer:  083
Problem: An angle $x$ is chosen at random from the interval $0^\circ < x < 90^\circ.$ Let $p$ be the probability that the numbers $\sin^2 x, \cos^2 x,$ and $\sin x \cos x$ are not the lengths of the sides of a triangle. Given that $p = d/n,$ where $d$ is the number of degrees in $\text{arctan}$ $m$ and $m$ and $n$ are positive integers with $m + n < 1000,$ find $m + n.$

Solution: Note that the three expressions are symmetric with respect to interchanging $\sin$ and $\cos$ , and so the probability is symmetric around $45^\circ$ .  Thus, take $0 < x < 45$ so that $\sin x < \cos x$ .  Then $\cos^2 x$ is the largest of the three given expressions and those three lengths not forming a triangle is equivalent to a violation of the triangle inequality
\[\cos^2 x > \sin^2 x + \sin x \cos x\]
This is equivalent to
\[\cos^2 x - \sin^2 x > \sin x \cos x\]
and, using some of our trigonometric identities , we can re-write this as $\cos 2x > \frac 12 \sin 2x$ . Since we've chosen $x \in (0, 45)$ $\cos 2x > 0$ so
\[2 > \tan 2x \Longrightarrow  x < \frac 12 \arctan 2.\]
The probability that $x$ lies in this range is $\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}$ so that $m = 2$ $n = 90$ and our answer is $\boxed{092}$

Answer:  092
Problem: In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$

Solution: By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$ ),
\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\] \[(AD^2 - BC^2) = 360(AD - BC) \cos A\] \[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\] \[(AD + BC) = 360 \cos A\] We know that $AD + BC = 640 - 360 = 280$ $\cos A = \dfrac{280}{360} =  \dfrac{7}{9} = 0.777 \ldots$
$\lfloor 1000 \cos  A \rfloor = \boxed{777}$

Answer:  777
Problem: In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$

Solution: Notice that $AB = CD$ , and $BD = DB$ , and $\angle{DAB} \cong \angle{BCD}$ , so we have side-side-angle matching on triangles $ABD$ and $CDB$ .  Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$ , we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Extend $AD$ to $C'$ so that $\triangle{ABC'}$ is isosceles with $AB = C'B$ .  Then notice that $\triangle{DC'B}$ has matching side-side-angle, and yet $\triangle{ADB} \not\cong \triangle{C'DB}$ because $\angle{ADB}$ is not right.  Therefore $\triangle{C'DB}$ is the unique triangle mentioned above, so $\triangle{CDB}$ is congruent, in some order of vertices, to $\triangle{C'DB}$ .  Since $\triangle{CDB} \cong \triangle{C'DB}$ would imply $\triangle{CDB} = \triangle{C'DB}$ , making quadrilateral $ABCD$ degenerate, we must have $\triangle{CDB} \cong \triangle{C'BD}$
Since the perimeter of $ABCD$ is $640$ $AD + BC = 640 - 180 - 180 = 280$ .  Hence $280 = AD + BC = AD + DC'$ .  Drop the altitude of $\triangle{ABC'}$ from $B$ and call the foot $P$ .  Then right triangle trigonometry on $\triangle{APB}$ shows that $\cos{A} = AP/AB = 140/180 = 7/9$ , so $\lfloor 1000 \cos A \rfloor = \boxed{777}$

Answer:  777
Problem: Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base- $2$ representation has more $1$ 's than $0$ 's. Find the remainder when $N$ is divided by $1000$

Solution: In base- $2$ representation, all positive numbers have a leftmost digit of $1$ . Thus there are ${n \choose k}$ numbers that have $n+1$ digits in base $2$ notation, with $k+1$ of the digits being $1$ 's.
In order for there to be more $1$ 's than $0$ 's, we must have $k+1 > \frac{n+1}{2} \implies k > \frac{n-1}{2} \implies k \ge \frac{n}{2}$ . Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle , from rows $0$ to $10$ (as $2003 < 2^{11}-1$ ). Since the sum of the elements of the $r$ th row is $2^r$ , it follows that the sum of all elements in rows $0$ through $10$ is $2^0 + 2^1 + \cdots + 2^{10} = 2^{11}-1 = 2047$ . The center elements are in the form ${2i \choose i}$ , so the sum of these elements is $\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351$
The sum of the elements on or to the right of the line of symmetry is thus $\frac{2047 + 351}{2} = 1199$ . However, we also counted the $44$ numbers from $2004$ to $2^{11}-1 = 2047$ . Indeed, all of these numbers have at least $6$ $1$ 's in their base- $2$ representation, as all of them are greater than $1984 = 11111000000_2$ , which has $5$ $1$ 's. Therefore, our answer is $1199 - 44 = 1155$ , and the remainder is $\boxed{155}$

Answer:  155
Problem: Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base- $2$ representation has more $1$ 's than $0$ 's. Find the remainder when $N$ is divided by $1000$

Solution: We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$ . If there are even digits, $2n$ , then the leftmost digit is $1$ , the rest, $2n-1$ , has odd number of digits. In order for the base-2 representation to have more $1$ 's, we will need more $1$ in the remaining $2n-1$ than $0$ 's. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$ 's at least as the number of $0$ 's. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 199$ . There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $199-44=\boxed{155}$

Answer:  155
Problem: In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: By the Angle Bisector Theorem, we know that $[CBD]=\frac{169}{289}[ABC]$ . Therefore, by finding the area of triangle $CBD$ , we see that \[\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].\] Solving for $BD$ yields \[BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.\] Furthermore, $\cos\frac{B}{2}=\frac{BD}{BF}$ , so \[BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.\] Now by the identity $2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B$ , we get \[BF=\frac{4[ABC]}{3\cdot289\sin B}.\] But then $[ABC]=\frac{360\cdot 507}{2}\sin B$ , so $BF=\frac{240}{289}\cdot 507$ . Thus $BF:FC=240:49$
Now by the Angle Bisector Theorem, $CD=\frac{169}{289}\cdot 780$ , and we know that $MC=\frac{1}{2}\cdot 780$ so $DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289$
We can now use mass points on triangle CBD. Assign a mass of $240\cdot 49$ to point $C$ . Then $D$ must have mass $240\cdot 289$ and $B$ must have mass $49\cdot 49$ . This gives $F$ a mass of $240\cdot 49+49\cdot 49=289\cdot 49$ . Therefore, $DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}$ , giving us an answer of $\boxed{289}.$

Answer:  289
Problem: In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: Let $\angle{DBM}=\theta$ and $\angle{DBC}=\alpha$ . Then because $BM$ is a median we have $360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}$ . Now we know \[\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}\] Expressing the area of $\triangle{BEF}$ in two ways we have \[\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD\] so \[\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}\] Plugging this in we have \[\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}\] so $\dfrac{DF+DE}{EF}=\dfrac{507}{360}$ . But $DF=DE+EF$ , so this simplifies to $1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}$ , and thus $\dfrac{DE}{EF}=\dfrac{49}{240}$ , and $m+n=\boxed{289}$

Answer:  289
Problem: In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: Firstly, angle bisector theorem yields $\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}$ . We're given that $AM=MC$ . Therefore, the cross ratio
\[(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}\]
We need a fourth point for this cross ratio to be useful, so reflect point $F$ over angle bisector $BD$ to a point $F'$ . Then $\triangle BFF'$ is isosceles and $BD$ is an altitude so $DF = DF'$ . Therefore,
\[(A,C;M,D) = (F,F';D,E) \implies \frac{FD(EF')}{EF(DF')} = \frac{EF'}{EF} = \frac{169}{120}\]
All that's left is to fiddle around with the ratios:
\[\frac{EF'}{EF} = \frac{ED+DF'}{EF} = \frac{EF+2DF}{EF} = 1\ +\ 2\left(\frac{DF}{EF}\right) \implies \frac{DF}{EF} = \frac{49}{240} \implies \boxed{289}\]

Answer:  289
Problem: In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution: Extend $DF$ to intersect with the extension of $AB$ at $G$ . Notice that $\triangle{BDF} \cong \triangle{BDG}$ , so $GD=DF$ . We now use Menelaus on $\triangle{GBF}$ , as $A$ $D$ , and $C$ are collinear; this gives us $\frac{GA}{BA} \cdot \frac{BC}{FC} \cdot \frac{DF}{GD}=1$ . As $GD=DF$ , we have $\frac{GA}{AB}=\frac{FC}{BC}$ , hence $\frac{GA}{120}=\frac{FC}{169}$ . Reflect $G$ over $A$ to $G'$ . Note that $\frac{G'A}{BA}=\frac{FC}{BC}$ , and reflexivity, hence $\triangle{ABC} \sim \triangle{BG'F}$ . It's easily concluded from this that $G'F \parallel AC$ , hence $G'F \parallel AD$ . As $GD=DF$ , we have $AD$ is a midsegment of $\triangle{GG'F}$ , thus $G'F = 2AD$ . We now focus on the ratio $\frac{BF}{BC}$ . From similarity, we have $\frac{BF}{BC}=\frac{G'F}{AC}=\frac{2AD}{AC}$ . By the angle bisector theorem, we have $AD:DC=120:169$ , hence $AD:AC=120:289$ , so $\frac{BF}{BC}=\frac{240}{289}$ . We now work out the ratio $\frac{DM}{MC}$ $\frac{DM}{MC}=\frac{CD-MC}{MC}=\frac{CD}{MC}-1=\frac{2CD}{AC}-1=\frac{338}{289}-1=\frac{49}{289}$ . We now use mass points on $\triangle{BDC}$ . We let the mass of $C$ be $240\cdot 49$ , so the mass of $B$ is $49 \cdot 49$ and the mass of $D$ is $289\cdot 240$ . Hence, the mass of $F$ is $289\cdot 49$ , so the ratio $\frac{DE}{EF}=\frac{49}{240}$ . Extracting gives $49+240=\boxed{289}.$

Answer:  289
Problem: The product $N$ of three positive integers is $6$ times their sum , and one of the integers is the sum of the other two. Find the sum of all possible values of $N$

Solution: Let the three integers be $a, b, c$ $N = abc = 6(a + b + c)$ and $c = a + b$ .  Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$ .  Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ is $12 \cdot (13 + 8 + 7) = 12(28) = \boxed{336}$

Answer:  336
Problem: Let $N$ be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when $N$ is divided by 1000?

Solution: We want a number with no digits repeating, so we can only use the digits $0-9$ once in constructing our number. To make the greatest number, we want the greatest digit to occupy the leftmost side and the least digit to occupy the rightmost side. Therefore, the last three digits of the greatest number should be an arrangement of the digits $0,1,2$ . Since the number has to be divisible by 8, the integer formed by the arrangement of $0,1,2$ is also divisible by 8. The only arrangement that works is $120$
Therefore, the remainder when the number is divided by $1000$ is $\boxed{120}$

Answer:  120
Problem: Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$ $B$ , and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$ $B$ is never immediately followed by $C$ , and $C$ is never immediately followed by $A$ . How many seven-letter good words are there?

Solution: There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is $3*2^6=192$
Therefore, there are $\boxed{192}$ seven-letter good words.

Answer:  192
Problem: Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$ $B$ , and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$ $B$ is never immediately followed by $C$ , and $C$ is never immediately followed by $A$ . How many seven-letter good words are there?

Solution: There are three choices for the first letter and two choices for each subsequent letter, so there are $3\cdot2^{n-1}\ n$ -letter good words. Substitute $n=7$ to find there are $3\cdot2^6=\boxed{192}$ seven-letter good words. ~ aopsav (Credit to AoPS Alcumus)

Answer:  192
Problem: Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$ $B$ , and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$ $B$ is never immediately followed by $C$ , and $C$ is never immediately followed by $A$ . How many seven-letter good words are there?

Solution: We solve this problem using recursion. Let $f(x)$ be the number of $x$ -letter good words. Thus $f(1) = 3$ (A, B or C) and the answer is just $f(7)$ .  The recurrence relation can be found by considering the last letter of one of the valid strings of length $x - 1$ .  There are $2$ possibilities for the next letter and thus $f(x) = 2 \cdot f(x-1)$ .  Now we can find a closed form as $f(x) = 3 \cdot 2 ^{x-1}$ (easy to prove by induction) and thus $f(7) = 64 * 3 = \boxed{192}$ seven-letter good words.  ~AK2006

Answer:  192
Problem: In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Embed the tetrahedron in 4-space to make calculations easier.
Its vertices are $(1,0,0,0)$ $(0,1,0,0)$ $(0,0,1,0)$ $(0,0,0,1)$
To get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0)$ $(\frac{1}{3}, \frac{1}{3},0, \frac{1}{3})$ $(\frac{1}{3},0, \frac{1}{3}, \frac{1}{3})$ $(0,\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$
The side length of the large tetrahedron is $\sqrt{2}$ by the distance formula.
The side length of the smaller tetrahedron is $\frac{\sqrt{2}}{3}$ by the distance formula.
Their ratio is $1:3$ , so the ratio of their volumes is $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$
$m+n = 1 + 27 = \boxed{28}$

Answer:  28
Problem: In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the large tetrahedron be $ABCD$ , and the small tetrahedron be $WXYZ$ , with $W$ on $ABC$ $X$ on $BCD$ $Y$ on $ACD$ , and $Z$ on $ABD$ . Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let $AB=1$ , for our convenience. Dropping an altitude from $W$ to $BC$ , and calling the foot $M$ , we have $WM=XM=\frac{\sqrt3}{6}$ . Since $\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3$ . By Law of Cosines, we have $WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3$ . Hence, the ratio of the volumes is $\left(\frac{1}{3}\right)^3=1/27$ $m+n=1+27=\boxed{028}$

Answer:  028
Problem: A cylindrical log has diameter $12$ inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a $45^\circ$ angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as $n\pi$ , where n is a positive integer. Find $n$

Solution: The volume of the wedge is half the volume of a cylinder with height $12$ and radius $6$ . (Imagine taking another identical wedge and sticking it to the existing one).  Thus, $V=\dfrac{6^2\cdot 12\pi}{2}=216\pi$ , so $n=\boxed{216}$

Answer:  216