di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Combinatorics


Question: In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.

It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.

| ARMLton |  |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |


| ARMLville |  |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |

The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.

A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.

Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
Let $n=\operatorname{pop}\left(\mathcal{F}_{d}\right)$. In terms of $n$, compute $d$.

Solution:
With $d$ dishes there are $\left(\begin{array}{l}d \\ 2\end{array}\right)=\frac{d^{2}-d}{2}$ possible pairs, so $n=\frac{d^{2}-d}{2}$. Then $2 n=d^{2}-d$, or $d^{2}-d-2 n=0$. Using the quadratic formula yields $d=\frac{1+\sqrt{1+8 n}}{2}$ (ignoring the negative value).

Final answer: $d=\frac{1+\sqrt{1+8 n}}{2}$
Subfield: Combinatorics


Question: In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.

It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.

| ARMLton |  |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |


| ARMLville |  |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |

The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.

A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.

Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
Let $T$ be a full town and let $D \in \operatorname{dish}(T)$. Let $T^{\prime}$ be the town consisting of all residents of $T$ who do not know how to make $D$. Prove that $T^{\prime}$ is full.

Solution:
The town $T^{\prime}$ consists of all residents of $T$ who do not know how to make $D$. Because $T$ is full, every pair of dishes $\left\{d_{i}, d_{j}\right\}$ in $\operatorname{dish}(T)$ can be made by some resident $r_{i j}$ in $T$. If $d_{i} \neq D$ and $d_{j} \neq D$, then $r_{i j} \in T^{\prime}$. So every pair of dishes in $\operatorname{dish}(T) \backslash\{D\}$ can be made by some resident of $T^{\prime}$. Hence $T^{\prime}$ is full.
Subfield: Combinatorics


Question: In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.

It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.

| ARMLton |  |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |


| ARMLville |  |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |

The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.

A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.

Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.

In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$.

For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups.
Show that gr(ARMLville $)=3$.

Solution:
Sally and Ross both make calzones, Ross and David both make pancakes, and Sally and David both make steak. So no two of these people can be in the same group, and $\operatorname{gr}($ ARMLville $)=3$.
Subfield: Number Theory


Question: A student computed the repeating decimal expansion of $\frac{1}{N}$ for some integer $N$, but inserted six extra digits into the repetend to get $.0 \overline{0231846597}$. Compute the value of $N$.

Solution:
Because the given repetend has ten digits, the original had four digits. If $\frac{1}{N}=.0 \underline{A} \underline{B} \underline{C} \underline{D}=$ $\frac{\underline{A} \underline{B} \underline{C} \underline{D}}{99990}$, then the numerator must divide $99990=10 \cdot 99 \cdot 101=2 \cdot 3^{2} \cdot 5 \cdot 11 \cdot 101$.

Note that all 3- and 4-digit multiples of 101 contain at least one digit which appears twice. Because the 10-digit string under the vinculum (i.e., 0231846597) contains no repeated digits, $\underline{A} \underline{B} \underline{C} \underline{D}$ cannot be a multiple of 101 . So $\underline{A} \underline{B} \underline{C} \underline{D}$ divides $2 \cdot 3^{2} \cdot 5 \cdot 11=990$. The only divisor of 990 that can be formed from four of the given digits (taken in order) is 0165, that is, 165. Hence $\frac{1}{N}=\frac{165}{99990}=\frac{1}{606} \Rightarrow N=\mathbf{6 0 6}$.

Final answer: 606
Subfield: Number Theory


Question: Let $n$ be a four-digit number whose square root is three times the sum of the digits of $n$. Compute $n$.

Solution:
Because $\sqrt{n}$ is a multiple of $3, n$ must be a multiple of 9 . Therefore the sum of the digits of $n$ is a multiple of 9 . Thus $\sqrt{n}$ must be a multiple of 27 , which implies that $n$ is a multiple of $27^{2}$. The only candidates to consider are $54^{2}(=2916)$ and $81^{2}(=6561)$, and only 2916 satisfies the desired conditions.

Final answer: 2916
Subfield: Number Theory


Question: Compute the sum of the reciprocals of the positive integer divisors of 24.

Solution:
The map $n \mapsto 24 / n$ establishes a one-to-one correspondence among the positive integer divisors of 24 . Thus

$$
\begin{aligned}
\sum_{\substack{n \mid 24 \\
n>0}} \frac{1}{n} & =\sum_{\substack{n \mid 24 \\
n>0}} \frac{1}{24 / n} \\
& =\frac{1}{24} \sum_{\substack{n \mid 24 \\
n>0}} n
\end{aligned}
$$

Because $24=2^{3} \cdot 3$, the sum of the positive divisors of 24 is $\left(1+2+2^{2}+2^{3}\right)(1+3)=15 \cdot 4=60$. Hence the sum is $60 / 24=\mathbf{5} / \mathbf{2}$.
Because $24=2^{3} \cdot 3$, any positive divisor of 24 is of the form $2^{a} 3^{b}$ where $a=0,1,2$, or 3 , and $b=0$ or 1 . So the sum of the positive divisors of 24 can be represented as the product $(1+2+4+8)(1+3)$. Similarly, the sum of their reciprocals can be represented as the product $\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)\left(\frac{1}{1}+\frac{1}{3}\right)$. The first sum is $\frac{15}{8}$ and the second is $\frac{4}{3}$, so the product is $\mathbf{5 / 2}$.

Final answer: $\frac{5}{2}$
Subfield: Number Theory


Question: There exists a digit $Y$ such that, for any digit $X$, the seven-digit number $\underline{1} \underline{2} \underline{3} \underline{X} \underline{5} \underline{Y} \underline{7}$ is not a multiple of 11. Compute $Y$.

Solution:
Consider the ordered pairs of digits $(X, Y)$ for which $\underline{1} \underline{2} \underline{3} \underline{X} \underline{5} \underline{Y} \underline{7}$ is a multiple of 11 . Recall that a number is a multiple of 11 if and only if the alternating sum of the digits is a multiple of 11 . Because $1+3+5+7=16$, the sum of the remaining digits, namely $2+X+Y$, must equal 5 or 16 . Thus $X+Y$ must be either 3 or 14 , making $X=3-Y$ (if $Y=0,1,2$, or 3 ) or $14-Y$ (if $Y=5,6,7,8$, or 9 ). Thus a solution $(X, Y)$ exists unless $Y=4$.

Final answer: 4
Subfield: Geometry


Question: A point is selected at random from the interior of a right triangle with legs of length $2 \sqrt{3}$ and 4 . Let $p$ be the probability that the distance between the point and the nearest vertex is less than 2. Then $p$ can be written in the form $a+\sqrt{b} \pi$, where $a$ and $b$ are rational numbers. Compute $(a, b)$.

Solution:
Label the triangle as $\triangle A B C$, with $A B=2 \sqrt{3}$ and $B C=4$. Let $D$ and $E$ lie on $\overline{A B}$ such that $D B=A E=2$. Let $F$ be the midpoint of $\overline{B C}$, so that $B F=F C=2$. Let $G$ and $H$ lie on $\overline{A C}$, with $A G=H C=2$. Now draw the arcs of radius 2 between $E$ and $G, D$ and $F$, and $F$ and $H$. Let the intersection of arc $D F$ and $\operatorname{arc} E G$ be $J$. Finally, let $M$ be the midpoint of $\overline{A B}$. The completed diagram is shown below.

<img_3875>

The region $R$ consisting of all points within $\triangle A B C$ that lie within 2 units of any vertex is the union of the three sectors $E A G, D B F$, and $F C H$. The angles of these sectors, being the angles $\angle A, \angle B$, and $\angle C$, sum to $180^{\circ}$, so the sum of their areas is $2 \pi$. Computing the area of $R$ requires subtracting the areas of all intersections of the three sectors that make up $R$.

The only sectors that intersect are $E A G$ and $D B F$. Half this area of intersection, the part above $\overline{M J}$, equals the difference between the areas of sector $D B J$ and of $\triangle M B J$. Triangle $M B J$ is a $1: \sqrt{3}: 2$ right triangle because $B M=\sqrt{3}$ and $B J=2$, so the area of $\triangle M B J$ is $\frac{\sqrt{3}}{2}$. Sector $D B J$ has area $\frac{1}{12}(4 \pi)=\frac{\pi}{3}$, because $\mathrm{m} \angle D B J=30^{\circ}$. Therefore the area of intersection of the sectors is $2\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)=\frac{2 \pi}{3}-\sqrt{3}$. Hence the total area of $R$ is $2 \pi-\left(\frac{2 \pi}{3}-\sqrt{3}\right)=\frac{4 \pi}{3}+\sqrt{3}$. The total area of $\triangle A B C$ is $4 \sqrt{3}$, therefore the desired probability is $\frac{\frac{4 \pi}{3}+\sqrt{3}}{4 \sqrt{3}}=\frac{\pi}{3 \sqrt{3}}+\frac{1}{4}$. Then $a=\frac{1}{4}$ and $b=\left(\frac{1}{3 \sqrt{3}}\right)^{2}=\frac{1}{27}$, hence the answer is $\left(\frac{1}{4}, \frac{1}{27}\right)$.

Final answer: $(\frac{1}{4}, \frac{1}{27})$
Subfield: Geometry


Question: The square $A R M L$ is contained in the $x y$-plane with $A=(0,0)$ and $M=(1,1)$. Compute the length of the shortest path from the point $(2 / 7,3 / 7)$ to itself that touches three of the four sides of square $A R M L$.

Solution:
Consider repeatedly reflecting square $A R M L$ over its sides so that the entire plane is covered by copies of $A R M L$. A path starting at $(2 / 7,3 / 7)$ that touches one or more sides and returns to $(2 / 7,3 / 7)$ corresponds to a straight line starting at $(2 / 7,3 / 7)$ and ending at the image of $(2 / 7,3 / 7)$ in one of the copies of $A R M L$. To touch three sides, the path must cross three lines, at least one of which must be vertical and at least one of which must be horizontal.

<img_3221>

If the path crosses two horizontal lines and the line $x=0$, it will have traveled a distance of 2 units vertically and $4 / 7$ units vertically for a total distance of $\sqrt{2^{2}+(4 / 7)^{2}}$ units. Similarly, the total distance traveled when crossing two horizontal lines and $x=1$ is $\sqrt{2^{2}+(10 / 7)^{2}}$, the total distance traveled when crossing two vertical lines and $y=0$ is $\sqrt{2^{2}+(6 / 7)^{2}}$, and the total distance traveled when crossing two vertical lines and $y=1$ is $\sqrt{2^{2}+(8 / 7)^{2}}$. The least of these is

$$
\sqrt{2^{2}+(4 / 7)^{2}}=\frac{2}{\mathbf{7}} \sqrt{\mathbf{5 3}}
$$

Final answer: $\frac{2}{7} \sqrt{53}$
Subfield: Algebra


Question: For each positive integer $k$, let $S_{k}$ denote the infinite arithmetic sequence of integers with first term $k$ and common difference $k^{2}$. For example, $S_{3}$ is the sequence $3,12,21, \ldots$ Compute the sum of all $k$ such that 306 is an element of $S_{k}$.

Solution:
If 306 is an element of $S_{k}$, then there exists an integer $m \geq 0$ such that $306=k+m k^{2}$. Thus $k \mid 306$ and $k^{2} \mid 306-k$. The second relation can be rewritten as $k \mid 306 / k-1$, which implies that $k \leq \sqrt{306}$ unless $k=306$. The prime factorization of 306 is $2 \cdot 3^{2} \cdot 17$, so the set of factors of 306 less than $\sqrt{306}$ is $\{1,2,3,6,9,17\}$. Check each in turn:

$$
\begin{aligned}
306-1 & =305, & & 1^{2} \mid 305 \\
306-2 & =304, & & 2^{2} \mid 304 \\
306-3 & =303, & & 3^{2} \nmid 303 \\
306-6 & =300, & & 6^{2} \nmid 300 \\
306-9 & =297, & & 9^{2} \nmid 297 \\
306-17 & =289, & & 17^{2} \mid 289 .
\end{aligned}
$$

Thus the set of possible $k$ is $\{1,2,17,306\}$, and the sum is $1+2+17+306=\mathbf{3 2 6}$.

Final answer: 326
Subfield: Algebra


Question: Compute the sum of all values of $k$ for which there exist positive real numbers $x$ and $y$ satisfying the following system of equations.

$$
\left\{\begin{aligned}
\log _{x} y^{2}+\log _{y} x^{5} & =2 k-1 \\
\log _{x^{2}} y^{5}-\log _{y^{2}} x^{3} & =k-3
\end{aligned}\right.
$$

Solution:
Let $\log _{x} y=a$. Then the first equation is equivalent to $2 a+\frac{5}{a}=2 k-1$, and the second equation is equivalent to $\frac{5 a}{2}-\frac{3}{2 a}=k-3$. Solving this system by eliminating $k$ yields the quadratic equation $3 a^{2}+5 a-8=0$, hence $a=1$ or $a=-\frac{8}{3}$. Substituting each of these values



of $a$ into either of the original equations and solving for $k$ yields $(a, k)=(1,4)$ or $\left(-\frac{8}{3},-\frac{149}{48}\right)$. Adding the values of $k$ yields the answer of $43 / 48$.
In terms of $a=\log _{x} y$, the two equations become $2 a+\frac{5}{a}=2 k-1$ and $\frac{5 a}{2}-\frac{3}{2 a}=k-3$. Eliminate $\frac{1}{a}$ to obtain $31 a=16 k-33$; substitute this into either of the original equations and clear denominators to get $96 k^{2}-86 k-1192=0$. The sum of the two roots is $86 / 96=\mathbf{4 3} / \mathbf{4 8}$.

Final answer: $\frac{43}{48}$
Subfield: Geometry


Question: Let $W=(0,0), A=(7,0), S=(7,1)$, and $H=(0,1)$. Compute the number of ways to tile rectangle $W A S H$ with triangles of area $1 / 2$ and vertices at lattice points on the boundary of WASH.

Solution:
Define a fault line to be a side of a tile other than its base. Any tiling of $W A S H$ can be represented as a sequence of tiles $t_{1}, t_{2}, \ldots, t_{14}$, where $t_{1}$ has a fault line of $\overline{W H}, t_{14}$ has a fault line of $\overline{A S}$, and where $t_{k}$ and $t_{k+1}$ share a fault line for $1 \leq k \leq 13$. Also note that to determine the position of tile $t_{k+1}$, it is necessary and sufficient to know the fault line that $t_{k+1}$ shares with $t_{k}$, as well as whether the base of $t_{k+1}$ lies on $\overline{W A}$ (abbreviated " $\mathrm{B}$ " for "bottom") or on $\overline{S H}$ (abbreviated "T" for "top"). Because rectangle $W A S H$ has width 7 , precisely 7 of the 14 tiles must have their bases on $\overline{W A}$. Thus any permutation of 7 B's and 7 T's determines a unique tiling $t_{1}, t_{2}, \ldots, t_{14}$, and conversely, any tiling $t_{1}, t_{2}, \ldots, t_{14}$ corresponds to a unique permutation of 7 B's and 7 T's. Thus the answer is $\left(\begin{array}{c}14 \\ 7\end{array}\right)=\mathbf{3 4 3 2}$.
Let $T(a, b)$ denote the number of ways to triangulate the polygon with vertices at $(0,0),(b, 0),(a, 1),(0,1)$, where each triangle has area $1 / 2$ and vertices at lattice points. The problem is to compute $T(7,7)$. It is easy to see that $T(a, 0)=T(0, b)=1$ for all $a$ and $b$. If $a$ and $b$ are both positive, then either one of the triangles includes the edge from $(a-1,1)$ to $(b, 0)$ or one of the triangles includes the edge from $(a, 1)$ to $(b-1,0)$, but not both. (In fact, as soon as there is an edge from $(a, 1)$ to $(x, 0)$ with $x<b$, there must be edges from $(a, 1)$ to $\left(x^{\prime}, 0\right)$ for all $x \leq x^{\prime}<b$.) If there is an edge from $(a-1,1)$ to $(b, 0)$, then the number of ways to complete the triangulation is $T(a-1, b)$; if there is an edge from $(a, 1)$ to $(b-1,0)$, then the number of ways to complete the triangulation is $T(a, b-1)$; thus $T(a, b)=T(a-1, b)+T(a, b-1)$. The recursion and the initial conditions describe Pascal's triangle, so $T(a, b)=\left(\begin{array}{c}a+b \\ a\end{array}\right)$. In particular, $T(7,7)=\left(\begin{array}{c}14 \\ 7\end{array}\right)=3432$.

Final answer: 3432
Subfield: Algebra


Question: Compute $\sin ^{2} 4^{\circ}+\sin ^{2} 8^{\circ}+\sin ^{2} 12^{\circ}+\cdots+\sin ^{2} 176^{\circ}$.

Solution:
Because $\cos 2 x=1-2 \sin ^{2} x, \sin ^{2} x=\frac{1-\cos 2 x}{2}$. Thus the desired sum can be rewritten as

$$
\frac{1-\cos 8^{\circ}}{2}+\frac{1-\cos 16^{\circ}}{2}+\cdots+\frac{1-\cos 352^{\circ}}{2}=\frac{44}{2}-\frac{1}{2}\left(\cos 8^{\circ}+\cos 16^{\circ}+\cdots+\cos 352^{\circ}\right) .
$$

If $\alpha=\cos 8^{\circ}+i \sin 8^{\circ}$, then $\alpha$ is a primitive $45^{\text {th }}$ root of unity, and $1+\alpha+\alpha^{2}+\alpha^{3}+\cdots+\alpha^{44}=0$. Hence $\alpha+\alpha^{2}+\cdots+\alpha^{44}=-1$, and because the real part of $\alpha^{n}$ is simply $\cos 8 n^{\circ}$,

$$
\cos 8^{\circ}+\cos 16^{\circ}+\cdots+\cos 352^{\circ}=-1 .
$$

Thus the desired sum is $22-(1 / 2)(-1)=\mathbf{4 5} / \mathbf{2}$.
The problem asks to simplify the sum

$$
\sin ^{2} a+\sin ^{2} 2 a+\sin ^{2} 3 a+\cdots+\sin ^{2} n a
$$



where $a=4^{\circ}$ and $n=44$. Because $\cos 2 x=1-2 \sin ^{2} x, \sin ^{2} x=\frac{1-\cos 2 x}{2}$. Thus the desired sum can be rewritten as

$$
\frac{1-\cos 2 a}{2}+\frac{1-\cos 4 a}{2}+\cdots+\frac{1-\cos 2 n a}{2}=\frac{n}{2}-\frac{1}{2}(\cos 2 a+\cos 4 a+\cdots+\cos 2 n a) .
$$

Let $Q=\cos 2 a+\cos 4 a+\cdots+\cos 2 n a$. By the sum-to-product identity,

$$
\begin{aligned}
\sin 3 a-\sin a & =2 \cos 2 a \sin a \\
\sin 5 a-\sin 3 a & =2 \cos 4 a \sin a \\
& \vdots \\
\sin (2 n+1) a-\sin (2 n-1) a & =2 \cos 2 n a \sin a
\end{aligned}
$$

Thus

$$
\begin{aligned}
Q \cdot 2 \sin a & =(\sin 3 a-\sin a)+(\sin 5 a-\sin 3 a)+\cdots+(\sin (2 n+1) a-\sin (2 n-1) a) \\
& =\sin (2 n+1) a-\sin a
\end{aligned}
$$

With $a=4^{\circ}$ and $n=44$, the difference on the right side becomes $\sin 356^{\circ}-\sin 4^{\circ}$; note that the terms in this difference are opposites, because of the symmetry of the unit circle. Hence

$$
\begin{aligned}
Q \cdot 2 \sin 4^{\circ} & =-2 \sin 4^{\circ}, \text { and } \\
Q & =-1 .
\end{aligned}
$$

Thus the original sum becomes $44 / 2-(1 / 2)(-1)=\mathbf{4 5} / \mathbf{2}$.

Final answer: $\frac{45}{2}$
Subfield: Geometry


Question: Compute the area of the region defined by $x^{2}+y^{2} \leq|x|+|y|$.

Solution:
Call the region $R$, and let $R_{q}$ be the portion of $R$ in the $q^{\text {th }}$ quadrant. Noting that the point $(x, y)$ is in $R$ if and only if $( \pm x, \pm y)$ is in $R$, it follows that $\left[R_{1}\right]=\left[R_{2}\right]=\left[R_{3}\right]=\left[R_{4}\right]$, and so $[R]=4\left[R_{1}\right]$. So it suffices to determine $\left[R_{1}\right]$.

In the first quadrant, the boundary equation is just $x^{2}+y^{2}=x+y \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$. This equation describes a circle of radius $\frac{\sqrt{2}}{2}$ centered at $\left(\frac{1}{2}, \frac{1}{2}\right)$. The portion of the circle's interior which is inside the first quadrant can be decomposed into a right isosceles triangle with side length 1 and half a circle of radius $\frac{\sqrt{2}}{2}$. Thus $\left[R_{1}\right]=\frac{1}{2}+\frac{\pi}{4}$, hence $[R]=\mathbf{2}+\boldsymbol{\pi}$.

Final answer: $2+\pi$