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


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836)

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that for all maps $M, C(M)<h(M)$.

Solution:
Let $n=h(M)$. The following strategy will always catch a Robber within two days using $n-1$ Cops, which proves that $C(M) \leq n-1$. Choose any subset $\mathcal{S}$ of $n-1$ hideouts and position $n-1$ Cops at the hideouts of $\mathcal{S}$ for 2 days. If the Robber is not caught on the first day, he must have been at the hideout not in $\mathcal{S}$, and therefore must move to a hideout in $\mathcal{S}$ on the following day.

Final answer: 证明题，略
Subfield: Combinatorics


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<img_3859>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that $C(M) \leq 3$ for the map below.

<img_3880>

Solution:
The following strategy guarantees capture using three Cops for four consecutive days, so $C(M) \leq 3$. Position three Cops at $\{B, E, H\}$ for two days, which will catch any Robber who starts out at $B, C, D, E, F, G$, or $H$, because a Robber at $C$ or $D$ would have to move to either $B$ or $E$, and a Robber at $F$ or $G$ would have to move to either $E$ or $H$. So if the Robber is not yet caught, he must have been at $I, J, K, L$, or $A$ for those two days. In this case, an analogous argument shows that placing Cops at $B, H, K$ for two consecutive days will guarantee a capture.
Subfield: Geometry


Question: Triangle $A B C$ has $\mathrm{m} \angle A>\mathrm{m} \angle B>\mathrm{m} \angle C$. The angle between the altitude and the angle bisector at vertex $A$ is $6^{\circ}$. The angle between the altitude and the angle bisector at vertex $B$ is $18^{\circ}$. Compute the degree measure of angle $C$.

Solution:
Let the feet of the altitudes from $A$ and $B$ be $E$ and $D$, respectively, and let $F$ and $G$ be the intersection points of the angle bisectors with $\overline{A C}$ and $\overline{B C}$, respectively, as shown below.

<img_3386>

Then $\mathrm{m} \angle G A E=6^{\circ}$ and $\mathrm{m} \angle D B F=18^{\circ}$. Suppose $\mathrm{m} \angle F B C=x^{\circ}$ and $\mathrm{m} \angle C A G=y^{\circ}$. So $\mathrm{m} \angle C A E=(y+6)^{\circ}$ and $\mathrm{m} \angle C B D=(x+18)^{\circ}$. Considering right triangle $B D C$, $\mathrm{m} \angle C=90^{\circ}-(x+18)^{\circ}=(72-x)^{\circ}$, while considering right triangle $A E C, \mathrm{~m} \angle C=$ $90^{\circ}-(y+6)^{\circ}=(84-y)^{\circ}$. Thus $84-y=72-x$ and $y-x=12$. Considering $\triangle A B E$, $\mathrm{m} \angle E A B=(y-6)^{\circ}$ and $\mathrm{m} \angle E B A=2 x^{\circ}$, so $(y-6)+2 x=90$, or $2 x+y=96$. Solving the system yields $x=28, y=40$. Therefore $\mathrm{m} \angle A=80^{\circ}$ and $\mathrm{m} \angle B=56^{\circ}$, so $\mathrm{m} \angle C=44^{\circ}$.
From right triangle $A B E, 90^{\circ}=\left(\frac{1}{2} A-6^{\circ}\right)+B$, and from right triangle $A B D, 90^{\circ}=\left(\frac{1}{2} B-18^{\circ}\right)+A$. Adding the two equations gives $180^{\circ}=\frac{3}{2}(A+B)-24^{\circ}$, so $A+B=\frac{2}{3} \cdot 204^{\circ}=136^{\circ}$ and $C=180^{\circ}-(A+B)=44^{\circ}$.

Final answer: $44^{\circ}$
Subfield: Algebra


Question: Compute the number of ordered pairs of integers $(b, c)$, with $-20 \leq b \leq 20,-20 \leq c \leq 20$, such that the equations $x^{2}+b x+c=0$ and $x^{2}+c x+b=0$ share at least one root.

Solution:
Let $r$ be the common root. Then $r^{2}+b r+c=r^{2}+c r+b \Rightarrow b r-c r=b-c$. So either $b=c$ or $r=1$. In the latter case, $1+b+c=0$, so $c=-1-b$.

There are 41 ordered pairs where $b=c$. If $c=-1-b$ and $-20 \leq b \leq 20$, then $-21 \leq c \leq 19$. Therefore there are 40 ordered pairs $(b,-1-b)$ where both terms are in the required intervals. Thus there are $41+40=\mathbf{8 1}$ solutions.

Final answer: 81
Subfield: Combinatorics


Question: A seventeen-sided die has faces numbered 1 through 17, but it is not fair: 17 comes up with probability $1 / 2$, and each of the numbers 1 through 16 comes up with probability $1 / 32$. Compute the probability that the sum of two rolls is either 20 or 12.

Solution:
The rolls that add up to 20 are $17+3,16+4,15+5,14+6,13+7,12+8,11+9$, and $10+10$. Accounting for order, the probability of $17+3$ is $\frac{1}{2} \cdot \frac{1}{32}+\frac{1}{32} \cdot \frac{1}{2}=2 \cdot \frac{1}{2} \cdot \frac{1}{32}=\frac{32}{1024}$. The combination $10+10$ has probability $\frac{1}{32} \cdot \frac{1}{32}=\frac{1}{1024}$; the other six combinations have probability $2 \cdot \frac{1}{32} \cdot \frac{1}{32}=\frac{2}{1024}$, for a total of $\frac{32+1+6 \cdot 2}{1024}=\frac{45}{1024}$ (again, accounting for two possible orders per combination). The rolls that add up to 12 are $1+11,2+10,3+9,4+8,5+7,6+6$, all



of which have probability $2 \cdot \frac{1}{32} \cdot \frac{1}{32}=\frac{2}{1024}$ except the last, which has probability $\left(\frac{1}{32}\right)^{2}$, for a total of $\frac{11}{1024}$. Thus the probability of either sum appearing is $\frac{45}{1024}+\frac{11}{1024}=\frac{56}{1024}=\frac{\mathbf{7}}{\mathbf{1 2 8}}$.

Final answer: $\frac{7}{128}$
Subfield: Geometry


Question: In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs.

![](https://cdn.mathpix.com/cropped/2023_12_21_78e54ff35f7d827e777fg-1.jpg?height=368&width=374&top_left_y=602&top_left_x=924)

Solution:
Because $A B=16, A C=B C=\frac{16}{\sqrt{2}}=8 \sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \sqrt{2}-8$. Each large arc has measure $45^{\circ}$ and the small arc has measure $90^{\circ}$. Therefore the area enclosed by each large arc is $\frac{45}{360} \cdot \pi \cdot 8^{2}=8 \pi$, and the area enclosed by the small arc is $\frac{90}{360} \cdot \pi \cdot(8 \sqrt{2}-8)^{2}=48 \pi-32 \pi \sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \pi-32 \pi \sqrt{2}$. On the other hand, the area of the triangle is $\frac{1}{2}(8 \sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \pi+32 \pi \sqrt{2}$.

Final answer: $\quad 64-64 \pi+32 \pi \sqrt{2}$
Subfield: Geometry


Question: In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs.

<img_3699>

Solution:
Because $A B=16, A C=B C=\frac{16}{\sqrt{2}}=8 \sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \sqrt{2}-8$. Each large arc has measure $45^{\circ}$ and the small arc has measure $90^{\circ}$. Therefore the area enclosed by each large arc is $\frac{45}{360} \cdot \pi \cdot 8^{2}=8 \pi$, and the area enclosed by the small arc is $\frac{90}{360} \cdot \pi \cdot(8 \sqrt{2}-8)^{2}=48 \pi-32 \pi \sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \pi-32 \pi \sqrt{2}$. On the other hand, the area of the triangle is $\frac{1}{2}(8 \sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \pi+32 \pi \sqrt{2}$.

Final answer: $\quad 64-64 \pi+32 \pi \sqrt{2}$
Subfield: Number Theory


Question: Compute the number of ordered pairs of integers $(a, b)$ such that $1<a \leq 50,1<b \leq 50$, and $\log _{b} a$ is rational.

Solution:
Begin by partitioning $\{2,3, \ldots, 50\}$ into the subsets

$$
\begin{aligned}
A & =\{2,4,8,16,32\} \\
B & =\{3,9,27\} \\
C & =\{5,25\} \\
D & =\{6,36\} \\
E & =\{7,49\} \\
F & =\text { all other integers between } 2 \text { and } 50, \text { inclusive. }
\end{aligned}
$$

If $\log _{b} a$ is rational, then either $a$ and $b$ are both members of one of the sets $A, B, C, D$, or $E$, or $a=b \in F$ (see note below for proof). Then the number of possible ordered pairs is

$$
\begin{aligned}
|A|^{2}+|B|^{2}+|C|^{2}+|D|^{2}+|E|^{2}+|F| & =25+9+4+4+4+35 \\
& =\mathbf{8 1}
\end{aligned}
$$

Final answer: 81
Subfield: Combinatorics


Question: Suppose that 5-letter "words" are formed using only the letters A, R, M, and L. Each letter need not be used in a word, but each word must contain at least two distinct letters. Compute the number of such words that use the letter A more than any other letter.

Solution:
Condition on the number $n$ of A's that appear in the word; $n$ is at least two, because of the requirement that $\mathbf{A}$ occur more often than any other letter, and $n$ is at most 4 , because of the requirement that there be at least two distinct letters. In the case $n=4$, there are 3 choices for the other letter, and 5 choices for where to place it, for a total of 15 possibilities. In the case $n=3$, there are two possibilities to consider: either a second letter occurs twice, or there are two distinct letters besides A. If a second letter occurs twice, there are 3 choices



for the other letter, and $\frac{5 !}{3 ! \cdot 2 !}=10$ ways to arrange the three A's and two non-A's, for their locations, for a total of 30 choices. If there are two distinct letters besides $A$, then there are $\left(\begin{array}{l}3 \\ 2\end{array}\right)=3$ ways to pick the two letters, and $\frac{5 !}{3 ! \cdot 1 ! \cdot 1 !}=20$ ways to arrange them, for a total of 60 words. Thus there are a combined total of 90 words when $n=3$. In the case $n=2$, no other letter can occur twice, so all the letters R, M, L, must appear in the word; they can be arranged in $\frac{5 !}{2 ! \cdot 1 ! \cdot 1 ! \cdot 1 !}=60$ ways. The total number of words satisfying the conditions is therefore $15+90+60=\mathbf{1 6 5}$.

Final answer: 165
Subfield: Algebra


Question: Positive integers $a_{1}, a_{2}, a_{3}, \ldots$ form an arithmetic sequence. If $a_{1}=10$ and $a_{a_{2}}=100$, compute $a_{a_{a_{3}}}$.

Solution:
Let $d$ be the common difference of the sequence. Then $a_{a_{2}}=a_{1}+\left(a_{2}-1\right) d=100 \Rightarrow\left(a_{2}-1\right) d=$ 90. But $a_{2}=a_{1}+d=10+d$, so $(9+d) d=90$. Solving the quadratic yields $d=-15$ or $d=6$, but the requirement that $a_{i}$ be positive for all $i$ rules out the negative value, so $d=6$ and $a_{n}=10+(n-1) \cdot 6$. Thus $a_{3}=10+2(6)=22$, and $a_{a_{3}}=a_{22}=10+21(6)=136$. Finally, $a_{a_{a_{3}}}=a_{136}=10+135(6)=\mathbf{8 2 0}$.

Final answer: 820
Subfield: Geometry


Question: The graphs of $y=x^{2}-|x|-12$ and $y=|x|-k$ intersect at distinct points $A, B, C$, and $D$, in order of increasing $x$-coordinates. If $A B=B C=C D$, compute $k$.

Solution:
First, note that both graphs are symmetric about the $y$-axis, so $C$ and $D$ must be reflections of $B$ and $A$, respectively, across the $y$-axis. Thus $x_{C}=-x_{B}$ and $y_{C}=y_{B}$, so $B C=2 x_{C}$. For $x<0$, the equations become $y=x^{2}+x-12$ and $y=-x-k$; setting the $x$-expressions equal to each other yields the equation $x^{2}+2 x+(k-12)=0$, from which $x=-1 \pm \sqrt{13-k}$. Therefore $x_{B}=-1+\sqrt{13-k}$ and $B C=2-2 \sqrt{13-k}$. (Note that the existence of two distinct negative values of $-1 \pm \sqrt{13-k}$ forces $12<k \leq 13$.)

Thus the $x$-coordinates of the four points are

$$
\begin{aligned}
& x_{A}=-1-\sqrt{13-k} \\
& x_{B}=-1+\sqrt{13-k} \\
& x_{C}=1-\sqrt{13-k} \\
& x_{D}=1+\sqrt{13-k} .
\end{aligned}
$$

To compute $y_{A}$, use the second equation $y=|x|-k$ to obtain $y_{A}=1+\sqrt{13-k}-k=$ $(1-k)+\sqrt{13-k}$; similarly, $y_{B}=(1-k)-\sqrt{13-k}$. Therefore

$$
\begin{aligned}
A B & =\sqrt{\left(x_{B}-x_{A}\right)^{2}+\left(y_{B}-y_{A}\right)^{2}} \\
& =\sqrt{(2 \sqrt{13-k})^{2}+(-2 \sqrt{13-k})^{2}} \\
& =2 \sqrt{2(13-k)}
\end{aligned}
$$

Because $A B=B C, 2 \sqrt{2(13-k)}=2-2 \sqrt{13-k}$. Let $u=\sqrt{13-k}$; then $2 \sqrt{2} u=2-2 u$, from which $u=\frac{2}{2+2 \sqrt{2}}=\frac{1}{1+\sqrt{2}}$, which equals $\sqrt{2}-1$ by rationalizing the denominator. Thus

$$
13-k=(\sqrt{2}-1)^{2}=3-2 \sqrt{2}, \text { so } k=\mathbf{1 0}+\mathbf{2} \sqrt{\mathbf{2}} \text {. }
$$

Because $10+2 \sqrt{2} \approx 12.8$, the value of $k$ determined algebraically satisfies the inequality $12<k \leq 13$ observed above.
Let $C=(a, b)$. Because $C$ and $D$ lie on a line with slope 1 , $D=(a+h, b+h)$ for some $h>0$. Because both graphs are symmetric about the $y$-axis, the other two points of intersection are $A=(-a-h, b+h)$ and $B=(-a, b)$, and $a>0$.

In terms of these coordinates, the distances are $A B=C D=\sqrt{2} h$ and $B C=2 a$. Thus the condition $A B=B C=C D$ holds if and only if $\sqrt{2} h=2 a$, or $h=\sqrt{2} a$.

The foregoing uses the condition that $C$ and $D$ lie on a line of slope 1 , so now use the remaining equation and subtract:

$$
\begin{aligned}
b & =a^{2}-a-12 \\
b+h & =(a+h)^{2}-(a+h)-12 \\
h & =2 a h+h^{2}-h
\end{aligned}
$$

Because the points are distinct, $h \neq 0$. Dividing by $h$ yields $2-2 a=h=\sqrt{2} a$. Thus $a=\frac{2}{2+\sqrt{2}}=2-\sqrt{2}$.

Finally, because $C$ lies on the two graphs, $b=a^{2}-a-12=-8-3 \sqrt{2}$ and $k=a-b=$ $10+2 \sqrt{2}$.

Final answer: $10+2 \sqrt{2}$
Subfield: Combinatorics


Question: The zeros of $f(x)=x^{6}+2 x^{5}+3 x^{4}+5 x^{3}+8 x^{2}+13 x+21$ are distinct complex numbers. Compute the average value of $A+B C+D E F$ over all possible permutations $(A, B, C, D, E, F)$ of these six numbers.

Solution:
There are $6 !=720$ permutations of the zeros, so the average value is the sum, $S$, divided by 720. Setting any particular zero as $A$ leaves $5 !=120$ ways to permute the other five zeros, so over the 720 permutations, each zero occupies the $A$ position 120 times. Similarly, fixing any ordered pair $(B, C)$ of zeros allows $4 !=24$ permutations of the other four zeros, and $B C=C B$ means that each value of $B C$ occurs 48 times. Finally, fixing any ordered triple $(D, E, F)$ allows $3 !=6$ permutations of the other variables, and there are $3 !=6$ equivalent arrangements within each product $D E F$, so that the product of any three zeros occurs 36 times within the sum. Let $S_{1}=A+B+C+D+E+F$ (i.e., the sum of the zeros taken singly), $S_{2}=A B+A C+\cdots+A F+B C+\cdots+E F$ (i.e., the sum of the zeros taken two at a time), and $S_{3}=A B C+A B D+\cdots+D E F$ be the sum of the zeros three at a time. Then $S=120 S_{1}+48 S_{2}+36 S_{3}$. Using the sums and products of roots formulas, $S_{1}=-2 / 1=-2$, $S_{2}=3 / 1=3$, and $S_{3}=-5 / 1=-5$. Thus $S=120(-2)+48(3)+36(-5)=-276$. The average value is thus $-\frac{276}{720}=-\frac{\mathbf{2 3}}{\mathbf{6 0}}$.

Final answer: $-\frac{23}{60}$
Subfield: Geometry


Question: Let $A B C$ be a triangle with $\mathrm{m} \angle B=\mathrm{m} \angle C=80^{\circ}$. Compute the number of points $P$ in the plane such that triangles $P A B, P B C$, and $P C A$ are all isosceles and non-degenerate. Note: the approximation $\cos 80^{\circ} \approx 0.17$ may be useful.

Solution:
Focus on $\triangle P B C$. Either $P B=P C$ or $P B=B C$ or $P C=B C$.

If $P B=P C$, then $P$ lies on the perpendicular bisector $l$ of side $\overline{B C}$. Considering now $\triangle P A B$, if $P A=P B$, then $P A=P C$, and $P$ must be the circumcenter of $\triangle A B C$; call this location $P_{1}$. If $P A=A B$, then $P A=A C$, and $P, B, C$ all lie on a circle with center $A$ and radius $A B$. There are two intersection points of that circle with $l$, one on each arc with endpoints $B$ and $C$; label the one on the major arc $P_{2}$ and on the minor $\operatorname{arc} P_{3}$. Finally, if $P B=A B$, then $P B=A C$ by the transitive property and $P C=A C$ by the perpendicular bisector theorem, so $P B A C$ is a rhombus; $P$ is the reflection of $A$ across $\overline{B C}$. Call this point $P_{4}$.

If $P B=B C$, then $P$ must lie on the circle centered at $B$ with radius $B C$. Considering $\triangle P A B$, if $P A=A B$, then $P$ lies on the circle centered at $A$ with radius $A B$. Now $\odot A$ and $\odot B$ intersect at two points, but one of them is $C$, so the other intersection must be the location of $P$, which is $P_{5}$. The condition $P B=A B$ is impossible, because it implies that $A B=B C$, which is false because in $\triangle A B C, \mathrm{~m} \angle C>\mathrm{m} \angle A=20^{\circ}$, so $A B>B C$. The third possibility for $\triangle P A B$ is that $P A=P B$, implying that the perpendicular bisector of $\overline{A B}$ intersects $\odot B$, which only occurs if $B C / A B \geq 1 / 2$ (although if $B C / A B=1 / 2$, the triangle is degenerate). But $B C / A B=2 \cos 80^{\circ}$, and the given approximation $\cos 80^{\circ} \approx 0.17$ implies that $B C / A B \approx 0.34$. Hence the perpendicular bisector of $\overline{A B}$ does not intersect $\odot B$. Thus the assumption $P B=B C$ yields only one additional location for $P, P_{5}$. Similarly, $P C=B C$ yields exactly one more location, $P_{6}$, for a total of $\mathbf{6}$ points. All six points, and their associated triangles, are pictured below.



<img_3810>

Final answer: 6
Subfield: Algebra


Question: If $x, y$, and $z$ are positive integers such that $x y=20$ and $y z=12$, compute the smallest possible value of $x+z$.

Solution:
Note that $x$ and $z$ can each be minimized by making $y$ as large as possible, so set $y=$ $\operatorname{lcm}(12,20)=4$. Then $x=5, z=3$, and $x+z=\mathbf{8}$.

Final answer: 8
Subfield: Geometry


Question: Let $T=8$. Let $A=(1,5)$ and $B=(T-1,17)$. Compute the value of $x$ such that $(x, 3)$ lies on the perpendicular bisector of $\overline{A B}$.

Solution:
The midpoint of $\overline{A B}$ is $\left(\frac{T}{2}, 11\right)$, and the slope of $\overleftrightarrow{A B}$ is $\frac{12}{T-2}$. Thus the perpendicular bisector of $\overline{A B}$ has slope $\frac{2-T}{12}$ and passes through the point $\left(\frac{T}{2}, 11\right)$. Thus the equation of the perpendicular bisector of $\overline{A B}$ is $y=\left(\frac{2-T}{12}\right) x+\left(11-\frac{2 T-T^{2}}{24}\right)$. Plugging $y=3$ into this equation and solving for $x$ yields $x=\frac{96}{T-2}+\frac{T}{2}$. With $T=8$, it follows that $x=\frac{96}{6}+\frac{8}{2}=16+4=\mathbf{2 0}$.

Final answer: 20
Subfield: Number Theory


Question: Let T be a rational number. Let $N$ be the smallest positive $T$-digit number that is divisible by 33 . Compute the product of the last two digits of $N$.

Solution:
The sum of the digits of $N$ must be a multiple of 3 , and the alternating sum of the digits must be a multiple of 11 . Because the number of digits of $N$ is fixed, the minimum $N$ will have the alternating sum of its digits equal to 0 , and therefore the sum of the digits of $N$ will be even, so it must be 6 . Thus if $T$ is even, then $N=1 \underbrace{0 \ldots .02}_{T-30^{\prime} \mathrm{s}}$, and if $T$ is odd, then $N=1 \underbrace{0 \ldots 0}_{T-30^{\prime} \mathrm{s}} 32$. Either way, the product of the last two digits of $N$ is 6 (independent of $T$ ).

Final answer: 6
Subfield: Geometry


Question: Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$.

<img_3165>

Solution:
Note that $D E S H$ is a trapezoid with height $\frac{T}{2}$. Because $\overline{A S}$ and $\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\frac{T}{6}$. Thus $S H=T-2 \cdot \frac{T}{6}=\frac{2 T}{3}$. Thus $[D E S H]=\frac{1}{2}\left(T+\frac{2 T}{3}\right) \cdot \frac{T}{2}=\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 .

Final answer: 15