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


Question: Eight students attend a Harper Valley ARML practice. At the end of the practice, they decide to take selfies to celebrate the event. Each selfie will have either two or three students in the picture. Compute the minimum number of selfies so that each pair of the eight students appears in exactly one selfie.

Solution:
The answer is 12 . To give an example in which 12 selfies is possible, consider regular octagon $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6} P_{7} P_{8}$. Each vertex of the octagon represents a student and each of the diagonals and sides of the octagon represents a pair of students. Construct eight triangles $P_{1} P_{2} P_{4}, P_{2} P_{3} P_{5}, P_{3} P_{4} P_{6}, \ldots, P_{8} P_{1} P_{3}$. Each of the segments in the forms of $\overline{P_{i} P_{i+1}}, \overline{P_{i} P_{i+2}}, \overline{P_{i} P_{i+3}}$ appears exactly once in these eight triangles. Taking 8 three-person selfies (namely $\left.\left\{P_{1}, P_{2}, P_{4}\right\},\left\{P_{2}, P_{3}, P_{5}\right\}, \ldots,\left\{P_{8}, P_{1}, P_{3}\right\}\right)$ and 4 two-person selfies (namely $\left.\left\{P_{1}, P_{5}\right\},\left\{P_{2}, P_{6}\right\},\left\{P_{3}, P_{7}\right\},\left\{P_{4}, P_{8}\right\}\right)$ gives a total of 12 selfies, completing the desired task.

A diagram of this construction is shown below. Each of the eight triangles is a different color, and each of the two-person selfies is represented by a dotted diameter.

<img_3543>

It remains to show fewer than 12 selfies is impossible. Assume that the students took $x$ three-person selfies and $y$ two-person selfies. Each three-person selfie counts 3 pairs of student appearances (in a selfie), and each two-person selfie counts 1 pair of student appearances (in a selfie). Together, these selfies count $3 x+y$ pairs of student appearances. There are $\left(\begin{array}{l}8 \\ 2\end{array}\right)=28$ pairs of student appearances. Hence $3 x+y=28$. The number of



selfies is $x+y=28-2 x$, so it is enough to show that $x \leq 8$.

Assume for contradiction there are $x \geq 9$ three-person selfies; then there are at least $3 \cdot 9=27$ (individual) student appearances on these selfies. Because there are 8 students, some student $s_{1}$ had at least $\lceil 27 / 8\rceil$ appearances; that is, $s_{1}$ appeared in at least 4 of these three-person selfies. There are $2 \cdot 4=8$ (individual) student appearances other than $s_{1}$ on these 4 selfies. Because there are only 7 students besides $s_{1}$, some other student $s_{2}$ had at least $[8 / 7\rceil$ (individual) appearances on these 4 selfies; that is, $s_{2}$ appeared (with $s_{1}$ ) in at least 2 of these 4 three-person selfies, violating the condition that each pair of the students appears in exactly one selfie. Thus the answer is $\mathbf{1 2}$.

Final answer: 12
Subfield: Algebra


Question: $\quad$ Compute the least positive value of $t$ such that

$$
\operatorname{Arcsin}(\sin (t)), \operatorname{Arccos}(\cos (t)), \operatorname{Arctan}(\tan (t))
$$

form (in some order) a three-term arithmetic progression with a nonzero common difference.

Solution:
For $0 \leq t<\pi / 2$, all three values are $t$, so the desired $t$ does not lie in this interval.

For $\pi / 2<t<\pi$,

$$
\begin{aligned}
\operatorname{Arcsin}(\sin (t)) & =\pi-t \in(0, \pi / 2) \\
\operatorname{Arccos}(\cos (t)) & =t \quad \in(\pi / 2, \pi) \\
\operatorname{Arctan}(\tan (t)) & =t-\pi \in(-\pi / 2,0)
\end{aligned}
$$

A graph of all three functions is shown below.

<img_3325>

Thus if the three numbers are to form an arithmetic progression, they should satisfy

$$
t-\pi<\pi-t<t
$$

The three numbers will be in arithmetic progression if and only if $t+(t-\pi)=2(\pi-t)$, which implies $t=\frac{\mathbf{3 \pi}}{\mathbf{4}}$. Note that if $t=\frac{3 \pi}{4}$, the arithmetic progression is $-\frac{\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}$, as required.

Final answer: $\frac{3 \pi}{4}$
Subfield: Geometry


Question: In non-right triangle $A B C$, distinct points $P, Q, R$, and $S$ lie on $\overline{B C}$ in that order such that $\angle B A P \cong \angle P A Q \cong \angle Q A R \cong \angle R A S \cong \angle S A C$. Given that the angles of $\triangle A B C$ are congruent to the angles of $\triangle A P Q$ in some order of correspondence, compute $\mathrm{m} \angle B$ in degrees.

Solution:
Let $\theta=\frac{1}{5} \mathrm{~m} \angle A$. Because $\mathrm{m} \angle P A Q=\theta<5 \theta=\mathrm{m} \angle A$, it follows that either $\mathrm{m} \angle B=\theta$ or $\mathrm{m} \angle C=\theta$. Thus there are two cases to consider.

If $\mathrm{m} \angle C=\theta$, then it follows that $\mathrm{m} \angle A Q P=\mathrm{m} \angle Q A C+\mathrm{m} \angle A C B=4 \theta$, and hence $\mathrm{m} \angle B=4 \theta$. So $\triangle A B C$ has angles of measures $5 \theta, 4 \theta, \theta$, and thus $\theta=18^{\circ}$. However, this implies $\mathrm{m} \angle A=5 \theta=90^{\circ}$, which is not the case.

<img_3197>

If instead $\mathrm{m} \angle B=\theta$, then it follows that $\mathrm{m} \angle A P Q=\mathrm{m} \angle B A P+\mathrm{m} \angle A B P=2 \theta$, and hence $\mathrm{m} \angle C=2 \theta$. So $\triangle A B C$ has angles of measures $5 \theta, 2 \theta, \theta$, and thus $\theta=22.5^{\circ}$. Hence $\mathrm{m} \angle B=\theta=\mathbf{2 2 . 5}$.

<img_3939>

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


Question: Consider the system of equations

$$
\begin{aligned}
& \log _{4} x+\log _{8}(y z)=2 \\
& \log _{4} y+\log _{8}(x z)=4 \\
& \log _{4} z+\log _{8}(x y)=5 .
\end{aligned}
$$

Given that $x y z$ can be expressed in the form $2^{k}$, compute $k$.

Solution:
Note that for $n>0, \log _{4} n=\log _{64} n^{3}$ and $\log _{8} n=\log _{64} n^{2}$. Adding together the three given equations and using both the preceding facts and properties of logarithms yields

$$
\begin{aligned}
& \log _{4}(x y z)+\log _{8}\left(x^{2} y^{2} z^{2}\right)=11 \\
\Longrightarrow & \log _{64}(x y z)^{3}+\log _{64}(x y z)^{4}=11 \\
\Longrightarrow & \log _{64}(x y z)^{7}=11 \\
\Longrightarrow & 7 \log _{64}(x y z)=11 .
\end{aligned}
$$

The last equation is equivalent to $x y z=64^{11 / 7}=2^{66 / 7}$, hence the desired value of $k$ is $\frac{\mathbf{6 6}}{\mathbf{7}}$.

Final answer: $\frac{66}{7}$
Subfield: Combinatorics


Question: A complex number $z$ is selected uniformly at random such that $|z|=1$. Compute the probability that $z$ and $z^{2019}$ both lie in Quadrant II in the complex plane.

Solution:
For convenience, let $\alpha=\pi / 4038$. Denote by

$$
0 \leq \theta<2 \pi=8076 \alpha
$$

the complex argument of $z$, selected uniformly at random from the interval $[0,2 \pi)$. Then $z$ itself lies in Quadrant II if and only if

$$
2019 \alpha=\frac{\pi}{2}<\theta<\pi=4038 \alpha
$$

On the other hand, $z^{2019}$ has argument 2019日, and hence it lies in Quadrant II if and only if there is some integer $k$ with

$$
\begin{gathered}
\frac{\pi}{2}+2 k \pi<2019 \theta<\pi+2 k \pi \\
\Longleftrightarrow(4 k+1) \cdot \frac{\pi}{2}<2019 \theta<(4 k+2) \cdot \frac{\pi}{2} \\
\Longleftrightarrow(4 k+1) \alpha<\theta<(4 k+2) \alpha .
\end{gathered}
$$

Because it is also true that $2019 \alpha<\theta<4038 \alpha$, the set of $\theta$ that satisfies the conditions of the problem is the union of intervals:

$$
(2021 \alpha, 2022 \alpha) \cup(2025 \alpha, 2026 \alpha) \cup \cdots \cup(4037 \alpha, 4038 \alpha)
$$

There are 505 such intervals, the $j^{\text {th }}$ interval consisting of $(4 j+2017) \alpha<\theta<(4 j+2018) \alpha$. Each interval has length $\alpha$, so the sum of the intervals has length $505 \alpha$. Thus the final answer is

$$
\frac{505 \alpha}{2 \pi}=\frac{505}{2 \cdot 4038}=\frac{\mathbf{5 0 5}}{\mathbf{8 0 7 6}} .
$$

Final answer: $\frac{505}{8076}$
Subfield: Number Theory


Question: Compute the least positive integer $n$ such that the sum of the digits of $n$ is five times the sum of the digits of $(n+2019)$.

Solution:
Let $S(n)$ denote the sum of the digits of $n$, so that solving the problem is equivalent to solving $S(n)=5 S(n+2019)$. Using the fact that $S(n) \equiv n(\bmod 9)$ for all $n$, it follows that

$$
\begin{aligned}
n & \equiv 5(n+2019) \equiv 5(n+3)(\bmod 9) \\
4 n & \equiv-15(\bmod 9) \\
n & \equiv 3(\bmod 9)
\end{aligned}
$$

Then $S(n+2019) \equiv 6(\bmod 9)$. In particular, $S(n+2019) \geq 6$ and $S(n) \geq 5 \cdot 6=30$. The latter inequality implies $n \geq 3999$, which then gives $n+2019 \geq 6018$. Thus if $n+2019$ were a four-digit number, then $S(n+2019) \geq 7$. Moreover, $S(n+2019)$ can only be 7, because otherwise, $S(n)=5 S(n+2019) \geq 40$, which is impossible (if $n$ has four digits, then $S(n)$ can be no greater than 36). So if $n+2019$ were a four-digit number, then $S(n+2019)=7$ and $S(n)=35$. But this would imply that the digits of $n$ are $8,9,9,9$ in some order, contradicting the assumption that $n+2019$ is a four-digit number. On the other hand, if $n+2019$ were a five-digit number such that $S(n+2019) \geq 6$, then the least such value of $n+2019$ is 10005 , and indeed, this works because it corresponds to $n=\mathbf{7 9 8 6}$, the least possible value of $n$.

Final answer: 7986
Subfield: Geometry


Question: $\quad$ Compute the greatest real number $K$ for which the graphs of

$$
(|x|-5)^{2}+(|y|-5)^{2}=K \quad \text { and } \quad(x-1)^{2}+(y+1)^{2}=37
$$

have exactly two intersection points.

Solution:
The graph of the second equation is simply the circle of radius $\sqrt{37}$ centered at $(1,-1)$. The first graph is more interesting, and its behavior depends on $K$.

- For small values of $K$, the first equation determines a set of four circles of radius $\sqrt{K}$ with centers at $(5,5),(5,-5),(-5,5)$, and $(-5,-5)$. Shown below are versions with $K=1, K=4$, and $K=16$.
<img_3421>
- However, when $K>25$, the graph no longer consists of four circles! As an example, for $K=36$, the value $x=5$ gives $(|y|-5)^{2}=36$; hence $|y|=-1$ or $|y|=6$. The first option is impossible; the graph ends up "losing" the portions of the upper-right circle that would cross the $x$ - or $y$-axes compared to the graph for $(x-5)^{2}+(y-5)^{2}=36$. The graph for $K=36$ is shown below.

<img_3535>

- As $K$ continues to increase, the "interior" part of the curve continues to shrink, until at $K=50$, it simply comprises the origin, and for $K>50$, it does not exist. As examples, the graphs with $K=50$ and $K=64$ are shown below.
<img_3577>



Overlay the graph of the circle of radius $\sqrt{37}$ centered at $(1,-1)$ with the given graphs. When $K=25$, this looks like the following graph.

<img_3905>

Note that the two graphs intersect at $(0,5)$ and $(-5,0)$, as well as four more points (two points near the positive $x$-axis and two points near the negative $y$-axis). When $K$ is slightly greater than 25 , this drops to four intersection points. The graph for $K=27$ is shown below.

<img_3751>

Thus for the greatest $K$ for which there are exactly two intersection points, those two intersection points should be along the positive $x$ - and negative $y$-axes. If the intersection point on the positive $x$-axis is at $(h, 0)$, then $(h-1)^{2}+(0+1)^{2}=37$ and $(h-5)^{2}+(0-5)^{2}=K$. Thus $h=7$ and $K=\mathbf{2 9}$

Final answer: 29
Subfield: Geometry


Question: Triangle $A B C$ is inscribed in circle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at point $T$. The tangent to $\omega$ at $A$ intersects the perpendicular bisector of $\overline{A T}$ at point $P$. Given that $A B=14, A C=30$, and $B C=40$, compute $[P B C]$.

Solution:
To begin, denote by $R$ the radius of $\omega$. The semiperimeter of triangle $A B C$ is 42 , and then applying Heron's formula yields

$$
[A B C]=\frac{14 \cdot 30 \cdot 40}{4 R}=\sqrt{42 \cdot 28 \cdot 12 \cdot 2}=168
$$

from which it follows that $R=\frac{14 \cdot 30 \cdot 40}{4 \cdot 168}=25$.

Now consider the point circle with radius zero centered at $T$ in tandem with the circle $\omega$. Because $P A=P T$, it follows that $P$ lies on the radical axis of these circles. Moreover, the midpoints of $\overline{T B}$ and $\overline{T C}$ lie on this radical axis as well. Thus $P$ lies on the midline of $\triangle T B C$ that is parallel to $\overline{B C}$.

<img_3211>

To finish, let $O$ denote the center of $\omega$ and $M$ the midpoint of $\overline{B C}$. By considering right triangle $T B O$ with altitude $\overline{B M}$, it follows that $M T \cdot M O=M B^{2}$, but also $M O=\sqrt{O B^{2}-M B^{2}}=\sqrt{25^{2}-20^{2}}=15$, so

$$
M T=\frac{M B^{2}}{M O}=\frac{400}{15}=\frac{80}{3}
$$

Thus the distance from $P$ to $\overline{B C}$ is $\frac{1}{2} M T=\frac{40}{3}$. Finally,

$$
[P B C]=\frac{1}{2} \cdot \frac{40}{3} \cdot B C=\frac{\mathbf{8 0 0}}{\mathbf{3}}
$$

Final answer: $\frac{800}{3}$
Subfield: Algebra


Question: Given that $a, b, c$, and $d$ are integers such that $a+b c=20$ and $-a+c d=19$, compute the greatest possible value of $c$.

Solution:
Adding the two given equations yields $b c+c d=c(b+d)=39$. The greatest possible value of $c$ therefore occurs when $c=\mathbf{3 9}$ and $b+d=1$.

Final answer: 39
Subfield: Combinatorics


Question: Let $T$ = 39. Emile randomly chooses a set of $T$ cards from a standard deck of 52 cards. Given that Emile's set contains no clubs, compute the probability that his set contains three aces.

Solution:
Knowing that 13 of the cards are not in Emile's set, there are $\left(\begin{array}{c}39 \\ T\end{array}\right)$ ways for him to have chosen a set of $T$ cards. Given that Emile's set contains no clubs, the suits of the three aces are fixed (i.e., diamonds, hearts, and spades). The number of possible sets of cards in which these three aces appear is therefore $\left(\begin{array}{c}36 \\ T-3\end{array}\right)$. The desired probability is therefore $\left(\begin{array}{c}36 \\ T-3\end{array}\right) /\left(\begin{array}{c}39 \\ T\end{array}\right)$. With $T=39$, this probability is $1 / 1=\mathbf{1}$, which is consistent with the fact that Emile's set contains all cards in the deck that are not clubs, hence he is guaranteed to have all three of the remaining aces.

Final answer: 1
Subfield: Geometry


Question: Let $T=1$. In parallelogram $A B C D, \frac{A B}{B C}=T$. Given that $M$ is the midpoint of $\overline{A B}$ and $P$ and $Q$ are the trisection points of $\overline{C D}$, compute $\frac{[A B C D]}{[M P Q]}$.

Solution:
Let $C D=3 x$ and let $h$ be the length of the altitude between bases $\overline{A B}$ and $\overline{C D}$. Then $[A B C D]=3 x h$ and $[M P Q]=\frac{1}{2} x h$. Hence $\frac{[A B C D]}{[M P Q]}=\mathbf{6}$. Both the position of $M$ and the ratio $\frac{A B}{B C}=T$ are irrelevant.

Final answer: 6
Subfield: Algebra


Question: Let $T=6$. Compute the value of $x$ such that $\log _{T} \sqrt{x-7}+\log _{T^{2}}(x-2)=1$.

Solution:
It can readily be shown that $\log _{a} b=\log _{a^{2}} b^{2}$. Thus it follows that $\log _{T} \sqrt{x-7}=\log _{T^{2}}(x-7)$. Hence the left-hand side of the given equation is $\log _{T^{2}}(x-7)(x-2)$ and the equation is equivalent to $(x-7)(x-2)=T^{2}$, which is equivalent to $x^{2}-9 x+14-T^{2}=0$. With $T=6$, this equation is $x^{2}-9 x-22=0 \Longrightarrow(x-11)(x+2)=0$. Plugging $x=-2$ into the given equation leads to the first term of the left-hand side having a negative radicand and the second term having an argument of 0 . However, one can easily check that $x=\mathbf{1 1}$ indeed satisfies the given equation.

Final answer: 11
Subfield: Number Theory


Question: Let $T=11$. Let $p$ be an odd prime and let $x, y$, and $z$ be positive integers less than $p$. When the trinomial $(p x+y+z)^{T-1}$ is expanded and simplified, there are $N$ terms, of which $M$ are always multiples of $p$. Compute $M$.

Solution:
A general term in the expansion of $(p x+y+z)^{T-1}$ has the form $K(p x)^{a} y^{b} z^{c}$, where $a, b$, and $c$ are nonnegative integers such that $a+b+c=T-1$. Using the "stars and bars" approach, the number of nonnegative integral solutions to $a+b+c=T-1$ is the number of arrangements of $T-1$ stars and 2 bars in a row (the bars act has separators and the " 2 " arises because it is one less than the number of variables in the equation). Thus there are $\left(\begin{array}{c}T+1 \\ 2\end{array}\right)$ solutions. Each term will be a multiple of $p$ unless $a=0$. In this case, the number of terms that are not multiples of $p$ is the number of nonnegative integral solutions to the equation $b+c=T-1$, which is $T$ ( $b$ can range from 0 to $T-1$ inclusive, and then $c$ is fixed). Hence $M=\left(\begin{array}{c}T+1 \\ 2\end{array}\right)-T=\frac{T^{2}-T}{2}$. With $T=11$, the answer is $\mathbf{5 5}$

Final answer: 55
Subfield: Algebra


Question: Let $T=55$. Compute the value of $K$ such that $20, T-5, K$ is an increasing geometric sequence and $19, K, 4 T+11$ is an increasing arithmetic sequence.

Solution:
The condition that $20, T-5, K$ is an increasing geometric sequence implies that $\frac{T-5}{20}=\frac{K}{T-5}$, hence $K=\frac{(T-5)^{2}}{20}$. The condition that $19, K, 4 T+11$ is an increasing arithmetic sequence implies that $K-19=4 T+11-K$, hence $K=2 T+15$. With $T=55$, each of these equations implies that $K=\mathbf{1 2 5}$. Note that the two equations can be combined and solved without being passed a value of $T$. A quadratic equation results, and its roots are $T=55$ or $T=-5$. However, with $T=-5$, neither of the given sequences is increasing.

Final answer: 125
Subfield: Geometry


Question: Let $T=125$. Cube $\mathcal{C}_{1}$ has volume $T$ and sphere $\mathcal{S}_{1}$ is circumscribed about $\mathcal{C}_{1}$. For $n \geq 1$, the sphere $\mathcal{S}_{n}$ is circumscribed about the cube $\mathcal{C}_{n}$ and is inscribed in the cube $\mathcal{C}_{n+1}$. Let $k$ be the least integer such that the volume of $\mathcal{C}_{k}$ is at least 2019. Compute the edge length of $\mathcal{C}_{k}$.

Solution:
In general, let cube $\mathcal{C}_{n}$ have edge length $x$. Then the diameter of sphere $\mathcal{S}_{n}$ is the space diagonal of $\mathcal{C}_{n}$, which has length $x \sqrt{3}$. This in turn is the edge length of cube $\mathcal{C}_{n+1}$. Hence the edge lengths of $\mathcal{C}_{1}, \mathcal{C}_{2}, \ldots$ form an increasing geometric sequence with common ratio $\sqrt{3}$ and volumes of $\mathcal{C}_{1}, \mathcal{C}_{2}, \ldots$ form an increasing geometric sequence with common ratio $3 \sqrt{3}$. With $T=125$, the edge length of $\mathcal{C}_{1}$ is 5 , so the sequence of edge lengths of the cubes is $5,5 \sqrt{3}, 15, \ldots$, and the respective sequence of the volumes of the cubes is $125,375 \sqrt{3}, 3375, \ldots$. Hence $k=3$, and the edge length of $\mathcal{C}_{3}$ is $\mathbf{1 5}$.

Final answer: 15
Subfield: Geometry


Question: Square $K E N T$ has side length 20 . Point $M$ lies in the interior of $K E N T$ such that $\triangle M E N$ is equilateral. Given that $K M^{2}=a-b \sqrt{3}$, where $a$ and $b$ are integers, compute $b$.

Solution:
Let $s$ be the side length of square $K E N T$; then $M E=s$. Let $J$ be the foot of the altitude from $M$ to $\overline{K E}$. Then $\mathrm{m} \angle J E M=30^{\circ}$ and $\mathrm{m} \angle E M J=60^{\circ}$. Hence $M J=\frac{s}{2}, J E=\frac{s \sqrt{3}}{2}$, and $K J=K E-J E=s-\frac{s \sqrt{3}}{2}$. Applying the Pythagorean Theorem to $\triangle K J M$ implies that $K M^{2}=\left(s-\frac{s \sqrt{3}}{2}\right)^{2}+\left(\frac{s}{2}\right)^{2}=2 s^{2}-s^{2} \sqrt{3}$. With $s=20$, the value of $b$ is therefore $s^{2}=\mathbf{4 0 0}$.

Final answer: 400
Subfield: Algebra


Question: Let $T$ be a rational number. Let $a, b$, and $c$ be the three solutions of the equation $x^{3}-20 x^{2}+19 x+T=0$. Compute $a^{2}+b^{2}+c^{2}$.

Solution:
According to Vieta's formulas, $a+b+c=-(-20)=20$ and $a b+b c+c a=19$. Noting that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)$, it follows that $a^{2}+b^{2}+c^{2}=20^{2}-2 \cdot 19=\mathbf{3 6 2}$. The value of $T$ is irrelevant.

Final answer: 362
Subfield: Algebra


Question: Let $T=362$ and let $K=\sqrt{T-1}$. Compute $\left|(K-20)(K+1)+19 K-K^{2}\right|$.

Solution:
The expression inside the absolute value bars simplifies to $K^{2}-19 K-20+19 K-K^{2}=-20$. Hence the answer is $\mathbf{2 0}$ and the value of $K(=\sqrt{361}=19)$ is not needed.

Final answer: 20
Subfield: Geometry


Question: Let $T=20$. In $\triangle L E O, \sin \angle L E O=\frac{1}{T}$. If $L E=\frac{1}{n}$ for some positive real number $n$, then $E O=$ $n^{3}-4 n^{2}+5 n$. As $n$ ranges over the positive reals, compute the least possible value of $[L E O]$.

Solution:
Note that $[L E O]=\frac{1}{2}(\sin \angle L E O) \cdot L E \cdot E O=\frac{1}{2} \cdot \frac{1}{T} \cdot \frac{1}{n} \cdot\left(n^{3}-4 n^{2}+5 n\right)=\frac{n^{2}-4 n+5}{2 T}$. Because $T$ is a constant, the least possible value of $[L E O]$ is achieved when the function $f(n)=n^{2}-4 n+5$ is minimized.



This occurs when $n=-(-4) /(2 \cdot 1)=2$, and the minimum value is $f(2)=1$. Hence the desired least possible value of $[L E O]$ is $\frac{1}{2 T}$, and with $T=20$, this is $\frac{1}{40}$.

Final answer: $\frac{1}{40}$
Subfield: Algebra


Question: Let $T=\frac{1}{40}$. Given that $x, y$, and $z$ are real numbers such that $x+y=5, x^{2}-y^{2}=\frac{1}{T}$, and $x-z=-7$, compute $x+z$

Solution:
Note that $x^{2}-y^{2}=(x+y)(x-y)=5(x-y)$, hence $x-y=\frac{1}{5 T}$. Then $x+z=(x+y)+(x-y)+(z-x)=$ $5+\frac{1}{5 T}+7=12+\frac{1}{5 T}$. With $T=\frac{1}{40}$, the answer is thus $12+8=\mathbf{2 0}$.

Final answer: 20
Subfield: Number Theory


Question: Let $T=20$. The product of all positive divisors of $2^{T}$ can be written in the form $2^{K}$. Compute $K$.

Solution:
When $n$ is a nonnegative integer, the product of the positive divisors of $2^{n}$ is $2^{0} \cdot 2^{1} \cdot \ldots \cdot 2^{n-1} \cdot 2^{n}=$ $2^{0+1+\cdots+(n-1)+n}=2^{n(n+1) / 2}$. Because $T=20$ is an integer, it follows that $K=\frac{T(T+1)}{2}=\mathbf{2 1 0}$.

Final answer: 210
Subfield: Combinatorics


Question: Let $T=210$. At the Westward House of Supper ("WHS"), a dinner special consists of an appetizer, an entrée, and dessert. There are 7 different appetizers and $K$ different entrées that a guest could order. There are 2 dessert choices, but ordering dessert is optional. Given that there are $T$ possible different orders that could be placed at the WHS, compute $K$.

Solution:
Because dessert is optional, there are effectively $2+1=3$ dessert choices. Hence, by the Multiplication Principle, it follows that $T=7 \cdot K \cdot 3$, thus $K=\frac{T}{21}$. With $T=210$, the answer is 10 .

Final answer: 10
Subfield: Algebra


Question: Let $S=15$ and let $M=10$ . Sam and Marty each ride a bicycle at a constant speed. Sam's speed is $S \mathrm{~km} / \mathrm{hr}$ and Marty's speed is $M \mathrm{~km} / \mathrm{hr}$. Given that Sam and Marty are initially $100 \mathrm{~km}$ apart and they begin riding towards one another at the same time, along a straight path, compute the number of kilometers that Sam will have traveled when Sam and Marty meet.

Solution:
In km/hr, the combined speed of Sam and Marty is $S+M$. Thus one can determine the total time they traveled and use this to determine the number of kilometers that Sam traveled. However, this is not needed, and there is a simpler approach. Suppose that Marty traveled a distance of $d$. Then because Sam's speed is $\frac{S}{M}$ of Marty's speed, Sam will have traveled a distance of $\frac{S}{M} \cdot d$. Thus, together, they traveled $d+\frac{S}{M} \cdot d$. Setting this equal to 100 and solving yields $d=\frac{100 M}{M+S}$. Thus Sam traveled $\frac{S}{M} \cdot d=\frac{100 S}{M+S}$. With $S=15$ and $M=10$, this is equal to $60 \mathrm{~km}$.

Final answer: 60
Subfield: Number Theory


Question: Compute the $2011^{\text {th }}$ smallest positive integer $N$ that gains an extra digit when doubled.

Solution:
Let $S$ be the set of numbers that gain an extra digit when doubled. First notice that the numbers in $S$ are precisely those whose first digit is at least 5 . Thus there are five one-digit numbers in $S, 50$ two-digit numbers in $S$, and 500 three-digit numbers in $S$. Therefore 5000 is the $556^{\text {th }}$ smallest number in $S$, and because all four-digit numbers greater than 5000 are in $S$, the $2011^{\text {th }}$ smallest number in $S$ is $5000+(2011-556)=\mathbf{6 4 5 5}$.

Final answer: 6455