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


Question: In triangle $A B C, C$ is a right angle and $M$ is on $\overline{A C}$. A circle with radius $r$ is centered at $M$, is tangent to $\overline{A B}$, and is tangent to $\overline{B C}$ at $C$. If $A C=5$ and $B C=12$, compute $r$.

Solution:
Let $N$ be the point of tangency of the circle with $\overline{A B}$ and draw $\overline{M B}$, as shown below.

<img_3520>

Because $\triangle B M C$ and $\triangle B M N$ are right triangles sharing a hypotenuse, and $\overline{M N}$ and $\overline{M C}$ are radii, $\triangle B M C \cong \triangle B M N$. Thus $B N=12$ and $A N=1$. Also $\triangle A N M \sim \triangle A C B$ because the right triangles share $\angle A$, so $\frac{N M}{A N}=\frac{C B}{A C}$. Therefore $\frac{r}{1}=\frac{12}{5}$, so $r=\frac{\mathbf{1 2}}{\mathbf{5}}$.
Let $r$ denote the radius of the circle, and let $D$ be the foot of the perpendicular from $O$ to $\overline{A B}$. Note that $\triangle A B C \sim \triangle A O D$. Thus $\frac{A B}{A O}=\frac{B C}{D O} \Longrightarrow \frac{13}{5-r}=\frac{12}{r}$, and $r=\frac{\mathbf{1 2}}{\mathbf{5}}$.

Final answer: $\frac{12}{5}$
Subfield: Number Theory


Question: The product of the first five terms of a geometric progression is 32 . If the fourth term is 17 , compute the second term.

Solution:
Let $a$ be the third term of the geometric progression, and let $r$ be the common ratio. Then the product of the first five terms is

$$
\left(a r^{-2}\right)\left(a r^{-1}\right)(a)(a r)\left(a r^{2}\right)=a^{5}=32
$$

so $a=2$. Because the fourth term is $17, r=\frac{17}{a}=\frac{17}{2}$. The second term is $a r^{-1}=\frac{2}{17 / 2}=\frac{4}{17}$.

Final answer: $\frac{4}{17}$
Subfield: Geometry


Question: Polygon $A_{1} A_{2} \ldots A_{n}$ is a regular $n$-gon. For some integer $k<n$, quadrilateral $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle of area 6 . If the area of $A_{1} A_{2} \ldots A_{n}$ is 60 , compute $n$.

Solution:
Because $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle, $n$ must be even, and moreover, $k=\frac{n}{2}$. Also, the rectangle's diagonals meet at the center $O$ of the circumscribing circle. $O$ is also the center of the $n$-gon. The diagram below shows the case $n=16$.



<img_3867>

Then $\left[A_{1} A_{2} O\right]=\frac{1}{4}\left[A_{1} A_{2} A_{k} A_{k+1}\right]=\frac{1}{n}\left[A_{1} A_{2} \ldots A_{n}\right]=60$. So $\frac{1}{4}(6)=\frac{1}{n}(60)$, and $n=40$.

Final answer: 40
Subfield: Combinatorics


Question: A bag contains 20 lavender marbles, 12 emerald marbles, and some number of orange marbles. If the probability of drawing an orange marble in one try is $\frac{1}{y}$, compute the sum of all possible integer values of $y$.

Solution:
Let $x$ be the number of orange marbles. Then the probability of drawing an orange marble is $\frac{x}{x+20+12}=\frac{x}{x+32}$. If this probability equals $\frac{1}{y}$, then $y=\frac{x+32}{x}=1+\frac{32}{x}$. This expression represents an integer only when $x$ is a factor of 32 , thus $x \in\{1,2,4,8,16,32\}$. The corresponding $y$-values are $33,17,9,5,3$, and 2 , and their sum is $\mathbf{6 9}$.

Final answer: 69
Subfield: Algebra


Question: Compute the number of ordered quadruples of integers $(a, b, c, d)$ satisfying the following system of equations:

$$
\left\{\begin{array}{l}
a b c=12,000 \\
b c d=24,000 \\
c d a=36,000
\end{array}\right.
$$

Solution:
From the first two equations, conclude that $d=2 a$. From the last two, $3 b=2 a$. Thus all solutions to the system will be of the form $(3 K, 2 K, c, 6 K)$ for some integer $K$. Substituting these expressions into the system, each equation now becomes $c K^{2}=2000=2^{4} \cdot 5^{3}$. So $K^{2}$ is of the form $2^{2 m} 5^{2 n}$. There are 3 choices for $m$ and 2 for $n$, so there are 6 values for $K^{2}$, which means there are 12 solutions overall, including negative values for $K$.

Although the problem does not require finding them, the twelve values of $K$ are $\pm 1, \pm 2, \pm 4$, $\pm 5, \pm 10, \pm 20$. These values yield the following quadruples $(a, b, c, d)$ :

$$
\begin{aligned}
& (3,2,2000,6),(-3,-2,2000,-6), \\
& (6,4,500,12),(-6,-4,500,-12), \\
& (12,8,125,24),(-12,-8,125,-24), \\
& (15,10,80,30),(-15,-10,80,-30), \\
& (30,20,20,60),(-30,-20,20,-60), \\
& (60,40,5,120),(-60,-40,5,-120) .
\end{aligned}
$$

Final answer: 12
Subfield: Algebra


Question: Let $n$ be a positive integer such that $\frac{3+4+\cdots+3 n}{5+6+\cdots+5 n}=\frac{4}{11}$. Compute $\frac{2+3+\cdots+2 n}{4+5+\cdots+4 n}$.

Solution:
In simplifying the numerator and denominator of the left side of the equation, notice that

$$
\begin{aligned}
k+(k+1)+\cdots+k n & =\frac{1}{2}(k n(k n+1)-k(k-1)) \\
& =\frac{1}{2}(k(n+1)(k n-k+1))
\end{aligned}
$$

This identity allows the given equation to be transformed:

$$
\begin{aligned}
\frac{3(n+1)(3 n-3+1)}{5(n+1)(5 n-5+1)} & =\frac{4}{11} \\
\frac{3(n+1)(3 n-2)}{5(n+1)(5 n-4)} & =\frac{4}{11} \\
\frac{3 n-2}{5 n-4} & =\frac{20}{33}
\end{aligned}
$$

Solving this last equation yields $n=14$. Using the same identity twice more, for $n=14$ and $k=2$ and $k=4$, the desired quantity is $\frac{2(2 n-1)}{4(4 n-3)}=\frac{\mathbf{2 7}}{\mathbf{1 0 6}}$.

Final answer: $\frac{27}{106}$
Subfield: Algebra


Question: The quadratic polynomial $f(x)$ has a zero at $x=2$. The polynomial $f(f(x))$ has only one real zero, at $x=5$. Compute $f(0)$.

Solution:
Let $f(x)=a(x-b)^{2}+c$. The graph of $f$ is symmetric about $x=b$, so the graph of $y=f(f(x))$ is also symmetric about $x=b$. If $b \neq 5$, then $2 b-5$, the reflection of 5 across $b$, must be a zero of $f(f(x))$. Because $f(f(x))$ has exactly one zero, $b=5$.

Because $f(2)=0$ and $f$ is symmetric about $x=5$, the other zero of $f$ is $x=8$. Because the zeros of $f$ are at 2 and 8 and $f(5)$ is a zero of $f$, either $f(5)=2$ or $f(5)=8$. The following argument shows that $f(5)=8$ is impossible. Because $f$ is continuous, if $f(5)=8$, then $f\left(x_{0}\right)=2$ for some $x_{0}$ in the interval $2<x_{0}<5$. In that case, $f\left(f\left(x_{0}\right)\right)=0$, so 5 would not be a unique zero of $f(f(x))$. Therefore $f(5)=2$ and $c=2$. Setting $f(2)=0$ yields the equation $a(2-5)^{2}+2=0$, so $a=-\frac{2}{9}$, and $f(0)=-\frac{\mathbf{3 2}}{\mathbf{9}}$.

Final answer: $-\frac{32}{9}$
Subfield: Combinatorics


Question: The Local Area Inspirational Math Exam comprises 15 questions. All answers are integers ranging from 000 to 999, inclusive. If the 15 answers form an arithmetic progression with the largest possible difference, compute the largest possible sum of those 15 answers.

Solution:
Let $a$ represent the middle $\left(8^{\text {th }}\right)$ term of the sequence, and let $d$ be the difference. Then the terms of the sequence are $a-7 d, a-6 d, \ldots, a+6 d, a+7 d$, their sum is $15 a$, and the difference between the largest and the smallest terms is $14 d$. The largest $d$ such that $14 d \leq 999$ is $d=71$. Thus the largest possible value for $a$ is $999-7 \cdot 71=502$. The maximal sum of the sequence is therefore $15 a=\mathbf{7 5 3 0}$.

Final answer: 7530
Subfield: Geometry


Question: Circle $\omega_{1}$ has center $O$, which is on circle $\omega_{2}$. The circles intersect at points $A$ and $C$. Point $B$ lies on $\omega_{2}$ such that $B A=37, B O=17$, and $B C=7$. Compute the area of $\omega_{1}$.

Solution:
The points $O, A, B, C$ all lie on $\omega_{2}$ in some order. There are two possible cases to consider: either $B$ is outside circle $\omega_{1}$, or it is inside the circle, as shown below.



<img_3962>

The following argument shows that the first case is impossible. By the Triangle Inequality on $\triangle A B O$, the radius $r_{1}$ of circle $\omega_{1}$ must be at least 20 . But because $B$ is outside $\omega_{1}, B O>r_{1}$, which is impossible, because $B O=17$. So $B$ must be inside the circle.

Construct point $D$ on minor arc $A O$ of circle $\omega_{2}$, so that $A D=O B$ (and therefore $\left.D O=B C\right)$.

<img_3873>

Because $A, D, O, B$ all lie on $\omega_{2}$, Ptolemy's Theorem applies to quadrilateral $A D O B$.



<img_3934>

Therefore $A D \cdot O B+O D \cdot A B=A O \cdot D B=r_{1}^{2}$. Substituting $A D=O B=17, D O=B C=7$, and $A B=37$ yields $r_{1}^{2}=37 \cdot 7+17^{2}=548$. Thus the area of $\omega_{1}$ is $\mathbf{5 4 8 \pi}$.

Final answer: $548 \pi$
Subfield: Algebra


Question: Compute the number of integers $n$ for which $2^{4}<8^{n}<16^{32}$.

Solution:
$8^{n}=2^{3 n}$ and $16^{32}=2^{128}$. Therefore $4<3 n<128$, and $2 \leq n \leq 42$. Thus there are 41 such integers $n$.

Final answer: 41
Subfield: Number Theory


Question: Let $T=41$. Compute the number of positive integers $b$ such that the number $T$ has exactly two digits when written in base $b$.

Solution:
If $T$ has more than one digit when written in base $b$, then $b \leq T$. If $T$ has fewer than three digits when written in base $b$, then $b^{2}>T$, or $b>\sqrt{T}$. So the desired set of bases $b$ is $\{b \mid \sqrt{T}<b \leq T\}$. When $T=41,\lfloor\sqrt{T}\rfloor=6$ and so $6<b \leq 41$. There are $41-6=\mathbf{3 5}$ such integers.

Final answer: 35
Subfield: Geometry


Question: Let $T=35$. Triangle $A B C$ has a right angle at $C$, and $A B=40$. If $A C-B C=T-1$, compute $[A B C]$, the area of $\triangle A B C$.

Solution:
Let $A C=b$ and $B C=a$. Then $a^{2}+b^{2}=1600$ and $|a-b|=T-1$. Squaring the second equation yields $a^{2}+b^{2}-2 a b=(T-1)^{2}$, so $1600-2 a b=(T-1)^{2}$. Hence the area of the triangle is $\frac{1}{2} a b=\frac{1600-(T-1)^{2}}{4}=400-\frac{(T-1)^{2}}{4}$ or $400-\left(\frac{T-1}{2}\right)^{2}$, which for $T=35$ yields $400-289=\mathbf{1 1 1}$.

Final answer: 111
Subfield: Algebra


Question: Let $x$ be a positive real number such that $\log _{\sqrt{2}} x=20$. Compute $\log _{2} \sqrt{x}$.

Solution:
The identity $\log _{b^{n}} x=\frac{1}{n} \log _{b} x$ yields $\log _{2} x=10$. Then $\log _{2} \sqrt{x}=\log _{2} x^{1 / 2}=\frac{1}{2} \log _{2} x=5$.
Use the definition of $\log$ to obtain $x=(\sqrt{2})^{20}=\left(2^{1 / 2}\right)^{20}=2^{10}$. Thus $\log _{2} \sqrt{x}=\log _{2} 2^{5}=\mathbf{5}$.
Use the change of base formula to obtain $\frac{\log x}{\log \sqrt{2}}=20$, so $\log x=$ $20 \log \sqrt{2}=20 \log 2^{1 / 2}=10 \log 2$. Thus $x=2^{10}$, and $\log _{2} \sqrt{x}=\log _{2} 2^{5}=5$.

Final answer: 5
Subfield: Combinatorics


Question: Let $T=5$. Hannah flips two fair coins, while Otto flips $T$ fair coins. Let $p$ be the probability that the number of heads showing on Hannah's coins is greater than the number of heads showing on Otto's coins. If $p=q / r$, where $q$ and $r$ are relatively prime positive integers, compute $q+r$.

Solution:
Because Hannah has only two coins, the only ways she can get more heads than Otto are if she gets 1 (and he gets 0 ), or she gets 2 (and he gets either 1 or 0 ).

The probability of Hannah getting exactly one head is $\frac{1}{2}$. The probability of Otto getting no heads is $\frac{1}{2^{T}}$. So the probability of both events occurring is $\frac{1}{2^{T+1}}$.

The probability of Hannah getting exactly two heads is $\frac{1}{4}$. The probability of Otto getting no heads is still $\frac{1}{2^{T}}$, but the probability of getting exactly one head is $\frac{T}{2^{T}}$, because there are $T$ possibilities for which coin is heads. So the probability of Otto getting either 0 heads or 1 head is $\frac{1+T}{2^{T}}$, and combining that with Hannah's result yields an overall probability of $\frac{1+T}{2^{T+2}}$.

Thus the probability that Hannah flips more heads than Otto is $\frac{1}{2^{T+1}}+\frac{1+T}{2^{T+2}}=\frac{3+T}{2^{T+2}}$. For $T=5$, the value is $\frac{8}{128}=\frac{1}{16}$, giving an answer of $1+16=\mathbf{1 7}$.

Final answer: 17
Subfield: Combinatorics


Question: Let $T=17$. In ARMLovia, the unit of currency is the edwah. Janet's wallet contains bills in denominations of 20 and 80 edwahs. If the bills are worth an average of $2 T$ edwahs each, compute the smallest possible value of the bills in Janet's wallet.

Solution:
Let $x$ be the number of twenty-edwah bills and $y$ be the number of eighty-edwah bills. Then

$$
\begin{aligned}
\frac{20 x+80 y}{x+y} & =2 T \\
20 x+80 y & =2 T x+2 T y \\
(80-2 T) y & =(2 T-20) x
\end{aligned}
$$



In the case where $T=17$ (and hence $2 T=34$ ), this equation reduces to $46 y=14 x$, or $23 y=7 x$. Because 23 and 7 are relatively prime, $23 \mid x$ and $7 \mid y$. Therefore the pair that yields the smallest possible value is $(x, y)=(23,7)$. Then there are $23+7=30$ bills worth a total of $23 \cdot 20+7 \cdot 80=460+560=1020$ edwahs, and $1020 / 30=34$, as required. The answer is $\mathbf{1 0 2 0}$.
Consider the equation $\frac{20 x+80 y}{x+y}=2 T$ derived in the first solution. The identity $\frac{20 x+80 y}{x+y}=20+\frac{60 y}{x+y}$ yields the following:

$$
\begin{aligned}
\frac{60 y}{x+y} & =2 T-20 \\
\frac{30 y}{x+y} & =T-10 \\
x+y & =\frac{30 y}{T-10} .
\end{aligned}
$$

For the smallest value of $x+y$, both $x+y$ and $y$ will be relatively prime. Thus the smallest value of $x+y$ is $\frac{30}{\operatorname{gcd}(T-10,30)}$, which occurs when $y=\frac{T-10}{\operatorname{gcd}(T-10,30)}$. Substituting $T=17$, the numbers $T-10=7$ and 30 are relatively prime, so $y=7$ and $x=23$, for a total of $\mathbf{1 0 2 0}$ edwahs.

Final answer: 1020
Subfield: Combinatorics


Question: The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473)

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$.

Solution:
$$
\begin{aligned}
& \mathrm{Pa}(1,1)+\mathrm{Pa}(2,1)+\mathrm{Pa}(3,1)+\mathrm{Pa}(4,1)+\mathrm{Pa}(5,1)=1+2+3+4+5=\mathbf{1 5} \\
& \mathrm{Pa}(2,2)+\mathrm{Pa}(3,2)+\mathrm{Pa}(4,2)+\mathrm{Pa}(5,2)+\mathrm{Pa}(6,2)=1+3+6+10+15=\mathbf{3 5} \\
& \mathrm{Pa}(3,3)+\mathrm{Pa}(4,3)+\mathrm{Pa}(5,3)+\mathrm{Pa}(6,3)+\mathrm{Pa}(7,3)=1+4+10+20+35=\mathbf{7 0} \\
& \mathrm{Pa}(4,4)+\mathrm{Pa}(5,4)+\mathrm{Pa}(6,4)+\mathrm{Pa}(7,4)+\mathrm{Pa}(8,4)=1+5+15+35+70=\mathbf{1 2 6}
\end{aligned}
$$

Final answer: 15, 35, 70, 126
Subfield: Combinatorics


Question: The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<img_3774>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$.

Solution:
$$
\begin{aligned}
& \mathrm{Pa}(1,1)+\mathrm{Pa}(2,1)+\mathrm{Pa}(3,1)+\mathrm{Pa}(4,1)+\mathrm{Pa}(5,1)=1+2+3+4+5=\mathbf{1 5} \\
& \mathrm{Pa}(2,2)+\mathrm{Pa}(3,2)+\mathrm{Pa}(4,2)+\mathrm{Pa}(5,2)+\mathrm{Pa}(6,2)=1+3+6+10+15=\mathbf{3 5} \\
& \mathrm{Pa}(3,3)+\mathrm{Pa}(4,3)+\mathrm{Pa}(5,3)+\mathrm{Pa}(6,3)+\mathrm{Pa}(7,3)=1+4+10+20+35=\mathbf{7 0} \\
& \mathrm{Pa}(4,4)+\mathrm{Pa}(5,4)+\mathrm{Pa}(6,4)+\mathrm{Pa}(7,4)+\mathrm{Pa}(8,4)=1+5+15+35+70=\mathbf{1 2 6}
\end{aligned}
$$

Final answer: 15, 35, 70, 126
Subfield: Combinatorics


Question: The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473)

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$.

Solution:
Notice that $\mathrm{Pa}(n, n)+\operatorname{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)=\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\mathrm{Pa}(n, n)$ and the right side is $\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(n+k+1, n+1)$, then adding $\mathrm{Pa}(n+k+1, n)$ to both sides yields $\mathrm{Pa}(n+k+1, n)+\mathrm{Pa}(n+k+1, n+1)=$ $\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\mathrm{Pa}$.

Final answer: $m=n+k+1$, $j=n+1$
Subfield: Combinatorics


Question: The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<img_3774>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$.

Solution:
Notice that $\mathrm{Pa}(n, n)+\operatorname{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)=\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\mathrm{Pa}(n, n)$ and the right side is $\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(n+k+1, n+1)$, then adding $\mathrm{Pa}(n+k+1, n)$ to both sides yields $\mathrm{Pa}(n+k+1, n)+\mathrm{Pa}(n+k+1, n+1)=$ $\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\mathrm{Pa}$.

Final answer: $m=n+k+1$, $j=n+1$
Subfield: Combinatorics


Question: The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

<img_3774>

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define

$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$

Prove that $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ for all nonnegative integers $n$.

Solution:
By definition of $\mathrm{Pa}, \operatorname{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$ for all nonnegative integers $n$, and this value is odd, so $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ by definition.