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


Question: Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below.
<img_3235>

Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also "adjacent" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$.
Determine the value of $f(7,3)$.

Solution:
Suppose that $p_{1}=1$.

Then $p_{2}=p_{7}=0$, so the string is of the form $10 p_{3} p_{4} p_{5} p_{6} 0$.

Since $k=3$, then 2 of $p_{3}, p_{4}, p_{5}, p_{6}$ equal 1 , but in such a way that no two adjacent digits are both 1 .

The possible strings in this case are 1010100, 1010010 and 1001010.

Suppose that $p_{1}=0$. Then $p_{2}$ can equal 1 or 0 .

If $p_{2}=1$, then $p_{3}=0$ as well. This means that the string is of the form $010 p_{4} p_{5} p_{6} p_{7}$, which is the same as the general string in the first case, but shifted by 1 position around the circle, so there are again 3 possibilities.

If $p_{2}=0$, then the string is of the form $00 p_{3} p_{4} p_{5} p_{6} p_{7}$ and 3 of the digits $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ equal 1 in such a way that no 2 adjacent digits equal 1.

There is only 1 way in which this can happen: 0010101.

Overall, this gives 7 possible configurations, so $f(7,3)=7$.

Final answer: 7
Subfield: Algebra


Question: Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below.
![](https://cdn.mathpix.com/cropped/2023_12_21_1bbe653ae31ae7a70c3cg-1.jpg?height=344&width=674&top_left_y=454&top_left_x=754)

Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also "adjacent" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$.
Determine the value of $f(7,3)$.

Solution:
Suppose that $p_{1}=1$.

Then $p_{2}=p_{7}=0$, so the string is of the form $10 p_{3} p_{4} p_{5} p_{6} 0$.

Since $k=3$, then 2 of $p_{3}, p_{4}, p_{5}, p_{6}$ equal 1 , but in such a way that no two adjacent digits are both 1 .

The possible strings in this case are 1010100, 1010010 and 1001010.

Suppose that $p_{1}=0$. Then $p_{2}$ can equal 1 or 0 .

If $p_{2}=1$, then $p_{3}=0$ as well. This means that the string is of the form $010 p_{4} p_{5} p_{6} p_{7}$, which is the same as the general string in the first case, but shifted by 1 position around the circle, so there are again 3 possibilities.

If $p_{2}=0$, then the string is of the form $00 p_{3} p_{4} p_{5} p_{6} p_{7}$ and 3 of the digits $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ equal 1 in such a way that no 2 adjacent digits equal 1.

There is only 1 way in which this can happen: 0010101.

Overall, this gives 7 possible configurations, so $f(7,3)=7$.

Final answer: 7
Subfield: Number Theory


Question: Determine the two pairs of positive integers $(a, b)$ with $a<b$ that satisfy the equation $\sqrt{a}+\sqrt{b}=\sqrt{50}$.

Solution:
First, we note that $\sqrt{50}=5 \sqrt{2}$.

Next, we note that $\sqrt{2}+4 \sqrt{2}=5 \sqrt{2}$ and $2 \sqrt{2}+3 \sqrt{2}=5 \sqrt{2}$.

From the first of these, we obtain $\sqrt{2}+\sqrt{32}=\sqrt{50}$.

From the second of these, we obtain $\sqrt{8}+\sqrt{18}=\sqrt{50}$.

Thus, $(a, b)=(2,32)$ and $(a, b)=(8,18)$ are solutions to the original equation.

(We are not asked to justify why these are the only two solutions.)

Final answer: (2,32), (8,18)
Subfield: Algebra


Question: Consider the system of equations:

$$
\begin{aligned}
c+d & =2000 \\
\frac{c}{d} & =k
\end{aligned}
$$

Determine the number of integers $k$ with $k \geq 0$ for which there is at least one pair of integers $(c, d)$ that is a solution to the system.

Solution:
From the second equation, we note that $d \neq 0$.

Rearranging this second equation, we obtain $c=k d$.

Substituting into the first equation, we obtain $k d+d=2000$ or $(k+1) d=2000$.

Since $k \geq 0$, note that $k+1 \geq 1$.

This means that if $(c, d)$ is a solution, then $k+1$ is a divisor of 2000 .

Also, if $k+1$ is a divisor of 2000 , then the equation $(k+1) d=2000$ gives us an integer value of $d$ (which is non-zero) from which we can find an integer value of $c$ using the first equation.

Therefore, the values of $k$ that we want to count correspond to the positive divisors of 2000.

Since $2000=10 \cdot 10 \cdot 20=2^{4} \cdot 5^{3}$, then 2000 has $(4+1)(3+1)=20$ positive divisors.

This comes from the fact that if $p$ and $q$ are distinct prime numbers then the positive integer $p^{a} \cdot q^{b}$ has $(a+1)(b+1)$ positive divisors.

We could list these divisors as

$$
1,2,4,5,8,10,16,20,25,40,50,80,100,125,200,250,400,500,1000,2000
$$



if we did not know the earlier formula.

Since 2000 has 20 positive divisors, then there are 20 values of $k$ for which the system of equations has at least one integer solution.

For example, if $k+1=8$, then $k=7$. This gives the system $c+d=2000$ and $\frac{c}{d}=7$ which has solution $(c, d)=(1750,250)$.

Final answer: 20
Subfield: Geometry


Question: In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant.)

<img_3667>

Solution:
Since the lengths of $A D, A B$ and $B C$ form a geometric sequence, we suppose that these lengths are $a$, ar and $a r^{2}$, respectively, for some real numbers $a>0$ and $r>0$.

Since the angles at $A$ and $B$ are both right angles, we assign coordinates to the diagram, putting $B$ at the origin ( 0,0$), C$ on the positive $x$-axis at $\left(a r^{2}, 0\right), A$ on the positive $y$-axis at $(0, a r)$, and $D$ at $(a, a r)$.

<img_3911>

Therefore, the slope of the line segment joining $B(0,0)$ and $D(a, a r)$ is $\frac{a r-0}{a-0}=r$.

Also, the slope of the line segment joining $A(0, a r)$ and $C\left(a r^{2}, 0\right)$ is $\frac{a r-0}{0-a r^{2}}=-\frac{1}{r}$.

Since the product of the slopes of these two line segments is -1 , then the segments are perpendicular, as required.
Subfield: Geometry


Question: In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant.)

![](https://cdn.mathpix.com/cropped/2023_12_21_17bd3319dc6e6567eee6g-1.jpg?height=271&width=401&top_left_y=819&top_left_x=1255)

Solution:
Since the lengths of $A D, A B$ and $B C$ form a geometric sequence, we suppose that these lengths are $a$, ar and $a r^{2}$, respectively, for some real numbers $a>0$ and $r>0$.

Since the angles at $A$ and $B$ are both right angles, we assign coordinates to the diagram, putting $B$ at the origin ( 0,0$), C$ on the positive $x$-axis at $\left(a r^{2}, 0\right), A$ on the positive $y$-axis at $(0, a r)$, and $D$ at $(a, a r)$.

![](https://cdn.mathpix.com/cropped/2023_12_21_edb67d8383f857f7729ag-1.jpg?height=366&width=591&top_left_y=451&top_left_x=867)

Therefore, the slope of the line segment joining $B(0,0)$ and $D(a, a r)$ is $\frac{a r-0}{a-0}=r$.

Also, the slope of the line segment joining $A(0, a r)$ and $C\left(a r^{2}, 0\right)$ is $\frac{a r-0}{0-a r^{2}}=-\frac{1}{r}$.

Since the product of the slopes of these two line segments is -1 , then the segments are perpendicular, as required.

Final answer: 证明题，略
Subfield: Algebra


Question: Determine all real numbers $x$ for which $2 \log _{2}(x-1)=1-\log _{2}(x+2)$.

Solution:
Using logarithm and exponent laws, we obtain the following equivalent equations:

$$
\begin{aligned}
2 \log _{2}(x-1) & =1-\log _{2}(x+2) \\
2 \log _{2}(x-1)+\log _{2}(x+2) & =1 \\
\log _{2}\left((x-1)^{2}\right)+\log _{2}(x+2) & =1 \\
\log _{2}\left((x-1)^{2}(x+2)\right) & =1 \\
(x-1)^{2}(x+2) & =2^{1} \\
\left(x^{2}-2 x+1\right)(x+2) & =2 \\
x^{3}-3 x+2 & =2 \\
x^{3}-3 x & =0 \\
x\left(x^{2}-3\right) & =0
\end{aligned}
$$

and so $x=0$ or $x=\sqrt{3}$ or $x=-\sqrt{3}$.

Note that if $x=0$, then $x-1=-1<0$ and so $\log _{2}(x-1)$ is not defined. Thus, $x \neq 0$. Note that if $x=-\sqrt{3}$, then $x-1=-\sqrt{3}-1<0$ and so $\log _{2}(x-1)$ is not defined. Thus, $x \neq-\sqrt{3}$.

If $x=\sqrt{3}$, we can verify that both logarithms in the original equation are defined and that the original equation is true. We could convince ourselves of this with a calculator or we could algebraically verify that raising 2 to the power of both sides gives the same number, so the expressions must actually be equal.

Therefore, $x=\sqrt{3}$ is the only solution.

Final answer: $\sqrt{3}$
Subfield: Algebra


Question: Consider the function $f(x)=x^{2}-2 x$. Determine all real numbers $x$ that satisfy the equation $f(f(f(x)))=3$.

Solution:
Let $a=f(f(x))$.

Thus, the equation $f(f(f(x)))=3$ is equivalent to $f(a)=3$.

Since $f(a)=a^{2}-2 a$, then we obtain the equation $a^{2}-2 a=3$ which gives $a^{2}-2 a-3=0$ and $(a-3)(a+1)=0$.

Thus, $a=3$ or $a=-1$ which means that $f(f(x))=3$ or $f(f(x))=-1$.

Let $b=f(x)$.

Thus, the equations $f(f(x))=3$ and $f(f(x))=-1$ become $f(b)=3$ and $f(b)=-1$.

If $f(b)=3$, then $b=f(x)=3$ or $b=f(x)=-1$ using similar reasoning to above when $f(a)=3$.

If $f(b)=-1$, then $b^{2}-2 b=-1$ and so $b^{2}-2 b+1=0$ or $(b-1)^{2}=0$ which means that $b=f(x)=1$.

Thus, $f(x)=3$ or $f(x)=-1$ or $f(x)=1$.

If $f(x)=3$, then $x=3$ or $x=-1$ as above.

If $f(x)=-1$, then $x=1$ as above.

If $f(x)=1$, then $x^{2}-2 x=1$ and so $x^{2}-2 x-1=0$.

By the quadratic formula,

$$
x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\frac{2 \pm \sqrt{8}}{2}=1 \pm \sqrt{2}
$$

Therefore, the solutions to the equation $f(f(f(x)))=3$ are $x=3,1,-1,1+\sqrt{2}, 1-\sqrt{2}$.

Final answer: $3,1,-1,1+\sqrt{2}, 1-\sqrt{2}$
Subfield: Geometry


Question: A circle has centre $O$ and radius 1. Quadrilateral $A B C D$ has all 4 sides tangent to the circle at points $P, Q, S$, and $T$, as shown. Also, $\angle A O B=\angle B O C=\angle C O D=\angle D O A$. If $A O=3$, determine the length of $D S$.

<img_4022>

Solution:
Since $\angle A O B=\angle B O C=\angle C O D=\angle D O A$ and these angles form a complete circle around $O$, then $\angle A O B=\angle B O C=\angle C O D=\angle D O A=\frac{1}{4} \cdot 360^{\circ}=90^{\circ}$.

Join point $O$ to $P, B, Q, C, S, D, T$, and $A$.

<img_3604>

Since $P, Q, S$, and $T$ are points of tangency, then the radii meet the sides of $A B C D$ at right angles at these points.

Since $A O=3$ and $O T=1$ and $\angle O T A=90^{\circ}$, then by the Pythagorean Theorem, $A T=\sqrt{A O^{2}-O T^{2}}=\sqrt{8}=2 \sqrt{2}$.

Since $\triangle O T A$ is right-angled at $T$, then $\angle T A O+\angle A O T=90^{\circ}$.

Since $\angle D O A=90^{\circ}$, then $\angle A O T+\angle D O T=90^{\circ}$.

Thus, $\angle T A O=\angle D O T$.

This means that $\triangle A T O$ is similar to $\triangle O T D$.

Thus, $\frac{D T}{O T}=\frac{O T}{A T}$ and so $D T=\frac{O T^{2}}{A T}=\frac{1}{2 \sqrt{2}}$.

Since $D S$ and $D T$ are tangents to the circle from the same point, then $D S=D T=\frac{1}{2 \sqrt{2}}$.

Final answer: $\frac{1}{2 \sqrt{2}}$
Subfield: Geometry


Question: A circle has centre $O$ and radius 1. Quadrilateral $A B C D$ has all 4 sides tangent to the circle at points $P, Q, S$, and $T$, as shown. Also, $\angle A O B=\angle B O C=\angle C O D=\angle D O A$. If $A O=3$, determine the length of $D S$.

![](https://cdn.mathpix.com/cropped/2023_12_21_17bd3319dc6e6567eee6g-1.jpg?height=279&width=412&top_left_y=1660&top_left_x=1298)

Solution:
Since $\angle A O B=\angle B O C=\angle C O D=\angle D O A$ and these angles form a complete circle around $O$, then $\angle A O B=\angle B O C=\angle C O D=\angle D O A=\frac{1}{4} \cdot 360^{\circ}=90^{\circ}$.

Join point $O$ to $P, B, Q, C, S, D, T$, and $A$.

![](https://cdn.mathpix.com/cropped/2023_12_21_e945ab0e1a14a95ca3e9g-1.jpg?height=274&width=464&top_left_y=1473&top_left_x=928)

Since $P, Q, S$, and $T$ are points of tangency, then the radii meet the sides of $A B C D$ at right angles at these points.

Since $A O=3$ and $O T=1$ and $\angle O T A=90^{\circ}$, then by the Pythagorean Theorem, $A T=\sqrt{A O^{2}-O T^{2}}=\sqrt{8}=2 \sqrt{2}$.

Since $\triangle O T A$ is right-angled at $T$, then $\angle T A O+\angle A O T=90^{\circ}$.

Since $\angle D O A=90^{\circ}$, then $\angle A O T+\angle D O T=90^{\circ}$.

Thus, $\angle T A O=\angle D O T$.

This means that $\triangle A T O$ is similar to $\triangle O T D$.

Thus, $\frac{D T}{O T}=\frac{O T}{A T}$ and so $D T=\frac{O T^{2}}{A T}=\frac{1}{2 \sqrt{2}}$.

Since $D S$ and $D T$ are tangents to the circle from the same point, then $D S=D T=\frac{1}{2 \sqrt{2}}$.

Final answer: $\frac{1}{2 \sqrt{2}}$
Subfield: Algebra


Question: Suppose that $x$ satisfies $0<x<\frac{\pi}{2}$ and $\cos \left(\frac{3}{2} \cos x\right)=\sin \left(\frac{3}{2} \sin x\right)$.

Determine all possible values of $\sin 2 x$, expressing your answers in the form $\frac{a \pi^{2}+b \pi+c}{d}$ where $a, b, c, d$ are integers.

Solution:
Since $0<x<\frac{\pi}{2}$, then $0<\cos x<1$ and $0<\sin x<1$.

This means that $0<\frac{3}{2} \cos x<\frac{3}{2}$ and $0<\frac{3}{2} \sin x<\frac{3}{2}$. Since $3<\pi$, then $0<\frac{3}{2} \cos x<\frac{\pi}{2}$ and $0<\frac{3}{2} \sin x<\frac{\pi}{2}$.

If $Y$ and $Z$ are angles with $0<Y<\frac{\pi}{2}$ and $0<Z<\frac{\pi}{2}$, then $\cos Y=\sin Z$ exactly when $Y+Z=\frac{\pi}{2}$. To see this, we could picture points $R$ and $S$ on the unit circle corresponding to the angles $Y$ and $Z$; the $x$-coordinate of $R$ is equal to the $y$-coordinate of $S$ exactly when the angles $Y$ and $Z$ are complementary.

Therefore, the following equations are equivalent:

$$
\begin{aligned}
\cos \left(\frac{3}{2} \cos x\right) & =\sin \left(\frac{3}{2} \sin x\right) \\
\frac{3}{2} \cos x+\frac{3}{2} \sin x & =\frac{\pi}{2} \\
\cos x+\sin x & =\frac{\pi}{3} \\
(\sin x+\cos x)^{2} & =\frac{\pi^{2}}{9} \\
\sin ^{2} x+2 \sin x \cos x+\sin ^{2} x & =\frac{\pi^{2}}{9} \\
2 \sin x \cos x+\left(\sin ^{2} x+\cos ^{2} x\right) & =\frac{\pi^{2}}{9} \\
\sin 2 x+1 & =\frac{\pi^{2}}{9} \\
\sin 2 x & =\frac{\pi^{2}-9}{9}
\end{aligned}
$$

Therefore, the only possible value of $\sin 2 x$ is $\frac{\pi^{2}-9}{9}$.

Final answer: $\frac{\pi^{2}-9}{9}$
Subfield: Number Theory


Question: For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.

For example, the value of $f(1,2)$ is 3 .
Determine the value of $f(2,5)$.

Solution:
By definition, $f(2,5)=\frac{2}{5}+\frac{5}{2}+\frac{1}{2 \cdot 5}=\frac{2 \cdot 2+5 \cdot 5+1}{2 \cdot 5}=\frac{4+25+1}{10}=\frac{30}{10}=3$.

Final answer: 3
Subfield: Number Theory


Question: For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.

For example, the value of $f(1,2)$ is 3 .
Determine all positive integers $a$ for which $f(a, a)$ is an integer.

Solution:
By definition, $f(a, a)=\frac{a}{a}+\frac{a}{a}+\frac{1}{a^{2}}=2+\frac{1}{a^{2}}$.

For $2+\frac{1}{a^{2}}$ to be an integer, it must be the case that $\frac{1}{a^{2}}$ is an integer.

For $\frac{1}{a^{2}}$ to be an integer and since $a^{2}$ is an integer, $a^{2}$ needs to be a divisor of 1 .

Since $a^{2}$ is positive, then $a^{2}=1$.

Since $a$ is a positive integer, then $a=1$.

Thus, the only positive integer $a$ for which $f(a, a)$ is an integer is $a=1$.

Final answer: 1
Subfield: Number Theory


Question: For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.

For example, the value of $f(1,2)$ is 3 .
If $a$ and $b$ are positive integers and $f(a, b)$ is an integer, prove that $f(a, b)$ must be a multiple of 3 .

Solution:
Suppose that $a$ and $b$ are positive integers for which $f(a, b)$ is an integer.

Assume that $k=f(a, b)$ is not a multiple of 3 .

We will show that there must be a contradiction, which will lead to the conclusion that $k$ must be a multiple of 3 .

By definition, $k=f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.

Multiplying by $a b$, we obtain $k a b=a^{2}+b^{2}+1$, which we re-write as $a^{2}-(k b) a+\left(b^{2}+1\right)=0$. We treat this as a quadratic equation in $a$ with coefficients in terms of the variables $b$ and $k$. Solving for $a$ in terms of $b$ and $k$ using the quadratic formula, we obtain

$$
a=\frac{k b \pm \sqrt{(-k b)^{2}-4(1)\left(b^{2}+1\right)}}{2}=\frac{k b \pm \sqrt{k^{2} b^{2}-4 b^{2}-4}}{2}
$$

Since $a$ is an integer, then the discriminant $k^{2} b^{2}-4 b^{2}-4$ must be a perfect square. Re-writing the discriminant, we obtain

$$
k^{2} b^{2}-4 b^{2}-4=b^{2}\left(k^{2}-4\right)-4=b^{2}(k-2)(k+2)-4
$$

Since $k$ is not a multiple of 3 , then it is either 1 more than a multiple of 3 or it is 2 more than a multiple of 3.

If $k$ is 1 more than a multiple of 3 , then $k+2$ is a multiple of 3 .

If $k$ is 2 more than a multiple of 3 , then $k-2$ is a multiple of 3 .

In either case, $(k-2)(k+2)$ is a multiple of 3 , say $(k-2)(k+2)=3 m$ for some integer $m$. This means that the discriminant can be re-written again as

$$
b^{2}(3 m)-4=3\left(b^{2} m-2\right)+2
$$

In other words, the discriminant is itself 2 more than a multiple of 3.

However, every perfect square is either a multiple of 3 or one more than a multiple of 3 :

Suppose that $r$ is an integer and consider $r^{2}$.

The integer $r$ can be written as one of $3 q, 3 q+1,3 q+2$, for some integer $q$.

These three cases give

$$
\begin{aligned}
(3 q)^{2} & =9 q^{2}=3\left(3 q^{2}\right) \\
(3 q+1)^{2} & =9 q^{2}+6 q+1=3\left(3 q^{2}+2 q\right)+1 \\
(3 q+2)^{2} & =9 q^{2}+12 q+4=3\left(3 q^{2}+4 q+1\right)+1
\end{aligned}
$$

and so $r^{2}$ is either a multiple of 3 or 1 more than a multiple of 3.



We have determined that the discriminant is a perfect square and is 2 more than a multiple of 3. This is a contradiction.

This means that our initial assumption must be incorrect, and so $k=f(a, b)$ must be a multiple of 3.
Subfield: Combinatorics


Question: Amir and Brigitte play a card game. Amir starts with a hand of 6 cards: 2 red, 2 yellow and 2 green. Brigitte starts with a hand of 4 cards: 2 purple and 2 white. Amir plays first. Amir and Brigitte alternate turns. On each turn, the current player chooses one of their own cards at random and places it on the table. The cards remain on the table for the rest of the game. A player wins and the game ends when they have placed two cards of the same colour on the table. Determine the probability that Amir wins the game.

Solution:
On her first two turns, Brigitte either chooses two cards of the same colour or two cards of different colours. If she chooses two cards of different colours, then on her third turn, she must choose a card that matches one of the cards that she already has.

Therefore, the game ends on or before Brigitte's third turn.

Thus, if Amir wins, he wins on his second turn or on his third turn. (He cannot win on his first turn.)

For Amir to win on his second turn, the second card he chooses must match the first card that he chooses.

On this second turn, there will be 5 cards in his hand, of which 1 matches the colour of the first card that he chose.

Therefore, the probability that Amir wins on his second turn is $\frac{1}{5}$.

Note that there is no restriction on the first card that he chooses or the first card that Brigitte chooses.

For Amir to win on his third turn, the following conditions must be true: (i) the colour of the second card that he chooses is different from the colour of the first card that he chooses, (ii) the colour of the second card that Brigitte chooses is different from the colour of the first card that she chooses, and (iii) the colour of the third card that Amir chooses matches the colour of one of the first two cards.

The probability of (i) is $\frac{4}{5}$, since he must choose any card other than the one that matches the first one.

The probability of (ii) is $\frac{2}{3}$, since Brigitte must choose either of the cards that does not match her first card.

The probability of (iii) is $\frac{2}{4}$, since Amir can choose either of the 2 cards that matches one of the first two cards that he chose.

Again, the cards that Amir and Brigitte choose on their first turns do not matter.

Thus, the probability that Amir wins on his third turn is $\frac{4}{5} \cdot \frac{2}{3} \cdot \frac{2}{4}$ which equals $\frac{4}{15}$.

Finally, the probabilty that Amir wins the game is thus $\frac{1}{5}+\frac{4}{15}$ which equals $\frac{7}{15}$.

Final answer: $\frac{7}{15}$
Subfield: Combinatorics


Question: Carlos has 14 coins, numbered 1 to 14. Each coin has exactly one face called "heads". When flipped, coins 1,2,3,..,13,14 land heads with probabilities $h_{1}, h_{2}, h_{3}, \ldots, h_{13}, h_{14}$, respectively. When Carlos flips each of the 14 coins exactly once, the probability that an even number of coins land heads is exactly $\frac{1}{2}$. Must there be a $k$ between 1 and 14, inclusive, for which $h_{k}=\frac{1}{2}$ ? Prove your answer.

Solution:
Suppose that, after flipping the first 13 coins, the probability that there is an even number of heads is $p$.

Then the probability that there is an odd number of heads is $1-p$.

When the 14th coin is flipped, the probability of heads is $h_{14}$ and the probability of not heads is $1-h_{14}$.

After the 14th coin is flipped, there can be an even number of heads if the first 13 include an even number of heads and the 14th is not heads, or if the first 13 include an odd number of heads and the 14th is heads.

The probability of this is $p\left(1-h_{14}\right)+(1-p) h_{14}$.

Therefore,

$$
\begin{aligned}
p\left(1-h_{14}\right)+(1-p) h_{14} & =\frac{1}{2} \\
2 p-2 p h_{14}+2 h_{14}-2 p h_{14} & =1 \\
0 & =4 p h_{14}-2 p-2 h_{14}+1 \\
0 & =2 p\left(2 h_{14}-1\right)-(2 h 14-1) \\
0 & =(2 p-1)\left(2 h_{14}-1\right)
\end{aligned}
$$

Therefore, either $h_{14}=\frac{1}{2}$ or $p=\frac{1}{2}$.

If $h_{14}=\frac{1}{2}$, we have proven the result.

If $h_{14} \neq \frac{1}{2}$, then $p=\frac{1}{2}$.

This would mean that the probability of getting an even number of heads when the first 13 coins are flipped is $\frac{1}{2}$.

We could repeat the argument above to conclude that either $h_{13}=\frac{1}{2}$ or the probability of obtaining an even number of heads when the first 12 coins are flipped is $\frac{1}{2}$.

Continuing in this way, either one of $h_{14}, h_{13}, \ldots, h_{3}, h_{2}$ will equal $\frac{1}{2}$, or the probability of obtaining an even number of heads when 1 coin is flipped is $\frac{1}{2}$.

This last statement is equivalent to saying that the probability of obtaining a head with the first coin is $\frac{1}{2}$ (that is, $h_{1}=\frac{1}{2}$ ).

Therefore, at least one of $h_{1}, h_{2}, \ldots, h_{13}, h_{14}$ must equal $\frac{1}{2}$.