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


Question: The function $f(x)$ satisfies the equation $f(x)=f(x-1)+f(x+1)$ for all values of $x$. If $f(1)=1$ and $f(2)=3$, what is the value of $f(2008)$ ?

Solution:
Since $f(x)=f(x-1)+f(x+1)$, then $f(x+1)=f(x)-f(x-1)$, and so

$$
\begin{aligned}
& f(1)=1 \\
& f(2)=3 \\
& f(3)=f(2)-f(1)=3-1=2 \\
& f(4)=f(3)-f(2)=2-3=-1 \\
& f(5)=f(4)-f(3)=-1-2=-3 \\
& f(6)=f(5)-f(4)=-3-(-1)=-2 \\
& f(7)=f(6)-f(5)=-2-(-3)=1=f(1) \\
& f(8)=f(7)-f(6)=1-(-2)=3=f(2)
\end{aligned}
$$

Since the value of $f$ at an integer depends only on the values of $f$ at the two previous integers, then the fact that the first several values form a cycle with $f(7)=f(1)$ and $f(8)=f(2)$ tells us that the values of $f$ will always repeat in sets of 6 .

Since 2008 is 4 more than a multiple of 6 (as $2008=4+2004=4+6(334)$ ), then $f(2008)=f(2008-6(334))=f(4)=-1$.

Final answer: -1
Subfield: Algebra


Question: The numbers $a, b, c$, in that order, form a three term arithmetic sequence (see below) and $a+b+c=60$.

The numbers $a-2, b, c+3$, in that order, form a three term geometric sequence. Determine all possible values of $a, b$ and $c$.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms.

A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)

Present your answer in the form of coordinates (e.g. (1, 2, 3) for a=1, b=2, c=3).

Solution:
Since $a, b, c$ form an arithmetic sequence, then we can write $a=b-d$ and $c=b+d$ for some real number $d$.

Since $a+b+c=60$, then $(b-d)+b+(b+d)=60$ or $3 b=60$ or $b=20$.

Therefore, we can write $a, b, c$ as $20-d, 20,20+d$.

(We could have written $a, b, c$ instead as $a, a+d, a+2 d$ and arrived at the same result.) Thus, $a-2=20-d-2=18-d$ and $c+3=20+d+3=23+d$, so we can write $a-2, b, c+3$ as $18-d, 20,23+d$.



Since these three numbers form a geometric sequence, then

$$
\begin{aligned}
\frac{20}{18-d} & =\frac{23+d}{20} \\
20^{2} & =(23+d)(18-d) \\
400 & =-d^{2}-5 d+414 \\
d^{2}+5 d-14 & =0 \\
(d+7)(d-2) & =0
\end{aligned}
$$

Therefore, $d=-7$ or $d=2$.

If $d=-7$, then $a=27, b=20$ and $c=13$.

If $d=2$, then $a=18, b=20$ and $c=22$.

(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)
Since $a, b, c$ form an arithmetic sequence, then $c-b=b-a$ or $a+c=2 b$.

Since $a+b+c=60$, then $2 b+b=60$ or $3 b=60$ or $b=20$.

Thus, $a+c=40$, so $a=40-c$.

Therefore, we can write $a, b, c$ as $40-c, 20, c$.

Also, $a-2=40-c-2=38-c$, so we can write $a-2, b, c+3$ as $38-c, 20, c+3$.

Since these three numbers form a geometric sequence, then

$$
\begin{aligned}
\frac{20}{38-c} & =\frac{c+3}{20} \\
20^{2} & =(38-c)(c+3) \\
400 & =-c^{2}+35 c+114 \\
c^{2}-35 d+286 & =0 \\
(c-13)(c-22) & =0
\end{aligned}
$$

Therefore, $c=13$ or $c=22$.

If $c=13$, then $a=27$, so $a=27, b=20$ and $c=13$.

If $c=22$, then $a=18$, so $a=18, b=20$ and $c=22$.

(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)

Final answer: $(27,20,13), (18,20,22)$
Subfield: Algebra


Question: The average of three consecutive multiples of 3 is $a$.

The average of four consecutive multiples of 4 is $a+27$.

The average of the smallest and largest of these seven integers is 42 .

Determine the value of $a$.

Solution:
Since the average of three consecutive multiples of 3 is $a$, then $a$ is the middle of these three integers, so the integers are $a-3, a, a+3$.

Since the average of four consecutive multiples of 4 is $a+27$, then $a+27$ is halfway in between the second and third of these multiples (which differ by 4), so the second and third of the multiples are $(a+27)-2=a+25$ and $(a+27)+2=a+29$, so the four integers are $a+21, a+25, a+29, a+33$.

(We have used in these two statements the fact that if a list contains an odd number of integers, then there is a middle integer in the list, and if the list contains an even number



of integers, then the "middle" integer is between two integers from the list.)

The smallest of these seven integers is $a-3$ and the largest is $a+33$.

The average of these two integers is $\frac{1}{2}(a-3+a+33)=\frac{1}{2}(2 a+30)=a+15$.

Since $a+15=42$, then $a=27$.

Final answer: 27
Subfield: Combinatorics


Question: Billy and Crystal each have a bag of 9 balls. The balls in each bag are numbered from 1 to 9. Billy and Crystal each remove one ball from their own bag. Let $b$ be the sum of the numbers on the balls remaining in Billy's bag. Let $c$ be the sum of the numbers on the balls remaining in Crystal's bag. Determine the probability that $b$ and $c$ differ by a multiple of 4 .

Solution:
Suppose that Billy removes the ball numbered $x$ from his bag and that Crystal removes the ball numbered $y$ from her bag.

Then $b=1+2+3+4+5+6+7+8+9-x=45-x$.

Also, $c=1+2+3+4+5+6+7+8+9-y=45-y$.

Hence, $b-c=(45-x)-(45-y)=y-x$.

Since $1 \leq x \leq 9$ and $1 \leq y \leq 9$, then $-8 \leq y-x \leq 8$.

(This is because $y-x$ is maximized when $y$ is largest (that is, $y=9$ ) and $x$ is smallest (that is, $x=1$ ), so $y-x \leq 9-1=8$. Similarly, $y-x \geq-8$.)

Since $b-c=y-x$ is between -8 and 8 , then for it to be a multiple of $4, b-c=y-x$ can be $-8,-4,0,4$, or 8 .

Since each of Billy and Crystal chooses 1 ball from 9 balls and each ball is equally likely to be chosen, then the probability of any specific ball being chosen from one of their bags is $\frac{1}{9}$. Thus, the probability of any specific pair of balls being chosen (one from each bag) is $\frac{1}{9} \times \frac{1}{9}=\frac{1}{81}$.

Therefore, to compute the desired probability, we must count the number of pairs $(x, y)$ where $y-x$ is $-8,-4,0,4,8$, and multiply this result by $\frac{1}{81}$.

Method 1 

If $y-x=-8$, then $(x, y)$ must be $(9,1)$.

If $y-x=8$, then $(x, y)$ must be $(1,9)$.

If $y-x=-4$, then $(x, y)$ can be $(5,1),(6,2),(7,3),(8,4),(9,5)$.

If $y-x=4$, then $(x, y)$ can be $(1,5),(2,6),(3,7),(4,8),(5,9)$.

If $y-x=0$, then $(x, y)$ can be $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)$.

There are thus 21 pairs $(x, y)$ that work, so the desired probability is $\frac{21}{81}=\frac{7}{27}$.

Method 2

If $x=9$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .

If $x=8$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .

If $x=7$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .

If $x=6$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .

If $x=5$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .

If $x=4$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .

If $x=3$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .

If $x=2$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .

If $x=1$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .



There are thus 21 pairs $(x, y)$ that work, so the desired probability is $\frac{21}{81}=\frac{7}{27}$.

Final answer: $\frac{7}{27}$
Subfield: Geometry


Question: In the diagram, $A B C$ is a right-angled triangle with $P$ and $R$ on $A B$. Also, $Q$ is on $A C$, and $P Q$ is parallel to $B C$. If $R P=2$, $B R=3, B C=4$, and the area of $\triangle Q R C$ is 5 , determine the length of $A P$.

![](https://cdn.mathpix.com/cropped/2023_12_21_16c4f36a80a838603da9g-1.jpg?height=334&width=499&top_left_y=1931&top_left_x=1255)

Solution:
Let $A P=x$ and $Q P=h$.

Since $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.

Consider trapezoid $C B P Q$. We can think of this as having parallel bases of lengths 4 and $h$ and height 5 . Thus, its area is $\frac{1}{2}(4+h)(5)$.

However, we can also compute its area by adding the areas of $\triangle C B R$ (which is $\left.\frac{1}{2}(4)(3)\right)$, $\triangle C R Q$ (which is given as 5), and $\triangle R P Q$ (which is $\left.\frac{1}{2}(2)(h)\right)$.

Thus,

$$
\begin{aligned}
\frac{1}{2}(4+h)(5) & =\frac{1}{2}(4)(3)+5+\frac{1}{2}(2)(h) \\
20+5 h & =12+10+2 h \\
3 h & =2 \\
h & =\frac{2}{3}
\end{aligned}
$$

Now, $\triangle A P Q$ is similar to $\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,

$$
\begin{aligned}
\frac{A P}{P Q} & =\frac{A B}{B C} \\
(A P)(B C) & =(P Q)(A B) \\
4 x & =\frac{2}{3}(x+5) \\
4 x & =\frac{2}{3} x+\frac{10}{3} \\
\frac{10}{3} x & =\frac{10}{3} \\
x & =1
\end{aligned}
$$



Therefore, $A P=x=1$.
Let $A P=x$ and $Q P=h$.

Since $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.

Since $\triangle A B C$ is right-angled at $B$, its area is $\frac{1}{2}(4)(5+x)=10+2 x$.

However, we can look at the area of the $\triangle A B C$ in terms of its four triangular pieces:

$\triangle C B R$ (which has area $\left.\frac{1}{2}(4)(3)\right), \triangle C R Q$ (which has area 5), $\triangle Q P R$ (which has area $\left.\frac{1}{2} h(2)\right)$, and $\triangle Q P A$ (which has area $\frac{1}{2} x h$ ).

Therefore, $10+2 x=6+5+h+\frac{1}{2} x h$ so $x h-4 x+2 h+2=0$.

Now, $\triangle A P Q$ is similar to $\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,

$$
\begin{aligned}
\frac{A P}{P Q} & =\frac{A B}{B C} \\
(A P)(B C) & =(P Q)(A B) \\
x(4) & =h(x+5) \\
4 x & =h x+5 h \\
-5 h & =h x-4 x
\end{aligned}
$$

Substituting this into the equation above, $x h+2 h-4 x+2=0$ becomes $-5 h+2 h+2=0$ or $3 h=2$ or $h=\frac{2}{3}$.

Lastly, we solve for $x$ by subsituting our value for $h$ : $-5\left(\frac{2}{3}\right)=\frac{2}{3} x-4 x$ or $-\frac{10}{3}=-\frac{10}{3} x$ and so $x=1$.

Therefore, $A P=x=1$.

Final answer: 1
Subfield: Geometry


Question: In the diagram, $A B C$ is a right-angled triangle with $P$ and $R$ on $A B$. Also, $Q$ is on $A C$, and $P Q$ is parallel to $B C$. If $R P=2$, $B R=3, B C=4$, and the area of $\triangle Q R C$ is 5 , determine the length of $A P$.

<img_3718>

Solution:
Let $A P=x$ and $Q P=h$.

Since $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.

Consider trapezoid $C B P Q$. We can think of this as having parallel bases of lengths 4 and $h$ and height 5 . Thus, its area is $\frac{1}{2}(4+h)(5)$.

However, we can also compute its area by adding the areas of $\triangle C B R$ (which is $\left.\frac{1}{2}(4)(3)\right)$, $\triangle C R Q$ (which is given as 5), and $\triangle R P Q$ (which is $\left.\frac{1}{2}(2)(h)\right)$.

Thus,

$$
\begin{aligned}
\frac{1}{2}(4+h)(5) & =\frac{1}{2}(4)(3)+5+\frac{1}{2}(2)(h) \\
20+5 h & =12+10+2 h \\
3 h & =2 \\
h & =\frac{2}{3}
\end{aligned}
$$

Now, $\triangle A P Q$ is similar to $\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,

$$
\begin{aligned}
\frac{A P}{P Q} & =\frac{A B}{B C} \\
(A P)(B C) & =(P Q)(A B) \\
4 x & =\frac{2}{3}(x+5) \\
4 x & =\frac{2}{3} x+\frac{10}{3} \\
\frac{10}{3} x & =\frac{10}{3} \\
x & =1
\end{aligned}
$$



Therefore, $A P=x=1$.
Let $A P=x$ and $Q P=h$.

Since $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.

Since $\triangle A B C$ is right-angled at $B$, its area is $\frac{1}{2}(4)(5+x)=10+2 x$.

However, we can look at the area of the $\triangle A B C$ in terms of its four triangular pieces:

$\triangle C B R$ (which has area $\left.\frac{1}{2}(4)(3)\right), \triangle C R Q$ (which has area 5), $\triangle Q P R$ (which has area $\left.\frac{1}{2} h(2)\right)$, and $\triangle Q P A$ (which has area $\frac{1}{2} x h$ ).

Therefore, $10+2 x=6+5+h+\frac{1}{2} x h$ so $x h-4 x+2 h+2=0$.

Now, $\triangle A P Q$ is similar to $\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,

$$
\begin{aligned}
\frac{A P}{P Q} & =\frac{A B}{B C} \\
(A P)(B C) & =(P Q)(A B) \\
x(4) & =h(x+5) \\
4 x & =h x+5 h \\
-5 h & =h x-4 x
\end{aligned}
$$

Substituting this into the equation above, $x h+2 h-4 x+2=0$ becomes $-5 h+2 h+2=0$ or $3 h=2$ or $h=\frac{2}{3}$.

Lastly, we solve for $x$ by subsituting our value for $h$ : $-5\left(\frac{2}{3}\right)=\frac{2}{3} x-4 x$ or $-\frac{10}{3}=-\frac{10}{3} x$ and so $x=1$.

Therefore, $A P=x=1$.

Final answer: 1
Subfield: Algebra


Question: The equation $2^{x+2} 5^{6-x}=10^{x^{2}}$ has two real solutions. Determine these two solutions.

Solution:
Rewriting the equation, we obtain

$$
\begin{aligned}
2^{x+2} 5^{6-x} & =2^{x^{2}} 5^{x^{2}} \\
1 & =2^{x^{2}} 2^{-2-x} 5^{x^{2}} 5^{x-6} \\
1 & =2^{x^{2}-x-2} 5^{x^{2}+x-6} \\
0 & =\left(x^{2}-x-2\right) \log _{10} 2+\left(x^{2}+x-6\right) \log _{10} 5 \\
0 & =(x-2)(x+1) \log _{10} 2+(x-2)(x+3) \log _{10} 5 \\
0 & =(x-2)\left[(x+1) \log _{10} 2+(x+3) \log _{10} 5\right] \\
0 & =(x-2)\left[\left(\log _{10} 2+\log _{10} 5\right) x+\left(\log _{10} 2+3 \log 105\right)\right] \\
0 & =(x-2)\left[\left(\log _{10} 10\right) x+\log _{10}\left(2 \cdot 5^{3}\right)\right] \\
0 & =(x-2)\left(x+\log _{10} 250\right)
\end{aligned}
$$

Therefore, $x=2$ or $x=-\log _{10} 250$.
We take base 10 logarithms of both sides:

$$
\begin{aligned}
\log _{10}\left(2^{x+2} 5^{6-x}\right) & =\log _{10}\left(10^{x^{2}}\right) \\
\log _{10}\left(2^{x+2}\right)+\log _{10}\left(5^{6-x}\right) & =x^{2} \\
(x+2) \log _{10} 2+(6-x) \log _{10} 5 & =x^{2} \\
x\left(\log _{10} 2-\log _{10} 5\right)+\left(2 \log _{10} 2+6 \log _{10} 5\right) & =x^{2} \\
x^{2}-x\left(\log _{10} 2-\log _{10} 5\right)-\left(2 \log _{10} 2+6 \log _{10} 5\right) & =0
\end{aligned}
$$

Now, $\log _{10} 2+\log _{10} 5=\log _{10} 10=1$ so $\log _{10} 5=1-\log _{10} 2$, so we can simplify the equation to

$$
x^{2}-x\left(2 \log _{10} 2-1\right)-\left(6-4 \log _{10} 2\right)=0
$$

This is a quadratic equation in $x$, so should have at most 2 real solutions.

By the quadratic formula,

$$
\begin{aligned}
x & =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{\left(2 \log _{10} 2-1\right)^{2}-4(1)\left(-\left(6-4 \log _{10} 2\right)\right)}}{2(1)} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{4\left(\log _{10} 2\right)^{2}-4\left(\log _{10} 2\right)+1+24-16 \log _{10} 2}}{2} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{4\left(\log _{10} 2\right)^{2}-20\left(\log _{10} 2\right)+25}}{2} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{\left(2 \log _{10} 2-5\right)^{2}}}{2} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm\left(5-2 \log _{10} 2\right)}{2}
\end{aligned}
$$

since $5-2 \log _{10} 2>0$.

Therefore,

$$
x=\frac{\left(2 \log _{10} 2-1\right)+\left(5-2 \log _{10} 2\right)}{2}=\frac{4}{2}=2
$$

or

$$
x=\frac{\left(2 \log _{10} 2-1\right)-\left(5-2 \log _{10} 2\right)}{2}=\frac{4 \log _{10} 2-6}{2}=2 \log _{10} 2-3
$$

(Note that at any point, we could have used a calculator to convert to decimal approximations and solve.)

Final answer: $2,-\log _{10} 250$
Subfield: Algebra


Question: Determine all real solutions to the system of equations

$$
\begin{aligned}
& x+\log _{10} x=y-1 \\
& y+\log _{10}(y-1)=z-1 \\
& z+\log _{10}(z-2)=x+2
\end{aligned}
$$

and prove that there are no more solutions.

Solution:
First, we rewrite the system as

$$
\begin{aligned}
& x+\log _{10} x=y-1 \\
& (y-1)+\log _{10}(y-1)=z-2 \\
& (z-2)+\log _{10}(z-2)=x
\end{aligned}
$$

Second, we make the substitution $a=x, b=y-1$ and $c=z-2$, allowing us to rewrite



the system as

$$
\begin{aligned}
a+\log _{10} a & =b \\
b+\log _{10} b & =c \\
c+\log _{10} c & =a
\end{aligned}
$$

Third, we observe that $(a, b, c)=(1,1,1)$ is a solution, since $1+\log _{10} 1=1+0=1$.

Next, if $a>1$, then $\log _{10} a>0$, so from (1),

$$
b=a+\log _{10} a>a+0=a>1
$$

so $\log _{10} b>0$, so from $(2)$,

$$
c=b+\log _{10} b>b+0=b>a>1
$$

so $\log _{10} c>0$, so from (3),

$$
a=c+\log _{10} c>c+0=c>b>a>1
$$

But this says that $a>c>b>a$, which is a contradiction.

Therefore, $a$ cannot be larger than 1 .

Lastly, if $0<a<1$ ( $a$ cannot be negative), then $\log _{10} a<0$, so from (1),

$$
b=a+\log _{10} a<a+0=a<1
$$

so $\log _{10} b<0$, so from $(2)$,

$$
c=b+\log _{10} b<b+0=b<a<1
$$

so $\log _{10} c<0$, so from (3),

$$
a=c+\log _{10} c>c+0=c<b<a<1
$$

But this says that $a<c<b<a$, which is a contradiction.

Therefore, $a$ cannot be smaller than 1 either.

Thus, $a$ must equal 1.

If $a=1$, then $b=a+\log _{10} a=1+\log _{10} 1=1+0=1$ from (1), which will similarly give $c=1$ from (2).

Thus, the only solution to the system is $(a, b, c)=(1,1,1)=(x, y-1, z-2)$ since $a$ cannot be either larger than or smaller than 1 , so $(x, y, z)=(1,2,3)$.

Final answer: $1,2,3$
Subfield: Number Theory


Question: The positive integers 34 and 80 have exactly two positive common divisors, namely 1 and 2. How many positive integers $n$ with $1 \leq n \leq 30$ have the property that $n$ and 80 have exactly two positive common divisors?

Solution:
Since $80=2^{4} \cdot 5$, its positive divisors are $1,2,4,5,8,10,16,20,40,80$.

For an integer $n$ to share exactly two positive common divisors with 80, these divisors must be either 1 and 2 or 1 and 5 . ( 1 is a common divisor of any two integers. The second common divisor must be a prime number since any composite divisor will cause there to be at least one more common divisor which is prime.)

Since $1 \leq n \leq 30$ and $n$ is a multiple of 2 or of 5 , then the possible values of $n$ come from the list

$$
2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26,28,30
$$

We remove the multiples of 4 from this list (since they would share at least the divisors $1,2,4$ with 80 ) and the multiples of 10 from this list (since they would share at least the divisors $1,2,5,10$ with 80 ).

This leaves the list

$$
2,5,6,14,15,18,22,25,26
$$

The common divisors of any number from this list and 80 are either 1 and 2 or 1 and 5 . There are 9 such integers.

Final answer: 9
Subfield: Algebra


Question: A function $f$ is defined so that

- $f(1)=1$,
- if $n$ is an even positive integer, then $f(n)=f\left(\frac{1}{2} n\right)$, and
- if $n$ is an odd positive integer with $n>1$, then $f(n)=f(n-1)+1$.

For example, $f(34)=f(17)$ and $f(17)=f(16)+1$.

Determine the value of $f(50)$.

Solution:
We start with $f(50)$ and apply the given rules for the function until we reach $f(1)$ :

$$
\begin{aligned}
f(50) & =f(25) \\
& =f(24)+1 \\
& =f(12)+1 \\
& =f(6)+1 \\
& =f(3)+1 \\
& =(f(2)+1)+1 \\
& =f(1)+1+1 \\
& =1+1+1 \\
& =3
\end{aligned}
$$

(since 50 is even and $\frac{1}{2}(50)=25$ )

(since 25 is odd and $25-1=24$ )

$$
\left(\frac{1}{2}(24)=12\right)
$$

$$
\begin{aligned}
\left(\frac{1}{2}(12)\right. & =6) \\
\left(\frac{1}{2}(6)\right. & =3) \\
(3-1 & =2) \\
\left(\frac{1}{2}(2)\right. & =1) \\
(f(1) & =1)
\end{aligned}
$$

Therefore, $f(50)=3$.

Final answer: 3
Subfield: Geometry


Question: The perimeter of equilateral $\triangle P Q R$ is 12. The perimeter of regular hexagon $S T U V W X$ is also 12. What is the ratio of the area of $\triangle P Q R$ to the area of $S T U V W X$ ?

Solution:
Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2 .

Since equilateral $\triangle P Q R$ has perimeter 12 , then its side length is 4 .

Consider equilateral triangles with side length 2.

Six of these triangles can be combined to form a regular hexagon with side length 2 and four of these can be combined to form an equilateral triangle with side length 4 .
<img_3579>

Note that the six equilateral triangles around the centre of the hexagon give a total central angle of $6 \cdot 60^{\circ}=360^{\circ}$ (a complete circle) and the three equilateral triangles along each side of the large equilateral triangle make a straight angle of $180^{\circ}\left(\right.$ since $3 \cdot 60^{\circ}=180^{\circ}$ ). Also, the length of each side of the hexagon is 2 and the measure of each internal angle is $120^{\circ}$, which means that the hexagon is regular. Similarly, the triangle is equilateral.

Since the triangle is made from four identical smaller triangles and the hexagon is made from six of these smaller triangles, the ratio of the area of the triangle to the hexagon is $4: 6$ which is equivalent to $2: 3$.

Final answer: $\frac{2}{3}$
Subfield: Geometry


Question: In the diagram, sector $A O B$ is $\frac{1}{6}$ of an entire circle with radius $A O=B O=18$. The sector is cut into two regions with a single straight cut through $A$ and point $P$ on $O B$. The areas of the two regions are equal. Determine the length of $O P$.

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Solution:
Since sector $A O B$ is $\frac{1}{6}$ of a circle with radius 18 , its area is $\frac{1}{6}\left(\pi \cdot 18^{2}\right)$ or $54 \pi$.

For the line $A P$ to divide this sector into two pieces of equal area, each piece has area $\frac{1}{2}(54 \pi)$ or $27 \pi$.

We determine the length of $O P$ so that the area of $\triangle P O A$ is $27 \pi$.

Since sector $A O B$ is $\frac{1}{6}$ of a circle, then $\angle A O B=\frac{1}{6}\left(360^{\circ}\right)=60^{\circ}$.

Drop a perpendicular from $A$ to $T$ on $O B$.

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The area of $\triangle P O A$ is $\frac{1}{2}(O P)(A T)$.

$\triangle A O T$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Since $A O=18$, then $A T=\frac{\sqrt{3}}{2}(A O)=9 \sqrt{3}$.

For the area of $\triangle P O A$ to equal $27 \pi$, we have $\frac{1}{2}(O P)(9 \sqrt{3})=27 \pi$ which gives $O P=\frac{54 \pi}{9 \sqrt{3}}=\frac{6 \pi}{\sqrt{3}}=2 \sqrt{3} \pi$.

(Alternatively, we could have used the fact that the area of $\triangle P O A$ is $\frac{1}{2}(O A)(O P) \sin (\angle P O A)$.)

Final answer: $2 \sqrt{3} \pi$
Subfield: Geometry


Question: In the diagram, sector $A O B$ is $\frac{1}{6}$ of an entire circle with radius $A O=B O=18$. The sector is cut into two regions with a single straight cut through $A$ and point $P$ on $O B$. The areas of the two regions are equal. Determine the length of $O P$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d6fc47bd20a6486a92ceg-1.jpg?height=273&width=288&top_left_y=1517&top_left_x=1271)

Solution:
Since sector $A O B$ is $\frac{1}{6}$ of a circle with radius 18 , its area is $\frac{1}{6}\left(\pi \cdot 18^{2}\right)$ or $54 \pi$.

For the line $A P$ to divide this sector into two pieces of equal area, each piece has area $\frac{1}{2}(54 \pi)$ or $27 \pi$.

We determine the length of $O P$ so that the area of $\triangle P O A$ is $27 \pi$.

Since sector $A O B$ is $\frac{1}{6}$ of a circle, then $\angle A O B=\frac{1}{6}\left(360^{\circ}\right)=60^{\circ}$.

Drop a perpendicular from $A$ to $T$ on $O B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4ac427a3237c2043180eg-1.jpg?height=268&width=276&top_left_y=1568&top_left_x=1033)

The area of $\triangle P O A$ is $\frac{1}{2}(O P)(A T)$.

$\triangle A O T$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Since $A O=18$, then $A T=\frac{\sqrt{3}}{2}(A O)=9 \sqrt{3}$.

For the area of $\triangle P O A$ to equal $27 \pi$, we have $\frac{1}{2}(O P)(9 \sqrt{3})=27 \pi$ which gives $O P=\frac{54 \pi}{9 \sqrt{3}}=\frac{6 \pi}{\sqrt{3}}=2 \sqrt{3} \pi$.

(Alternatively, we could have used the fact that the area of $\triangle P O A$ is $\frac{1}{2}(O A)(O P) \sin (\angle P O A)$.)

Final answer: $2 \sqrt{3} \pi$