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


Question: The circle $(x-p)^{2}+y^{2}=r^{2}$ has centre $C$ and the circle $x^{2}+(y-p)^{2}=r^{2}$ has centre $D$. The circles intersect at two distinct points $A$ and $B$, with $x$-coordinates $a$ and $b$, respectively.
Prove that $a+b=p$ and $a^{2}+b^{2}=r^{2}$.

Solution:
We have $(x-p)^{2}+y^{2}=r^{2}$ and $x^{2}+(y-p)^{2}=r^{2}$, so at the points of intersection,

$$
\begin{aligned}
(x-p)^{2}+y^{2} & =x^{2}+(y-p)^{2} \\
x^{2}-2 p x+p^{2}+y^{2} & =x^{2}+y^{2}-2 p y+p^{2} \\
-2 p x & =-2 p y
\end{aligned}
$$

and so $x=y$ (since we may assume that $p \neq 0$ otherwise the two circles would coincide). Therefore, $a$ and $b$ are the two solutions of the equation $(x-p)^{2}+x^{2}=r^{2}$ or $2 x^{2}-2 p x+$ $\left(p^{2}-r^{2}\right)=0$ or $x^{2}-p x+\frac{1}{2}\left(p^{2}-r^{2}\right)=0$.

Using the relationship between the sum and product of roots of a quadratic equation and its coefficients, we obtain that $a+b=p$ and $a b=\frac{1}{2}\left(p^{2}-r^{2}\right)$.

(We could have solved for $a$ and $b$ using the quadratic formula and calculated these directly.)

So we know that $a+b=p$.

Lastly, $a^{2}+b^{2}=(a+b)^{2}-2 a b=p^{2}-2\left(\frac{1}{2}\left(p^{2}-r^{2}\right)\right)=r^{2}$, as required.
Since the circles are reflections of one another in the line $y=x$, then the two points of intersection must both lie on the line $y=x$, ie. $A$ has coordinates $(a, a)$ and $B$ has coordinates $(b, b)$.

Therefore, $(a-p)^{2}+a^{2}=r^{2}$ and $(b-p)^{2}+b^{2}=r^{2}$, since these points lie on both circles.



Subtracting the two equations, we get

$$
\begin{aligned}
(b-p)^{2}-(a-p)^{2}+b^{2}-a^{2} & =0 \\
((b-p)-(a-p))((b-p)+(a-p))+(b-a)(b+a) & =0 \\
(b-a)(a+b-2 p)+(b-a)(b+a) & =0 \\
(b-a)(a+b-2 p+b+a) & =0 \\
2(b-a)(a+b-p) & =0
\end{aligned}
$$

Since $a \neq b$, then we must have $a+b=p$, as required.

Since $a+b=p$, then $a-p=-b$, so substituting back into $(a-p)^{2}+a^{2}=r^{2}$ gives $(-b)^{2}+a^{2}=r^{2}$, or $a^{2}+b^{2}=r^{2}$, as required.
Subfield: Geometry


Question: The circle $(x-p)^{2}+y^{2}=r^{2}$ has centre $C$ and the circle $x^{2}+(y-p)^{2}=r^{2}$ has centre $D$. The circles intersect at two distinct points $A$ and $B$, with $x$-coordinates $a$ and $b$, respectively.
If $r$ is fixed and $p$ is then found to maximize the area of quadrilateral $C A D B$, prove that either $A$ or $B$ is the origin.

Solution:
We first draw a diagram.

<img_3773>

We know that $C$ has coordinates $(p, 0)$ and $D$ has coordinates $(0, p)$.

Thus, the slope of line segment $C D$ is -1 .

Since the points $A$ and $B$ both lie on the line $y=x$, then the slope of line segment $A B$ is 1 .

Therefore, $A B$ is perpendicular to $C D$, so $C A D B$ is a kite, and so its area is equal to $\frac{1}{2}(A B)(C D)$.

(We could derive this by breaking quadrilateral $C A D B$ into $\triangle C A B$ and $\triangle D A B$.)

Since $C$ has coordinates $(p, 0)$ and $D$ has coordinates $(0, p)$, then $C D=\sqrt{p^{2}+(-p)^{2}}=$ $\sqrt{2 p^{2}}$.

(We do not know if $p$ is positive, so this is not necessarily equal to $\sqrt{2} p$.)

We know that $A$ has coordinates $(a, a)$ and $B$ has coordinates $(b, b)$, so

$$
\begin{aligned}
A B & =\sqrt{(a-b)^{2}+(a-b)^{2}} \\
& =\sqrt{2 a^{2}-4 a b+2 b^{2}} \\
& =\sqrt{2\left(a^{2}+b^{2}\right)-4 a b} \\
& =\sqrt{2 r^{2}-4\left(\frac{1}{2}\left(p^{2}-r^{2}\right)\right)} \\
& =\sqrt{4 r^{2}-2 p^{2}}
\end{aligned}
$$

Therefore, the area of quadrilateral $C A D B$ is

$$
\frac{1}{2}(A B)(C D)=\frac{1}{2} \sqrt{4 r^{2}-2 p^{2}} \sqrt{2 p^{2}}=\sqrt{2 r^{2} p^{2}-p^{4}}
$$



To maximize this area, we must maximize $2 r^{2} p^{2}-p^{4}=2 r^{2}\left(p^{2}\right)-\left(p^{2}\right)^{2}$.

Since $r$ is fixed, we can consider this as a quadratic polynomial in $p^{2}$. Since the coefficient of $\left(p^{2}\right)^{2}$ is negative, then this is a parabola opening downwards, so we find its maximum value by finding its vertex.

The vertex of $2 r^{2}\left(p^{2}\right)-\left(p^{2}\right)^{2}$ is at $p^{2}=-\frac{2 r^{2}}{2(-1)}=r^{2}$.

So the maximum area of the quadrilateral occurs when $p$ is chosen so that $p^{2}=r^{2}$.

Since $p^{2}=r^{2}$, then $(a+b)^{2}=p^{2}=r^{2}$ so $a^{2}+2 a b+b^{2}=r^{2}$.

Since $a^{2}+b^{2}=r^{2}$, then $2 a b=0$ so either $a=0$ or $b=0$, and so either $A$ has coordinates $(0,0)$ or $B$ has coordinates $(0,0)$, ie. either $A$ is the origin or $B$ is the origin.
Subfield: Combinatorics


Question: A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below:

![](https://cdn.mathpix.com/cropped/2023_12_21_7e10dbc5f8ae70c21b8eg-1.jpg?height=449&width=1283&top_left_y=1719&top_left_x=429)
Determine $f(50)$.

Solution:
We proceed directly.

On the first pass from left to right, Josephine closes all of the even numbered lockers, leaving the odd ones open.

The second pass proceeds from right to left. Before the pass, the lockers which are open are $1,3, \ldots, 47,49$.

On the second pass, she shuts lockers 47, 43, 39, .., 3 .

The third pass proceeds from left to right. Before the pass, the lockers which are open are $1,5, \ldots, 45,49$.

On the third pass, she shuts lockers $5,13, \ldots, 45$.

This leaves lockers 1, 9, 17, 25, 33, 41, 49 open.

On the fourth pass, from right to left, lockers 41, 25 and 9 are shut, leaving 1, 17, 33, 49.

On the fifth pass, from left to right, lockers 17 and 49 are shut, leaving 1 and 33 open.

On the sixth pass, from right to left, locker 1 is shut, leaving 33 open.

Thus, $f(50)=33$.

Final answer: 33
Subfield: Combinatorics


Question: A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below:

<img_3864>
Determine $f(50)$.

Solution:
We proceed directly.

On the first pass from left to right, Josephine closes all of the even numbered lockers, leaving the odd ones open.

The second pass proceeds from right to left. Before the pass, the lockers which are open are $1,3, \ldots, 47,49$.

On the second pass, she shuts lockers 47, 43, 39, .., 3 .

The third pass proceeds from left to right. Before the pass, the lockers which are open are $1,5, \ldots, 45,49$.

On the third pass, she shuts lockers $5,13, \ldots, 45$.

This leaves lockers 1, 9, 17, 25, 33, 41, 49 open.

On the fourth pass, from right to left, lockers 41, 25 and 9 are shut, leaving 1, 17, 33, 49.

On the fifth pass, from left to right, lockers 17 and 49 are shut, leaving 1 and 33 open.

On the sixth pass, from right to left, locker 1 is shut, leaving 33 open.

Thus, $f(50)=33$.

Final answer: 33
Subfield: Algebra


Question: What is the sum of the digits of the integer equal to $\left(10^{3}+1\right)^{2}$ ?

Solution:
Using a calculator, we see that

$$
\left(10^{3}+1\right)^{2}=1001^{2}=1002001
$$

The sum of the digits of this integer is $1+2+1$ which equals 4 .

To determine this integer without using a calculator, we can let $x=10^{3}$.

Then

$$
\begin{aligned}
\left(10^{3}+1\right)^{2} & =(x+1)^{2} \\
& =x^{2}+2 x+1 \\
& =\left(10^{3}\right)^{2}+2\left(10^{3}\right)+1 \\
& =1002001
\end{aligned}
$$

Final answer: 1002001
Subfield: Algebra


Question: A bakery sells small and large cookies. Before a price increase, the price of each small cookie is $\$ 1.50$ and the price of each large cookie is $\$ 2.00$. The price of each small cookie is increased by $10 \%$ and the price of each large cookie is increased by $5 \%$. What is the percentage increase in the total cost of a purchase of 2 small cookies and 1 large cookie?

Solution:
Before the price increase, the total cost of 2 small cookies and 1 large cookie is $2 \cdot \$ 1.50+\$ 2.00=\$ 5.00$.

$10 \%$ of $\$ 1.50$ is $0.1 \cdot \$ 1.50=\$ 0.15$. After the price increase, 1 small cookie costs $\$ 1.50+\$ 0.15=\$ 1.65$.

$5 \%$ of $\$ 2.00$ is $0.05 \cdot \$ 2.00=\$ 0.10$. After the price increase, 1 large cookie costs $\$ 2.00+\$ 0.10=\$ 2.10$.

After the price increase, the total cost of 2 small cookies and 1 large cookie is $2 \cdot \$ 1.65+\$ 2.10=\$ 5.40$.

The percentage increase in the total cost is $\frac{\$ 5.40-\$ 5.00}{\$ 5.00} \times 100 \%=\frac{40}{500} \times 100 \%=8 \%$.

Final answer: $8 \%$
Subfield: Algebra


Question: Qing is twice as old as Rayna. Qing is 4 years younger than Paolo. The average age of Paolo, Qing and Rayna is 13. Determine their ages.

Solution:
Suppose that Rayna's age is $x$ years.

Since Qing is twice as old as Rayna, Qing's age is $2 x$ years.

Since Qing is 4 years younger than Paolo, Paolo's age is $2 x+4$ years.

Since the average of their ages is 13 years, we obtain

$$
\frac{x+(2 x)+(2 x+4)}{3}=13
$$

This gives $5 x+4=39$ and so $5 x=35$ or $x=7$.

Therefore, Rayna is 7 years old, Qing is 14 years old, and Paolo is 18 years old.

(Checking, the average of 7,14 and 18 is $\frac{7+14+18}{3}=\frac{39}{3}=13$.)

Final answer: 7,14,18
Subfield: Geometry


Question: In the diagram, $P Q R S$ is a quadrilateral. What is its perimeter?

<img_3956>

Solution:
The length of $P Q$ is equal to $\sqrt{(0-5)^{2}+(12-0)^{2}}=\sqrt{(-5)^{2}+12^{2}}=13$.

In a similar way, we can see that $Q R=R S=S P=13$.

Therefore, the perimeter of $P Q R S$ is $4 \cdot 13=52$.

(We can also see that if $O$ is the origin, then $\triangle P O Q, \triangle P O S, \triangle R O Q$, and $\triangle R O S$ are congruent because $O Q=O S$ and $O P=O R$, which means that $P Q=Q R=R S=S P$.)

Final answer: 52
Subfield: Geometry


Question: In the diagram, $P Q R S$ is a quadrilateral. What is its perimeter?

![](https://cdn.mathpix.com/cropped/2023_12_21_620dad4369d1ef5c24f3g-1.jpg?height=442&width=498&top_left_y=191&top_left_x=1258)

Solution:
The length of $P Q$ is equal to $\sqrt{(0-5)^{2}+(12-0)^{2}}=\sqrt{(-5)^{2}+12^{2}}=13$.

In a similar way, we can see that $Q R=R S=S P=13$.

Therefore, the perimeter of $P Q R S$ is $4 \cdot 13=52$.

(We can also see that if $O$ is the origin, then $\triangle P O Q, \triangle P O S, \triangle R O Q$, and $\triangle R O S$ are congruent because $O Q=O S$ and $O P=O R$, which means that $P Q=Q R=R S=S P$.)

Final answer: 52
Subfield: Geometry


Question: In the diagram, $A$ has coordinates $(0,8)$. Also, the midpoint of $A B$ is $M(3,9)$ and the midpoint of $B C$ is $N(7,6)$. What is the slope of $A C$ ?

<img_4040>

Solution:
Suppose that $B$ has coordinates $(r, s)$ and $C$ has coordinates $(t, u)$.

Since $M(3,9)$ is the midpoint of $A(0,8)$ and $B(r, s)$, then 3 is the average of 0 and $r$ (which gives $r=6)$ and 9 is the average of 8 and $s$ (which gives $s=10$ ).

Since $N(7,6)$ is the midpoint of $B(6,10)$ and $C(t, u)$, then 7 is the average of 6 and $t$ (which gives $t=8$ ) and 6 is the average of 10 and $u$ (which gives $u=2$ ).

The slope of the line segment joining $A(0,8)$ and $C(8,2)$ is $\frac{8-2}{0-8}$ which equals $-\frac{3}{4}$.
Since $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$, then $M N$ is parallel to $A C$. Therefore, the slope of $A C$ equals the slope of the line segment joining $M(3,9)$ to $N(7,6)$, which is $\frac{9-6}{3-7}$ or $-\frac{3}{4}$.

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


Question: In the diagram, $A$ has coordinates $(0,8)$. Also, the midpoint of $A B$ is $M(3,9)$ and the midpoint of $B C$ is $N(7,6)$. What is the slope of $A C$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_620dad4369d1ef5c24f3g-1.jpg?height=440&width=415&top_left_y=669&top_left_x=1256)

Solution:
Suppose that $B$ has coordinates $(r, s)$ and $C$ has coordinates $(t, u)$.

Since $M(3,9)$ is the midpoint of $A(0,8)$ and $B(r, s)$, then 3 is the average of 0 and $r$ (which gives $r=6)$ and 9 is the average of 8 and $s$ (which gives $s=10$ ).

Since $N(7,6)$ is the midpoint of $B(6,10)$ and $C(t, u)$, then 7 is the average of 6 and $t$ (which gives $t=8$ ) and 6 is the average of 10 and $u$ (which gives $u=2$ ).

The slope of the line segment joining $A(0,8)$ and $C(8,2)$ is $\frac{8-2}{0-8}$ which equals $-\frac{3}{4}$.
Since $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$, then $M N$ is parallel to $A C$. Therefore, the slope of $A C$ equals the slope of the line segment joining $M(3,9)$ to $N(7,6)$, which is $\frac{9-6}{3-7}$ or $-\frac{3}{4}$.

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


Question: The parabola with equation $y=-2 x^{2}+4 x+c$ has vertex $V(1,18)$. The parabola intersects the $y$-axis at $D$ and the $x$-axis at $E$ and $F$. Determine the area of $\triangle D E F$.

Solution:
Since $V(1,18)$ is on the parabola, then $18=-2\left(1^{2}\right)+4(1)+c$ and so $c=18+2-4=16$.

Thus, the equation of the parabola is $y=-2 x^{2}+4 x+16$.

The $y$-intercept occurs when $x=0$, and so $y=16$. Thus, $D$ has coordinates $(0,16)$.

The $x$-intercepts occur when $y=0$. Here,

$$
\begin{array}{r}
-2 x^{2}+4 x+16=0 \\
-2\left(x^{2}-2 x-8\right)=0 \\
-2(x-4)(x+2)=0
\end{array}
$$



and so $x=4$ and $x=-2$.

This means that $E$ and $F$, in some order, have coordinates $(4,0)$ and $(-2,0)$.

Therefore, $\triangle D E F$ has base $E F$ of length $4-(-2)=6$ and height 16 (vertical distance from the $x$-axis to the point $D$ ).

Finally, the area of $\triangle D E F$ is $\frac{1}{2} \cdot 6 \cdot 16=48$.

Final answer: 48
Subfield: Algebra


Question: If $3\left(8^{x}\right)+5\left(8^{x}\right)=2^{61}$, what is the value of the real number $x$ ?

Solution:
We obtain successively

$$
\begin{aligned}
3\left(8^{x}\right)+5\left(8^{x}\right) & =2^{61} \\
8\left(8^{x}\right) & =2^{61} \\
8^{x+1} & =2^{61} \\
\left(2^{3}\right)^{x+1} & =2^{61} \\
2^{3(x+1)} & =2^{61}
\end{aligned}
$$

Thus, $3(x+1)=61$ and so $3 x+3=61$ which gives $3 x=58$ or $x=\frac{58}{3}$.

Final answer: $\frac{58}{3}$
Subfield: Number Theory


Question: For some real numbers $m$ and $n$, the list $3 n^{2}, m^{2}, 2(n+1)^{2}$ consists of three consecutive integers written in increasing order. Determine all possible values of $m$.

Solution:
Since the list $3 n^{2}, m^{2}, 2(n+1)^{2}$ consists of three consecutive integers written in increasing order, then

$$
\begin{aligned}
2(n+1)^{2}-3 n^{2} & =2 \\
2 n^{2}+4 n+2-3 n^{2} & =2 \\
-n^{2}+4 n & =0 \\
-n(n-4) & =0
\end{aligned}
$$

and so $n=0$ or $n=4$.

If $n=0$, the list becomes $0, m^{2}, 2$. This means that $m^{2}=1$ and so $m= \pm 1$.

If $n=4$, we have $3 n^{2}=3 \cdot 16=48$ and $2(n+1)^{2}=2 \cdot 25=50$ giving the list $48, m^{2}, 50$. This means that $m^{2}=49$ and so $m= \pm 7$.

Thus, the possible values for $m$ are $1,-1,7,-7$.

Final answer: 1,-1,7,-7
Subfield: Combinatorics


Question: Chinara starts with the point $(3,5)$, and applies the following three-step process, which we call $\mathcal{P}$ :

Step 1: Reflect the point in the $x$-axis.

Step 2: Translate the resulting point 2 units upwards.

Step 3: Reflect the resulting point in the $y$-axis.

As she does this, the point $(3,5)$ moves to $(3,-5)$, then to $(3,-3)$, and then to $(-3,-3)$.

Chinara then starts with a different point $S_{0}$. She applies the three-step process $\mathcal{P}$ to the point $S_{0}$ and obtains the point $S_{1}$. She then applies $\mathcal{P}$ to $S_{1}$ to obtain the point $S_{2}$. She applies $\mathcal{P}$ four more times, each time using the previous output of $\mathcal{P}$ to be the new input, and eventually obtains the point $S_{6}(-7,-1)$. What are the coordinates of the point $S_{0}$ ?

Solution:
Suppose that $S_{0}$ has coordinates $(a, b)$.

Step 1 moves $(a, b)$ to $(a,-b)$.

Step 2 moves $(a,-b)$ to $(a,-b+2)$.

Step 3 moves $(a,-b+2)$ to $(-a,-b+2)$.

Thus, $S_{1}$ has coordinates $(-a,-b+2)$.

Step 1 moves $(-a,-b+2)$ to $(-a, b-2)$.

Step 2 moves $(-a, b-2)$ to $(-a, b)$.

Step 3 moves $(-a, b)$ to $(a, b)$.

Thus, $S_{2}$ has coordinates $(a, b)$, which are the same coordinates as $S_{0}$.

Continuing this process, $S_{4}$ will have the same coordinates as $S_{2}$ (and thus as $S_{0}$ ) and $S_{6}$ will have the same coordinates as $S_{4}, S_{2}$ and $S_{0}$.

Since the coordinates of $S_{6}$ are $(-7,-1)$, the coordinates of $S_{0}$ are also $(-7,-1)$.
We work backwards from $S_{6}(-7,-1)$.

To do this, we undo the Steps of the process $\mathcal{P}$ by applying them in reverse order.



Since Step 3 reflects a point in the $y$-axis, its inverse does the same.

Since Step 2 translates a point 2 units upwards, its inverse translates a point 2 units downwards.

Since Step 1 reflects a point in the $x$-axis, its inverse does the same.

Applying these inverse steps to $S_{6}(-7,-1)$, we obtain $(7,-1)$, then $(7,-3)$, then $(7,3)$.

Thus, $S_{5}$ has coordinates $(7,3)$.

Applying the inverse steps to $S_{5}(7,3)$, we obtain $(-7,3)$, then $(-7,1)$, then $(-7,-1)$.

Thus, $S_{4}$ has coordinates $(-7,-1)$, which are the same coordinates as $S_{6}$.

If we apply these steps two more times, we will see that $S_{2}$ is the same point as $S_{4}$.

Two more applications tell us that $S_{0}$ is the same point as $S_{2}$.

Therefore, the coordinates of $S_{0}$ are the same as the coordinates of $S_{6}$, which are $(-7,-1)$.

Final answer: (-7,-1)
Subfield: Geometry


Question: In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_3c4a0213f9fdb308210bg-1.jpg?height=244&width=428&top_left_y=274&top_left_x=1355)
Determine the length of $A B$ in terms of $x$.

Solution:
We begin by determining the length of $A B$ in terms of $x$.

Since $A B D E$ is a rectangle, $B D=A E=2 x$.

Since $\triangle B C D$ is equilateral, $\angle D B C=60^{\circ}$.

Join $A$ to $D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4720ccb222a99329fc9eg-1.jpg?height=247&width=434&top_left_y=955&top_left_x=951)

Since $A D$ and $B C$ are parallel, $\angle A D B=\angle D B C=60^{\circ}$.

Consider $\triangle A D B$. This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle since $\angle A B D$ is a right angle.

Using ratios of side lengths, $\frac{A B}{B D}=\frac{\sqrt{3}}{1}$ and so $A B=\sqrt{3} B D=2 \sqrt{3} x$

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


Question: In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$.

<img_3702>
Determine the length of $A B$ in terms of $x$.

Solution:
We begin by determining the length of $A B$ in terms of $x$.

Since $A B D E$ is a rectangle, $B D=A E=2 x$.

Since $\triangle B C D$ is equilateral, $\angle D B C=60^{\circ}$.

Join $A$ to $D$.

<img_3330>

Since $A D$ and $B C$ are parallel, $\angle A D B=\angle D B C=60^{\circ}$.

Consider $\triangle A D B$. This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle since $\angle A B D$ is a right angle.

Using ratios of side lengths, $\frac{A B}{B D}=\frac{\sqrt{3}}{1}$ and so $A B=\sqrt{3} B D=2 \sqrt{3} x$

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


Question: In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$.

<img_3702>
Determine positive integers $r$ and $s$ for which

$$
\frac{A C}{A D}=\sqrt{\frac{r}{s}}
$$

Solution:
We begin by determining the length of $A B$ in terms of $x$.

Since $A B D E$ is a rectangle, $B D=A E=2 x$.

Since $\triangle B C D$ is equilateral, $\angle D B C=60^{\circ}$.

Join $A$ to $D$.

<img_3330>

Since $A D$ and $B C$ are parallel, $\angle A D B=\angle D B C=60^{\circ}$.

Consider $\triangle A D B$. This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle since $\angle A B D$ is a right angle.

Using ratios of side lengths, $\frac{A B}{B D}=\frac{\sqrt{3}}{1}$ and so $A B=\sqrt{3} B D=2 \sqrt{3} x$

Next, we determine $\frac{A C}{A D}$.

Now, $\frac{A D}{B D}=\frac{2}{1}$ and so $A D=2 B D=4 x$.

Suppose that $M$ is the midpoint of $A E$ and $N$ is the midpoint of $B D$.

Since $A E=B D=2 x$, then $A M=M E=B N=N D=x$.

Join $M$ to $N$ and $N$ to $C$ and $A$ to $C$.

<img_3952>

Since $A B D E$ is a rectangle, then $M N$ is parallel to $A B$ and so $M N$ is perpendicular to both $A E$ and $B D$.

Also, $M N=A B=2 \sqrt{3} x$.

Since $\triangle B C D$ is equilateral, its median $C N$ is perpendicular to $B D$.

Since $M N$ and $N C$ are perpendicular to $B D, M N C$ is actually a straight line segment and so $M C=M N+N C$.

Now $\triangle B N C$ is also a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $N C=\sqrt{3} B N=\sqrt{3} x$.

This means that $M C=2 \sqrt{3} x+\sqrt{3} x=3 \sqrt{3} x$.



Finally, $\triangle A M C$ is right-angled at $M$ and so

$$
A C=\sqrt{A M^{2}+M C^{2}}=\sqrt{x^{2}+(3 \sqrt{3} x)^{2}}=\sqrt{x^{2}+27 x^{2}}=\sqrt{28 x^{2}}=2 \sqrt{7} x
$$

since $x>0$.

This means that $\frac{A C}{A D}=\frac{2 \sqrt{7} x}{4 x}=\frac{\sqrt{7}}{2}=\sqrt{\frac{7}{4}}$, which means that the integers $r=7$ and $s=4$ satisfy the conditions.

Final answer: 7,4
Subfield: Geometry


Question: In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_3c4a0213f9fdb308210bg-1.jpg?height=244&width=428&top_left_y=274&top_left_x=1355)
Determine positive integers $r$ and $s$ for which

$$
\frac{A C}{A D}=\sqrt{\frac{r}{s}}
$$

Solution:
We begin by determining the length of $A B$ in terms of $x$.

Since $A B D E$ is a rectangle, $B D=A E=2 x$.

Since $\triangle B C D$ is equilateral, $\angle D B C=60^{\circ}$.

Join $A$ to $D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4720ccb222a99329fc9eg-1.jpg?height=247&width=434&top_left_y=955&top_left_x=951)

Since $A D$ and $B C$ are parallel, $\angle A D B=\angle D B C=60^{\circ}$.

Consider $\triangle A D B$. This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle since $\angle A B D$ is a right angle.

Using ratios of side lengths, $\frac{A B}{B D}=\frac{\sqrt{3}}{1}$ and so $A B=\sqrt{3} B D=2 \sqrt{3} x$

Next, we determine $\frac{A C}{A D}$.

Now, $\frac{A D}{B D}=\frac{2}{1}$ and so $A D=2 B D=4 x$.

Suppose that $M$ is the midpoint of $A E$ and $N$ is the midpoint of $B D$.

Since $A E=B D=2 x$, then $A M=M E=B N=N D=x$.

Join $M$ to $N$ and $N$ to $C$ and $A$ to $C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4720ccb222a99329fc9eg-1.jpg?height=250&width=447&top_left_y=1840&top_left_x=945)

Since $A B D E$ is a rectangle, then $M N$ is parallel to $A B$ and so $M N$ is perpendicular to both $A E$ and $B D$.

Also, $M N=A B=2 \sqrt{3} x$.

Since $\triangle B C D$ is equilateral, its median $C N$ is perpendicular to $B D$.

Since $M N$ and $N C$ are perpendicular to $B D, M N C$ is actually a straight line segment and so $M C=M N+N C$.

Now $\triangle B N C$ is also a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $N C=\sqrt{3} B N=\sqrt{3} x$.

This means that $M C=2 \sqrt{3} x+\sqrt{3} x=3 \sqrt{3} x$.



Finally, $\triangle A M C$ is right-angled at $M$ and so

$$
A C=\sqrt{A M^{2}+M C^{2}}=\sqrt{x^{2}+(3 \sqrt{3} x)^{2}}=\sqrt{x^{2}+27 x^{2}}=\sqrt{28 x^{2}}=2 \sqrt{7} x
$$

since $x>0$.

This means that $\frac{A C}{A D}=\frac{2 \sqrt{7} x}{4 x}=\frac{\sqrt{7}}{2}=\sqrt{\frac{7}{4}}$, which means that the integers $r=7$ and $s=4$ satisfy the conditions.

Final answer: 7,4