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


Question: 
Prove that for every real number $t$ such that $0<t<\frac{1}{2}$ there exists a positive integer $n$ with the following property: for every set $S$ of $n$ positive integers there exist two different elements $x$ and $y$ of $S$, and a non-negative integer $m$ (i.e. $m \geq 0$ ), such that

$$
|x-m y| \leq t y .
$$

Solution:
Let $n$ be any positive integer such that

$$
(1+t)^{n-1} \geq \frac{1}{t} \tag{Q6.1}
$$

(this inequality is actually true for every large enough $n$ due to Bernoulli's inequality).

Let $S$ be any set of $n$ distinct positive integers, which we denote by

$$
s_{1}<s_{2}<\ldots<s_{n}
$$

We distinguish two cases.

- If $s_{i+1} \leq(1+t) s_{i}$ for some $i \in\{1, \ldots, n-1\}$, then

$$
\left|s_{i+1}-s_{i}\right|=s_{i+1}-s_{i} \leq t s_{i}
$$

and therefore the required inequality is satisfied with $x=s_{i+1}, y=s_{i}$, and $m=1$.

- If $s_{i+1}>(1+t) s_{i}$ for every $i \in\{1, \ldots, n-1\}$, then by induction we obtain that

$$
s_{n}>(1+t)^{n-1} s_{1}
$$

As a consequence, from (Q6.1) it follows that

$$
\left|s_{1}\right|=s_{1}<\frac{1}{(1+t)^{n-1}} \cdot s_{n} \leq t s_{n}
$$

and therefore the required inequality is satisfied with $x=s_{1}, y=s_{n}$, and $m=0$.
Subfield: Algebra


Question: Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that

(1) $a_{0}+a_{1}=-\frac{1}{n}$, and

(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.

Solution:
$\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ is equivalent to $\left(a_{k}+a_{k-1}+1\right)\left(a_{k}+a_{k+1}-1\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \ldots$ the following way: $b_{0}=-\frac{1}{n}$ and $\left(b_{k-1}+1\right)\left(b_{k}-1\right)=-1$. There is a proper sequence $b_{0}, b_{1}, \ldots, b_{N-1}$ if and only if there is proper sequence $a_{0}, a_{1}, \ldots, a_{N}$, because from a a proper $\left(a_{k}\right)$ sequence we can get a proper $\left(b_{k}\right)$ sequence with $b_{k}=a_{k}+a_{k+1}$ for $k=0,1, \ldots, N-1$ and from a proper $\left(b_{k}\right)$ sequence we can get a proper $\left(a_{k}\right)$ sequence by arbitrarily setting $a_{0}$ and then inductively defining $a_{k}=b_{k-1}-a_{k-1}$ for $k=1,2, \ldots, N$.

We prove by induction that $b_{k}=-\frac{1}{n-k}$ for $k<n$. This is true for $k=0$, as $b_{0}=-\frac{1}{n}$ and

$$
b_{k}=1-\frac{1}{b_{k-1}+1}=1-\frac{1}{1-\frac{1}{n-k+1}}=-\frac{1}{n-k}
$$

for $k<n$. Thus there is a proper sequence $b_{0}, b_{1}, \ldots, b_{n-1}$, but it can't be continued, because $b_{n-1}+1=$ 0 so there is no $b_{n}$ for which $\left(b_{n-1}+1\right)\left(b_{n}-1\right)=-1$.

Therefore the longest proper sequence $\left(b_{k}\right)$ is $n$-long, so the longest proper sequence $\left(a_{k}\right)$ is $n+1$ long, so $N=n$.
The required maximum is $N=n$.

To rule out the case $N \geq n+1$, it is clearly sufficient to rule out the case $N=n+1$.

Assume for contradiction that $a_{0}, a_{1}, \ldots, a_{n+1}$ are real numbers satisfying both conditions in the statement. It is sufficient to show that $a_{k}+a_{k+1}=0$ for some $k \leq n$, because then $a_{k-1}-a_{k+1}=0$ so $a_{k+1}=a_{k-1}$, therefore $a_{k-1}+a_{k}=0$ and so on, by backwards recursion we get that $a_{j}+a_{j+1}=0$ for all $0 \leq j \leq k$, but this is a contradiction with $a_{0}+a_{1}=-\frac{1}{n}$.

To prove that $a_{k}+a_{k+1}=0$ for some $k \leq n$, assume that $a_{k}+a_{k+1} \neq 0$ for all $k \leq n$, to rewrite the second condition in the statement in the form

$$
\frac{1}{a_{k}+a_{k+1}}-\frac{1}{a_{k-1}+a_{k}}=1, \quad k=1, \ldots, n
$$

and sum both sides over the full range from $k=1$ to $n$. This gives

$$
\frac{1}{a_{n}+a_{n+1}}-\frac{1}{a_{0}+a_{1}}=n
$$

As $a_{0}+a_{1}=-\frac{1}{n}$, this means that $\frac{1}{a_{n}+a_{n+1}}=0$, which is a contradiction. Consequently, $N \leq n$.

To provide $n+1$ real numbers satisfying both conditions in the statement, fix $a_{0}$ and go through the telescoping procedure above to obtain

$$
a_{k}=(-1)^{k} a_{0}+\sum_{j=1}^{k} \frac{(-1)^{k-j+1}}{n-j+1}, \quad k=1, \ldots, n
$$

This concludes the proof.

Final answer: $N=n$
Subfield: Combinatorics


Question: Let $n$ be a positive integer.
Prove that there exists a set $S$ of $6 n$ pairwise different positive integers, such that the least common multiple of any two elements of $S$ is no larger than $32 n^{2}$.

Solution:
Let the set $A$ consist of the $4 n$ integers $1,2, \ldots, 4 n$ and let the set $B$ consist of the $2 n$ even integers $4 n+2,4 n+4, \ldots, 8 n$. We claim that the $6 n$-element set $S=A \cup B$ has the desired property.

Indeed, the least common multiple of two (even) elements of $B$ is no larger than $8 n \cdot(8 n / 2)=32 n^{2}$, and the least common multiple of some element of $A$ and some element of $A \cup B$ is at most their product, which is at most $4 n \cdot 8 n=32 n^{2}$.
Subfield: Combinatorics


Question: Let $n$ be a positive integer.
Prove that every set $T$ of $6 n$ pairwise different positive integers contains two elements the least common multiple of which is larger than $9 n^{2}$.

Solution:
We prove the following lemma: "If a set $U$ contains $m+1$ integers, where $m \geqslant 2$, that are all not less than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$."

Let the elements of $U$ be $u_{1}>u_{2}>\cdots>u_{m+1} \geqslant m$. Note that $1 / u_{1} \leqslant 1 / u_{i} \leqslant 1 / m$ for $1 \leqslant i \leqslant m+1$. We partition the interval $\left[1 / u_{1} ; 1 / m\right]$ into $m$ subintervals of equal length. By the pigeonhole principle, there exist indices $i, j$ with $1 \leqslant i<j \leqslant m+1$ such that $1 / u_{i}$ and $1 / u_{j}$ belong to the same subinterval. Hence

$$
0<\frac{1}{u_{j}}-\frac{1}{u_{i}} \leqslant \frac{1}{m}\left(\frac{1}{m}-\frac{1}{u_{1}}\right)<\frac{1}{m^{2}}
$$

Now $1 / u_{j}-1 / u_{i}$ is a positive fraction with denominator $\operatorname{lcm}\left(u_{i}, u_{j}\right)$. The above thus yields the lower bound $\operatorname{lcm}\left(u_{i}, u_{j}\right)>m^{2}$, completing the proof of the lemma.

Applying the lemma with $m=3 n$ to the $3 n+1$ largest elements of $T$, which are all not less than $3 n$, we arrive at the desired statement.

A Variant. Alternatively, for part (b), we prove the following lemma: "If a set $U$ contains $m \geqslant 2$ integers that all are greater than $m$, then some two of its elements have least common multiple strictly larger than $\mathrm{m}^{2} . "$

Let $u_{1}>u_{2}>\cdots>u_{m}$ be the elements of $U$. Since $u_{m}>m=m^{2} / m$, there exists a smallest index $k$ such that $u_{k}>m^{2} / k$. If $k=1$, then $u_{1}>m^{2}$, and the least common multiple of $u_{1}$ and $u_{2}$ is strictly larger than $m^{2}$. So let us suppose $k>1$ from now on, so that we have $u_{k}>m^{2} / k$ and $u_{k-1} \leqslant m^{2} /(k-1)$. The greatest common divisor $d$ of $u_{k-1}$ and $u_{k}$ satisfies

$$
d \leqslant u_{k-1}-u_{k}<\frac{m^{2}}{k-1}-\frac{m^{2}}{k}=\frac{m^{2}}{(k-1) k}
$$



This implies $m^{2} /(d k)>k-1$ and $u_{k} / d>k-1$, and hence $u_{k} / d \geqslant k$. But then the least common multiple of $u_{k-1}$ and $u_{k}$ equals

$$
\frac{u_{k-1} u_{k}}{d} \geqslant u_{k} \cdot \frac{u_{k}}{d}>\frac{m^{2}}{k} \cdot k=m^{2}
$$

and the proof of the lemma is complete.

If we remove the $3 n$ smallest elements from set $T$ and apply the lemma with $m=3 n$ to the remaining elements, we arrive at the desired statement.
Subfield: Algebra


Question: Let $n, m$ be integers greater than 1 , and let $a_{1}, a_{2}, \ldots, a_{m}$ be positive integers not greater than $n^{m}$. Prove that there exist positive integers $b_{1}, b_{2}, \ldots, b_{m}$ not greater than $n$, such that

$$
\operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right)<n
$$

where $\operatorname{gcd}\left(x_{1}, x_{2}, \ldots, x_{m}\right)$ denotes the greatest common divisor of $x_{1}, x_{2}, \ldots, x_{m}$.

Solution:
Suppose without loss of generality that $a_{1}$ is the smallest of the $a_{i}$. If $a_{1} \geq n^{m}-1$, then the problem is simple: either all the $a_{i}$ are equal, or $a_{1}=n^{m}-1$ and $a_{j}=n^{m}$ for some $j$. In the first case, we can take (say) $b_{1}=1, b_{2}=2$, and the rest of the $b_{i}$ can be arbitrary, and we have

$$
\operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right) \leq \operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}\right)=1
$$

In the second case, we can take $b_{1}=1, b_{j}=1$, and the rest of the $b_{i}$ arbitrary, and again

$$
\operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right) \leq \operatorname{gcd}\left(a_{1}+b_{1}, a_{j}+b_{j}\right)=1
$$

So from now on we can suppose that $a_{1} \leq n^{m}-2$.

Now, let us suppose the desired $b_{1}, \ldots, b_{m}$ do not exist, and seek a contradiction. Then, for any choice of $b_{1}, \ldots, b_{m} \in\{1, \ldots, n\}$, we have

$$
\operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right) \geq n \text {. }
$$

Also, we have

$$
\operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right) \leq a_{1}+b_{1} \leq n^{m}+n-2
$$

Thus there are at most $n^{m}-1$ possible values for the greatest common divisor. However, there are $n^{m}$ choices for the $m$-tuple $\left(b_{1}, \ldots, b_{m}\right)$. Then, by the pigeonhole principle, there are two $m$-tuples that yield the same values for the greatest common divisor, say $d$. But since $d \geq n$, for each $i$ there can be at most one choice of $b_{i} \in\{1,2, \ldots, n\}$ such that $a_{i}+b_{i}$ is divisible by $d$ - and therefore there can be at most one $m$-tuple $\left(b_{1}, b_{2}, \ldots, b_{m}\right)$ yielding $d$ as the greatest common divisor. This is the desired contradiction.
We will prove stronger version of this problem:

For $m, n>1$, let $a_{1}, \ldots, a_{m}$ be positive integers with at least one $a_{i} \leq n^{2^{m-1}}$. Then there are integers $b_{1}, \ldots, b_{m}$, each equal to 1 or 2 , such that $\operatorname{gcd}\left(a_{1}+b_{1}, \ldots, a_{m}+b_{m}\right)<n$.

Proof: Suppose otherwise. Then the $2^{m-1}$ integers $\operatorname{gcd}\left(a_{1}+b_{1}, \ldots, a_{m}+b_{m}\right)$ with $b_{1}=1$ and $b_{i}=1$ or 2 for $i>1$ are all pairwise coprime, since for any two of them, there is some $i>1$ with $a_{i}+1$ appearing in one and $a_{i}+2$ in the other. Since each of these $2^{m-1}$ integers divides $a_{1}+1$, and each is $\geq n$ with at most one equal to $n$, it follows that $a_{1}+1 \geq n(n+1)^{2^{m-1}-1}$ so $a_{1} \geq n^{2^{m-1}}$. The same is true for each $a_{i}, i=1, \ldots, n$, a contradiction.
Subfield: Algebra


Question: In an increasing sequence of numbers with an odd number of terms, the difference between any two consecutive terms is a constant $d$, and the middle term is 302 . When the last 4 terms are removed from the sequence, the middle term of the resulting sequence is 296. What is the value of $d$ ?

Solution:
Let the number of terms in the sequence be $2 k+1$.

We label the terms $a_{1}, a_{2}, \ldots, a_{2 k+1}$.

The middle term here is $a_{k+1}=302$.

Since the difference between any two consecutive terms in this increasing sequence is $d$, $a_{m+1}-a_{m}=d$ for $m=1,2, \ldots, 2 k$.

When the last 4 terms are removed, the last term is now $a_{2 k-3}$ so the middle term is then $a_{k-1}=296$. (When four terms are removed from the end, the middle term shifts two terms to the left.)

Now $6=a_{k+1}-a_{k-1}=\left(a_{k+1}-a_{k}\right)+\left(a_{k}-a_{k-1}\right)=d+d=2 d$.

Therefore $d=3$.
If the last four terms are removed from the sequence this results in 302 shifting 2 terms to the left in the new sequence meaning that $302-296=2 d, d=3$.

Final answer: 3
Subfield: Algebra


Question: There are two increasing sequences of five consecutive integers, each of which have the property that the sum of the squares of the first three integers in the sequence equals the sum of the squares of the last two. Determine these two sequences.

Solution:
Let $n$ be the smallest integer in one of these sequences.

So we want to solve the equation $n^{2}+(n+1)^{2}+(n+2)^{2}=(n+3)^{2}+(n+4)^{2}$ (translating the given problem into an equation).

Thus $n^{2}+n^{2}+2 n+1+n^{2}+4 n+4=n^{2}+6 n+9+n^{2}+8 n+16$



$$
\begin{array}{r}
n^{2}-8 n-20=0 \\
(n-10)(n+2)=0
\end{array}
$$

So $n=10$ or $n=-2$.

Therefore, the sequences are 10, 11, 12, 13, 14 and $-2,-1,0,1,2$.

Verification 

$(-2)^{2}+(-1)^{2}+0^{2}=1^{2}+2^{2}=5$ and $10^{2}+11^{2}+12^{2}=13^{2}+14^{2}=365$

Final answer: 10,11,12,13,14,-2,-1,0,1,2
Subfield: Algebra


Question: If $f(t)=\sin \left(\pi t-\frac{\pi}{2}\right)$, what is the smallest positive value of $t$ at which $f(t)$ attains its minimum value?

Solution:
Since $t>0, \pi t-\frac{\pi}{2}>-\frac{\pi}{2}$. So $\sin \left(\pi t-\frac{\pi}{2}\right)$ first attains its minimum value when

$$
\begin{aligned}
\pi t-\frac{\pi}{2} & =\frac{3 \pi}{2} \\
t & =2 .
\end{aligned}
$$
Rewriting $f(t)$ as, $f(t)=\sin \left[\pi\left(t-\frac{1}{2}\right)\right]$.

Thus $f(t)$ has a period $\frac{2 \pi}{\pi}=2$ and appears in the diagram at the right.

Thus $f(t)$ attains its minimum at $t=2$. Note that $f(t)$ attains a minimum value at $t=0$ but since $t>0$, the required answer is $t=2$.

<img_3914>

Final answer: 2
Subfield: Geometry


Question: In the diagram, $\angle A B F=41^{\circ}, \angle C B F=59^{\circ}, D E$ is parallel to $B F$, and $E F=25$. If $A E=E C$, determine the length of $A E$, to 2 decimal places.

![](https://cdn.mathpix.com/cropped/2023_12_21_3d20c72c65b4fe186184g-1.jpg?height=626&width=331&top_left_y=305&top_left_x=1469)

Solution:
Let the length of $A E=E C$ be $x$.

Then $A F=x-25$.

In, $\triangle B C F, \frac{x+25}{B F}=\tan \left(59^{\circ}\right)$.

In $\triangle A B F, \frac{x-25}{B F}=\tan \left(41^{\circ}\right)$.

Solving for $B F$ in these two equations and equating,

$$
B F=\frac{x+25}{\tan 59^{\circ}}=\frac{x-25}{\tan 41^{\circ}}
$$

so $\quad\left(\tan 41^{\circ}\right)(x+25)=\left(\tan 59^{\circ}\right)(x-25)$

$$
\begin{aligned}
25\left(\tan 59^{\circ}+\tan 41^{\circ}\right) & =x\left(\tan 59^{\circ}-\tan 41^{\circ}\right) \\
x & =\frac{25\left(\tan 59^{\circ}+\tan 41^{\circ}\right)}{\tan 59^{\circ}-\tan 41^{\circ}} \\
x & \doteq 79.67 .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2023_12_21_204c58d31fdb1aa1cd53g-1.jpg?height=691&width=426&top_left_y=1064&top_left_x=1397)

Therefore the length of $A E$ is 79.67 .

Final answer: 79.67
Subfield: Geometry


Question: In the diagram, $\angle A B F=41^{\circ}, \angle C B F=59^{\circ}, D E$ is parallel to $B F$, and $E F=25$. If $A E=E C$, determine the length of $A E$, to 2 decimal places.

<img_3276>

Solution:
Let the length of $A E=E C$ be $x$.

Then $A F=x-25$.

In, $\triangle B C F, \frac{x+25}{B F}=\tan \left(59^{\circ}\right)$.

In $\triangle A B F, \frac{x-25}{B F}=\tan \left(41^{\circ}\right)$.

Solving for $B F$ in these two equations and equating,

$$
B F=\frac{x+25}{\tan 59^{\circ}}=\frac{x-25}{\tan 41^{\circ}}
$$

so $\quad\left(\tan 41^{\circ}\right)(x+25)=\left(\tan 59^{\circ}\right)(x-25)$

$$
\begin{aligned}
25\left(\tan 59^{\circ}+\tan 41^{\circ}\right) & =x\left(\tan 59^{\circ}-\tan 41^{\circ}\right) \\
x & =\frac{25\left(\tan 59^{\circ}+\tan 41^{\circ}\right)}{\tan 59^{\circ}-\tan 41^{\circ}} \\
x & \doteq 79.67 .
\end{aligned}
$$

<img_3420>

Therefore the length of $A E$ is 79.67 .

Final answer: 79.67
Subfield: Algebra


Question: Determine all integer values of $x$ such that $\left(x^{2}-3\right)\left(x^{2}+5\right)<0$.

Solution:
Since $x^{2} \geq 0$ for all $x, x^{2}+5>0$. Since $\left(x^{2}-3\right)\left(x^{2}+5\right)<0, x^{2}-3<0$, so $x^{2}<3$ or $-\sqrt{3}<x<\sqrt{3}$. Thus $x=-1,0,1$.

Final answer: -1,0,1
Subfield: Algebra


Question: At present, the sum of the ages of a husband and wife, $P$, is six times the sum of the ages of their children, $C$. Two years ago, the sum of the ages of the husband and wife was ten times the sum of the ages of the same children. Six years from now, it will be three times the sum of the ages of the same children. Determine the number of children.

Solution:
Let $n$ be the number of children.

At the present, $P=6 C$, where $P$ and $C$ are as given.  (1)

Two years ago, the sum of the ages of the husband and wife was $P-4$, since they were each two years younger.

Similarly, the sum of the ages of the children was $C-n(2)$ ( $n$ is the number of children).

So two years ago, $P-4=10(C-2 n)$    (2) , from the given condition.

Similarly, six years from now, $P+12=3(C+6 n)$       (3), from the given condition.

We want to solve for $n$.

Substituting (1) into each of (2) and (3),

$$
\begin{aligned}
& 6 C-4=10(C-2 n) \quad \text { or } \quad 20 n-4 C=4 \quad \text { or } \quad 5 n-C=1 \\
& 6 C+12=3(C+6 n) \quad \text { or } \quad-18 n+3 C=-12 \quad \text { or } \quad-6 n+C=-4
\end{aligned}
$$

Adding these two equations, $-n=-3$, so $n=3$.

Therefore, there were three children.

Final answer: 3
Subfield: Geometry


Question: In triangle $A B C, A B=B C=25$ and $A C=30$. The circle with diameter $B C$ intersects $A B$ at $X$ and $A C$ at $Y$. Determine the length of $X Y$.

![](https://cdn.mathpix.com/cropped/2023_12_21_3d20c72c65b4fe186184g-1.jpg?height=347&width=450&top_left_y=1881&top_left_x=1399)

Solution:
Join $B Y$. Since $B C$ is a diameter, then $\angle B Y C=90^{\circ}$. Since $A B=B C, \triangle A B C$ is isosceles and $B Y$ is an altitude in $\triangle A B C$, then $A Y=Y C=15$.

Let $\angle B A C=\theta$.

Since $\triangle A B C$ is isosceles, $\angle B C A=\theta$.

Since $B C Y X$ is cyclic, $\angle B X Y=180-\theta$ and so $\angle A X Y=\theta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_5a97850acc9642377b17g-1.jpg?height=390&width=521&top_left_y=759&top_left_x=1363)

Thus $\triangle A X Y$ is isosceles and so $X Y=A Y=15$.

Therefore $X Y=15$.
Join $B Y . \angle B Y C=90^{\circ}$ since it is inscribed in a semicircle.

$\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.

Therefore $B Y=\sqrt{25^{2}-15^{2}}=20$.

Join $C X . \angle C X B=90^{\circ}$ since it is also inscribed in a semicircle.

![](https://cdn.mathpix.com/cropped/2023_12_21_5a97850acc9642377b17g-1.jpg?height=398&width=523&top_left_y=1362&top_left_x=1365)

The area of $\triangle A B C$ is

$$
\begin{aligned}
\frac{1}{2}(A C)(B Y) & =\frac{1}{2}(A B)(C X) \\
\frac{1}{2}(30)(20) & =\frac{1}{2}(25)(C X) \\
C X & =\frac{600}{25}=24 .
\end{aligned}
$$

From $\triangle A B Y$ we conclude that $\cos \angle A B Y=\frac{B Y}{A B}=\frac{20}{25}=\frac{4}{5}$.

In $\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \cos \angle X B Y$.

Now (by Pythagoras $\triangle B X C$ ),



$$
\begin{aligned}
B X^{2} & =B C^{2}-C X^{2} \\
& =25^{2}-24^{2} \\
& =49 \\
B X & =7 .
\end{aligned}
$$

Therefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \frac{4}{5}$

$$
=49+400-224
$$

$$
=225 \text {. }
$$

Therefore $X Y=15$.

Final answer: 15
Subfield: Geometry


Question: In triangle $A B C, A B=B C=25$ and $A C=30$. The circle with diameter $B C$ intersects $A B$ at $X$ and $A C$ at $Y$. Determine the length of $X Y$.

<img_3898>

Solution:
Join $B Y$. Since $B C$ is a diameter, then $\angle B Y C=90^{\circ}$. Since $A B=B C, \triangle A B C$ is isosceles and $B Y$ is an altitude in $\triangle A B C$, then $A Y=Y C=15$.

Let $\angle B A C=\theta$.

Since $\triangle A B C$ is isosceles, $\angle B C A=\theta$.

Since $B C Y X$ is cyclic, $\angle B X Y=180-\theta$ and so $\angle A X Y=\theta$.

<img_3475>

Thus $\triangle A X Y$ is isosceles and so $X Y=A Y=15$.

Therefore $X Y=15$.
Join $B Y . \angle B Y C=90^{\circ}$ since it is inscribed in a semicircle.

$\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.

Therefore $B Y=\sqrt{25^{2}-15^{2}}=20$.

Join $C X . \angle C X B=90^{\circ}$ since it is also inscribed in a semicircle.

<img_3739>

The area of $\triangle A B C$ is

$$
\begin{aligned}
\frac{1}{2}(A C)(B Y) & =\frac{1}{2}(A B)(C X) \\
\frac{1}{2}(30)(20) & =\frac{1}{2}(25)(C X) \\
C X & =\frac{600}{25}=24 .
\end{aligned}
$$

From $\triangle A B Y$ we conclude that $\cos \angle A B Y=\frac{B Y}{A B}=\frac{20}{25}=\frac{4}{5}$.

In $\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \cos \angle X B Y$.

Now (by Pythagoras $\triangle B X C$ ),



$$
\begin{aligned}
B X^{2} & =B C^{2}-C X^{2} \\
& =25^{2}-24^{2} \\
& =49 \\
B X & =7 .
\end{aligned}
$$

Therefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \frac{4}{5}$

$$
=49+400-224
$$

$$
=225 \text {. }
$$

Therefore $X Y=15$.

Final answer: 15
Subfield: Algebra


Question: What is the value of $x$ such that $\log _{2}\left(\log _{2}(2 x-2)\right)=2$ ?

Solution:
$$
\begin{aligned}
\log _{2}\left(\log _{2}(2 x-2)\right) & =2 \\
\log _{2}(2 x-2) & =2^{2} \\
2 x-2 & =2^{\left(2^{2}\right)} \\
2 x-2 & =2^{4} \\
2 x-2 & =16 \\
2 x & =18 \\
x & =9
\end{aligned}
$$

Final answer: 9
Subfield: Algebra


Question: Let $f(x)=2^{k x}+9$, where $k$ is a real number. If $f(3): f(6)=1: 3$, determine the value of $f(9)-f(3)$.

Solution:
From the given condition,

$$
\begin{aligned}
\frac{f(3)}{f(6)}=\frac{2^{3 k}+9}{2^{6 k}+9} & =\frac{1}{3} \\
3\left(2^{3 k}+9\right) & =2^{6 k}+9 \\
0 & =2^{6 k}-3\left(2^{3 k}\right)-18 .
\end{aligned}
$$

We treat this as a quadratic equation in the variable $x=2^{3 k}$, so

$$
\begin{aligned}
& 0=x^{2}-3 x-18 \\
& 0=(x-6)(x+3)
\end{aligned}
$$

Therefore, $2^{3 k}=6$ or $2^{3 k}=-3$. Since $2^{a}>0$ for any $a$, then $2^{3 k} \neq-3$.

So $2^{3 k}=6$. We could solve for $k$ here, but this is unnecessary.



We calculate $f(9)-f(3)=\left(2^{9 k}+9\right)-\left(2^{3 k}+9\right)$

$$
\begin{aligned}
& =2^{9 k}-2^{3 k} \\
& =\left(2^{3 k}\right)^{3}-2^{3 k} \\
& =6^{3}-6 \\
& =210 .
\end{aligned}
$$

Therefore $f(9)-f(3)=210$.

Final answer: 210
Subfield: Algebra


Question: Determine, with justification, all values of $k$ for which $y=x^{2}-4$ and $y=2|x|+k$ do not intersect.

Solution:
Since each of these two graphs is symmetric about the $y$-axis (i.e. both are even functions), then we only need to find $k$ so that there are no points of intersection with $x \geq 0$.

So let $x \geq 0$ and consider the intersection between $y=2 x+k$ and $y=x^{2}-4$.

Equating, we have, $2 x+k=x^{2}-4$.

Rearranging, we want $x^{2}-2 x-(k+4)=0$ to have no solutions.



For no solutions, the discriminant is negative, i.e.

$$
\begin{aligned}
20+4 k & <0 \\
4 k & <-20 \\
k & <-5 .
\end{aligned}
$$

So $y=x^{2}-4$ and $y=2|x|+k$ have no intersection points when $k<-5$.



Final answer: (-\infty,-5)
Subfield: Geometry


Question: In the diagram, $\angle E A D=90^{\circ}, \angle A C D=90^{\circ}$, and $\angle A B C=90^{\circ}$. Also, $E D=13, E A=12$, $D C=4$, and $C B=2$. Determine the length of $A B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_be0b235a4c1314f85f7dg-1.jpg?height=271&width=556&top_left_y=2336&top_left_x=1251)

Solution:
By the Pythagorean Theorem in $\triangle E A D$, we have $E A^{2}+A D^{2}=E D^{2}$ or $12^{2}+A D^{2}=13^{2}$, and so $A D=\sqrt{169-144}=5$, since $A D>0$.

By the Pythagorean Theorem in $\triangle A C D$, we have $A C^{2}+C D^{2}=A D^{2}$ or $A C^{2}+4^{2}=5^{2}$, and so $A C=\sqrt{25-16}=3$, since $A C>0$.

(We could also have determined the lengths of $A D$ and $A C$ by recognizing 3-4-5 and 5-12-13 right-angled triangles.)

By the Pythagorean Theorem in $\triangle A B C$, we have $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+2^{2}=3^{2}$, and so $A B=\sqrt{9-4}=\sqrt{5}$, since $A B>0$.

Final answer: $\sqrt{5}$