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


Question: In the diagram, $A B$ and $B C$ are chords of the circle with $A B<B C$. If $D$ is the point on the circle such that $A D$ is perpendicular to $B C$ and $E$ is the point on the circle such that $D E$ is parallel to $B C$, carefully prove, explaining all steps, that $\angle E A C+\angle A B C=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_32eb610075db476eabecg-1.jpg?height=442&width=455&top_left_y=375&top_left_x=1255)

Solution:
Join $A$ to $E$ and $C$, and $B$ to $E$.

![](https://cdn.mathpix.com/cropped/2023_12_21_f51119572293c3cfd163g-1.jpg?height=439&width=450&top_left_y=751&top_left_x=943)

Since $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\angle A D E=90^{\circ}$.

Therefore, $A E$ is a diameter.

Now $\angle E A C=\angle E B C$ since both are subtended by $E C$.

Therefore, $\angle E A C+\angle A B C=\angle E B C+\angle A B C=\angle E B A$ which is indeed equal to $90^{\circ}$ as required, since $A E$ is a diameter.
Join $A$ to $E$ and $C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_f51119572293c3cfd163g-1.jpg?height=450&width=444&top_left_y=1686&top_left_x=946)

Since $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\angle A D E=90^{\circ}$.

Therefore, $A E$ is a diameter.

Thus, $\angle E C A=90^{\circ}$.

Now $\angle A B C=\angle A E C$ since both are subtended by $A C$.

Now $\angle E A C+\angle A B C=\angle E A C+\angle A E C=180^{\circ}-\angle E C A$ using the sum of the angles in $\triangle A E C$.

But $\angle E C A=90^{\circ}$, so $\angle E A C+\angle A E C=90^{\circ}$.
Join $A$ to $E$ and $C$, and $C$ to $D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_3adcbcafa02c5a2fde09g-1.jpg?height=447&width=444&top_left_y=281&top_left_x=946)

Since $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\angle A D E=90^{\circ}$.

Therefore, $A E$ is a diameter.

Now $\angle A B C=\angle A D C$ since both are subtended by $A C$.

Also $\angle E A C=\angle E D C$ since both are subtended by $E C$.

So $\angle E A C+\angle A B C=\angle E D C+\angle A D C=\angle A D E=90^{\circ}$.

Final answer: 证明题，略
Subfield: Geometry


Question: In the diagram, $A B$ and $B C$ are chords of the circle with $A B<B C$. If $D$ is the point on the circle such that $A D$ is perpendicular to $B C$ and $E$ is the point on the circle such that $D E$ is parallel to $B C$, carefully prove, explaining all steps, that $\angle E A C+\angle A B C=90^{\circ}$.

<img_3926>

Solution:
Join $A$ to $E$ and $C$, and $B$ to $E$.

<img_3770>

Since $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\angle A D E=90^{\circ}$.

Therefore, $A E$ is a diameter.

Now $\angle E A C=\angle E B C$ since both are subtended by $E C$.

Therefore, $\angle E A C+\angle A B C=\angle E B C+\angle A B C=\angle E B A$ which is indeed equal to $90^{\circ}$ as required, since $A E$ is a diameter.
Join $A$ to $E$ and $C$.

<img_3932>

Since $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\angle A D E=90^{\circ}$.

Therefore, $A E$ is a diameter.

Thus, $\angle E C A=90^{\circ}$.

Now $\angle A B C=\angle A E C$ since both are subtended by $A C$.

Now $\angle E A C+\angle A B C=\angle E A C+\angle A E C=180^{\circ}-\angle E C A$ using the sum of the angles in $\triangle A E C$.

But $\angle E C A=90^{\circ}$, so $\angle E A C+\angle A E C=90^{\circ}$.
Join $A$ to $E$ and $C$, and $C$ to $D$.

<img_3930>

Since $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\angle A D E=90^{\circ}$.

Therefore, $A E$ is a diameter.

Now $\angle A B C=\angle A D C$ since both are subtended by $A C$.

Also $\angle E A C=\angle E D C$ since both are subtended by $E C$.

So $\angle E A C+\angle A B C=\angle E D C+\angle A D C=\angle A D E=90^{\circ}$.
Subfield: Algebra


Question: Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
Determine all real numbers $k$ for which $f(x)$ is constant for all values of $x$.

Solution:
Since $\sin ^{2} x+\cos ^{2} x=1$, then $\cos ^{2} x=1-\sin ^{2} x$, so

$$
\begin{aligned}
f(x) & =\sin ^{6} x+\left(1-\sin ^{2} x\right)^{3}+k\left(\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}\right) \\
& =\sin ^{6} x+1-3 \sin ^{2} x+3 \sin ^{4} x-\sin ^{6} x+k\left(\sin ^{4} x+1-2 \sin ^{2} x+\sin ^{4} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x+(3+2 k) \sin ^{4} x
\end{aligned}
$$

Therefore, if $3+2 k=0$ or $k=-\frac{3}{2}$, then $f(x)=1+k=-\frac{1}{2}$ for all $x$ and so is constant. (If $k \neq-\frac{3}{2}$, then we get

$$
\begin{aligned}
f(0) & =1+k \\
f\left(\frac{1}{4} \pi\right) & =(1+k)-(3+2 k)\left(\frac{1}{2}\right)+(3+2 k)\left(\frac{1}{4}\right)=\frac{1}{4}+\frac{1}{2} k \\
f\left(\frac{1}{6} \pi\right) & =(1+k)-(3+2 k)\left(\frac{1}{4}\right)+(3+2 k)\left(\frac{1}{16}\right)=\frac{7}{16}+\frac{5}{8} k
\end{aligned}
$$

which cannot be all equal for any single value of $k$, so $f(x)$ is not constant if $k \neq-\frac{3}{2}$.)
Since $\sin ^{2} x+\cos ^{2} x=1$, then

$$
\begin{aligned}
f(x) & =\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)+k\left(\sin ^{4} x+\cos ^{4} x\right) \\
& =\left(\sin ^{4}+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-3 \sin ^{2} x \cos ^{2} x\right) \\
& \quad k\left(\sin ^{4} x+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-2 \sin ^{2} x \cos ^{2} x\right) \\
& =\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x\right)+k\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right) \\
& =1-3 \sin ^{2} x \cos ^{2} x+k\left(1-2 \sin ^{2} x \cos ^{2} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x \cos ^{2} x
\end{aligned}
$$

Therefore, if $3+2 k=0$ or $k=-\frac{3}{2}$, then $f(x)=1+k=-\frac{1}{2}$ for all $x$ and so is constant.
For $f(x)$ to be constant, we need $f^{\prime}(x)=0$ for all values of $x$.

Calculating using the Chain Rule,

$$
\begin{aligned}
f^{\prime}(x) & =6 \sin ^{5} x \cos x-6 \cos ^{5} x \sin x+k\left(4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x\right) \\
& =2 \sin x \cos x\left(3\left(\sin ^{4} x-\cos ^{4} x\right)+2 k\left(\sin ^{2} x-\cos ^{2} x\right)\right) \\
& =2 \sin x \cos x\left(\sin ^{2} x-\cos ^{2} x\right)\left(3\left(\sin ^{2} x+\cos ^{2} x\right)+2 k\right) \\
& =2 \sin x \cos x\left(\sin ^{2} x-\cos ^{2} x\right)(3+2 k)
\end{aligned}
$$

If $3+2 k=0$ or $k=-\frac{3}{2}$, then $f^{\prime}(x)=0$ for all $x$, so $f(x)$ is constant.

(If $3+2 k \neq 0$, then choosing $x=\frac{1}{6} \pi$ for example gives $f^{\prime}(x) \neq 0$ so $f(x)$ is not constant.)

Final answer: $-\frac{3}{2}$
Subfield: Algebra


Question: Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
If $k=-0.7$, determine all solutions to the equation $f(x)=0$.

Solution:
Since $\sin ^{2} x+\cos ^{2} x=1$, then $\cos ^{2} x=1-\sin ^{2} x$, so

$$
\begin{aligned}
f(x) & =\sin ^{6} x+\left(1-\sin ^{2} x\right)^{3}+k\left(\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}\right) \\
& =\sin ^{6} x+1-3 \sin ^{2} x+3 \sin ^{4} x-\sin ^{6} x+k\left(\sin ^{4} x+1-2 \sin ^{2} x+\sin ^{4} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x+(3+2 k) \sin ^{4} x
\end{aligned}
$$

Now, we have

$$
f(x)=(1+k)-(3+2 k) \sin ^{2} x+(3+2 k) \sin ^{4} x
$$

and so we want to solve

$$
\begin{array}{r}
0.3-(1.6) \sin ^{2} x+(1.6) \sin ^{4} x=0 \\
16 \sin ^{4} x-16 \sin ^{2} x+3=0 \\
\left(4 \sin ^{2} x-3\right)\left(4 \sin ^{2} x-1\right)=0
\end{array}
$$

Therefore, $\sin ^{2} x=\frac{1}{4}, \frac{3}{4}$, and so $\sin x= \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}$.

Therefore,

$$
x=\frac{1}{6} \pi+2 \pi k, \frac{5}{6} \pi+2 \pi k, \frac{7}{6} \pi+2 \pi k, \frac{11}{6} \pi+2 \pi k, \frac{1}{3} \pi+2 \pi k, \frac{2}{3} \pi+2 \pi k, \frac{4}{3} \pi+2 \pi k, \frac{5}{3} \pi+2 \pi k
$$

for $k \in \mathbb{Z}$.
Since $\sin ^{2} x+\cos ^{2} x=1$, then

$$
\begin{aligned}
f(x) & =\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)+k\left(\sin ^{4} x+\cos ^{4} x\right) \\
& =\left(\sin ^{4}+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-3 \sin ^{2} x \cos ^{2} x\right) \\
& \quad k\left(\sin ^{4} x+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-2 \sin ^{2} x \cos ^{2} x\right) \\
& =\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x\right)+k\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right) \\
& =1-3 \sin ^{2} x \cos ^{2} x+k\left(1-2 \sin ^{2} x \cos ^{2} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x \cos ^{2} x
\end{aligned}
$$

Now, we have

$$
f(x)=(1+k)-(3+2 k) \sin ^{2} x \cos ^{2} x
$$

Using the fact that $\sin 2 x=2 \sin x \cos x$, we can further simplify $f(x)$ to

$$
f(x)=(1+k)-\frac{1}{4}(3+2 k) \sin ^{2} 2 x
$$



and so we want to solve

$$
\begin{aligned}
0.3-\frac{1}{4}(1.6) \sin ^{2} 2 x & =0 \\
4 \sin ^{2} 2 x & =3 \\
\sin ^{2} 2 x & =\frac{3}{4}
\end{aligned}
$$

and so $\sin 2 x= \pm \frac{\sqrt{3}}{2}$.

Therefore,

$$
2 x=\frac{1}{3} \pi+2 \pi k, \frac{2}{3} \pi+2 \pi k, \frac{4}{3} \pi+2 \pi k, \frac{5}{3} \pi+2 \pi k
$$

for $k \in \mathbb{Z}$, and so

$$
x=\frac{1}{6} \pi+\pi k, \frac{1}{3} \pi+\pi k, \frac{2}{3} \pi+\pi k, \frac{5}{6} \pi+\pi k
$$

for $k \in \mathbb{Z}$.

Final answer: $x=\frac{1}{6} \pi+\pi k, \frac{1}{3} \pi+\pi k, \frac{2}{3} \pi+\pi k, \frac{5}{6} \pi+\pi k$
Subfield: Geometry


Question: Hexagon $A B C D E F$ has vertices $A(0,0), B(4,0), C(7,2), D(7,5), E(3,5)$, $F(0,3)$. What is the area of hexagon $A B C D E F$ ?

Solution:
Let $P$ be the point with coordinates $(7,0)$ and let $Q$ be the point with coordinates $(0,5)$.

<img_4025>

Then $A P D Q$ is a rectangle with width 7 and height 5 , and so it has area $7 \cdot 5=35$.

Hexagon $A B C D E F$ is formed by removing two triangles from rectangle $A P D Q$, namely $\triangle B P C$ and $\triangle E Q F$.

Each of $\triangle B P C$ and $\triangle E Q F$ is right-angled, because each shares an angle with rectangle $A P D Q$.

Each of $\triangle B P C$ and $\triangle E Q F$ has a base of length 3 and a height of 2.

Thus, their combined area is $2 \cdot \frac{1}{2} \cdot 3 \cdot 2=6$.

This means that the area of hexagon $A B C D E F$ is $35-6=29$.

Final answer: 29
Subfield: Geometry


Question: In the diagram, $\triangle P Q S$ is right-angled at $P$ and $\triangle Q R S$ is right-angled at $Q$. Also, $P Q=x, Q R=8, R S=x+8$, and $S P=x+3$ for some real number $x$. Determine all possible values of the perimeter of quadrilateral $P Q R S$.

<img_3302>

Solution:
Since $\triangle P Q S$ is right-angled at $P$, then by the Pythagorean Theorem,

$$
S Q^{2}=S P^{2}+P Q^{2}=(x+3)^{2}+x^{2}
$$

Since $\triangle Q R S$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain

$$
\begin{aligned}
R S^{2} & =S Q^{2}+Q R^{2} \\
(x+8)^{2} & =\left((x+3)^{2}+x^{2}\right)+8^{2} \\
x^{2}+16 x+64 & =x^{2}+6 x+9+x^{2}+64 \\
0 & =x^{2}-10 x+9 \\
0 & =(x-1)(x-9)
\end{aligned}
$$

and so $x=1$ or $x=9$.

(We can check that if $x=1, \triangle P Q S$ has sides of lengths 4,1 and $\sqrt{17}$ and $\triangle Q R S$ has sides of lengths $\sqrt{17}, 8$ and 9 , both of which are right-angled, and if $x=9, \triangle P Q S$ has sides of lengths 12,9 and 15 and $\triangle Q R S$ has sides of lengths 15,8 and 17 , both of which are right-angled.)

In terms of $x$, the perimeter of $P Q R S$ is $x+8+(x+8)+(x+3)=3 x+19$.

Thus, the possible perimeters of $P Q R S$ are 22 (when $x=1$ ) and 46 (when $x=9$ ).

Final answer: 22,46
Subfield: Geometry


Question: In the diagram, $\triangle P Q S$ is right-angled at $P$ and $\triangle Q R S$ is right-angled at $Q$. Also, $P Q=x, Q R=8, R S=x+8$, and $S P=x+3$ for some real number $x$. Determine all possible values of the perimeter of quadrilateral $P Q R S$.

![](https://cdn.mathpix.com/cropped/2023_12_21_1ca8be37cce6eab426b2g-1.jpg?height=336&width=458&top_left_y=1244&top_left_x=1256)

Solution:
Since $\triangle P Q S$ is right-angled at $P$, then by the Pythagorean Theorem,

$$
S Q^{2}=S P^{2}+P Q^{2}=(x+3)^{2}+x^{2}
$$

Since $\triangle Q R S$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain

$$
\begin{aligned}
R S^{2} & =S Q^{2}+Q R^{2} \\
(x+8)^{2} & =\left((x+3)^{2}+x^{2}\right)+8^{2} \\
x^{2}+16 x+64 & =x^{2}+6 x+9+x^{2}+64 \\
0 & =x^{2}-10 x+9 \\
0 & =(x-1)(x-9)
\end{aligned}
$$

and so $x=1$ or $x=9$.

(We can check that if $x=1, \triangle P Q S$ has sides of lengths 4,1 and $\sqrt{17}$ and $\triangle Q R S$ has sides of lengths $\sqrt{17}, 8$ and 9 , both of which are right-angled, and if $x=9, \triangle P Q S$ has sides of lengths 12,9 and 15 and $\triangle Q R S$ has sides of lengths 15,8 and 17 , both of which are right-angled.)

In terms of $x$, the perimeter of $P Q R S$ is $x+8+(x+8)+(x+3)=3 x+19$.

Thus, the possible perimeters of $P Q R S$ are 22 (when $x=1$ ) and 46 (when $x=9$ ).

Final answer: 22,46
Subfield: Algebra


Question: A list $a_{1}, a_{2}, a_{3}, a_{4}$ of rational numbers is defined so that if one term is equal to $r$, then the next term is equal to $1+\frac{1}{1+r}$. For example, if $a_{3}=\frac{41}{29}$, then $a_{4}=1+\frac{1}{1+(41 / 29)}=\frac{99}{70}$. If $a_{3}=\frac{41}{29}$, what is the value of $a_{1} ?$

Solution:
If $r$ is a term in the sequence and $s$ is the next term, then $s=1+\frac{1}{1+r}$.

This means that $s-1=\frac{1}{1+r}$ and so $\frac{1}{s-1}=1+r$ which gives $r=\frac{1}{s-1}-1$.

Therefore, since $a_{3}=\frac{41}{29}$, then

$$
a_{2}=\frac{1}{a_{3}-1}-1=\frac{1}{(41 / 29)-1}-1=\frac{1}{12 / 29}-1=\frac{29}{12}-1=\frac{17}{12}
$$

Further, since $a_{2}=\frac{17}{12}$, then

$$
a_{1}=\frac{1}{a_{2}-1}-1=\frac{1}{(17 / 12)-1}-1=\frac{1}{5 / 12}-1=\frac{12}{5}-1=\frac{7}{5}
$$

Final answer: $\frac{7}{5}$
Subfield: Geometry


Question: A hollow cylindrical tube has a radius of $10 \mathrm{~mm}$ and a height of $100 \mathrm{~mm}$. The tube sits flat on one of its circular faces on a horizontal table. The tube is filled with water to a depth of $h \mathrm{~mm}$. A solid cylindrical rod has a radius of $2.5 \mathrm{~mm}$ and a height of $150 \mathrm{~mm}$. The rod is inserted into the tube so that one of its circular faces sits flat on the bottom of the tube. The height of the water in the tube is now $64 \mathrm{~mm}$. Determine the value of $h$.

Solution:
Initially, the water in the hollow tube forms a cylinder with radius $10 \mathrm{~mm}$ and height $h \mathrm{~mm}$. Thus, the volume of the water is $\pi(10 \mathrm{~mm})^{2}(h \mathrm{~mm})=100 \pi h \mathrm{~mm}^{3}$.

After the rod is inserted, the level of the water rises to $64 \mathrm{~mm}$. Note that this does not overflow the tube, since the tube's height is $100 \mathrm{~mm}$.

Up to the height of the water, the tube is a cylinder with radius $10 \mathrm{~mm}$ and height 64 mm.

Thus, the volume of the tube up to the height of the water is

$$
\pi(10 \mathrm{~mm})^{2}(64 \mathrm{~mm})=6400 \pi \mathrm{mm}^{3}
$$

This volume consists of the water that is in the tube (whose volume, which has not changed, is $100 \pi h \mathrm{~mm}^{3}$ ) and the rod up to a height of $64 \mathrm{~mm}$.
<img_3180>

Since the radius of the rod is $2.5 \mathrm{~mm}$, the volume of the rod up to a height of $64 \mathrm{~mm}$ is $\pi(2.5 \mathrm{~mm})^{2}(64 \mathrm{~mm})=400 \pi \mathrm{mm}^{3}$.

Comparing volumes, $6400 \pi \mathrm{mm}^{3}=100 \pi h \mathrm{~mm}^{3}+400 \pi \mathrm{mm}^{3}$ and so $100 h=6000$ which gives $h=60$.

Final answer: 60
Subfield: Algebra


Question: A function $f$ has the property that $f\left(\frac{2 x+1}{x}\right)=x+6$ for all real values of $x \neq 0$. What is the value of $f(4) ?$

Solution:
We note that $\frac{2 x+1}{x}=\frac{2 x}{x}+\frac{1}{x}=2+\frac{1}{x}$.

Therefore, $\frac{2 x+1}{x}=4$ exactly when $2+\frac{1}{x}=4$ or $\frac{1}{x}=2$ and so $x=\frac{1}{2}$.

Alternatively, we could solve $\frac{2 x+1}{x}=4$ directly to obtain $2 x+1=4 x$, which gives $2 x=1$ and so $x=\frac{1}{2}$.

Thus, to determine the value of $f(4)$, we substitute $x=\frac{1}{2}$ into the given equation $f\left(\frac{2 x+1}{x}\right)=x+6$ and obtain $f(4)=\frac{1}{2}+6=\frac{13}{2}$.

Final answer: $\frac{13}{2}$
Subfield: Algebra


Question: Determine all real numbers $a, b$ and $c$ for which the graph of the function $y=\log _{a}(x+b)+c$ passes through the points $P(3,5), Q(5,4)$ and $R(11,3)$.

Solution:
Since the graph passes through $(3,5),(5,4)$ and $(11,3)$, we can substitute these three points and obtain the following three equations:

$$
\begin{aligned}
& 5=\log _{a}(3+b)+c \\
& 4=\log _{a}(5+b)+c \\
& 3=\log _{a}(11+b)+c
\end{aligned}
$$

Subtracting the second equation from the first and the third equation from the second, we obtain:

$$
\begin{aligned}
& 1=\log _{a}(3+b)-\log _{a}(5+b) \\
& 1=\log _{a}(5+b)-\log _{a}(11+b)
\end{aligned}
$$

Equating right sides and manipulating, we obtain the following equivalent equations:

$$
\begin{aligned}
\log _{a}(5+b)-\log _{a}(11+b) & =\log _{a}(3+b)-\log _{a}(5+b) \\
2 \log _{a}(5+b) & =\log _{a}(3+b)+\log _{a}(11+b) \\
\log _{a}\left((5+b)^{2}\right) & =\log _{a}((3+b)(11+b)) \quad(\text { using log laws }) \\
(5+b)^{2} & =(3+b)(11+b) \quad \text { (raising both sides to the power of } a) \\
25+10 b+b^{2} & =33+14 b+b^{2} \quad \\
-8 & =4 b \\
b & =-2
\end{aligned}
$$

Since $b=-2$, the equation $1=\log _{a}(3+b)-\log _{a}(5+b)$ becomes $1=\log _{a} 1-\log _{a} 3$.

Since $\log _{a} 1=0$ for every admissible value of $a$, then $\log _{a} 3=-1$ which gives $a=3^{-1}=\frac{1}{3}$.

Finally, the equation $5=\log _{a}(3+b)+c$ becomes $5=\log _{1 / 3}(1)+c$ and so $c=5$.

Therefore, $a=\frac{1}{3}, b=-2$, and $c=5$, which gives $y=\log _{1 / 3}(x-2)+5$.

Checking:

- When $x=3$, we obtain $y=\log _{1 / 3}(3-2)+5=\log _{1 / 3} 1+5=0+5=5$.
- When $x=5$, we obtain $y=\log _{1 / 3}(5-2)+5=\log _{1 / 3} 3+5=-1+5=4$.
- When $x=11$, we obtain $y=\log _{1 / 3}(11-2)+5=\log _{1 / 3} 9+5=-2+5=3$.

Final answer: $\frac{1}{3},-2,5$
Subfield: Algebra


Question: A computer is programmed to choose an integer between 1 and 99, inclusive, so that the probability that it selects the integer $x$ is equal to $\log _{100}\left(1+\frac{1}{x}\right)$. Suppose that the probability that $81 \leq x \leq 99$ is equal to 2 times the probability that $x=n$ for some integer $n$. What is the value of $n$ ?

Solution:
The probability that the integer $n$ is chosen is $\log _{100}\left(1+\frac{1}{n}\right)$.

The probability that an integer between 81 and 99 , inclusive, is chosen equals the sum of the probabilities that the integers $81,82, \ldots, 98,99$ are selected, which equals

$$
\log _{100}\left(1+\frac{1}{81}\right)+\log _{100}\left(1+\frac{1}{82}\right)+\cdots+\log _{100}\left(1+\frac{1}{98}\right)+\log _{100}\left(1+\frac{1}{99}\right)
$$

Since the second probability equals 2 times the first probability, the following equations are equivalent:

$$
\begin{aligned}
\log _{100}\left(1+\frac{1}{81}\right)+\log _{100}\left(1+\frac{1}{82}\right)+\cdots+\log _{100}\left(1+\frac{1}{98}\right)+\log _{100}\left(1+\frac{1}{99}\right) & =2 \log _{100}\left(1+\frac{1}{n}\right) \\
\log _{100}\left(\frac{82}{81}\right)+\log _{100}\left(\frac{83}{82}\right)+\cdots+\log _{100}\left(\frac{99}{98}\right)+\log _{100}\left(\frac{100}{99}\right) & =2 \log _{100}\left(1+\frac{1}{n}\right)
\end{aligned}
$$

Using logarithm laws, these equations are further equivalent to

$$
\begin{aligned}
\log _{100}\left(\frac{82}{81} \cdot \frac{83}{82} \cdots \cdot \frac{99}{98} \cdot \frac{100}{99}\right) & =\log _{100}\left(1+\frac{1}{n}\right)^{2} \\
\log _{100}\left(\frac{100}{81}\right) & =\log _{100}\left(1+\frac{1}{n}\right)^{2}
\end{aligned}
$$

Since logarithm functions are invertible, we obtain $\frac{100}{81}=\left(1+\frac{1}{n}\right)^{2}$.

Since $n>0$, then $1+\frac{1}{n}=\sqrt{\frac{100}{81}}=\frac{10}{9}$, and so $\frac{1}{n}=\frac{1}{9}$, which gives $n=9$.

Final answer: 9
Subfield: Geometry


Question: In the diagram, $\triangle A B D$ has $C$ on $B D$. Also, $B C=2, C D=1, \frac{A C}{A D}=\frac{3}{4}$, and $\cos (\angle A C D)=-\frac{3}{5}$. Determine the length of $A B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_31c9558111b63654ac46g-1.jpg?height=271&width=417&top_left_y=859&top_left_x=1258)

Solution:
Since $\frac{A C}{A D}=\frac{3}{4}$, then we let $A C=3 t$ and $A D=4 t$ for some real number $t>0$.

![](https://cdn.mathpix.com/cropped/2023_12_21_8df3bd851dc8f5cdf24eg-1.jpg?height=274&width=418&top_left_y=1592&top_left_x=951)

Using the cosine law in $\triangle A C D$, the following equations are equivalent:

$$
\begin{aligned}
A D^{2} & =A C^{2}+C D^{2}-2 \cdot A C \cdot C D \cdot \cos (\angle A C D) \\
(4 t)^{2} & =(3 t)^{2}+1^{2}-2(3 t)(1)\left(-\frac{3}{5}\right) \\
16 t^{2} & =9 t^{2}+1+\frac{18}{5} t \\
80 t^{2} & =45 t^{2}+5+18 t \\
35 t^{2}-18 t-5 & =0 \\
(7 t-5)(5 t+1) & =0
\end{aligned}
$$

Since $t>0$, then $t=\frac{5}{7}$.

Thus, $A C=3 t=\frac{15}{7}$.

Using the cosine law in $\triangle A C B$ and noting that

$$
\cos (\angle A C B)=\cos \left(180^{\circ}-\angle A C D\right)=-\cos (\angle A C D)=\frac{3}{5}
$$



the following equations are equivalent:

$$
\begin{aligned}
A B^{2} & =A C^{2}+B C^{2}-2 \cdot A C \cdot B C \cdot \cos (\angle A C B) \\
& =\left(\frac{15}{7}\right)^{2}+2^{2}-2\left(\frac{15}{7}\right)(2)\left(\frac{3}{5}\right) \\
& =\frac{225}{49}+4-\frac{36}{7} \\
& =\frac{225}{49}+\frac{196}{49}-\frac{252}{49} \\
& =\frac{169}{49}
\end{aligned}
$$

Since $A B>0$, then $A B=\frac{13}{7}$.

Final answer: $\frac{13}{7}$
Subfield: Geometry


Question: In the diagram, $\triangle A B D$ has $C$ on $B D$. Also, $B C=2, C D=1, \frac{A C}{A D}=\frac{3}{4}$, and $\cos (\angle A C D)=-\frac{3}{5}$. Determine the length of $A B$.

<img_4045>

Solution:
Since $\frac{A C}{A D}=\frac{3}{4}$, then we let $A C=3 t$ and $A D=4 t$ for some real number $t>0$.

<img_3309>

Using the cosine law in $\triangle A C D$, the following equations are equivalent:

$$
\begin{aligned}
A D^{2} & =A C^{2}+C D^{2}-2 \cdot A C \cdot C D \cdot \cos (\angle A C D) \\
(4 t)^{2} & =(3 t)^{2}+1^{2}-2(3 t)(1)\left(-\frac{3}{5}\right) \\
16 t^{2} & =9 t^{2}+1+\frac{18}{5} t \\
80 t^{2} & =45 t^{2}+5+18 t \\
35 t^{2}-18 t-5 & =0 \\
(7 t-5)(5 t+1) & =0
\end{aligned}
$$

Since $t>0$, then $t=\frac{5}{7}$.

Thus, $A C=3 t=\frac{15}{7}$.

Using the cosine law in $\triangle A C B$ and noting that

$$
\cos (\angle A C B)=\cos \left(180^{\circ}-\angle A C D\right)=-\cos (\angle A C D)=\frac{3}{5}
$$



the following equations are equivalent:

$$
\begin{aligned}
A B^{2} & =A C^{2}+B C^{2}-2 \cdot A C \cdot B C \cdot \cos (\angle A C B) \\
& =\left(\frac{15}{7}\right)^{2}+2^{2}-2\left(\frac{15}{7}\right)(2)\left(\frac{3}{5}\right) \\
& =\frac{225}{49}+4-\frac{36}{7} \\
& =\frac{225}{49}+\frac{196}{49}-\frac{252}{49} \\
& =\frac{169}{49}
\end{aligned}
$$

Since $A B>0$, then $A B=\frac{13}{7}$.

Final answer: $\frac{13}{7}$
Subfield: Number Theory


Question: Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=151&width=891&top_left_y=453&top_left_x=644)

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
Determine the sum of the terms in the $(4,2)$-sawtooth sequence.

Solution:
The $(4,2)$-sawtooth sequence consists of the terms

$$
1, \quad 2,3,4,3,2,1, \quad 2,3,4,3,2,1
$$

whose sum is 31 .

Final answer: 31
Subfield: Number Theory


Question: Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

<img_3354>

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
Determine the sum of the terms in the $(4,2)$-sawtooth sequence.

Solution:
The $(4,2)$-sawtooth sequence consists of the terms

$$
1, \quad 2,3,4,3,2,1, \quad 2,3,4,3,2,1
$$

whose sum is 31 .

Final answer: 31