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


Question: A snail's shell is formed from six triangular sections, as shown. Each triangle has interior angles of $30^{\circ}, 60^{\circ}$ and $90^{\circ}$. If $A B$ has a length of $1 \mathrm{~cm}$, what is the length of $A H$, in $\mathrm{cm}$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_6612ae75d8ef169f4be5g-1.jpg?height=288&width=523&top_left_y=1905&top_left_x=1256)

Solution:
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, the ratio of the side opposite the $90^{\circ}$ to the side opposite the $60^{\circ}$ angle is $2: \sqrt{3}$.

Note that each of $\triangle A B C, \triangle A C D, \triangle A D E, \triangle A E F, \triangle A F G$, and $\triangle A G H$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $\frac{A H}{A G}=\frac{A G}{A F}=\frac{A F}{A E}=\frac{A E}{A D}=\frac{A D}{A C}=\frac{A C}{A B}=\frac{2}{\sqrt{3}}$.

Thus, $A H=\frac{2}{\sqrt{3}} A G=\left(\frac{2}{\sqrt{3}}\right)^{2} A F=\left(\frac{2}{\sqrt{3}}\right)^{3} A E=\left(\frac{2}{\sqrt{3}}\right)^{4} A D=\left(\frac{2}{\sqrt{3}}\right)^{5} A C=\left(\frac{2}{\sqrt{3}}\right)^{6} A B$.

(In other words, to get from $A B=1$ to the length of $A H$, we multiply by the "scaling factor" $\frac{2}{\sqrt{3}}$ six times.)

Therefore, $A H=\left(\frac{2}{\sqrt{3}}\right)^{6}=\frac{64}{27}$.

Final answer: $\frac{64}{27}$
Subfield: Geometry


Question: In rectangle $A B C D$, point $E$ is on side $D C$. Line segments $A E$ and $B D$ are perpendicular and intersect at $F$. If $A F=4$ and $D F=2$, determine the area of quadrilateral $B C E F$.

<img_4004>

Solution:
Since $\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,

$$
A D=\sqrt{A F^{2}+F D^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2 \sqrt{5}
$$

since $A D>0$.

Let $\angle F A D=\beta$.

Since $A B C D$ is a rectangle, then $\angle B A F=90^{\circ}-\beta$.

Since $\triangle A F D$ is right-angled at $F$, then $\angle A D F=90^{\circ}-\beta$.

Since $A B C D$ is a rectangle, then $\angle B D C=90^{\circ}-\left(90^{\circ}-\beta\right)=\beta$.



<img_3979>

Therefore, $\triangle B F A, \triangle A F D$, and $\triangle D F E$ are all similar as each is right-angled and has either an angle of $\beta$ or an angle of $90^{\circ}-\beta$ (and hence both of these angles).

Therefore, $\frac{A B}{A F}=\frac{D A}{D F}$ and so $A B=\frac{4(2 \sqrt{5})}{2}=4 \sqrt{5}$.

Also, $\frac{F E}{F D}=\frac{F D}{F A}$ and so $F E=\frac{2(2)}{4}=1$.

Since $A B C D$ is a rectangle, then $B C=A D=2 \sqrt{5}$, and $D C=A B=4 \sqrt{5}$.

Finally, the area of quadrilateral $B C E F$ equals the area of $\triangle D C B$ minus the area $\triangle D F E$. Thus, the required area is

$$
\frac{1}{2}(D C)(C B)-\frac{1}{2}(D F)(F E)=\frac{1}{2}(4 \sqrt{5})(2 \sqrt{5})-\frac{1}{2}(2)(1)=20-1=19
$$
Since $\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,

$$
A D=\sqrt{A F^{2}+F D^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2 \sqrt{5}
$$

since $A D>0$.

Let $\angle F A D=\beta$.

Since $A B C D$ is a rectangle, then $\angle B A F=90^{\circ}-\beta$. Since $\triangle B A F$ is right-angled at $F$, then $\angle A B F=\beta$.

Since $\triangle A F D$ is right-angled at $F$, then $\angle A D F=90^{\circ}-\beta$.

Since $A B C D$ is a rectangle, then $\angle B D C=90^{\circ}-\left(90^{\circ}-\beta\right)=\beta$.

<img_3412>

Looking at $\triangle A F D$, we see that $\sin \beta=\frac{F D}{A D}=\frac{2}{2 \sqrt{5}}=\frac{1}{\sqrt{5}}, \cos \beta=\frac{A F}{A D}=\frac{4}{2 \sqrt{5}}=\frac{2}{\sqrt{5}}$, and $\tan \beta=\frac{F D}{A F}=\frac{2}{4}=\frac{1}{2}$.

Since $A F=4$ and $\angle A B F=\beta$, then $A B=\frac{A F}{\sin \beta}=\frac{4}{\frac{1}{\sqrt{5}}}=4 \sqrt{5}$.

Since $F D=2$ and $\angle F D E=\beta$, then $F E=F D \tan \beta=2 \cdot \frac{1}{2}=1$.

Since $A B C D$ is a rectangle, then $B C=A D=2 \sqrt{5}$, and $D C=A B=4 \sqrt{5}$.

Finally, the area of quadrilateral $E F B C$ equals the area of $\triangle D C B$ minus the area $\triangle D F E$. Thus, the required area is

$$
\frac{1}{2}(D C)(C B)-\frac{1}{2}(D F)(F E)=\frac{1}{2}(4 \sqrt{5})(2 \sqrt{5})-\frac{1}{2}(2)(1)=20-1=19
$$

Final answer: 19
Subfield: Geometry


Question: In rectangle $A B C D$, point $E$ is on side $D C$. Line segments $A E$ and $B D$ are perpendicular and intersect at $F$. If $A F=4$ and $D F=2$, determine the area of quadrilateral $B C E F$.

![](https://cdn.mathpix.com/cropped/2023_12_21_6612ae75d8ef169f4be5g-1.jpg?height=339&width=483&top_left_y=2243&top_left_x=1252)

Solution:
Since $\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,

$$
A D=\sqrt{A F^{2}+F D^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2 \sqrt{5}
$$

since $A D>0$.

Let $\angle F A D=\beta$.

Since $A B C D$ is a rectangle, then $\angle B A F=90^{\circ}-\beta$.

Since $\triangle A F D$ is right-angled at $F$, then $\angle A D F=90^{\circ}-\beta$.

Since $A B C D$ is a rectangle, then $\angle B D C=90^{\circ}-\left(90^{\circ}-\beta\right)=\beta$.



![](https://cdn.mathpix.com/cropped/2023_12_21_98e0ff93ea7be169163dg-1.jpg?height=331&width=548&top_left_y=165&top_left_x=889)

Therefore, $\triangle B F A, \triangle A F D$, and $\triangle D F E$ are all similar as each is right-angled and has either an angle of $\beta$ or an angle of $90^{\circ}-\beta$ (and hence both of these angles).

Therefore, $\frac{A B}{A F}=\frac{D A}{D F}$ and so $A B=\frac{4(2 \sqrt{5})}{2}=4 \sqrt{5}$.

Also, $\frac{F E}{F D}=\frac{F D}{F A}$ and so $F E=\frac{2(2)}{4}=1$.

Since $A B C D$ is a rectangle, then $B C=A D=2 \sqrt{5}$, and $D C=A B=4 \sqrt{5}$.

Finally, the area of quadrilateral $B C E F$ equals the area of $\triangle D C B$ minus the area $\triangle D F E$. Thus, the required area is

$$
\frac{1}{2}(D C)(C B)-\frac{1}{2}(D F)(F E)=\frac{1}{2}(4 \sqrt{5})(2 \sqrt{5})-\frac{1}{2}(2)(1)=20-1=19
$$
Since $\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,

$$
A D=\sqrt{A F^{2}+F D^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2 \sqrt{5}
$$

since $A D>0$.

Let $\angle F A D=\beta$.

Since $A B C D$ is a rectangle, then $\angle B A F=90^{\circ}-\beta$. Since $\triangle B A F$ is right-angled at $F$, then $\angle A B F=\beta$.

Since $\triangle A F D$ is right-angled at $F$, then $\angle A D F=90^{\circ}-\beta$.

Since $A B C D$ is a rectangle, then $\angle B D C=90^{\circ}-\left(90^{\circ}-\beta\right)=\beta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_98e0ff93ea7be169163dg-1.jpg?height=328&width=544&top_left_y=1663&top_left_x=888)

Looking at $\triangle A F D$, we see that $\sin \beta=\frac{F D}{A D}=\frac{2}{2 \sqrt{5}}=\frac{1}{\sqrt{5}}, \cos \beta=\frac{A F}{A D}=\frac{4}{2 \sqrt{5}}=\frac{2}{\sqrt{5}}$, and $\tan \beta=\frac{F D}{A F}=\frac{2}{4}=\frac{1}{2}$.

Since $A F=4$ and $\angle A B F=\beta$, then $A B=\frac{A F}{\sin \beta}=\frac{4}{\frac{1}{\sqrt{5}}}=4 \sqrt{5}$.

Since $F D=2$ and $\angle F D E=\beta$, then $F E=F D \tan \beta=2 \cdot \frac{1}{2}=1$.

Since $A B C D$ is a rectangle, then $B C=A D=2 \sqrt{5}$, and $D C=A B=4 \sqrt{5}$.

Finally, the area of quadrilateral $E F B C$ equals the area of $\triangle D C B$ minus the area $\triangle D F E$. Thus, the required area is

$$
\frac{1}{2}(D C)(C B)-\frac{1}{2}(D F)(F E)=\frac{1}{2}(4 \sqrt{5})(2 \sqrt{5})-\frac{1}{2}(2)(1)=20-1=19
$$

Final answer: 19
Subfield: Algebra


Question: Determine all real values of $x$ for which $3^{(x-1)} 9^{\frac{3}{2 x^{2}}}=27$.

Solution:
Using the facts that $9=3^{2}$ and $27=3^{3}$, and the laws for manipulating exponents, we have

$$
\begin{aligned}
3^{x-1} 9^{\frac{3}{2 x^{2}}} & =27 \\
3^{x-1}\left(3^{2}\right)^{\frac{3}{2 x^{2}}} & =3^{3} \\
3^{x-1} 3^{\frac{3}{x^{2}}} & =3^{3} \\
3^{x-1+\frac{3}{x^{2}}} & =3^{3}
\end{aligned}
$$

When two powers of 3 are equal, their exponents must be equal so

$$
\begin{aligned}
x-1+\frac{3}{x^{2}} & =3 \\
x^{3}-x^{2}+3 & \left.=3 x^{2} \quad \text { (multiplying by } x^{2}\right) \\
x^{3}-4 x^{2}+3 & =0
\end{aligned}
$$

Since $x=1$ satisfies the equation, then $x-1$ is a factor of the left side. Using long division or synthetic division, we can factor this out to get $(x-1)\left(x^{2}-3 x-3\right)=0$.

Using the quadratic formula, the quadratic equation $x^{2}-3 x-3=0$ has roots

$$
x=\frac{3 \pm \sqrt{(-3)^{2}-4(1)(-3)}}{2}=\frac{3 \pm \sqrt{21}}{2}
$$

Therefore, the solutions to the original equation are $x=1$ and $x=\frac{3 \pm \sqrt{21}}{2}$.

Final answer: $1$,$\frac{3 + \sqrt{21}}{2}$,$\frac{3 - \sqrt{21}}{2}$
Subfield: Algebra


Question: Determine all points $(x, y)$ where the two curves $y=\log _{10}\left(x^{4}\right)$ and $y=\left(\log _{10} x\right)^{3}$ intersect.

Solution:
To determine the points of intersection, we equate $y$ values of the two curves and obtain $\log _{10}\left(x^{4}\right)=\left(\log _{10} x\right)^{3}$.

Since $\log _{10}\left(a^{b}\right)=b \log _{10} a$, the equation becomes $4 \log _{10} x=\left(\log _{10} x\right)^{3}$.

We set $u=\log _{10} x$ and so the equation becomes $4 u=u^{3}$, or $u^{3}-4 u=0$.

We can factor the left side as $u^{3}-4 u=u\left(u^{2}-4\right)=u(u+2)(u-2)$.

Therefore, $u(u+2)(u-2)=0$, and so $u=0$ or $u=-2$ or $u=2$.

Therefore, $\log _{10} x=0$ or $\log _{10} x=-2$ or $\log _{10} x=2$.

Therefore, $x=1$ or $x=\frac{1}{100}$ or $x=100$.

Finally, we must calculate the $y$-coordinates of the points of intersection. Since one of the original curves is $y=\left(\log _{10} x\right)^{3}$, we can calculate the corresponding values of $y$ by using the fact that $y=u^{3}$.

The corresponding values of $y$ are $y=0^{3}=0$ and $y=(-2)^{3}=-8$ and $y=2^{3}=8$.

Therefore, the points of intersection are $(1,0),\left(\frac{1}{100},-8\right)$ and $(100,8)$.

Final answer: $(1,0),(\frac{1}{100},-8),(100,8)$
Subfield: Combinatorics


Question: Oi-Lam tosses three fair coins and removes all of the coins that come up heads. George then tosses the coins that remain, if any. Determine the probability that George tosses exactly one head.

Solution:
If Oi-Lam tosses 3 heads, then George has no coins to toss, so cannot toss exactly 1 head. If Oi-Lam tosses 2, 1 or 0 heads, then George has at least one coin to toss, so can toss exactly 1 head.

Therefore, the following possibilities exist:

* Oi-Lam tosses 2 heads out of 3 coins and George tosses 1 head out of 1 coin
* Oi-Lam tosses 1 head out of 3 coins and George tosses 1 head out of 2 coins
* Oi-Lam tosses 0 heads out of 3 coins and George tosses 1 head out of 3 coins

We calculate the various probabilities.

If 3 coins are tossed, there are 8 equally likely possibilities: $\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{TTH}$, THT, HTT, TTT. Each of these possibilities has probability $\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$. Therefore,



* the probability of tossing 0 heads out of 3 coins is $\frac{1}{8}$
* the probability of tossing 1 head out of 3 coins is $\frac{3}{8}$
* the probability of tossing 2 heads out of 3 coins is $\frac{3}{8}$
* the probability of tossing 3 heads out of 3 coins is $\frac{1}{8}$

If 2 coins are tossed, there are 4 equally likely possibilities: HH, HT, TH, TT. Each of these possibilities has probability $\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$. Therefore, the probability of tossing 1 head out of 2 coins is $\frac{2}{4}=\frac{1}{2}$.

If 1 coin is tossed, the probability of tossing 1 head is $\frac{1}{2}$.

To summarize, the possibilities are

* Oi-Lam tosses 2 heads out of 3 coins (with probability $\frac{3}{8}$ ) and George tosses 1 head out of 1 coin (with probability $\frac{1}{2}$ )
* Oi-Lam tosses 1 head out of 3 coins (with probability $\frac{3}{8}$ ) and George tosses 1 head out of 2 coins (with probability $\frac{1}{2}$ )
* Oi-Lam tosses 0 heads out of 3 coins (with probability $\frac{1}{8}$ ) and George tosses 1 head out of 3 coins (with probability $\frac{3}{8}$ )

Therefore, the overall probability is $\frac{3}{8} \cdot \frac{1}{2}+\frac{3}{8} \cdot \frac{1}{2}+\frac{1}{8} \cdot \frac{3}{8}=\frac{27}{64}$.

Final answer: $\frac{27}{64}$
Subfield: Geometry


Question: In the diagram, points $B, P, Q$, and $C$ lie on line segment $A D$. The semi-circle with diameter $A C$ has centre $P$ and the semi-circle with diameter $B D$ has centre $Q$. The two semi-circles intersect at $R$. If $\angle P R Q=40^{\circ}$, determine the measure of $\angle A R D$.

<img_3416>

Solution:
Suppose $\angle P A R=x^{\circ}$ and $\angle Q D R=y^{\circ}$.

<img_3198>

Since $P R$ and $P A$ are radii of the larger circle, then $\triangle P A R$ is isosceles.

Thus, $\angle P R A=\angle P A R=x^{\circ}$.

Since $Q D$ and $Q R$ are radii of the smaller circle, then $\triangle Q R D$ is isosceles.

Thus, $\angle Q R D=\angle Q D R=y^{\circ}$.

In $\triangle A R D$, the sum of the angles is $180^{\circ}$, so $x^{\circ}+\left(x^{\circ}+40^{\circ}+y^{\circ}\right)+y^{\circ}=180^{\circ}$ or $2 x+2 y=140$ or $x+y=70$.

Therefore, $\angle C P D=x^{\circ}+40^{\circ}+y^{\circ}=(x+y+40)^{\circ}=110^{\circ}$.

Final answer: $110^{\circ}$
Subfield: Geometry


Question: In the diagram, points $B, P, Q$, and $C$ lie on line segment $A D$. The semi-circle with diameter $A C$ has centre $P$ and the semi-circle with diameter $B D$ has centre $Q$. The two semi-circles intersect at $R$. If $\angle P R Q=40^{\circ}$, determine the measure of $\angle A R D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_8305addb4aaa1260f985g-1.jpg?height=271&width=545&top_left_y=683&top_left_x=1256)

Solution:
Suppose $\angle P A R=x^{\circ}$ and $\angle Q D R=y^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_96c935933a9e5a2ec6ebg-1.jpg?height=267&width=555&top_left_y=1127&top_left_x=885)

Since $P R$ and $P A$ are radii of the larger circle, then $\triangle P A R$ is isosceles.

Thus, $\angle P R A=\angle P A R=x^{\circ}$.

Since $Q D$ and $Q R$ are radii of the smaller circle, then $\triangle Q R D$ is isosceles.

Thus, $\angle Q R D=\angle Q D R=y^{\circ}$.

In $\triangle A R D$, the sum of the angles is $180^{\circ}$, so $x^{\circ}+\left(x^{\circ}+40^{\circ}+y^{\circ}\right)+y^{\circ}=180^{\circ}$ or $2 x+2 y=140$ or $x+y=70$.

Therefore, $\angle C P D=x^{\circ}+40^{\circ}+y^{\circ}=(x+y+40)^{\circ}=110^{\circ}$.

Final answer: $110^{\circ}$
Subfield: Algebra


Question: If $\theta$ is an angle whose measure is not an integer multiple of $90^{\circ}$, prove that

$$
\cot \theta-\cot 2 \theta=\frac{1}{\sin 2 \theta}
$$

Solution:
$$
\begin{aligned}
\mathrm{LS} & =\cot \theta-\cot 2 \theta \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{\sin 2 \theta} \\
& =\frac{\sin 2 \theta \cos \theta-\cos 2 \theta \sin \theta}{\sin \theta \sin 2 \theta} \\
& =\frac{\sin (2 \theta-\theta)}{\sin \theta \sin 2 \theta} \\
& =\frac{\sin \theta}{\sin \theta \sin 2 \theta} \\
& =\frac{1}{\sin 2 \theta} \\
& =\mathrm{RS}
\end{aligned}
$$

as required.
$$
\begin{aligned}
\mathrm{LS} & =\cot \theta-\cot 2 \theta \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{\sin 2 \theta} \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos ^{2} \theta-\cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos ^{2} \theta-\left(2 \cos ^{2} \theta-1\right)}{\sin 2 \theta} \\
& =\frac{1}{\sin 2 \theta} \\
& =\mathrm{RS}
\end{aligned}
$$

as required.
Subfield: Algebra


Question: Ross starts with an angle of measure $8^{\circ}$ and doubles it 10 times until he obtains $8192^{\circ}$. He then adds up the reciprocals of the sines of these 11 angles. That is, he calculates

$$
S=\frac{1}{\sin 8^{\circ}}+\frac{1}{\sin 16^{\circ}}+\frac{1}{\sin 32^{\circ}}+\cdots+\frac{1}{\sin 4096^{\circ}}+\frac{1}{\sin 8192^{\circ}}
$$

Determine, without using a calculator, the measure of the acute angle $\alpha$ so that $S=\frac{1}{\sin \alpha}$.

Solution:
We first prove Lemma(i): If $\theta$ is an angle whose measure is not an integer multiple of $90^{\circ}$, then
$$
\cot \theta-\cot 2 \theta=\frac{1}{\sin 2 \theta}
$$

Proof. 
$$
\begin{aligned}
\mathrm{LS} & =\cot \theta-\cot 2 \theta \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{\sin 2 \theta} \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos ^{2} \theta-\cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos ^{2} \theta-\left(2 \cos ^{2} \theta-1\right)}{\sin 2 \theta} \\
& =\frac{1}{\sin 2 \theta} \\
& =\mathrm{RS}
\end{aligned}
$$

as required.

We use (i) to note that $\frac{1}{\sin 8^{\circ}}=\cot 4^{\circ}-\cot 8^{\circ}$ and $\frac{1}{\sin 16^{\circ}}=\cot 8^{\circ}-\cot 16^{\circ}$ and so on. Thus,

$$
\begin{aligned}
S= & \frac{1}{\sin 8^{\circ}}+\frac{1}{\sin 16^{\circ}}+\frac{1}{\sin 32^{\circ}}+\cdots+\frac{1}{\sin 4096^{\circ}}+\frac{1}{\sin 8192^{\circ}} \\
= & \left(\cot 4^{\circ}-\cot 8^{\circ}\right)+\left(\cot 8^{\circ}-\cot 16^{\circ}\right)+\left(\cot 16^{\circ}-\cot 32^{\circ}\right)+ \\
& \cdots+\left(\cot 2048^{\circ}-\cot 4096^{\circ}\right)+\left(\cot 4096^{\circ}-\cot 8192^{\circ}\right) \\
= & \cot 4^{\circ}-\cot 8192^{\circ}
\end{aligned}
$$

since the sum "telescopes".

Since the cotangent function has a period of $180^{\circ}$, and $8100^{\circ}$ is a multiple of $180^{\circ}$, then $\cot 8192^{\circ}=\cot 92^{\circ}$.

Therefore,

$$
\begin{aligned}
S & =\cot 4^{\circ}-\cot 92^{\circ} \\
& =\frac{\cos 4^{\circ}}{\sin 4^{\circ}}-\frac{\cos 92^{\circ}}{\sin 92^{\circ}} \\
& =\frac{\cos 4^{\circ}}{\sin 4^{\circ}}-\frac{-\sin 2^{\circ}}{\cos 2^{\circ}} \\
& =\frac{\cos 4^{\circ}}{2 \sin 2^{\circ} \cos 2^{\circ}}+\frac{\sin 2^{\circ}}{\cos 2^{\circ}} \\
& =\frac{\cos 4^{\circ}+2 \sin ^{2} 2^{\circ}}{2 \sin 2^{\circ} \cos 2^{\circ}} \\
& =\frac{\left(1-2 \sin ^{2} 2^{\circ}\right)+2 \sin ^{2} 2^{\circ}}{\sin 4^{\circ}} \\
& =\frac{1}{\sin 4^{\circ}}
\end{aligned}
$$

Therefore, $\alpha=4^{\circ}$.

Final answer: $4^{\circ}$
Subfield: Combinatorics


Question: For each positive integer $n$, let $T(n)$ be the number of triangles with integer side lengths, positive area, and perimeter $n$. For example, $T(6)=1$ since the only such triangle with a perimeter of 6 has side lengths 2,2 and 2 .
Determine the values of $T(10), T(11)$ and $T(12)$.

Solution:
Denote the side lengths of a triangle by $a, b$ and $c$, with $0<a \leq b \leq c$.

In order for these lengths to form a triangle, we need $c<a+b$ and $b<a+c$ and $a<b+c$. Since $0<a \leq b \leq c$, then $b<a+c$ and $a<b+c$ follow automatically, so only $c<a+b$ ever needs to be checked.

Instead of directly considering triangles and sets of triangle, we can consider triples $(a, b, c)$ and sets of triples $(a, b, c)$ with the appropriate conditions.

For each positive integer $k \geq 3$, we use the notation $S_{k}$ to denote the set of triples of positive integers $(a, b, c)$ with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=k$.

In this case, $c<a+b$ and $a+b+c=k$, so $c+c<a+b+c=k$, so $2 c<k$ or $c<\frac{1}{2} k$.

Also, if $0<a \leq b \leq c$ and $a+b+c=k$, then $k=a+b+c \leq c+c+c$, so $3 c \geq k$ or $c \geq \frac{1}{3} k$.


Consider $T(10)$, which is the number of elements in $S_{10}$.

We want to find all possible triples $(a, b, c)$ of integers with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=10$.

We need $c<\frac{10}{2}=5$ and $c \geq \frac{10}{3}$. Thus, $c=4$.

Therefore, we need $0<a \leq b \leq 4$ and $a+b=6$.

There are two possibilities: $(a, b, c)=(2,4,4)$ or $(a, b, c)=(3,3,4)$.

Therefore, $T(10)=2$.

Consider $T(11)$. We want to find all possible triples $(a, b, c)$ of integers with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=11$.

We need $c<\frac{11}{2}$ and $c \geq \frac{11}{3}$. Thus, $c=4$ or $c=5$.

If $c=4$, we need $0<a \leq b \leq 4$ and $a+b=7$.

There is only one possibility: $(a, b, c)=(3,4,4)$.

If $c=5$, we need $0<a \leq b \leq 5$ and $a+b=6$.

There are three possibilities: $(a, b, c)=(1,5,5)$ or $(a, b, c)=(2,4,5)$ or $(a, b, c)=(3,3,5)$.

Therefore, $T(11)=4$.

Consider $T(12)$. We want to find all possible triples $(a, b, c)$ of integers with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=12$.

We need $c<\frac{12}{2}$ and $c \geq \frac{12}{3}$. Thus, $c=4$ or $c=5$.

If $c=4$, we need $0<a \leq b \leq 4$ and $a+b=8$.

There is only one possibility: $(a, b, c)=(4,4,4)$.



If $c=5$, we need $0<a \leq b \leq 5$ and $a+b=7$.

There are two possibilities: $(a, b, c)=(2,5,5)$ or $(a, b, c)=(3,4,5)$.

Therefore, $T(12)=3$.

Final answer: 2,4,3
Subfield: Geometry


Question: In the diagram, a line is drawn through points $P, Q$ and $R$. If $P Q=Q R$, what are the coordinates of $R$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=363&width=564&top_left_y=840&top_left_x=1255)

Solution:
To get from $P$ to $Q$, we move 3 units right and 4 units up.

Since $P Q=Q R$ and $R$ lies on the line through $Q$, then we must use the same motion to get from $Q$ to $R$.

Therefore, to get from $Q(0,4)$ to $R$, we move 3 units right and 4 units up, so the coordinates of $R$ are $(3,8)$.
The line through $P(-3,0)$ and $Q(0,4)$ has slope $\frac{4-0}{0-(-3)}=\frac{4}{3}$ and $y$-intercept 4 , so has equation $y=\frac{4}{3} x+4$.

Thus, $R$ has coordinates $\left(a, \frac{4}{3} a+4\right)$ for some $a>0$.

Since $P Q=Q R$, then $P Q^{2}=Q R^{2}$, so

$$
\begin{aligned}
(-3)^{2}+4^{2} & =a^{2}+\left(\frac{4}{3} a+4-4\right)^{2} \\
25 & =a^{2}+\frac{16}{9} a^{2} \\
\frac{25}{9} a^{2} & =25 \\
a^{2} & =9
\end{aligned}
$$

so $a=3$ since $a>0$.

Thus, $R$ has coordinates $\left(3, \frac{4}{3}(3)+4\right)=(3,8)$.

Final answer: (3,8)
Subfield: Geometry


Question: In the diagram, a line is drawn through points $P, Q$ and $R$. If $P Q=Q R$, what are the coordinates of $R$ ?

<img_4009>

Solution:
To get from $P$ to $Q$, we move 3 units right and 4 units up.

Since $P Q=Q R$ and $R$ lies on the line through $Q$, then we must use the same motion to get from $Q$ to $R$.

Therefore, to get from $Q(0,4)$ to $R$, we move 3 units right and 4 units up, so the coordinates of $R$ are $(3,8)$.
The line through $P(-3,0)$ and $Q(0,4)$ has slope $\frac{4-0}{0-(-3)}=\frac{4}{3}$ and $y$-intercept 4 , so has equation $y=\frac{4}{3} x+4$.

Thus, $R$ has coordinates $\left(a, \frac{4}{3} a+4\right)$ for some $a>0$.

Since $P Q=Q R$, then $P Q^{2}=Q R^{2}$, so

$$
\begin{aligned}
(-3)^{2}+4^{2} & =a^{2}+\left(\frac{4}{3} a+4-4\right)^{2} \\
25 & =a^{2}+\frac{16}{9} a^{2} \\
\frac{25}{9} a^{2} & =25 \\
a^{2} & =9
\end{aligned}
$$

so $a=3$ since $a>0$.

Thus, $R$ has coordinates $\left(3, \frac{4}{3}(3)+4\right)=(3,8)$.

Final answer: (3,8)
Subfield: Geometry


Question: In the diagram, $O A=15, O P=9$ and $P B=4$. Determine the equation of the line through $A$ and $B$. Explain how you got your answer.

<img_4005>

Solution:
Since $O P=9$, then the coordinates of $P$ are $(9,0)$.

Since $O P=9$ and $O A=15$, then by the Pythagorean Theorem,

$$
A P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144
$$

so $A P=12$.

Since $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.

Since $P B=4$, then $B$ has coordinates $(13,0)$.

The line through $A(9,12)$ and $B(13,0)$ has slope $\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$.

Final answer: $y=-3 x+39$
Subfield: Geometry


Question: In the diagram, $O A=15, O P=9$ and $P B=4$. Determine the equation of the line through $A$ and $B$. Explain how you got your answer.

![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=366&width=444&top_left_y=1259&top_left_x=1250)

Solution:
Since $O P=9$, then the coordinates of $P$ are $(9,0)$.

Since $O P=9$ and $O A=15$, then by the Pythagorean Theorem,

$$
A P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144
$$

so $A P=12$.

Since $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.

Since $P B=4$, then $B$ has coordinates $(13,0)$.

The line through $A(9,12)$ and $B(13,0)$ has slope $\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$.

Final answer: $y=-3 x+39$