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


Question: In the diagram, $\triangle X Y Z$ is isosceles with $X Y=X Z=a$ and $Y Z=b$ where $b<2 a$. A larger circle of radius $R$ is inscribed in the triangle (that is, the circle is drawn so that it touches all three sides of the triangle). A smaller circle of radius $r$ is drawn so that it touches $X Y, X Z$ and the larger circle. Determine an expression for $\frac{R}{r}$ in terms of $a$ and $b$.

![](https://cdn.mathpix.com/cropped/2023_12_21_ff40fe20a2d5d1ed9823g-1.jpg?height=474&width=333&top_left_y=1991&top_left_x=1254)

Solution:
Suppose that $M$ is the midpoint of $Y Z$.

Suppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $X Y$ at $C$ and $X Z$ at $D$, and that the larger circle touches $X Y$ at $E$ and $X Z$ at $F$.

Join $O C, O D$ and $P E$.

Since $O C$ and $P E$ are radii that join the centres of circles to points of tangency, then $O C$ and $P E$ are perpendicular to $X Y$.

Join $X M$. Since $\triangle X Y Z$ is isosceles, then $X M$ (which is a median by construction) is an altitude (that is, $X M$ is perpendicular to $Y Z)$ and an angle bisector (that is, $\angle M X Y=\angle M X Z$ ).

Now $X M$ passes through $O$ and $P$. (Since $X C$ and $X D$ are tangents from $X$ to the same circle, then $X C=X D$. This means that $\triangle X C O$ is congruent to $\triangle X D O$ by side-side-side. This means that $\angle O X C=\angle O X D$ and so $O$ lies on the angle bisector of $\angle C X D$, and so $O$ lies on $X M$. Using a similar argument, $P$ lies on $X M$.)

![](https://cdn.mathpix.com/cropped/2023_12_21_04e57b63898bbe86db6ag-1.jpg?height=474&width=333&top_left_y=454&top_left_x=1579)
Draw a perpendicular from $O$ to $T$ on $P E$. Note that $O T$ is parallel to $X Y$ (since each is perpendicular to $P E$ ) and that $O C E T$ is a rectangle (since it has three right angles).

Consider $\triangle X M Y$ and $\triangle O T P$.

Each triangle is right-angled (at $M$ and at $T$ ).

Also, $\angle Y X M=\angle P O T$. (This is because $O T$ is parallel to $X Y$, since both are perpendicular to $P E$.)

Therefore, $\triangle X M Y$ is similar to $\triangle O T P$.

Thus, $\frac{X Y}{Y M}=\frac{O P}{P T}$.

Now $X Y=a$ and $Y M=\frac{1}{2} b$.

Also, $O P$ is the line segment joining the centres of two tangent circles, so $O P=r+R$.

Lastly, $P T=P E-E T=R-r$, since $P E=R, E T=O C=r$, and $O C E T$ is a rectangle. Therefore,

$$
\begin{aligned}
\frac{a}{b / 2} & =\frac{R+r}{R-r} \\
\frac{2 a}{b} & =\frac{R+r}{R-r} \\
2 a(R-r) & =b(R+r) \\
2 a R-b R & =2 a r+b r \\
R(2 a-b) & =r(2 a+b) \\
\frac{R}{r} & =\frac{2 a+b}{2 a-b} \quad(\text { since } 2 a>b \text { so } 2 a-b \neq 0, \text { and } r>0)
\end{aligned}
$$

Therefore, $\frac{R}{r}=\frac{2 a+b}{2 a-b}$.

Final answer: $\frac{2 a+b}{2 a-b}$
Subfield: Geometry


Question: In the diagram, $\triangle X Y Z$ is isosceles with $X Y=X Z=a$ and $Y Z=b$ where $b<2 a$. A larger circle of radius $R$ is inscribed in the triangle (that is, the circle is drawn so that it touches all three sides of the triangle). A smaller circle of radius $r$ is drawn so that it touches $X Y, X Z$ and the larger circle. Determine an expression for $\frac{R}{r}$ in terms of $a$ and $b$.

<img_3482>

Solution:
Suppose that $M$ is the midpoint of $Y Z$.

Suppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $X Y$ at $C$ and $X Z$ at $D$, and that the larger circle touches $X Y$ at $E$ and $X Z$ at $F$.

Join $O C, O D$ and $P E$.

Since $O C$ and $P E$ are radii that join the centres of circles to points of tangency, then $O C$ and $P E$ are perpendicular to $X Y$.

Join $X M$. Since $\triangle X Y Z$ is isosceles, then $X M$ (which is a median by construction) is an altitude (that is, $X M$ is perpendicular to $Y Z)$ and an angle bisector (that is, $\angle M X Y=\angle M X Z$ ).

Now $X M$ passes through $O$ and $P$. (Since $X C$ and $X D$ are tangents from $X$ to the same circle, then $X C=X D$. This means that $\triangle X C O$ is congruent to $\triangle X D O$ by side-side-side. This means that $\angle O X C=\angle O X D$ and so $O$ lies on the angle bisector of $\angle C X D$, and so $O$ lies on $X M$. Using a similar argument, $P$ lies on $X M$.)

<img_3625>
Draw a perpendicular from $O$ to $T$ on $P E$. Note that $O T$ is parallel to $X Y$ (since each is perpendicular to $P E$ ) and that $O C E T$ is a rectangle (since it has three right angles).

Consider $\triangle X M Y$ and $\triangle O T P$.

Each triangle is right-angled (at $M$ and at $T$ ).

Also, $\angle Y X M=\angle P O T$. (This is because $O T$ is parallel to $X Y$, since both are perpendicular to $P E$.)

Therefore, $\triangle X M Y$ is similar to $\triangle O T P$.

Thus, $\frac{X Y}{Y M}=\frac{O P}{P T}$.

Now $X Y=a$ and $Y M=\frac{1}{2} b$.

Also, $O P$ is the line segment joining the centres of two tangent circles, so $O P=r+R$.

Lastly, $P T=P E-E T=R-r$, since $P E=R, E T=O C=r$, and $O C E T$ is a rectangle. Therefore,

$$
\begin{aligned}
\frac{a}{b / 2} & =\frac{R+r}{R-r} \\
\frac{2 a}{b} & =\frac{R+r}{R-r} \\
2 a(R-r) & =b(R+r) \\
2 a R-b R & =2 a r+b r \\
R(2 a-b) & =r(2 a+b) \\
\frac{R}{r} & =\frac{2 a+b}{2 a-b} \quad(\text { since } 2 a>b \text { so } 2 a-b \neq 0, \text { and } r>0)
\end{aligned}
$$

Therefore, $\frac{R}{r}=\frac{2 a+b}{2 a-b}$.

Final answer: $\frac{2 a+b}{2 a-b}$
Subfield: Combinatorics


Question: For each positive integer $n \geq 1$, let $C_{n}$ be the set containing the $n$ smallest positive integers; that is, $C_{n}=\{1,2, \ldots, n-1, n\}$. For example, $C_{4}=\{1,2,3,4\}$. We call a set, $F$, of subsets of $C_{n}$ a Furoni family of $C_{n}$ if no element of $F$ is a subset of another element of $F$.
Consider $A=\{\{1,2\},\{1,3\},\{1,4\}\}$. Note that $A$ is a Furoni family of $C_{4}$. Determine the two Furoni families of $C_{4}$ that contain all of the elements of $A$ and to which no other subsets of $C_{4}$ can be added to form a new (larger) Furoni family.

Solution:
The subsets of $C_{4}$ are:

|  | \{\} | $\{1\}$ | $\{2\}$ | $\{3\}$ | $\{4\}$ |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\{1,2\}$ | $\{1,3\}$ | $\{1,4\}$ | $\{2,3\}$ | $\{2,4\}$ | $\{3,4\}$ |  |
| $\{1,2,3\}$ | $\{1,2,4\}$ | $\{1,3,4\}$ | $\{2,3,4\}$ | $\{1,2,3,4\}$ |  |  |

(There are 16 such subsets including the empty set \{\} and the complete set $C_{4}=\{1,2,3,4\}$.) Consider the Furoni family $A=\{\{1,2\},\{1,3\},\{1,4\}\}$.

Each of the following subsets of $C_{4}$ is already an element of $A:\{1,2\},\{1,3\},\{1,4\}$.

Each of the following subsets of $C_{4}$ is a subset of one or more of the elements of $A$ : \{\}$,\{1\},\{2\},\{3\},\{4\}$.

Each of the following subsets of $C_{4}$ has the property that one or more of the elements of $A$ is a subset of it: $\{1,2,3\},\{1,2,4\},\{1,3,4\},\{1,2,3,4\}$.

Since a Furoni family of $C_{4}$ cannot contain two subsets of $C_{4}$ one of which is a subset of the other, none of the subsets in either of these two lists can be added to $A$ to form a larger Furoni family.

This leaves the following subsets of $C_{4}$ to consider as possible elements to add to $A$ : $\{2,3\},\{2,4\},\{3,4\},\{2,3,4\}$.

If $\{2,3,4\}$ is added to $A$ to form $A^{\prime}=\{\{1,2\},\{1,3\},\{1,4\},\{2,3,4\}\}$, then $A^{\prime}$ is still a Furoni family of $C_{4}$ and none of $\{2,3\},\{2,4\},\{3,4\}$ can be added, since each is a subset of $\{2,3,4\}$. Therefore, $A^{\prime}$ is a Furoni family of $C_{4}$ to which no other subset can be added. If any of $\{2,3\},\{2,4\},\{3,4\}$ is added to $A$, then $\{2,3,4\}$ cannot be added (since each of these three two elements sets is a subset of $\{2,3,4\}$ ) but each of the remaining two element sets can be still added without violating the conditions for being a Furoni family.

Thus, $A^{\prime \prime}=\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$ is a Furoni family of $C_{4}$ to which no other subset can be added.

Therefore, the two Furoni families of $C_{4}$ that contain all of the elements of $A$ and to which no further subsets of $C_{4}$ can be added are

$$
A^{\prime}=\{\{1,2\},\{1,3\},\{1,4\},\{2,3,4\}\} \quad A^{\prime \prime}=\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}
$$

Final answer: $\{\{1,2\},\{1,3\},\{1,4\},\{2,3,4\}\},\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$
Subfield: Geometry


Question: In the diagram, what is the area of figure $A B C D E F$ ?

<img_3643>

Solution:
Because all of the angles in the figure are right angles, then $B C=D E=4$.

Thus, we can break up the figure into a 4 by 8 rectangle and a 4 by 4 square, by extending $B C$ to hit $F E$. Therefore, the area of the figure is $(8)(4)+(4)(4)=48$.

Final answer: 48
Subfield: Geometry


Question: In the diagram, what is the area of figure $A B C D E F$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=364&width=338&top_left_y=680&top_left_x=1403)

Solution:
Because all of the angles in the figure are right angles, then $B C=D E=4$.

Thus, we can break up the figure into a 4 by 8 rectangle and a 4 by 4 square, by extending $B C$ to hit $F E$. Therefore, the area of the figure is $(8)(4)+(4)(4)=48$.

Final answer: 48
Subfield: Geometry


Question: In the diagram, $A B C D$ is a rectangle with $A E=15, E B=20$ and $D F=24$. What is the length of $C F$ ?

<img_3192>

Solution:
By the Pythagorean Theorem in triangle $A B E$, $A B^{2}=15^{2}+20^{2}=625$, so $A B=25$.

Since $A B C D$ is a rectangle, $C D=A B=25$, so by the Pythagorean Theorem in triangle $C F D$, we have $625=25^{2}=24^{2}+C F^{2}$, so $C F^{2}=625-576=49$, or $C F=7$.

Final answer: 7
Subfield: Geometry


Question: In the diagram, $A B C D$ is a rectangle with $A E=15, E B=20$ and $D F=24$. What is the length of $C F$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=314&width=374&top_left_y=1079&top_left_x=1361)

Solution:
By the Pythagorean Theorem in triangle $A B E$, $A B^{2}=15^{2}+20^{2}=625$, so $A B=25$.

Since $A B C D$ is a rectangle, $C D=A B=25$, so by the Pythagorean Theorem in triangle $C F D$, we have $625=25^{2}=24^{2}+C F^{2}$, so $C F^{2}=625-576=49$, or $C F=7$.

Final answer: 7
Subfield: Geometry


Question: In the diagram, $A B C D$ is a square of side length 6. Points $E, F, G$, and $H$ are on $A B, B C, C D$, and $D A$, respectively, so that the ratios $A E: E B, B F: F C$, $C G: G D$, and $D H: H A$ are all equal to $1: 2$.

What is the area of $E F G H$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=361&width=372&top_left_y=1432&top_left_x=1386)

Solution:
Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.

Thus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.

Then the area of $E F G H$ is equal to the area of square $A B C D$ minus the combined area of the four triangles, or $6^{2}-4\left[\frac{1}{2}(2)(4)\right]=36-16=20$ square units.
Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.

Thus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.

By the Pythagorean Theorem,

$E F=F G=G H=H E=\sqrt{2^{2}+4^{2}}=\sqrt{20}$.

Since the two triangles $H A E$ and $E B F$ are congruent (we know the lengths of all three sides of each), then $\angle A H E=\angle B E F$. But $\angle A H E+\angle A E H=90^{\circ}$, so $\angle B E F+\angle A E H=90^{\circ}$, so $\angle H E F=90^{\circ}$.

In a similar way, we can show that each of the four angles of $E F G H$ is a right-angle, and so $E F G H$ is a square of side length $\sqrt{20}$.

Therefore, the area of $E F G H$ is $(\sqrt{20})^{2}=20$ square units.

Final answer: 20
Subfield: Geometry


Question: In the diagram, $A B C D$ is a square of side length 6. Points $E, F, G$, and $H$ are on $A B, B C, C D$, and $D A$, respectively, so that the ratios $A E: E B, B F: F C$, $C G: G D$, and $D H: H A$ are all equal to $1: 2$.

What is the area of $E F G H$ ?

<img_3566>

Solution:
Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.

Thus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.

Then the area of $E F G H$ is equal to the area of square $A B C D$ minus the combined area of the four triangles, or $6^{2}-4\left[\frac{1}{2}(2)(4)\right]=36-16=20$ square units.
Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.

Thus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.

By the Pythagorean Theorem,

$E F=F G=G H=H E=\sqrt{2^{2}+4^{2}}=\sqrt{20}$.

Since the two triangles $H A E$ and $E B F$ are congruent (we know the lengths of all three sides of each), then $\angle A H E=\angle B E F$. But $\angle A H E+\angle A E H=90^{\circ}$, so $\angle B E F+\angle A E H=90^{\circ}$, so $\angle H E F=90^{\circ}$.

In a similar way, we can show that each of the four angles of $E F G H$ is a right-angle, and so $E F G H$ is a square of side length $\sqrt{20}$.

Therefore, the area of $E F G H$ is $(\sqrt{20})^{2}=20$ square units.

Final answer: 20
Subfield: Geometry


Question: In the diagram, line $A$ has equation $y=2 x$. Line $B$ is obtained by reflecting line $A$ in the $y$-axis. Line $C$ is perpendicular to line $B$. What is the slope of line $C$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=626&width=613&top_left_y=1942&top_left_x=1206)

Solution:
When line $A$ with equation $y=2 x$ is reflected in the $y$-axis, the resulting line (line $B$ ) has equation $y=-2 x$. (Reflecting a line in the $y$-axis changes the sign of the slope.)

Since the slope of line $B$ is -2 and line $C$ is perpendicular to line $B$, then the slope of line $C$ is $\frac{1}{2}$ (the slopes of perpendicular lines are negative reciprocals).

Final answer: $\frac{1}{2}$
Subfield: Geometry


Question: In the diagram, line $A$ has equation $y=2 x$. Line $B$ is obtained by reflecting line $A$ in the $y$-axis. Line $C$ is perpendicular to line $B$. What is the slope of line $C$ ?

<img_3231>

Solution:
When line $A$ with equation $y=2 x$ is reflected in the $y$-axis, the resulting line (line $B$ ) has equation $y=-2 x$. (Reflecting a line in the $y$-axis changes the sign of the slope.)

Since the slope of line $B$ is -2 and line $C$ is perpendicular to line $B$, then the slope of line $C$ is $\frac{1}{2}$ (the slopes of perpendicular lines are negative reciprocals).

Final answer: $\frac{1}{2}$
Subfield: Geometry


Question: Three squares, each of side length 1 , are drawn side by side in the first quadrant, as shown. Lines are drawn from the origin to $P$ and $Q$. Determine, with explanation, the length of $A B$.

<img_3679>

Solution:
Consider the line through $O$ and $P$. To get from $O$ to $P$, we go right 2 and up 1. Since $B$ lies on this line and to get from $O$ to $B$ we go over 1, then we must go up $\frac{1}{2}$, to keep the ratio constant.

Consider the line through $O$ and $Q$. To get from $O$ to $Q$, we go right 3 and up 1. Since $A$ lies on this line and to get from $O$ to $A$ we go over 1, then we must go up $\frac{1}{3}$, to keep the ratio constant.

Therefore, since $A$ and $B$ lie on the same vertical line, then $A B=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.
Since the line through $P$ passes through the origin, then its equation is of the form $y=m x$. Since it passes through the point $(2,1)$, then $1=2 m$, so the line has equation $y=\frac{1}{2} x$. Since $B$ has $x$-coordinate 1, then $y=\frac{1}{2}(1)=\frac{1}{2}$, so $B$ has coordinates $\left(1, \frac{1}{2}\right)$. Similarly, we can determine that the equation of the line through $Q$ is $y=\frac{1}{3} x$, and so $A$ has coordinates $\left(1, \frac{1}{3}\right)$.

Therefore, since $A$ and $B$ lie on the same vertical line, then $A B=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.

Final answer: $\frac{1}{6}$
Subfield: Geometry


Question: Three squares, each of side length 1 , are drawn side by side in the first quadrant, as shown. Lines are drawn from the origin to $P$ and $Q$. Determine, with explanation, the length of $A B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b2aa85be1fe1a84ee805g-1.jpg?height=423&width=615&top_left_y=260&top_left_x=1189)

Solution:
Consider the line through $O$ and $P$. To get from $O$ to $P$, we go right 2 and up 1. Since $B$ lies on this line and to get from $O$ to $B$ we go over 1, then we must go up $\frac{1}{2}$, to keep the ratio constant.

Consider the line through $O$ and $Q$. To get from $O$ to $Q$, we go right 3 and up 1. Since $A$ lies on this line and to get from $O$ to $A$ we go over 1, then we must go up $\frac{1}{3}$, to keep the ratio constant.

Therefore, since $A$ and $B$ lie on the same vertical line, then $A B=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.
Since the line through $P$ passes through the origin, then its equation is of the form $y=m x$. Since it passes through the point $(2,1)$, then $1=2 m$, so the line has equation $y=\frac{1}{2} x$. Since $B$ has $x$-coordinate 1, then $y=\frac{1}{2}(1)=\frac{1}{2}$, so $B$ has coordinates $\left(1, \frac{1}{2}\right)$. Similarly, we can determine that the equation of the line through $Q$ is $y=\frac{1}{3} x$, and so $A$ has coordinates $\left(1, \frac{1}{3}\right)$.

Therefore, since $A$ and $B$ lie on the same vertical line, then $A B=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.

Final answer: $\frac{1}{6}$
Subfield: Combinatorics


Question: Two fair dice, each having six faces numbered 1 to 6 , are thrown. What is the probability that the product of the two numbers on the top faces is divisible by 5 ?

Solution:
There are 36 possibilities for the pair of numbers on the faces when the dice are thrown. For the product of the two numbers, each of which is between 1 and 6 , to be divisible by 5 , one of the two numbers must be equal to 5 .

Therefore, the possible pairs for the faces are

$$
(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)
$$

ie. there are 11 possibilities.

Thus, the probability is $\frac{11}{36}$.
For the product of the two numbers, each of which is between 1 and 6 , to be divisible by 5 , one of the two numbers must be equal to 5 .

When the two dice are thrown, the probability that the first die has a 5 on the top face and any number appears on the second die has any number on the top face is $\frac{1}{6} \times 1=\frac{1}{6}$.

Also, the probability that any number appears on the first die and a 5 appears on the second die is $1 \times \frac{1}{6}=\frac{1}{6}$.



If we consider the sum of these probabilities, we have double-counted the possibility that a 5 occurs on both dice, which happens with probability $\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$.

Therefore, the required probability is $\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$.

Final answer: $\frac{11}{36}$
Subfield: Algebra


Question: If $f(x)=x^{2}-x+2, g(x)=a x+b$, and $f(g(x))=9 x^{2}-3 x+2$, determine all possible ordered pairs $(a, b)$ which satisfy this relationship.

Solution:
First, we compute an expression for the composition of the two given functions:

$$
\begin{aligned}
f(g(x)) & =f(a x+b) \\
& =(a x+b)^{2}-(a x+b)+2 \\
& =a^{2} x^{2}+2 a b x+b^{2}-a x-b+2 \\
& =a^{2} x^{2}+(2 a b-a) x+\left(b^{2}-b+2\right)
\end{aligned}
$$

But we already know that $f(g(x))=9 x^{2}-3 x+2$, so comparing coefficients, we see that

$$
\begin{aligned}
a^{2} & =9 \\
2 a b-a & =-3 \\
b^{2}-b+2 & =2
\end{aligned}
$$

From the first equation, $a=3$ or $a=-3$.

From the third equation, $b^{2}-b=b(b-1)=0$ so $b=0$ or $b=1$.

There are thus 4 possible pairs $(a, b)$ which could solve the problem. We will check which pairs work by looking at the second equation.

From the second equation, $a(2 b-1)=-3$, so if $a=3$ then $b=0$, and if $a=-3$ then $b=1$. Therefore, the possible ordered pairs $(a, b)$ are $(3,0)$ and $(-3,1)$.

Final answer: $(3,0),(-3,1)$
Subfield: Geometry


Question: In the diagram, the parabola with equation $y=x^{2}+t x-2$ intersects the $x$-axis at points $P$ and $Q$.

Also, the line with equation $y=3 x+3$ intersects the parabola at points $P$ and $R$. Determine the value of $t$ and the area of triangle $P Q R$.

<img_3829>

Solution:
Point $P$ is the point where the line $y=3 x+3$ crosses the $x$ axis, and so has coordinates $(-1,0)$.

Therefore, one of the roots of the parabola $y=x^{2}+t x-2$ is $x=-1$, so

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

The parabola now has equation $y=x^{2}-x-2=(x+1)(x-2)$ (we already knew one of the roots so this helped with the factoring) and so its two $x$-intercepts are -1 and 2 , ie. $P$ has coordinates $(-1,0)$ and $Q$ has coordinates $(2,0)$.

We now have to find the coordinates of the point $R$. We know that $R$ is one of the two points of intersection of the line and the parabola, so we equate their equations:

$$
\begin{aligned}
3 x+3 & =x^{2}-x-2 \\
0 & =x^{2}-4 x-5 \\
0 & =(x+1)(x-5)
\end{aligned}
$$

(Again, we already knew one of the solutions to this equation $(x=-1)$ so this made factoring easier.) Since $R$ does not have $x$-coordinate -1 , then $R$ has $x$-coordinate $x=5$. Since $R$ lies on the line, then $y=3(5)+3=18$, so $R$ has coordinates $(5,18)$.

We can now calculate the area of triangle $P Q R$. This triangle has base of length 3 (from $P$ to $Q$ ) and height of length 18 (from the $x$-axis to $R$ ), and so has area $\frac{1}{2}(3)(18)=27$.

Thus, $t=-1$ and the area of triangle $P Q R$ is 27 .

Final answer: -1,27
Subfield: Geometry


Question: In the diagram, the parabola with equation $y=x^{2}+t x-2$ intersects the $x$-axis at points $P$ and $Q$.

Also, the line with equation $y=3 x+3$ intersects the parabola at points $P$ and $R$. Determine the value of $t$ and the area of triangle $P Q R$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b2aa85be1fe1a84ee805g-1.jpg?height=721&width=556&top_left_y=1488&top_left_x=1245)

Solution:
Point $P$ is the point where the line $y=3 x+3$ crosses the $x$ axis, and so has coordinates $(-1,0)$.

Therefore, one of the roots of the parabola $y=x^{2}+t x-2$ is $x=-1$, so

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

The parabola now has equation $y=x^{2}-x-2=(x+1)(x-2)$ (we already knew one of the roots so this helped with the factoring) and so its two $x$-intercepts are -1 and 2 , ie. $P$ has coordinates $(-1,0)$ and $Q$ has coordinates $(2,0)$.

We now have to find the coordinates of the point $R$. We know that $R$ is one of the two points of intersection of the line and the parabola, so we equate their equations:

$$
\begin{aligned}
3 x+3 & =x^{2}-x-2 \\
0 & =x^{2}-4 x-5 \\
0 & =(x+1)(x-5)
\end{aligned}
$$

(Again, we already knew one of the solutions to this equation $(x=-1)$ so this made factoring easier.) Since $R$ does not have $x$-coordinate -1 , then $R$ has $x$-coordinate $x=5$. Since $R$ lies on the line, then $y=3(5)+3=18$, so $R$ has coordinates $(5,18)$.

We can now calculate the area of triangle $P Q R$. This triangle has base of length 3 (from $P$ to $Q$ ) and height of length 18 (from the $x$-axis to $R$ ), and so has area $\frac{1}{2}(3)(18)=27$.

Thus, $t=-1$ and the area of triangle $P Q R$ is 27 .

Final answer: -1,27