di-zhang-fdu/OlympiadExtended huggingface dataset:
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 $A B$.

<img_3333>

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: $\sqrt{5}$
Subfield: Algebra


Question: If $2 \leq x \leq 5$ and $10 \leq y \leq 20$, what is the maximum value of $15-\frac{y}{x}$ ?

Solution:
Since we want to make $15-\frac{y}{x}$ as large as possible, then we want to subtract as little as possible from 15.

In other words, we want to make $\frac{y}{x}$ as small as possible.

To make a fraction with positive numerator and denominator as small as possible, we make the numerator as small as possible and the denominator as large as possible.

Since $2 \leq x \leq 5$ and $10 \leq y \leq 20$, then we make $x=5$ and $y=10$.

Therefore, the maximum value of $15-\frac{y}{x}$ is $15-\frac{10}{5}=13$.
Since $y$ is positive and $2 \leq x \leq 5$, then $15-\frac{y}{x} \leq 15-\frac{y}{5}$ for any $x$ with $2 \leq x \leq 5$ and positive $y$.

Since $10 \leq y \leq 20$, then $15-\frac{y}{5} \leq 15-\frac{10}{5}$ for any $y$ with $10 \leq y \leq 20$.

Therefore, for any $x$ and $y$ in these ranges, $15-\frac{y}{x} \leq 15-\frac{10}{5}=13$, and so the maximum possible value is 13 (which occurs when $x=5$ and $y=10$ ).

Final answer: 13
Subfield: Algebra


Question: The functions $f$ and $g$ satisfy

$$
\begin{aligned}
& f(x)+g(x)=3 x+5 \\
& f(x)-g(x)=5 x+7
\end{aligned}
$$

for all values of $x$. Determine the value of $2 f(2) g(2)$.

Solution:
First, we add the two given equations to obtain

$$
(f(x)+g(x))+(f(x)-g(x))=(3 x+5)+(5 x+7)
$$

or $2 f(x)=8 x+12$ which gives $f(x)=4 x+6$.

Since $f(x)+g(x)=3 x+5$, then $g(x)=3 x+5-f(x)=3 x+5-(4 x+6)=-x-1$.

(We could also find $g(x)$ by subtracting the two given equations or by using the second of the given equations.)

Since $f(x)=4 x+6$, then $f(2)=14$.

Since $g(x)=-x-1$, then $g(2)=-3$.

Therefore, $2 f(2) g(2)=2 \times 14 \times(-3)=-84$.
Since the two given equations are true for all values of $x$, then we can substitute $x=2$ to obtain

$$
\begin{aligned}
& f(2)+g(2)=11 \\
& f(2)-g(2)=17
\end{aligned}
$$

Next, we add these two equations to obtain $2 f(2)=28$ or $f(2)=14$.

Since $f(2)+g(2)=11$, then $g(2)=11-f(2)=11-14=-3$.

(We could also find $g(2)$ by subtracting the two equations above or by using the second of these equations.)

Therefore, $2 f(2) g(2)=2 \times 14 \times(-3)=-84$.

Final answer: -84
Subfield: Combinatorics


Question: Three different numbers are chosen at random from the set $\{1,2,3,4,5\}$.

The numbers are arranged in increasing order.

What is the probability that the resulting sequence is an arithmetic sequence?

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9 is an arithmetic sequence with four terms.)

Solution:
We consider choosing the three numbers all at once.

We list the possible sets of three numbers that can be chosen:

$$
\{1,2,3\}\{1,2,4\}\{1,2,5\} \quad\{1,3,4\} \quad\{1,3,5\} \quad\{1,4,5\} \quad\{2,3,4\} \quad\{2,3,5\} \quad\{2,4,5\} \quad\{3,4,5\}
$$

We have listed each in increasing order because once the numbers are chosen, we arrange them in increasing order.

There are 10 sets of three numbers that can be chosen.

Of these 10, the 4 sequences 1,2,3 and 1,3,5 and 2,3,4 and 3,4,5 are arithmetic sequences. Therefore, the probability that the resulting sequence is an arithmetic sequence is $\frac{4}{10}$ or $\frac{2}{5}$.

Final answer: $\frac{2}{5}$
Subfield: Geometry


Question: In the diagram, $A B C D$ is a quadrilateral with $A B=B C=C D=6, \angle A B C=90^{\circ}$, and $\angle B C D=60^{\circ}$. Determine the length of $A D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_e4d45cb8a520c4aa5010g-1.jpg?height=455&width=477&top_left_y=1057&top_left_x=1258)

Solution:
Join $B$ to $D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4e662a5ccdbfc2f8eac7g-1.jpg?height=453&width=477&top_left_y=283&top_left_x=932)

Consider $\triangle C B D$.

Since $C B=C D$, then $\angle C B D=\angle C D B=\frac{1}{2}\left(180^{\circ}-\angle B C D\right)=\frac{1}{2}\left(180^{\circ}-60^{\circ}\right)=60^{\circ}$.

Therefore, $\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.

Consider $\triangle D B A$.

Note that $\angle D B A=90^{\circ}-\angle C B D=90^{\circ}-60^{\circ}=30^{\circ}$.

Since $B D=B A=6$, then $\angle B D A=\angle B A D=\frac{1}{2}\left(180^{\circ}-\angle D B A\right)=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$.

We calculate the length of $A D$.

Method 1

By the Sine Law in $\triangle D B A$, we have $\frac{A D}{\sin (\angle D B A)}=\frac{B A}{\sin (\angle B D A)}$.

Therefore, $A D=\frac{6 \sin \left(30^{\circ}\right)}{\sin \left(75^{\circ}\right)}=\frac{6 \times \frac{1}{2}}{\sin \left(75^{\circ}\right)}=\frac{3}{\sin \left(75^{\circ}\right)}$.

Method 2

If we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\triangle B D A$ is isosceles. Thus, $A D=2 A P$.

Also, $B P$ bisects $\angle D B A$, so $\angle A B P=15^{\circ}$.

Now, $A P=B A \sin (\angle A B P)=6 \sin \left(15^{\circ}\right)$.

Therefore, $A D=2 A P=12 \sin \left(15^{\circ}\right)$.

Method 3

By the Cosine Law in $\triangle D B A$,

$$
\begin{aligned}
A D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \cos (\angle A B D) \\
& =6^{2}+6^{2}-2(6)(6) \cos \left(30^{\circ}\right) \\
& =72-72\left(\frac{\sqrt{3}}{2}\right) \\
& =72-36 \sqrt{3}
\end{aligned}
$$

Therefore, $A D=\sqrt{36(2-\sqrt{3})}=6 \sqrt{2-\sqrt{3}}$ since $A D>0$.
Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a94f238e941925a9b44ag-1.jpg?height=466&width=482&top_left_y=285&top_left_x=927)

Then $C Q=C D \cos (\angle D C Q)=6 \cos \left(60^{\circ}\right)=6 \times \frac{1}{2}=3$.

Also, $D Q=C D \sin (\angle D C Q)=6 \sin \left(60^{\circ}\right)=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}$.

Since $B C=6$, then $B Q=B C-C Q=6-3=3$.

Now quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.

Thus, $R D=B Q=3$ and $R B=D Q=3 \sqrt{3}$.

Since $A B=6$, then $A R=A B-R B=6-3 \sqrt{3}$.

Since $\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain

$$
A D=\sqrt{R D^{2}+A R^{2}}=\sqrt{3^{2}+(6-3 \sqrt{3})^{2}}=\sqrt{9+36-36 \sqrt{3}+27}=\sqrt{72-36 \sqrt{3}}
$$

which we can rewrite as $A D=\sqrt{36(2-\sqrt{3})}=6 \sqrt{2-\sqrt{3}}$.

Final answer: $6\sqrt{2-\sqrt{3}}$
Subfield: Geometry


Question: In the diagram, $A B C D$ is a quadrilateral with $A B=B C=C D=6, \angle A B C=90^{\circ}$, and $\angle B C D=60^{\circ}$. Determine the length of $A D$.

<img_3208>

Solution:
Join $B$ to $D$.

<img_3655>

Consider $\triangle C B D$.

Since $C B=C D$, then $\angle C B D=\angle C D B=\frac{1}{2}\left(180^{\circ}-\angle B C D\right)=\frac{1}{2}\left(180^{\circ}-60^{\circ}\right)=60^{\circ}$.

Therefore, $\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.

Consider $\triangle D B A$.

Note that $\angle D B A=90^{\circ}-\angle C B D=90^{\circ}-60^{\circ}=30^{\circ}$.

Since $B D=B A=6$, then $\angle B D A=\angle B A D=\frac{1}{2}\left(180^{\circ}-\angle D B A\right)=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$.

We calculate the length of $A D$.

Method 1

By the Sine Law in $\triangle D B A$, we have $\frac{A D}{\sin (\angle D B A)}=\frac{B A}{\sin (\angle B D A)}$.

Therefore, $A D=\frac{6 \sin \left(30^{\circ}\right)}{\sin \left(75^{\circ}\right)}=\frac{6 \times \frac{1}{2}}{\sin \left(75^{\circ}\right)}=\frac{3}{\sin \left(75^{\circ}\right)}$.

Method 2

If we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\triangle B D A$ is isosceles. Thus, $A D=2 A P$.

Also, $B P$ bisects $\angle D B A$, so $\angle A B P=15^{\circ}$.

Now, $A P=B A \sin (\angle A B P)=6 \sin \left(15^{\circ}\right)$.

Therefore, $A D=2 A P=12 \sin \left(15^{\circ}\right)$.

Method 3

By the Cosine Law in $\triangle D B A$,

$$
\begin{aligned}
A D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \cos (\angle A B D) \\
& =6^{2}+6^{2}-2(6)(6) \cos \left(30^{\circ}\right) \\
& =72-72\left(\frac{\sqrt{3}}{2}\right) \\
& =72-36 \sqrt{3}
\end{aligned}
$$

Therefore, $A D=\sqrt{36(2-\sqrt{3})}=6 \sqrt{2-\sqrt{3}}$ since $A D>0$.
Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.

<img_3835>

Then $C Q=C D \cos (\angle D C Q)=6 \cos \left(60^{\circ}\right)=6 \times \frac{1}{2}=3$.

Also, $D Q=C D \sin (\angle D C Q)=6 \sin \left(60^{\circ}\right)=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}$.

Since $B C=6$, then $B Q=B C-C Q=6-3=3$.

Now quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.

Thus, $R D=B Q=3$ and $R B=D Q=3 \sqrt{3}$.

Since $A B=6$, then $A R=A B-R B=6-3 \sqrt{3}$.

Since $\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain

$$
A D=\sqrt{R D^{2}+A R^{2}}=\sqrt{3^{2}+(6-3 \sqrt{3})^{2}}=\sqrt{9+36-36 \sqrt{3}+27}=\sqrt{72-36 \sqrt{3}}
$$

which we can rewrite as $A D=\sqrt{36(2-\sqrt{3})}=6 \sqrt{2-\sqrt{3}}$.

Final answer: $6\sqrt{2-\sqrt{3}}$
Subfield: Combinatorics


Question: What is the largest two-digit number that becomes $75 \%$ greater when its digits are reversed?

Solution:
Let $n$ be the original number and $N$ be the number when the digits are reversed. Since we are looking for the largest value of $n$, we assume that $n>0$.

Since we want $N$ to be $75 \%$ larger than $n$, then $N$ should be $175 \%$ of $n$, or $N=\frac{7}{4} n$.

Suppose that the tens digit of $n$ is $a$ and the units digit of $n$ is $b$. Then $n=10 a+b$.

Also, the tens digit of $N$ is $b$ and the units digit of $N$ is $a$, so $N=10 b+a$.

We want $10 b+a=\frac{7}{4}(10 a+b)$ or $4(10 b+a)=7(10 a+b)$ or $40 b+4 a=70 a+7 b$ or $33 b=66 a$, and so $b=2 a$.

This tells us that that any two-digit number $n=10 a+b$ with $b=2 a$ has the required property.

Since both $a$ and $b$ are digits then $b<10$ and so $a<5$, which means that the possible values of $n$ are 12, 24, 36, and 48 .

The largest of these numbers is 48.

Final answer: 48
Subfield: Geometry


Question: A triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0<k<4$. If the area of the triangle is 8 , determine the value of $k$.

![](https://cdn.mathpix.com/cropped/2023_12_21_e4d45cb8a520c4aa5010g-1.jpg?height=371&width=540&top_left_y=1701&top_left_x=1256)

Solution:
We "complete the rectangle" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a94f238e941925a9b44ag-1.jpg?height=369&width=533&top_left_y=2255&top_left_x=899)



Since $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.

Since $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .

Since $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .

Therefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \times 5=20$.

Now rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .

Let us examine each of these triangles:

- $\triangle A B C$ has area 8 (given information)
- $\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\frac{1}{2} \times 4 \times 3=6$.
- $\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\frac{1}{2} \times k \times 2=k$.
- $\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\frac{1}{2} \times(4-k) \times 5=10-\frac{5}{2} k$.

Since the sum of the areas of these triangles is 20 , then $8+6+k+10-\frac{5}{2} k=20$ or $4=\frac{3}{2} k$ and so $k=\frac{8}{3}$.

Final answer: $\frac{8}{3}$
Subfield: Geometry


Question: A triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0<k<4$. If the area of the triangle is 8 , determine the value of $k$.

<img_3619>

Solution:
We "complete the rectangle" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.

<img_3215>



Since $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.

Since $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .

Since $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .

Therefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \times 5=20$.

Now rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .

Let us examine each of these triangles:

- $\triangle A B C$ has area 8 (given information)
- $\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\frac{1}{2} \times 4 \times 3=6$.
- $\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\frac{1}{2} \times k \times 2=k$.
- $\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\frac{1}{2} \times(4-k) \times 5=10-\frac{5}{2} k$.

Since the sum of the areas of these triangles is 20 , then $8+6+k+10-\frac{5}{2} k=20$ or $4=\frac{3}{2} k$ and so $k=\frac{8}{3}$.

Final answer: $\frac{8}{3}$
Subfield: Geometry


Question: Square $O P Q R$ has vertices $O(0,0), P(0,8), Q(8,8)$, and $R(8,0)$. The parabola with equation $y=a(x-2)(x-6)$ intersects the sides of the square $O P Q R$ at points $K, L, M$, and $N$. Determine all the values of $a$ for which the area of the trapezoid $K L M N$ is 36 .

Solution:
First, we note that $a \neq 0$. (If $a=0$, then the "parabola" $y=a(x-2)(x-6)$ is actually the horizontal line $y=0$ which intersects the square all along $O R$.)

Second, we note that, regardless of the value of $a \neq 0$, the parabola has $x$-intercepts 2 and 6 , and so intersects the $x$-axis at $(2,0)$ and $(6,0)$, which we call $K(2,0)$ and $L(6,0)$. This gives $K L=4$.

Third, we note that since the $x$-intercepts of the parabola are 2 and 6 , then the axis of symmetry of the parabola has equation $x=\frac{1}{2}(2+6)=4$.

Since the axis of symmetry of the parabola is a vertical line of symmetry, then if the parabola intersects the two vertical sides of the square, it will intersect these at the same height, and if the parabola intersects the top side of the square, it will intersect it at two points that are symmetrical about the vertical line $x=4$.

Fourth, we recall that a trapezoid with parallel sides of lengths $a$ and $b$ and height $h$ has area $\frac{1}{2} h(a+b)$.

We now examine three cases.



Case 1: $a<0$

Here, the parabola opens downwards.

Since the parabola intersects the square at four points, it must intersect $P Q$ at points $M$ and $N$. (The parabola cannot intersect the vertical sides of the square since it gets "narrower" towards the vertex.)

<img_3541>

Since the parabola opens downwards, then $M N<K L=4$.

Since the height of the trapezoid equals the height of the square (or 8), then the area of the trapezoid is $\frac{1}{2} h(K L+M N)$ which is less than $\frac{1}{2}(8)(4+4)=32$.

But the area of the trapezoid must be 36 , so this case is not possible.

Case 2: $a>0 ; M$ and $N$ on $P Q$

We have the following configuration:

<img_4039>

Here, the height of the trapezoid is $8, K L=4$, and $M$ and $N$ are symmetric about $x=4$. Since the area of the trapezoid is 36 , then $\frac{1}{2} h(K L+M N)=36$ or $\frac{1}{2}(8)(4+M N)=36$ or $4+M N=9$ or $M N=5$.

Thus, $M$ and $N$ are each $\frac{5}{2}$ units from $x=4$, and so $N$ has coordinates $\left(\frac{3}{2}, 8\right)$.

Since this point lies on the parabola with equation $y=a(x-2)(x-6)$, then $8=a\left(\frac{3}{2}-2\right)\left(\frac{3}{2}-6\right)$ or $8=a\left(-\frac{1}{2}\right)\left(-\frac{9}{2}\right)$ or $8=\frac{9}{4} a$ or $a=\frac{32}{9}$.



Case 3: $a>0 ; M$ and $N$ on $Q R$ and $P O$

We have the following configuration:

<img_3268>

Here, $K L=4, M N=8$, and $M$ and $N$ have the same $y$-coordinate.

Since the area of the trapezoid is 36 , then $\frac{1}{2} h(K L+M N)=36$ or $\frac{1}{2} h(4+8)=36$ or $6 h=36$ or $h=6$.

Thus, $N$ has coordinates $(0,6)$.

Since this point lies on the parabola with equation $y=a(x-2)(x-6)$, then $6=a(0-2)(0-6)$ or $6=12 a$ or $a=\frac{1}{2}$.

Therefore, the possible values of $a$ are $\frac{32}{9}$ and $\frac{1}{2}$.

Final answer: $\frac{32}{9}$,$\frac{1}{2}$
Subfield: Algebra


Question: A 75 year old person has a $50 \%$ chance of living at least another 10 years.

A 75 year old person has a $20 \%$ chance of living at least another 15 years. An 80 year old person has a $25 \%$ chance of living at least another 10 years. What is the probability that an 80 year old person will live at least another 5 years?

Solution:
Consider a population of 100 people, each of whom is 75 years old and who behave according to the probabilities given in the question.

Each of the original 100 people has a $50 \%$ chance of living at least another 10 years, so there will be $50 \% \times 100=50$ of these people alive at age 85 .

Each of the original 100 people has a $20 \%$ chance of living at least another 15 years, so there will be $20 \% \times 100=20$ of these people alive at age 90 .

Since there is a $25 \%$ ( or $\frac{1}{4}$ ) chance that an 80 year old person will live at least another 10 years (that is, to age 90), then there should be 4 times as many of these people alive at age 80 than at age 90 .

Since there are 20 people alive at age 90 , then there are $4 \times 20=80$ of the original 100 people alive at age 80 .

In summary, of the initial 100 people of age 75, there are 80 alive at age 80,50 alive at age 85 , and 20 people alive at age 90 .

Because 50 of the 80 people alive at age 80 are still alive at age 85 , then the probability that an 80 year old person will live at least 5 more years (that is, to age 85 ) is $\frac{50}{80}=\frac{5}{8}$, or $62.5 \%$.
Suppose that the probability that a 75 year old person lives to 80 is $p$, the probability that an 80 year old person lives to 85 is $q$, and the probability that an 85 year old person lives to 90 is $r$.

We want to the determine the value of $q$.

For a 75 year old person to live at least another 10 years, they must live another 5 years (to age 80) and then another 5 years (to age 85). The probability of this is equal to $p q$. We are told in the question that this is equal to $50 \%$ or 0.5 .

Therefore, $p q=0.5$.



For a 75 year old person to live at least another 15 years, they must live another 5 years (to age 80), then another 5 years (to age 85), and then another 5 years (to age 90). The probability of this is equal to $p q r$. We are told in the question that this is equal to $20 \%$ or 0.2 .

Therefore, $p q r=0.2$

Similarly, since the probability that an 80 year old person will live another 10 years is $25 \%$, then $q r=0.25$.

Since $p q r=0.2$ and $p q=0.5$, then $r=\frac{p q r}{p q}=\frac{0.2}{0.5}=0.4$.

Since $q r=0.25$ and $r=0.4$, then $q=\frac{q r}{r}=\frac{0.25}{0.4}=0.625$.

Therefore, the probability that an 80 year old man will live at least another 5 years is 0.625 , or $62.5 \%$.

Final answer: 62.5%
Subfield: Algebra


Question: Determine all values of $x$ for which $2^{\log _{10}\left(x^{2}\right)}=3\left(2^{1+\log _{10} x}\right)+16$.

Solution:
Using logarithm rules, the given equation is equivalent to $2^{2 \log _{10} x}=3\left(2 \cdot 2^{\log _{10} x}\right)+16$ or $\left(2^{\log _{10} x}\right)^{2}=6 \cdot 2^{\log _{10} x}+16$.

Set $u=2^{\log _{10} x}$. Then the equation becomes $u^{2}=6 u+16$ or $u^{2}-6 u-16=0$.

Factoring, we obtain $(u-8)(u+2)=0$ and so $u=8$ or $u=-2$.

Since $2^{a}>0$ for any real number $a$, then $u>0$ and so we can reject the possibility that $u=-2$.

Thus, $u=2^{\log _{10} x}=8$ which means that $\log _{10} x=3$.

Therefore, $x=1000$.

Final answer: 1000
Subfield: Algebra


Question: The Sieve of Sundaram uses the following infinite table of positive integers:

| 4 | 7 | 10 | 13 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 12 | 17 | 22 | $\cdots$ |
| 10 | 17 | 24 | 31 | $\cdots$ |
| 13 | 22 | 31 | 40 | $\cdots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |

The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown.
Determine the number in the 50th row and 40th column.

Solution:
First, we determine the first entry in the 50th row.

Since the first column is an arithmetic sequence with common difference 3, then the 50th entry in the first column (the first entry in the 50th row) is $4+49(3)=4+147=151$.

Second, we determine the common difference in the 50th row by determining the second entry in the 50th row.

Since the second column is an arithmetic sequence with common difference 5 , then the 50 th entry in the second column (that is, the second entry in the 50th row) is $7+49(5)$ or $7+245=252$.

Therefore, the common difference in the 50th row must be $252-151=101$.

Thus, the 40th entry in the 50th row (that is, the number in the 50th row and the 40th column) is $151+39(101)=151+3939=4090$.

Final answer: 4090
Subfield: Algebra


Question: The Sieve of Sundaram uses the following infinite table of positive integers:

| 4 | 7 | 10 | 13 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 12 | 17 | 22 | $\cdots$ |
| 10 | 17 | 24 | 31 | $\cdots$ |
| 13 | 22 | 31 | 40 | $\cdots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |

The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown.
Determine a formula for the number in the $R$ th row and $C$ th column.

Solution:
First, we determine the first entry in the $R$ th row.

Since the first column is an arithmetic sequence with common difference 3 , then the $R$ th entry in the first column (that is, the first entry in the $R$ th row) is $4+(R-1)(3)$ or $4+3 R-3=3 R+1$.

Second, we determine the common difference in the $R$ th row by determining the second entry in the $R$ th row.

Since the second column is an arithmetic sequence with common difference 5 , then the $R$ th entry in the second column (that is, the second entry in the $R$ th row) is $7+(R-1)(5)$ or $7+5 R-5=5 R+2$.

Therefore, the common difference in the $R$ th row must be $(5 R+2)-(3 R+1)=2 R+1$. Thus, the $C$ th entry in the $R$ th row (that is, the number in the $R$ th row and the $C$ th column) is

$$
3 R+1+(C-1)(2 R+1)=3 R+1+2 R C+C-2 R-1=2 R C+R+C
$$

Final answer: $2RC+R+C$