Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E;  //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E);  //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW);  //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

Solution: Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ $[\triangle CDB]$ . So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$ . Therefore, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{4}$

Answer: B 4
Problem: In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E;  //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E);  //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW);  //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

Solution: Since $AE \perp AB$ and $BC \perp AB$ $AE \parallel BC$ . By alternate interior angles and $AA\sim$ , we find that $\triangle ADE \sim \triangle CDB$ , with side length ratio $\frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$ . Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{4}$

Answer: B 4
Problem: In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E;  //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E);  //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW);  //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

Solution: Let $[X]$ represent the area of figure $X$ . Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{4}$

Answer: B 4
Problem: In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E;  //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E);  //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW);  //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

Solution: We want to figure out $Area(\triangle ADE) - Area(\triangle BDC)$ .
Notice that $\triangle ABC$ and $\triangle BAE$ "intersect" and form $\triangle ADB$
This means that $Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$ because $Area(\triangle ADB)$ cancels out, which can be seen easily in the diagram.
$Area(\triangle BAE) = 0.5 * 4 * 8 = 16$
$Area(\triangle ABC) = 0.5 * 4 * 16 = 12$
$Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{4}$

Answer: B 4
Problem: A company sells peanut butter in cylindrical jars.  Marketing research suggests that using wider jars will increase sales.  If the diameter of the jars is increased by $25\%$ without altering the volume , by what percent must the height be decreased?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60$

Solution: When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$

Answer: C 36
Problem: Henry's Hamburger Haven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions.  A customer can choose one, two,or three meat patties and any collection of condiments.  How many different kinds of hamburgers can be ordered?
$\text{(A) \ } 24 \qquad \text{(B) \ } 256 \qquad \text{(C) \ } 768 \qquad \text{(D) \ } 40,320 \qquad \text{(E) \ } 120,960$

Solution: For each condiment, a customer may either choose to order it or not.  There are $8$ total condiments to choose from.  Therefore, there are $2^8=256$ ways to order the condiments. There are also $3$ choices for the meat, making a total of $256\times3=768$ possible hamburgers. $\boxed{768}$

Answer: C 768
Problem: At a party, each man danced with exactly three women and each woman danced with exactly two men.  Twelve men attended the party.  How many women attended the party?
$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$

Solution: If each man danced with $3$ women, then there will be a total of $3\times12=36$ pairs of men and women.  However, each woman only danced with $2$ men, so there must have been $\frac{36}2 \Longrightarrow \boxed{18}$ women.

Answer: D 18
Problem: At a party, each man danced with exactly three women and each woman danced with exactly two men.  Twelve men attended the party.  How many women attended the party?
$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$

Solution: Consider drawing out a diagram. Let a circle represent a man, and let a shaded circle represent a woman.
Then, we know that for every 2 men, there will be 3 woman using our diagram. Therefore, the ratio between the number of men and women is 2:3.
Hence, we know that:
$\frac{2}{3} = \frac{12}{x} \implies x = 18 \implies \boxed{18}.$

Answer: D 18
Problem: The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$

Solution: Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{0}$ dimes.

Answer: A 0
Problem: The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$

Solution: If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\boxed{0}.$

Answer: A 0
Problem: Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$

Solution: Rewrite $\frac{(x+y)}x$ as $\frac{x}x+\frac{y}x=1+\frac{y}x$
We also know that $\frac{y}x<0$ because $x$ and $y$ are of opposite sign.
Therefore, $1+\frac{y}x$ is maximized when $|\frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.
This occurs at $(-4,2)$ , so $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{12}$

Answer: D 12
Problem: Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$

Solution: If the answer choice is valid, then it must satisfy $\frac{(x+y)}x$ . We use answer choices from greatest to least since the question asks for the greatest value.
Answer choice $\text{(E)}$ . We see that if $\frac{(x+y)}x = 1$ then
$x+y=x$ and $y=0$ . However, $0$ is not in the domain of $y$ , so $\text{(E)}$ is incorrect.
Answer choice $\text{(D)}$ , however, we can find a value that satisfies $\frac{x+y}{x}=\frac{1}{2}$ which simplifies to $x+2y=0$ , such as $(-4,2)$
Therefore, $\boxed{12}$ is the greatest.

Answer: D 12
Problem: Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$

Solution: As $-4\leq x\leq-2$ , we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or $x+y$ must also be as small as possible.
So we pick our smallest value for $y$ , which is $2$
Now if we if set our value of $x$ to its lowest, our expression becomes $\frac{(-2+2)}{-2} = \frac{0}{2}$ As we decrease our $x$ value, we see that our numerator decrease from $0$ and our denominator decrease from $-2$ at the same rate. So decreasing our $x$ value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number.
So we also pick the smallest value for $x$ , which is $-4$ . We know have $\frac{(-4+2)}{-4} = \frac{-2}{-4} = \frac{1}{2}$
Therefore, $\boxed{12}$ is our greatest possible value.

Answer: D 12
Problem: The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?
2004 AMC 10A problem 16.png
$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20$

Solution: Since there are five types of squares: $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,$ and $5 \times 5.$ We must find how many of each square contain the black shaded square in the center.
If we list them, we get that
Thus, the answer is $1+4+9+4+1=19\Rightarrow\boxed{19}$

Answer: D 19
Problem: The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?
2004 AMC 10A problem 16.png
$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20$

Solution: We use complementary counting. There are only $2\times2$ and $1\times1$ squares that do not contain the black square. Counting, there are $12$ $2\times2$ squares, and $25-1 = 24$ $1\times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible $(25$ $1\times1$ squares $16$ $2\times2$ squares $9$ $3\times3$ squares $4$ $4\times4$ squares and $1$ $5\times5$ square), therefore there are $55-36 = 19$ squares that contain the black square, which is $\boxed{19}$

Answer: D 19
Problem: Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points.  They first meet after Brenda has run 100 meters.  They next meet after Sally has run 150 meters past their first meeting point.  Each girl runs at a constant speed.  What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ }  400\qquad \mathrm{(E) \ } 500$

Solution: Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350\Longrightarrow\boxed{350}$

Answer: C 350
Problem: Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points.  They first meet after Brenda has run 100 meters.  They next meet after Sally has run 150 meters past their first meeting point.  Each girl runs at a constant speed.  What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ }  400\qquad \mathrm{(E) \ } 500$

Solution: The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{350}$

Answer: C 350
Problem: sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{1}$

Answer: A 1
Problem: sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ .  Thus, we get the following equations:
$9r=11+d\Rightarrow d=9r-11$
$9r^2=29+2d$
Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ .  If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ .  Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{1}$

Answer: A 1
Problem: sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: Let the three numbers be, in increasing order, $9,y,z$
Hence, we have that $9-y=y-z\implies 9+z=2y$
Also, from the second part of information given, we get that
$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2$
Plugging back in...
$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$
Simplifying, we get that $z^2-10z-551=0$
Applying the quadratic formula, we get that $z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}$
Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$
However, the problem asks for the minimum value of the third term in a geometric progression.
Hence, the answer is $-19+20=\boxed{1}$

Answer: A 1
Problem: sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1}$

Answer: A 1
Problem: sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\boxed{1}$

Answer: A 1
Problem: sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\boxed{1}$

Answer: A 1
Problem: A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?
[asy]  size(250);defaultpen(linewidth(0.8)); draw(ellipse(origin, 3, 1)); fill((3,0)--(3,2)--(-3,2)--(-3,0)--cycle, white); draw((3,0)--(3,16)^^(-3,0)--(-3,16)); draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); filldraw(ellipse((0, 16), 3, 1), white, black); draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((-3,2)--(0,-1)^^(-3,1)--(-1,-0.89)); draw((0,-1)--(0,15), dashed); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4)); draw((-7,0)--(-5,0)^^(-7,16)--(-5,16)); draw((3,-3)--(-3,-3), Arrows(6)); draw((-6,0)--(-6,16), Arrows(6)); draw((-2,9)--(-1,9), Arrows(3)); label("$3$", (-1.375,9.05), dir(260), fontsize(7)); label("$A$", (0,15), N); label("$B$", (0,-1), NE); label("$30$", (0, -3), S); label("$80$", (-6, 8), W);[/asy]
$\mathrm{(A) \ } 120 \qquad \mathrm{(B) \ } 180 \qquad \mathrm{(C) \ } 240 \qquad \mathrm{(D) \ } 360 \qquad \mathrm{(E) \ } 480$

Solution: The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base $3$ and height $80$ . Thus, we get $3\times80=240\Rightarrow\boxed{240}$

Answer: C 240
Problem: Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$

Solution: Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=x\sqrt{2}$ . If we go angle chasing, we find out that $\angle AEB=75^{\circ}$ , thus $\angle ABE=15^{\circ}$ $\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ . Thus $\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ , or $AE=\frac{x(\sqrt{3}-1)}{2}$ . Thus $AB=\frac{x(\sqrt{3}+1)}{2}$ , and $[ABE]=\frac{x^2}{4}$ , and $[DEF]=\frac{x^2}{2}$ . Thus the ratio of the areas is $\boxed{2}$

Answer: D 2
Problem: Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$

Solution: WLOG, let the side length of $ABCD$ be 1. Let $DE = x$ . It suffices that $AE = 1 - x$ . Then triangles $ABE$ and $CBF$ are congruent by HL, so $CF = AE$ and $DE = DF$ . We find that $BE = EF = x \sqrt{2}$ , and so, by the Pythagorean Theorem, we have $(1 - x)^2 + 1 = 2x^2.$ This yields $x^2 + 2x = 2$ , so $x^2 = 2 - 2x$ . Thus, the desired ratio of areas is \[\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{2}.\]

Answer: D 2
Problem: Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$

Solution: Assume $AB=1$ . Then, $FC$ is $x$ and $ED$ is $1-x$ . We see that using $HL$ $FCB$ is congruent to EAB. Using Pythagoras of triangles $FCB$ and $FDE$ we get $2{(1-x)}^2=x^2+1$ . Expanding, we get $2x^2-4x+2=x^2+1$ . Simplifying gives $x^2-4x+1=0$ solving using completing the square (or other methods) gives 2 answers: $2-\sqrt{3}$ and $2+\sqrt{3}$ . Because $x < 1$ $x=2-\sqrt{3}$ . Using the areas, the answer is $\boxed{2}$

Answer: D 2