Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
$\text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2}$

Solution: Assume Moe, Loki, and Nick each give Ott $$ 1$ . Therefore, Moe has $$ 5$ , Loki has $$ 4$ , and Nick has $$ 3$ . After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original; which is $$ 12$ . Ott gets $$ 1$ $+$ $$ 1$ $+$ $$ 1$ $=$ $$ 3$ . Thus, the answer is $\frac{3}{12}=\boxed{14}$

Answer: B 14
Problem: Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will he need to do his job?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$

Solution: It will take him $300\cdot2=600$ seconds to paint all the dimples.
This is equivalent to $\frac{600}{60}=10$ minutes $\Rightarrow \boxed{10}$

Answer: D 10
Problem: I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Solution: Let the numbers be $x$ and $y$ . Then we have $x+y=11$ and $xy=24$ . Solving for $x$ in the first equation yields $x=11-y$ , and substituting this into the second equation gives $(11-y)(y)=24$ . Simplifying this gives $-y^2+11y=24$ , or $y^2-11y+24=0$ . This factors as $(y-3)(y-8)=0$ , so $y=3$ or $y=8$ , and the corresponding $x$ values are $x=8$ and $x=3$ . These are essentially the same answer: one number is $3$ and one number is $8$ , so the largest number is $8, \boxed{8}$

Answer: D 8
Problem: I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Solution: Use the answers to attempt to "reverse engineer" an appropriate pair of numbers.  Looking at option $A$ , guess that one of the numbers is $3$ .  If the sum of two numbers is $11$ and one is $3$ , then other must be $11 - 3 = 8$ .  The product of those numbers is $3\cdot 8 = 24$ , which is the second condition of the problem, so our number are $3$ and $8$
However, $3$ is the smaller of the two numbers, so the answer is $8$ or $\boxed{8}$

Answer: D 8
Problem: Granny Smith has $63. Elberta has $2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?
$\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$

Solution: Since Anjou has $\frac{1}{3}$ the amount of money as Granny Smith and Granny Smith has $ $63$ , Anjou has $\frac{1}{3}\times63=21$ dollars. Elberta has $ $2$ more than this, so she has $ $23$ , or $\boxed{23}$

Answer: E 23
Problem: The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 9$

Solution: Since the number is even, the last digit must be $2$ or $4$ . To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is $1$ . Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is decided, so the thousands digit is $2$ . Similarly, the hundreds digit needs to be the next smallest number, so it is $3$ . However, for the tens digit, we can't use $4$ , since we already used $2$ and the number must be even, so the units digit must be $4$ and the tens digit is $9, \boxed{9}$ (The number is $12394$ .)

Answer: E 9
Problem: Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is 60 feet. What is the distance in feet between the first and last trees?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 105 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 140$

Solution: There are $3$ spaces between the 1st and 4th trees, so each of these spaces has $\frac{60}{3}=20$ feet. Between the first and last trees there are $5$ spaces, so the distance between them is $20\times5=100$ feet, $\boxed{100}$

Answer: B 100
Problem: To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
What is the number of square inches in the area of the small kite?
$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

Solution: The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$ , so the area is $\frac{(6)(7)}{2}=21, \boxed{21}$

Answer: A 21
Problem: To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
What is the number of square inches in the area of the small kite?
$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

Solution: Drawing in the diagonals of the kite will form four right triangles on the "inside" part of the grid.  Drawing in the border of the 7 by 6 grid will form four right triangles on the "outside" part of the grid.  Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside, the area of the kite is half the total area of the grid, or $\frac{(6)(7)}{2}=21, \boxed{21}$

Answer: A 21
Problem: To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
What is the number of square inches in the area of the small kite?
$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

Solution: Pick's Theorem states: \[\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1\] as the area of a figure on a grid. Counting, we see there are $4$ boundary points and $20$ interior points. Therefore, we have \[\frac{4}{2}+20-1\implies 20+1\implies 21.\] Hence, the answer is $\boxed{21}$

Answer: A 21
Problem: To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] size(85); for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 38 \qquad \text{(E)}\ 39$

Solution: Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in each direction. The diagonals of the small kite are $6$ and $7$ , so the diagonals of the large kite are $18$ and $21$ , and the amount of bracing Genevieve needs is the sum of these lengths, which is $39, \boxed{39}$

Answer: E 39
Problem: To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]
The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?
$\textbf{(A)}\ 63 \qquad \textbf{(B)}\ 72 \qquad \textbf{(C)}\ 180 \qquad \textbf{(D)}\ 189 \qquad \textbf{(E)}\ 264$

Solution: The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\times3(7)$ , or $18\times21$ , so the area is $(18)(21)=378$ . The area of the kite is half of the area of the rectangle as you can see, so the area of the waste material is also half the area of the rectangle.  Thus, the area of the waste material is $378/2=\boxed{189}$

Answer: D 189
Problem: A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$

Solution: $2000\%$ is equivalent to $20\times100\%$ . Therefore, $2000\%$ of a number is the same as $20$ times that number. $4$ quarters is $1$ dollar, so Bryden will get $20\times1={20}$ dollars, $\boxed{20}$

Answer: A 20
Problem: A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$

Solution: Since $2000\%$ is just $\frac{2000}{100}$ , we can multiply that by $100$ , because four quarters is a $100$ cents. After the multiplication, we get $2000$ . Since our answer is in cents right now, we need to convert it to dollars, which would be $\boxed{20}$ dollars.

Answer: A 20
Problem: Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is
[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } }  draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy]
$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

Solution: [asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } }  draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } { draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewidth(1)); }  label("$A$",(3,2),NE); label("$B$",(3,-2),SE); label("$C$",(-3,-2),SW); label("$D$",(-3,0),NW); [/asy]
This quadrilateral is a trapezoid, because $AB\parallel CD$ but $BC$ is not parallel to $AD$ . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are $AB$ and $CD$ , which have lengths $2$ and $4$ , respectively, so the length of the median is $\frac{2+4}{2}=3$ $CB$ is perpendicular to the bases, so it is the height, and has length $6$ . Therefore, the area of the trapezoid is $(3)(6)=18, \boxed{18}$

Answer: C 18
Problem: Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is
[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } }  draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy]
$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

Solution: Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle.  Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle.
$A_{trap} = A_{tri} + A_{rect}$
$A_{trap} = \frac{1}{2}bh + lw$
$A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2$
$A_{trap}  = 6 + 12 = 18 \rightarrow \boxed{18}$

Answer: C 18
Problem: If $a\otimes b = \dfrac{a + b}{a - b}$ , then $(6\otimes 4)\otimes 3 =$
$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$

Solution: $6\otimes4=\frac{6+4}{6-4}=5$ $5\otimes3=\frac{5+3}{5-3}=4, \boxed{4}$

Answer: A 4
Problem: Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 72$

Solution: There are $36$ students in the class: $12$ prefer chocolate pie, $8$ prefer apple pie, and $6$ prefer blueberry pie. Therefore, $36-12-8-6=10$ students prefer cherry pie or lemon pie. Half of these prefer each, so $5$ students prefer cherry pie. This means that $\frac{5}{36}$ of the students prefer cherry pie, so $\frac{5}{36}$ of the full $360^\circ$ should be used for cherry pie. This is $(\frac{5}{36})(360^\circ)=50^\circ, \boxed{50}$

Answer: D 50
Problem: Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?
$\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$

Solution: There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, we counted 'baked beans and corn' and 'corn and baked beans' as $2$ different possibilities, so the total possibilites for the two vegetables is $\frac{4\times3}{2}=6$ , and the total number of possibilites is $12\times6=\boxed{72}.$

Answer: C 72
Problem: Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled?
$\text{(A)}\ 20 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 40$

Solution: After the $4$ minutes of Homer peeling alone, he had peeled $4\times3=12$ potatoes. This means that there are $44-12=32$ potatoes left. Once Christen joins him, the two are peeling potatoes at a rate of $3+5=8$ potatoes per minute. So, they finish peeling after another $\frac{32}{8}=4$ minutes. In these $4$ minutes, Christen peeled $4\times5=20$ potatoes, $\boxed{20}$

Answer: A 20
Problem: The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$

Solution: Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than $18$ must be positive and distinct, so the smallest possible numbers for these are $1$ and $2$ . The number right after $18$ also needs to be as small as possible, so it must be $19$ . This means that the remaining number, the maximum possible value for a number in the set, is $75-1-2-18-19=35, \boxed{35}$

Answer: D 35
Problem: On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$

Solution: The highest possible score is if you get every answer right, to get $5(20)=100$ . The second highest possible score is if you get $19$ questions right and leave the remaining one blank, to get a $5(19)+1(1)=96$ . Therefore, no score between $96$ and $100$ , exclusive, is possible, so $97$ is not possible, $\boxed{97}$

Answer: E 97
Problem: On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$

Solution: We can equivalently construct the following rules: You have 100 point at first, but if you give the wrong answer, you will lose 5 points, if you don't answer a question you will lose 4 points. Obviously, you can lose 10 points, 9 points, 8 points, 5 points or 4 points, but you cannot lose 3 points. The answer is $\boxed{97}$

Answer: E 97
Problem: Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution: There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.
Case 1: Triangles congruent to $\triangle RST$ There is obviously only $1$ of these: $\triangle RST$ itself.
Case 2: Triangles congruent to $\triangle SYZ$ There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$
Case 3: Triangles congruent to $\triangle RSX$ There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$
Case 4: Triangles congruent to $\triangle SYX$ There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$
However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$ . Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{4}$

Answer: D 4
Problem: Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution: We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is $3+1 = \boxed{4}$

Answer: D 4
Problem: Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution: Notice that 20 is obviously too high (There are only 20 ways to choose 3 of the points to form a triangle or a line in total!) and you can count 4 distinct triangles quickly: $\triangle RYX$ $\triangle RYT$ $\triangle RYZ$ $\triangle RST$ . So the answer is $\boxed{4}$

Answer: D 4
Problem: Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?
[asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$

Solution: Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue triangles on each side, so there is $1$ red triangle, $2$ blue triangles, and $8$ white triangles remaining on each half. Also, we have $2$ red-white pairs. This obviously can't use $2$ red triangles on one side, since there is only $1$ on each side, so we must use $1$ red triangle and $1$ white triangle per side, leaving $2$ blue triangles and $7$ white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than $3$ blue pairs, so the remaining blue triangles must be paired with white triangles, yielding $4$ blue-white pairs, one for each of the remaining blue triangles. This uses $2$ blue triangles and $2$ white triangles on each side, leaving $5$ white triangles apiece, which must be paired with each other, so there are $5$ white-white pairs, $\boxed{5}$

Answer: B 5