Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $a$ and $b$ be real numbers greater than $1$ for which there exists a positive real number $c,$ different from $1$ , such that
\[2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.\]
Find the largest possible value of $\log_a b.$
$\text{(A) }\sqrt{2} \qquad \text{(B) }\sqrt{3} \qquad \text{(C) }2 \qquad \text{(D) }\sqrt{6} \qquad \text{(E) }3$

Solution: We may rewrite the given equation as \[2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}\] Since $c \neq 1$ , we have $\log c \neq 0$ , so we may divide by $\log c$ on both sides. After making the substitutions $x = \log a$ and $y = \log b$ , our equation becomes \[\frac {2}{x} + \frac {2}{y} = \frac {9}{x+y}\]
Rewriting the left-hand side gives \[\frac {2(x+y)}{xy} = \frac {9}{x+y}\]
Cross-multiplying gives $2(x+y)^2 = 9xy$ or \[2x^2 - 5xy + 2y^2 = 0\]
Factoring gives $(2x-y)(x-2y) = 0$ or $\frac {x}{y} = 2, \frac {1}{2}$
Recall that $\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b$ . Therefore, the maximum value of $\log_{a} b$ is $\boxed{2}$

Answer: C 2
Problem: Under the new AMC $10, 12$ scoring method, $6$ points are given for each correct answer, $2.5$ points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between $0$ and $150$ can be obtained in only one way, for example, a score of $104.5$ can be obtained with $17$ correct answers, $1$ unanswered question, and $7$ incorrect answers, and also with $12$ correct answers and $13$ unanswered questions. There are scores that can be obtained in exactly three ways. What is their sum?
$\text{(A) }175 \qquad \text{(B) }179.5 \qquad \text{(C) }182 \qquad \text{(D) }188.5 \qquad \text{(E) }201$

Solution: $\boxed{188.5}$

Answer: D 188.5
Problem: The equation $z(z+i)(z+3i)=2002i$ has a zero of the form $a+bi$ , where $a$ and $b$ are positive real numbers. Find $a.$
$\text{(A) }\sqrt{118} \qquad \text{(B) }\sqrt{210} \qquad \text{(C) }2 \sqrt{210} \qquad \text{(D) }\sqrt{2002} \qquad \text{(E) }100 \sqrt{2}$

Solution: According to Wolfram-Alpha, the answer is $\boxed{118}$

Answer: A 118
Problem: Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits
of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a
two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$
$\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)}\ 9$

Solution: Denote $a$ and $b$ as the tens and units digit of $N$ , respectively. Then $N = 10a+b$ . It follows that $10a+b=ab+a+b$ , which implies that $9a=ab$ . Since $a\neq0$ $b=9$ . So the units digit of $N$ is $\boxed{9}$

Answer: E 9
Problem: Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$

Solution: Letting $x = 500$ and $y = \dfrac65$ in the given equation, we get $f(500\cdot\frac65) = \frac3{\frac65} = \frac52$ , or $f(600) = \boxed{52}$

Answer: C 52
Problem: Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$

Solution: The only function that satisfies the given condition is $y = \frac{k}{x}$ , for some constant $k$ . Thus, the answer is $\frac{500 \cdot 3}{600} = \boxed{52}$

Answer: C 52
Problem: Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$

Solution: Note that the equation given above is symmetric, so we have $x \cdot f(x)=y \cdot f(y)$ . Plugging in $x=500$ and $y=600$ gives $f(y)=\boxed{52}$

Answer: C 52
Problem: Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$ , how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$

Solution: Each of the $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$ $A_2A_5A_8$ , and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles.

Answer:  66
Problem: An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)
$\text{(A) }\frac {1}{2} \sqrt {3} \qquad \text{(B) }1 \qquad \text{(C) }\sqrt {2} \qquad \text{(D) }\frac {3}{2} \qquad \text{(E) }2$

Solution: Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. Its length is the same as the length of the tetrahedron's edge, i.e. $\boxed{1}$

Answer:  1
Problem: A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W);  filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$

Solution: [asy] unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$S$",S,N);  filldraw( circle(S,4/9), lightgray, black ); dot(S); draw( rightanglemark(A,C,B) ); draw( S -- A ); draw( S -- B ); [/asy]
In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$ , thus by the Pythagorean theorem we have $AC=4$
Let $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $S$ to $\overline{AC}$ . Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$
We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$ . Hence we get the following two equations:
\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*}
Simplifying both, we get
\begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*}
As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$
Now there are two possibilities: either $\frac{4-s}s=-2$ , or $\frac{4-s}s=2$
In the first case clearly $s<0$ , which puts the center on the wrong side of $A$ , so this is not the correct case.
(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$ , this circle turns out to have the same radius as circle $B$ , with center directly left of center $B$ , and tangent to $B$ directly above center $A$ .)
The second case solves to $s=\frac 43$ . We then have $4r = s^2 = \frac {16}9$ , hence $r = \boxed{49}$

Answer:  49
Problem: A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W);  filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$

Solution: The horizontal line is the equivalent of a circle of curvature $0$ , thus we can apply Descartes' Circle Formula
The four circles have curvatures $0, 1, \frac 14$ , and $\frac 1r$
We have $2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2$
Simplifying, we get $\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}$
\[\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0\] \[\frac{16}{r^2}-\frac{40}{r}+9=0\] \[\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0\]
Obviously $r$ cannot equal $4$ , therefore $r = \boxed{49}$

Answer:  49
Problem: Points $A = (3,9)$ $B = (1,1)$ $C = (5,3)$ , and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$ . The quadrilateral formed by joining the midpoints of $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$
$\text{(A) }7 \qquad \text{(B) }9 \qquad \text{(C) }10 \qquad \text{(D) }12 \qquad \text{(E) }16$

Solution: [asy] pair A=(3,9), B=(1,1), C=(5,3), D=(7,3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,E); pair AB = (A + B)/2, BC = (B + C)/2, CD = (C + D)/2, DA = (D + A)/2; draw(AB--BC--CD--DA--cycle); [/asy]
We already know two vertices of the square: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$
There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$ , or they are $(0,1)$ and $(-1,4)$ . The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$ , we can compute $D=(7,3)$ , and $7+3=\boxed{10}$

Answer:  10
Problem: Four positive integers $a$ $b$ $c$ , and $d$ have a product of $8!$ and satisfy:
\[\begin{array}{rl} ab + a + b & = 524 \\  bc + b + c & = 146 \\  cd + c + d & = 104 \end{array}\]
What is $a-d$
$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$

Solution: Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
\begin{align*} (a+1)(b+1) & = 525 \\  (b+1)(c+1) & = 147 \\  (c+1)(d+1) & = 105 \end{align*}
Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$ . We get:
\begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\  fg & = 3\cdot 7\cdot 7 \\  gh & = 3\cdot 5\cdot 7 \end{align*}
Clearly $7^2$ divides $fg$ . On the other hand, $7^2$ can not divide $f$ , as it then would divide $ef$ . Similarly, $7^2$ can not divide $g$ . Hence $7$ divides both $f$ and $g$ . This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$
The first case solves to $(e,f,g,h)=(75,7,21,5)$ , which gives us $(a,b,c,d)=(74,6,20,4)$ , but then $abcd \not= 8!$ . We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$ . (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)
The second case solves to $(e,f,g,h)=(25,21,7,15)$ , which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$ , and we have $a-d=24-14 =\boxed{10}$

Answer:  10
Problem: In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution: [asy] unitsize(0.5cm); defaultpen(0.8); pair A=(0,0), B=(10,0), C=(10,7), D=(0,7), E=(C+D)/2, F=(2*A+B)/3, G=(A+2*B)/3; pair H = intersectionpoint(A--C,E--F); pair J = intersectionpoint(A--C,E--G); draw(A--B--C--D--cycle); draw(G--E--F); draw(A--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,S); label("$G$",G,S); label("$H$",H,SE); label("$J$",J,ESE); filldraw(E--H--J--cycle,lightgray,black); draw(H--D, dashed); [/asy] Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac {AH}{HC} = \frac{AF}{EC} = \frac 23$
Also, triangles $AGJ$ and $CEJ$ are similar, hence $\frac {AJ}{JC} = \frac {AG}{EC} = \frac 43$
We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$ . We have:
Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}$

Answer:  3
Problem: In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution: As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$
We can then compute that $HJ = AC \cdot \left( \frac 47 - \frac 25 \right) = AC \cdot \frac{6}{35}$
As $E$ is the midpoint of $CD$ , the height from $E$ onto $AC$ is $1/2$ of the height from $D$ onto $AC$ . Therefore we have $[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}$

Answer:  3
Problem: In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution: Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are $7$ and $10$ (or any other two positive real numbers that multiply to 70). We can find $H$ and $J$ by intersecting lines, and then we calculate the area of $EHJ$ using shoelace formula. This yields $\boxed{3}$

Answer:  3
Problem: In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution: Note that triangle $AFH$ is similar to triangle $CEH$ with ratio $\frac{2}{3}$ . Similarly, triangle $AGJ$ is similar to triangle $ECJ$ with ratio $\frac{4}{3}$ . Thus, if $AC = a$ then we know that $AH = \frac{2}{5}a$ and $JC = \frac{3}{7}a$ meaning $HJ = \frac{6}{35}a$ and thus the ratio of $HJ$ to $JC$ is $\frac{\frac{6}{35}}{\frac{3}{7}} = \frac{2}{5}$ which equals the ratio of the areas of $HJE$ to $JEC$ . If $y = AD, x = DC$ , then we know that $JEC = \text{(altitude from J to EC)} \cdot EC = \frac{3}{7}y \cdot \frac{1}{2}x \cdot \frac{1}{2}$ and since $xy = 70$ and we want to find $\frac{2}{5}$ of this, we get our answer is $\frac{2}{5} \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot 70 \cdot \frac{1}{2} = \boxed{3}$ . -SuperJJ

Answer:  3
Problem: In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution: $[CEF] = \frac{[ABCD]}{4} = \frac{35}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{HE}{HF} = \frac{CE}{AF} = \frac{3}{2}$ $\frac{HE}{EF} = \frac{3}{5}$
$[CEH] = \frac{HE}{EF} \cdot [CEF] = \frac{3}{5} \cdot \frac{35}{2} = \frac{21}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{AH}{HC} = \frac{AF}{CE} = \frac{2}{3}$ $\frac{AH}{AC} = \frac{2}{5}$ $\frac{CH}{AC} = \frac{3}{5}$
$\triangle CEJ \sim \triangle AGJ$ $\frac{AJ}{JC} = \frac{AG}{CE} = \frac{4}{3}$ $\frac{AJ}{AC} = \frac{4}{7}$
$\frac{HJ}{AC} = \frac{AJ}{AC} - \frac{AH}{AC} = \frac{4}{7} - \frac{2}{5} = \frac{6}{35}$
$\frac{HJ}{CH} = \frac{HJ}{AC} \cdot \frac{AC}{CH} = \frac{6}{35} \cdot \frac{5}{3} = \frac{2}{7}$
$[EHJ] = \frac{HJ}{CH} \cdot [CEH] = \frac{2}{7} \cdot \frac{21}{2} = \boxed{3}$

Answer: C 3
Problem: In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution: We need one more pair of ratios to fully define our mass point system.  Let's use $\triangle AFH \sim \triangle CEH\implies EH:HF = 3:2$ and now do mass points on $\triangle AEG$
[asy] size(250); pair A = (0,0), B = (10,0), C = (10,7), D = (0,7); pair F = (10/3,0), G = (20/3,0), E = (5,7); pair H = intersectionpoint(A--C, E--F); pair J = intersectionpoint(A--C, E--G); filldraw(A--E--G--cycle, rgb(1,1,1)+opacity(0.3), red+2bp); draw(A--B--C--D--cycle); draw(A--C); draw(E--F); draw(E--G); draw(A--E, dashed); draw(E--B, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, N); dot("$F$", F, S); dot("$G$", G, S); dot("$H$", H, ESE); dot("$J$", J, W);  // mass point labels pair mass = A + SW; label(scale(0.8)*"$3$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = F + S; label(scale(0.8)*"$6$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = G + SE; label(scale(0.8)*"$3$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = J + .7*ESE; label(scale(0.8)*"$7$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = E + N; label(scale(0.8)*"$4$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = H + .7*WNW; label(scale(0.8)*"$10$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); [/asy]
Now it's just a standard "Area Reduction by Ratios"™ problem going from:
\[[ABCD]\xrightarrow[]{\frac{1}{2}}[AEB]\xrightarrow[]{\frac{2}{3}}[AEG]\xrightarrow[]{\frac{3}{7}}[AEJ]\xrightarrow[]{\frac{3}{10}}[HEJ]\]
or,
\[70 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{7}\cdot \frac{3}{10} = \boxed{3}\]

Answer: C 3
Problem: In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$

Solution: Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $\angle ABD = 45^\circ$ and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$ . Notice that in $\triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $\triangle{ADH}$ , we obtain $DH=\frac{\sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:
$kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $\tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $\sin(x)^2+\cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $\boxed{75}$

Answer:  75
Problem: In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$

Solution: Without loss of generality, we can assume that $BD = 1$ and $CD = 2$ . As above, we are able to find that $\angle ADC = 60^\circ$ and $\angle ADB = 120^\circ$
Using Law of Sines on triangle $ADB$ , we find that \[\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}.\] Since we know that \[\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},\] \[\sin 45^\circ = \frac{\sqrt{2}}{2},\] \[\sin 120^\circ = \frac{\sqrt{3}}{2},\] we can compute $AD$ to equal $1+\sqrt{3}$ and $AB$ to be $\frac{3\sqrt{2}+\sqrt{6}}{2}$
Next, we apply Law of Cosines to triangle $ADC$ to see that \[AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ).\] Simplifying the right side, we get $AC^2 = 6$ , so $AC = \sqrt{6}$
Now, we apply Law of Sines to triangle $ABC$ to see that \[\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}.\] After rearranging and noting that $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , we get \[\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}.\]
Dividing the right side by $\sqrt{3}$ , we see that \[\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4},\] so $\angle ACB$ is either $75^\circ$ or $105^\circ$ . Since $105^\circ$ is not a choice, we know $\angle ACB = \boxed{75}$

Answer:  75