Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

Solution: \[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]
There are $3 \times 3$ squares in the corners of this $7 \times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)
\[\begin{tabular}{|c|c|c|} \hline A & B & C \\ \hline B & D & E \\ \hline C & E & F \\ \hline \end{tabular}\] (Note that coloring one $3 \times 3$ square will also determine the colorings of the other $3$ because the large $7 \times 7$ square must look the same when it is rotated by $90^\circ$
There are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \times 64 = 1024$ . Going back to the question, we see that "there must be at least one square of each color in this grid of $49$ squares." We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \boxed{1022}$ -hansenhe

Answer: B 1022
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\boxed{12}$

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$
Since we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \frac{x^2}{2} + \frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\left(0, \frac{1}{2}\right)$ . Hence, we know that the minimum is $\frac{1}{2}$ , which further implies that $\boxed{12}$

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: [asy] Label f;  f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2));  yaxis(-2,2,Ticks(f, 0.2));  real g(real x)  {  return x^2-1;  }  draw(graph(g, 1.7, -1.7)); real h(real x)  {  return sqrt(1-x^2);  }  draw(graph(h, 1, -1)); real j(real x)  {  return -sqrt(1-x^2);  }  draw(graph(j, 1, -1)); [/asy]
Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{12}$ as the right answer.

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\boxed{12}$ is correct.

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: Now, let's graph these two equations. We want the blue parabola to be inside this red circle. [asy] import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x)  {  return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*"$a$",(0,5),NE); dot(scale(.7)*"$-a$",(0,-5),N); dot(scale(.7)*"$a$",(5,0),NE); dot(scale(.7)*"$-a$",(-5,0),SE); [/asy] Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\boxed{12}$

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,
\[b^2 - 4ac > 0 \rightarrow 0 - 4(6)(-(2a - 1)) > 0 \rightarrow a > \frac{1}{2}\] The answer is $\boxed{12}$ and we are done.

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \[4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}\] Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{12}$

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: We can see that if $a = 1$ , we know that the points where the two curves intersect are $(0, -1), (1, 0)$ and $(-1, 0)$ .Because there are only $3$ intersections and $a > 1/2$ , as well as $a > 1/4$ we know that either $\textbf{(D)}$ or $\textbf{(E)}$ is the correct answer. Then we can test a number from $(1/2, 1/4)$ to eliminate the remaining answer. So $\boxed{12}$ is the correct answer.

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: Simply plug in $a = \frac{1}{2}, \frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\boxed{12}$

Answer: E 12
Problem: Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution: An ideal solution come to mind is where they intersect at the $x$ -axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\sqrt{a}$ , set them equal we get $a=\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\boxed{12}$

Answer: E 12
Problem: Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution: The GCD information tells us that $24$ divides $a$ , both $24$ and $36$ divide $b$ , both $36$ and $54$ divide $c$ , and $54$ divides $d$ . Note that we have the prime factorizations: \begin{align*} 24 &= 2^3\cdot 3,\\ 36 &= 2^2\cdot 3^2,\\ 54 &= 2\cdot 3^3. \end{align*}
Hence we have \begin{align*} a &= 2^3\cdot 3\cdot w\\ b &= 2^3\cdot 3^2\cdot x\\ c &= 2^2\cdot 3^3\cdot y\\ d &= 2\cdot 3^3\cdot z \end{align*} for some positive integers $w,x,y,z$ . Now if $3$ divides $w$ , then $\gcd(a,b)$ would be at least $2^3\cdot 3^2$ which is too large, hence $3$ does not divide $w$ . Similarly, if $2$ divides $z$ , then $\gcd(c,d)$ would be at least $2^2\cdot 3^3$ which is too large, so $2$ does not divide $z$ . Therefore, \[\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)\] where neither $2$ nor $3$ divide $\gcd(w,z)$ . In other words, $\gcd(w,z)$ is divisible only by primes that are at least $5$ . The only possible value of $\gcd(a,d)$ between $70$ and $100$ and which fits this criterion is $78=2\cdot3\cdot13$ , so the answer is $\boxed{13}$

Answer: D 13
Problem: Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution: We can say that $a$ and $b$ 'have' $2^3 \cdot 3$ , that $b$ and $c$ have $2^2 \cdot 3^2$ , and that $c$ and $d$ have $3^3 \cdot 2$ . Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 \cdot 3^2$ , and thus $a$ has $2^3 \cdot 3$ (and no more powers of $3$ because otherwise $\gcd(a,b)$ would be different). In addition, $c$ has $3^3 \cdot 2^2$ , and thus $d$ has $3^3 \cdot 2$ (similar to $a$ , we see that $d$ cannot have any other powers of $2$ ). We now assume the simplest scenario, where $a = 2^3 \cdot 3$ and $d = 3^3 \cdot 2$ . According to this base case, we have $\gcd(a, d) = 2 \cdot 3 = 6$ . We want an extra factor between the two such that this number is between $70$ and $100$ , and this new factor cannot be divisible by $2$ or $3$ . Checking through, we see that $6 \cdot 13$ is the only one that works. Therefore the answer is $\boxed{13}$

Answer: D 13
Problem: Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution: First off, note that $24$ $36$ , and $54$ are all of the form $2^x\times3^y$ . The prime factorizations are $2^3\times 3^1$ $2^2\times 3^2$ and $2^1\times 3^3$ , respectively. Now, let $a_2$ and $a_3$ be the number of times $2$ and $3$ go into $a$ , respectively. Define $b_2$ $b_3$ $c_2$ , and $c_3$ similarly. Now, translate the $lcm$ s into the following: \[1) \min(a_2,b_2)=3\] \[2) \min(a_3,b_3)=1\] \[3) \min(b_2,c_2)=2\] \[4) \min(b_3,c_3)=2\] \[5) \min(c_2,d_2)=1\] \[6) \min(c_3,d_3)=3\]
From $4)$ , we see that $b_3 \geq 2$ , thus from $2)$ $a_3 = 1$ . Similarly, from $3)$ $c_2 \geq 2$ , thus from $5)$ $d_2 = 1$
Note also that $d_3 \geq 3$ and $a_2 \geq 3$ . Therefore \[\min(a_2, d_2) = 1\] \[\min(a_3, d_3) = 1\] Thus, $\gcd(a, d) = 2 \times 3 \times k$ for some $k$ having no factors of $2$ or $3$
Since $70 < \gcd(a, d) < 100$ , the only values for k are $12, 13, 14, 15, 16$ , but all have either factors of $2$ or $3$ , except $\boxed{13}$ . And we're done.

Answer: D 13
Problem: Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution: Notice that $\gcd (a,b,c,d)=\gcd(\gcd(a,b),\gcd(b,c),\gcd(c,d))=\gcd(24,36,54)=6$ , so $\gcd(d,a)$ must be a multiple of $6$ . The only answer choice that gives a value between $70$ and $100$ when multiplied by $6$ is $\boxed{13}$ . - mathleticguyyy + einstein

Answer: D 13
Problem: Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution: [asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z, W; real R; path quad;  //Variable Definitions R = 1; X = R*dir(45); Y = R*dir(135); Z = R*dir(-135); W = R*dir(-45); quad = X--Y--Z--W--cycle;  //Diagram draw(quad); label("$b$",X,NE); label("$a=2^3 \cdot 3 \cdot p$",Y,NW); label("$d=2 \cdot 3^3 \cdot q$",Z,SW); label("$c$",W,SE); label("$2^3 \cdot 3$",X--Y); label("$2^2 \cdot 3^2$",X--W); label("$2 \cdot 3^3$",Z--W); label("$2 \cdot 3 \cdot k$",Z--Y); [/asy]
The relationship of $a$ $b$ $c$ , and $d$ is shown in the above diagram. $gcd(a,d)=2 \cdot 3 \cdot k$ $70 < 6k < 100$ $12 \le k \le 16$ $k=\boxed{13}$

Answer: D 13
Problem: Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution: Just as in solution $1$ , we prime factorize $a, b, c$ and $d$ to observe that
$a=2^3\cdot{3}\cdot{w}$
$b=2^3\cdot{3^2}\cdot{x}$
$c=2^2\cdot{3^3}\cdot{y}$
$d=2\cdot{3^3}\cdot{z}.$
Substituting these expressions for $a$ and $d$ into the final given,
$70<\text{gcd}(2\cdot{3^3}\cdot{z}, 2^3\cdot{3}\cdot{w})<100.$
The greatest common divisor of these two numbers is already $6$ . If $k$ is what we wish to multiply $6$ by to obtain the gcd of these two numbers, then
$70<6k<100$ . Testing the answer choices, only $13$ works for $k$ (in order for the compound inequality to hold). so our gcd is $78$ , which means that $\boxed{13}$ must divide $a$

Answer: D 13
Problem: Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution: Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \[\frac{50}{x} = \frac{10}{y},\] where $y = -x+a$ . Therefore substituting we have that $BG=\frac{5a}{6}$ . By similar triangles, we have that $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ . Then, we have that $\frac{ah}{2}=120$ . We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$ , and we have that it is $\boxed{75}$ by substituting.

Answer: D 75
Problem: Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution: For this problem, we have $\triangle{ADE}\sim\triangle{ABC}$ because of SAS and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$

Answer: D 75
Problem: Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution: The ratio of the $\overline{BG}$ to $\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\bigtriangleup ABG$ to the area of $\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\bigtriangleup ABG$ is $\frac{5}{5+1}\times120=100$ . Since $\overline{DE}$ is the midsegment of $\bigtriangleup ABC$ , so $\overline{DF}$ is the midsegment of $\bigtriangleup ABG$ . Thus, the ratio of the area of $\bigtriangleup ADF$ to the area of $\bigtriangleup ABG$ is $1:4$ , so the area of $\bigtriangleup ACG$ is $\frac{1}{4}\times100=25$ . Therefore, the area of quadrilateral $FDBG$ is $[ABG]-[ADF]=100-25=\boxed{75}$

Answer: D 75
Problem: Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution: The area of quadrilateral $FDBG$ is the area of $\bigtriangleup ABG$ minus the area of $\bigtriangleup ADF$ . Notice, $\overline{DE} || \overline{BC}$ , so $\bigtriangleup ABG \sim \bigtriangleup ADF$ , and since $\overline{AD}:\overline{AB}=1:2$ , the area of $\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4$ . Given that the area of $\bigtriangleup ABC$ is $120$ , using $\frac{bh}{2}$ on side $AB$ yields $\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}$ . Using the Angle Bisector Theorem, $\overline{BG}:\overline{BC}=50:(10+50)=5:6$ , so the height of $\bigtriangleup ABG: \bigtriangleup ACB=5:6$ . Therefore our answer is $\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}$

Answer: D 75