Answer: 12.0

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

Solution: Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields $6$ more possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Thus, our answer is $6 + 6 = \boxed{12}$

Solution:
Burnside's Lemma can be applied to the problem. Number the disks 1-6 from top to bottom and left to right. First, we count the number of group actions. There are $6$ in total: $3$ rotations, $0$ °, $120$ °, and $240$ °. Additionally, there are $3$ different reflections, which are over lines passing through the middle of disks $1$ $4$ , and $6$ . The $0$ ° rotation has the total number of diagrams fixed, which is $6!/3!\cdot2!=60$ . The $120$ ° and $240$ ° rotations each have $0$ fixed diagrams, as one would need the same color for disks $1, 4, 6$ and the same color for disks $2, 3, 5,$ which isn't possible. Consider the reflection over the line passing through disk $1$ .
The green can be placed in $2$ places, disk $1$ or $5$ , and then, the blue must go in either disks $2$ and $3$ or disks $4$ and $6$ , so the number of fixed diagrams is $2\cdot2=4$ . The other two reflections also have $4$ unique fixed diagrams. Therefore, applying Burnside's Lemma, we get $(60+4+4+4)/6 = \boxed{12}$


Answer: 12.0

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution: First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$ . For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \boxed{12}$

Solution:
Denote the $6$ discs as in the first solution. Ignoring reflections or rotations, there are $\binom{6}{3} \cdot \binom{3}{2} = 60$ colorings. Now we need to count the number of fixed points under possible transformations:
1. The identity transformation. Since this doesn't change anything, there are $60$ fixed points.
2. Reflect on a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$ . Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection.
3. Rotate by $120^\circ$ counterclockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$ $4$ , and $6$ will be the same. Similarly, the colors of circles $2$ $3$ , and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation.
By Burnside's Lemma , the total number of colorings is $(1 \cdot 60+3 \cdot 4+2 \cdot 0)/(1+3+2) = \boxed{12}$

Solution:
Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as

Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are $\dbinom{5}{2}=10$ of these. However, due to reflection we must divide this by two. But, in two of these cases, the reflection is itself, so we must subtract these out before dividing by 2, and add them back afterwards, giving $\frac{10-2}{2}+2=6$ arrangements in this case.
Now, look at the second case. We again must pick two of the five remaining circles, and like in the first case, two of the reflections give the same arrangement. Thus, there are also $6$ arrangements in this case.
In total, we have $6+6=\boxed{12}$

Solution:
We note that the group $G$ acting on the possible colorings is $D_3 = \{e, r, r^2, s, sr, sr^2\}$ , where $r$ is a $120^\circ$ rotation and $s$ is a reflection. In particular, the possible actions are the identity, the $120^\circ$ and $240^\circ$ rotations, and the three reflections.
We will calculate the number of colorings that are fixed under each action. Every coloring is fixed under the identity, so we count $\dfrac{6!}{3!2!1!} = 60$ fixed colorings. Note that no colorings are fixed under the rotations, since then the outer three and inner three circle must be the same color, which is impossible in our situation.
Finally, consider the reflection with a line of symmetry going through the top circle. Every fixed coloring is determined by the color of the top circle (either green or blue), and the color of the middle circles (either blue or red). Hence, there are $2\cdot 2 = 4$ colorings fixed under this reflection action. The other two actions are symmetric, so they also have $4$ fixed colorings. Hence, by Burnside's lemma, the number of unique colorings up to reflections and rotations is \[\dfrac{1}{|D_3|} (1\cdot 60 + 2\cdot 0 + 3\cdot 4) = \dfrac{1}{6}\cdot 72 = \boxed{12}.\]


Answer: 19.0

In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. The ratio $A:B$ is $1:18$ . What is $N$
[asy] draw((0,0)--(18,0)); draw(arc((9,0),9,0,180)); filldraw(arc((1,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((3,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((5,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((7,0),1,0,180)--cycle,gray(0.8)); label("...",(9,0.5)); filldraw(arc((11,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((13,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((15,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((17,0),1,0,180)--cycle,gray(0.8)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 36$

Solution: Use the answer choices and calculate them. The one that works is $\bold{\boxed{19}$

Solution:
Let the number of semicircles be $n$ and let the radius of each semicircle be $r$ . To find the total area of all of the small semicircles, we have $n \cdot \frac{\pi \cdot r^2}{2}$
Next, we have to find the area of the larger semicircle. The radius of the large semicircle can be deduced to be $n \cdot r$ . So, the area of the larger semicircle is $\frac{\pi \cdot n^2 \cdot r^2}{2}$
Now that we have found the area of both A and B, we can find the ratio. $\frac{A}{B}=\frac{1}{18}$ , so part-to-whole ratio is $1:19$ . When we divide the area of the small semicircles combined by the area of the larger semicircles, we get $\frac{1}{n}$ . This is equal to $\frac{1}{19}$ . By setting them equal, we find that $n = 19$ . This is our answer, which corresponds to choice $\bold{\boxed{19}$


Answer: 19.0

Solution: Each small semicircle is $\frac{1}{N^2}$ of the large semicircle. Since $N$ small semicircles make $\frac{1}{19}$ of the large one, $\frac{N}{N^2} = \frac1{19}$ . Solving this, we get $\boxed{19}$ . ...


Answer: 7.6

In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\overline{HE}$ and $\overline{FG}$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt));  pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5);  draw(A--B--C--D--cycle); draw(E--F--G--H--cycle);  label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);  label("$E$",E,N); label("$F$",(10.8,3)); label("$G$",G,S); label("$H$",H,W);  label("$4$",A--E,N); label("$6$",B--E,N); label("$5$",(10.8,5.5)); label("$3$",(10.8,1.5)); label("$4$",G--C,S); label("$6$",G--D,S); label("$5$",D--H,W); label("$3$",A--H,W);  draw((3,7.25)--(7.56,1.17)); label("$d$",(3,7.25)--(7.56,1.17), NE);  [/asy]
$\textbf{(A)}\ 6.8\qquad \textbf{(B)}\ 7.1\qquad \textbf{(C)}\ 7.6\qquad \textbf{(D)}\ 7.8\qquad \textbf{(E)}\ 8.1$

Solution: The area of the parallelogram can be found in two ways. The first is by taking rectangle $ABCD$ and subtracting the areas of the triangles cut out to create parallelogram $EFGH$ . This is \[(4+6)(5+3) - 2 \cdot \frac12 \cdot 6 \cdot 5 - 2 \cdot \frac12 \cdot 3 \cdot 4 = 80 - 30 - 12 = 38\] The second way is by multiplying the base of the parallelogram such as $\overline{FG}$ with its altitude $d$ , which is perpendicular to both bases. $\triangle FGC$ is a $3-4-5$ triangle so $\overline{FG} = 5$ . Set these two representations of the area together. \[5d = 38 \rightarrow d = \boxed{7.6}\]


Answer: 4.0

In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$ $AB=4$ , and $DC=10$ . The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$ . Then $CF=$
[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy]
$\textrm{(A)}\ 3.25\qquad\textrm{(B)}\ 3.5\qquad\textrm{(C)}\ 3.75\qquad\textrm{(D)}\ 4.0\qquad\textrm{(E)}\ 4.25$

Solution: Drop perpendiculars from $A$ and $B$ ; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length.  $XY=AB=4$ and $DX=YC=3$ , hence $DY=7\implies DF=14\implies CF=\boxed{4.0}$


Answer: 36.0

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$ . What is the area of the inner square $EFGH$
AMC102005Aq.png
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$

Solution:
We see that side $BE$ , which we know is $1$ , is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, $AH = 1$ . Then $HB = HE + BE = HE + 1$ , and $HE$ is one of the sides of the square whose area we want to find. So:
\[1^2 + (HE+1)^2=\sqrt{50}^2\]
\[1 + (HE+1)^2=50\]
\[(HE+1)^2=49\]
\[HE+1=7\]
\[HE=6\] So, the area of the square is $6^2=\boxed{36}$


Answer: 12.0

In the five-sided star shown, the letters $A$ $B$ $C$ $D$ and $E$ are replaced by the
numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
numbers at the ends of the line segments $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ form an
arithmetic sequence, although not necessarily in that order. What is the middle
term of the arithmetic sequence?
[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("$A$",(0.5,1.54),N); label("$B$",(1,0),SE); label("$C$",(-0.31,0.95),W); label("$D$",(1.31,0.95),E); label("$E$",(0,0),SW); [/asy]
$(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13$

Solution:
Let the terms in the arithmetic sequence be $a$ $a + d$ $a + 2d$ $a + 3d$ , and $a + 4d$ . We seek the middle term $a + 2d$
These five terms are $A + B$ $B + C$ $C + D$ $D + E$ , and $E + A$ , in some order. The numbers $A$ $B$ $C$ $D$ , and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \[A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.\] Hence, the sum of the five terms is \[(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.\] But adding all five numbers, we also get $a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d$ , so \[5a + 10d = 60.\] Dividing both sides by 5, we get $a + 2d = \boxed{12}$ , which is the middle term. The answer is (D).


Answer: 12.0

In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9$ , although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$ $BC$ $CD$ $DE$ , and $EA$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?
2005amc10a17.gif
$\textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13$

Solution:
Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ( $A$ goes to $AE$ and $AB$ $D$ goes to $DC$ and $DE$ ). The sum of every term is equal to $2(3+5+6+7+9)=60$
Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=\boxed{12}$


Answer: 12.0

Solution: We know that the smallest number in this sequence must be $3 + 5 = 8$ , and the biggest number must be $7 + 9 = 16$ . Since there are $5$ terms in this sequence, we know that $8 + 4d = 16$ , or that $d = 2$ . Thus, we know that the middle term must be $8 + 2 \cdot 2 = \boxed{12}.$ ~yk2007 (Daniel K.)


Answer: 21.0

In the following equation, each of the letters represents uniquely a different digit in base ten:
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$

Solution:
The right side of the equation can be rewritten as $111T = 37 \cdot 3T$ .  With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$
The only possibility that works is $37 \cdot 27 = 999$ .  That means $E+M+T+Y = \boxed{21}$


Answer: 142.0

In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$

Solution: The $x$ th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$ . This is approximately the number divided by $\sqrt{2}$ $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{142}$ ~Lopkiloinm

Solution:
We can look at answer choice $\textbf{(C)}$ , which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$
The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{142}.$

Solution:
We can arrange the numbers in the following pattern: \[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]
We can see this as a isosceles right triangle, with legs of length $200.$
Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.
Solving for $x$ yields: \begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*} We see that $\boxed{142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer.


Answer: 142.0

Solution: The $x$ th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$ . This is approximately the number divided by $\sqrt{2}$ $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{142}$ ~Lopkiloinm

Solution:
We can look at answer choice $\textbf{(C)}$ , which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$
The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{142}.$

Solution:
We can arrange the numbers in the following pattern: \[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]
We can see this as a isosceles right triangle, with legs of length $200.$
Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.
Solving for $x$ yields: \begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*} We see that $\boxed{142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer.


Answer: 130.0

In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

Solution:
We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$ . When an inscribed angle contains the diameter , the inscribed angle is a right angle . Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$
$\angle ABE = \angle BED$ because they are alternate interior angles and $\overline{AB} \parallel \overline{ED}$ . Opposite angles in a cyclic quadrilateral are supplementary , so $\angle BED + \angle BCD = 180$ . Use substitution to get $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{130}$

Solution: Note $\angle ABE = \angle BED=50$ as before. The sum of the interior angles for quadrilateral $EBCD$ is $360$ . Denote the center of the circle as $P$ $\angle PDE = \angle PED = 50$ . Denote $\angle PDC = \angle PCD = x$ and $\angle PBC = \angle PCB = y$ . We wish to find $\angle BCD = x+y$ . Our equation is $(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360$ . Our final equation becomes $2(x+y)+100 = 360$ . After subtracting $100$ and dividing by $2$ , our answer becomes $x+y=\boxed{130}$

Solution:
Note that $\overset{\Large\frown} {BE}$ intercepts $\angle BAE$ . Since, $\overset{\Large\frown} {BE}=180$ , thus $\angle BAE=90°$ (courtesy of the Inscribed Angles Theorem).
Since we know that $\angle BAE=90°$ , then $\angle AEB + \angle ABE = 90°$ , (courtesy of the Triangle Sum Theorem) and also $5\angle AEB = 4\angle ABE$ . By solving this variation, $\angle AEB = 40$ and $\angle ABE = 50$ . After that, due to the Alternate Interior Angles Theorem, $\angle ABE \cong \angle BED$ , which means $\angle BED = 50$
After doing some angle chasing, then these following facts should be true, $\overset{\Large\frown} {AB}=80$ $\overset{\Large\frown} {BD}=100$ $\overset{\Large\frown} {AE}=100$
Note that the arcs have to equal 360, so, $360=\overset{\Large\frown} {AB}+\overset{\Large\frown} {BD}+\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}$
$360=80+100+100+\overset{\Large\frown} {DE}$
$\overset{\Large\frown} {DE}=80$
Notice how $\overset{\Large\frown} {DB}$ intercepts $\angle BCD$ and that $\overset{\Large\frown} {DB}=\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}+\overset{\Large\frown} {AB}$
$\overset{\Large\frown} {DB}=80+100+80$
$\overset{\Large\frown} {DB}=260$
According to the Inscribed Angles Theorem, $2\angle BCD=\overset{\Large\frown} {DB}$ , therefore the answer is $\frac{260}{2}= \boxed{130}$


Answer: 12.0

Solution:
Clearly, since $\overline{FE}$ is a side of a square with area $32$ $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$ . Now, since $\overline{FE} = \overline{BC}$ , we have $\overline{BC} = 4 \sqrt{2}$
Now, $\overline{JB}$ is a side of a square with area $18$ , so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$ . Since $\Delta JBK$ is equilateral, $\overline{BK} = 3 \sqrt{2}$
Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$ , so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$ . Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{12}$

Solution: Since $\overline{FE} = \sqrt{32}$ , and $\overline{FE} = \overline{BC}$ $\overline{BC} = 4\sqrt{2}$ . Meanwhile, $\overline{JB} = 3\sqrt{2}$ , and since $\triangle JBK$ is equilateral, $\overline{BK} = 3\sqrt{2}$ . If $ABCDEF$ is equiangular, $\angle ABC = \frac{180 \cdot (n-2)}{n} = 120^{\circ}$ , where $n$ is the number of sides of the shape. Adding all the angles around $B$ gives $270^{\circ}$ , so $\angle KBC = 360 - 90 = 270^{\circ}$ . Because $\triangle KBC$ is right, the area of $\triangle KBC = \frac{4\sqrt{2} \cdot 3\sqrt{2}} {2} = 12$ . Therefore, the answer is $\boxed{12}$ . ~strongstephen


Answer: 46.0

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ $w$ $x$ $y$ , and $z$ . Find $y + z$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$

Solution:
We know that $y+z=2v$ , so we could find one variable rather than two.
$v+24+w=43+v$
$24+w=43$
$w=19$

$44+x=24+x+z \implies z=20$

The sum per row is $25+21+20=66$
Thus $66-18-25=66-43=v=23$
Since we needed $2v$ and we know $v=23$ $23 \times 2 = \boxed{46}$

Solution:
$v+24+w=43+v$
$24+w=43$
$w=19$

$44+x=24+x+z \implies z=20$

The magic sum is determined by the bottom row. $25+20+21=66$
Solving for $y$
$y=66-19-21=66-40=26$
To find our answer, we need to find $y+z$ $y+z=20+26 = \boxed{46}$

Solution:
A nice thing to know is that any $3$ numbers that go through the middle form an arithmetic sequence.
Using this, we know that $x=(24+z)/2$ , or $2x=24+z$ because $x$ would be the average.
We also know that because $x$ is the average the magic sum would be $3x$ , so we can also write the equation $3x-46=z$ using the bottom row.
Solving for x in this system we get $x=22$ , so now using the arithmetic sequence knowledge we find that $y=26$ and $z=20$
Adding these we get $\boxed{46}$

Solution:
Create an equation for every row, column, and diagonal. Let $e$ be the sum of the rows, columns, and diagonals. \[w+v+24=e\] \[x+y+18=e\] \[z+46=e\] \[v+43=e\] \[x+z+24=e\] \[w+y+21=e\] \[x+w+25=e\] \[x+v+21=e\]
Notice that $z+46=e$ and $x+z+24=e$ both have $z$ . Equate them and you get that $x=22$ .
Using that same strategy, we use $v+43=e$ instead. $w+v+24=e$ is good for our purposes. It turns out that $w=19$ . Since we already know those numbers, and $x+w+25=e$ , We can say that $e$ will be $66$ . We are now able to solve: $x+z+24=e$ $w+y+21=e$ $x+v+21=e$ , and $x+y+18=e$ . Respectively, $v=23$ $w=19$ $x=22$ $y=26$ , and $z=20$ . We only require The sum of $y+z$ , which is $26+20=46$ .
We get that the sum of $y$ and $z$ respectively is $\boxed{46}$


Answer: 731.0

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Let the intersection of the highways be at the origin $O$ , and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\frac{x}{50}$ hours, or $d=rt=14t=1.4-\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$ . All these circles are homothetic with respect to a center at $(5,0)$
Now consider the circle at $(0,0)$ . Draw a line tangent to it at $A$ and passing through $B (5,0)$ . By the Pythagorean Theorem $AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$ . Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$ , so the slope of line $AB$ is $\frac{-7}{24}$ . Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$
This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\left(\frac{35}{31},\frac{35}{31}\right)$ . The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$ . By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\frac{700}{31}$ so the answer is $700+31=\boxed{731}$


Answer: 24.0

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$

Solution:
We first connect point $B$ with point $D$

We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$ , and the area of $\triangle BCD$ is $\frac{3\cdot 4}{2}$ . Thus, the area of quadrilateral $ABCD$ is $30-6 = \boxed{24}.$

Solution: $\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12,and 13. The area of $\triangle ABD$ $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.$ Thus, the area of $\triangle ABD$ is 30, so the area of $ABCD$ is $30-6 = \boxed{24}.$ ---LarryFlora


Answer: 2.0

In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$ , and $d$ such that for all positive integers $n$ $a_n=b\lfloor \sqrt{n+c} \rfloor +d$ ,
where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . The sum $b+c+d$ equals
$\text{(A)} \ 0 \qquad  \text{(B)} \ 1 \qquad  \text{(C)} \ 2 \qquad  \text{(D)} \ 3 \qquad  \text{(E)} \ 4$

Solution:
Solution by e_power_pi_times_i
Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$ . However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$ $a = 1$ . Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$ , and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{2}$


Answer: 100000.0

In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$

Solution: The digit $9$ is $5$ places to the left of the digit $3$ . Thus, it has a place value that is $10^5 = 100,000$ times greater. $\boxed{100,000}$

Solution: The digit $9$ is in the $100$ s place.  The digit $3$ is in the $\frac{1}{1000}$ ths place.  The ratio of these two numbers is $\frac{100}{\frac{1}{1000}} = 100\times1000 = 100,000$ , and the answer is $\boxed{100,000}$


Answer: 4.0

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, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{4}$

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, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{4}$

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

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: 4.0

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, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{4}$

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, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{4}$

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

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: 7.0

In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution:
The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:

where the light gray boxes are the ones we have filled. Counting these, we get $\boxed{7}$ total boxes.

Solution:
We label the three shaded unit squares $A,B,$ and $C,$ then construct the two lines of symmetry of the resulting figure, as shown below:  Note that:
The shaded unit squares contributed by $A,B,$ and $C$ are all distinct, so we need to shade at least $4+4+2-3=\boxed{7} draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); label("$A'$",(3.5,3.5)); label("$A'$",(1.5,0.5)); label("$A'$",(3.5,0.5)); label("$B'$",(2.5,2.5)); label("$C'$",(0.5,0.5)); label("$C'$",(0.5,3.5)); label("$C'$",(4.5,3.5)); [/asy] ~MRENTHUSIASM


Answer: 100.0

Solution:
Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$ . We break the figure into squares as shown below.

We see that there are a total of $2(1+3+5)+7=25$ squares with side length $2$ . Therefore, the total area is $4\cdot 25=\boxed{100.}$


Answer: 86.0

In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$

Solution:
Using Vieta's formulas, we have:
\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , and so $ac+ad+bc+bd=k-30$ . The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$ , so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$
Let $p=a+b$ and $q=c+d$ . Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$ , we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$ . Moreover, the first Vieta equation, $a+b+c+d=18$ , gives $p+q=18$ . Thus we have two linear equations in $p$ and $q$ , which we solve to obtain $p=4$ and $q=14$
Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$ , yielding $k=4\cdot 14+30 = \boxed{86}$

Solution:
We start as before: $ab=-32$ and $cd=62$ . We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$ , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$
Now
\begin{align*}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+62)\\  =& x^4+(r+s)x^3+(62-32+rs)x^2\\ &+(62s-32r)x-1984.\end{align*}
Equating the coefficients of $x^3$ and $x$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of $x^2$ and get $k=\boxed{86}.$

Solution: Let the roots of the equation be $a,b,c,$ and $d$ . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\  abc+abd+acd+bcd &=-200\\  abcd &=-1984.\\  \end{align*} Since $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that \[abc + abd + acd + bcd = -200\] can be factored into \[ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).\] From the first equation, $c+d=18-a-b$ . Substituting it back into the equation, \[-32(18-a-b)+62(a+b)=-200\] Expanding, \[-576+32a+32b+62a+62b=-200 \implies 94a+94b=376\] So, $a+b=4$ and $c+d=14$ . Notice that \[ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)\] Plugging all our values in, \[-32+62+4(14)=\boxed{86}.\]

Solution:
Since two of the roots have product $-32,$ the equation can be factored in the form \[x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\] Expanding, we get \[x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\] Matching coefficients, we get
\begin{align*}
a + b &= -18, \\
ab + c - 32 &= k, \\
ac - 32b &= 200, \\
-32c &= -1984.
\end{align*}Then $c = \frac{-1984}{-32} = 62,$ so $62a - 32b = 200.$ With $a + b = -18,$ we can solve to find $a = -4$ and $b = -14.$ Then \[k = ab + c - 32 = \boxed{86}.\]


Answer: 8.0

In the product shown, $\text{B}$ is a digit.  The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution:
Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, $2 \times \text{B}$ has a units digit of $6$ , so $\text{B}$ is either $3$ or $8$ . If $\text{B}=3$ , then the product is $32\times 73$ , which is clearly too small, so $\text{B}=8$
$\boxed{8}$


Answer: 2.0

In the rectangular parallelepiped shown, $AB$ $3$ $BC$ $1$ , and $CG$ $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$
[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G = (5.95,4.2); pair C = (5.95,1.2); pair D = (1.2,1.2); pair H= (1.2,4.2); pair M = ((4.75+5.95)/2,3.6); draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H); draw(B--C); draw(F--G); draw(A--D--H--C--D,dashed); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,W); label("$E$",E1,W); label("$F$",F,SW); label("$G$",G,NE); label("$H$",H,NW); label("$M$",M,N); dot(A); dot(B); dot(E1); dot(F); dot(G); dot(C); dot(D); dot(H); dot(M); label("3",A/2+B/2,S); label("2",C/2+G/2,E); label("1",C/2+B/2,SE); [/asy]
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2$

Solution: Consider the cross-sectional plane and label its area $b$ . Note that the volume of the triangular prism that encloses the pyramid is $\frac{bh}{2}=3$ , and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is $\frac{bh}{3}$ , so the answer is $\boxed{2}$ .uwu (AOPS12142015)

Solution:
You can calculate the volume of the rectangular pyramid by using the formula, $\frac{Ah}{3}$ $A$ is the area of the base, $BCHE$ , and is equal to $BC * BE$ . The height, $h$ , is equal to the height of triangle $FBE$ drawn from $F$ to $BE$
$BE=\sqrt{BF^2 + EF^2}=\sqrt{13}$ Area of $BCHE = BC * BE = \sqrt{13}$
$h = 2 *$ Area of $FBE / BE$ (since Area $= \frac{1}{2}bh$ ).
Area of $FBE = \frac{1}{2} * FB * FE = 3$
$h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}$
Volume of pyramid $=\frac{1}{3} * \sqrt{13} * \frac{6}{\sqrt{13}} = 2$
Answer is $\boxed{2}$

Solution:
We can start by identifying the information we need. We need to find the area of rectangle $EHCB$ and the height of rectangular prism $EHCBM$
In order to find the area of $EHCB,$ we can use the Pythagorean Theorem. We find that $EB = \sqrt{13}$ , so the area of rectangle $EHCB = \sqrt{13}$ . We shall refer to this as $x$
In order to find the height of rectangular prism $EHCBM$ , we can examine triangle $EFB$ . We can use the Geometric Mean Theorem to find that when an altitude is dropped from point $F,$ $\overline{EB}$ is split into segments of length $\dfrac{4 \cdot \sqrt{13}}{13}$ and $\dfrac{9 \cdot \sqrt{13}}{13}$ . Taking the geometric mean of these numbers, we find that the altitude has length $\dfrac{6 \cdot \sqrt{13}}{13}$ . This is also the height of the rectangular prism, which we shall refer to as $y$
Plugging $x$ and $y$ into the formula $V = \dfrac{b \cdot h}{3},$ we find that the volume is $\boxed{2}$

Solution: We start by setting the formula for the volume of a rectangular pyramid: $\frac{1}{3}Bh$ . By the Pythagorean Theorem, we know that $BE = \sqrt{BF^2 + EF^2} = \sqrt{13}$ . Therefore, the area of the base is $1 \times \sqrt{13} = \sqrt{13}$ . Next, we would like to know the height of the pyramid. We can observe that the altitude from point $F$ in $\triangle EFB$ is parallel to the height of the pyramid and therefore congruent because those two altitudes are on the same plane of base $EBCH$ . From this, we only need to find the altitude from point $F$ in $\triangle EFB$ and plug it into our formula for the volume of a rectangular pyramid. This is easy because we already know the area of $\triangle EFB$ and the base from point $F$ , so all we need to do is divide: $\frac{2 \times 3}{\sqrt{13}} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}$ . Now all we need to do is plug in all our known values into the volume formula: $\frac{1}{3}Bh = \frac{\sqrt{13} \times \frac{6\sqrt{13}}{13}}{3} = \boxed{2}$

Solution:
AMC 10B 10 2018.jpg
Using the Pythagorean Theorem, we can easily find that $EB = \sqrt{2^2 + 3^2} = \sqrt{13}$ . Quickly computing, we find the area of the base, $BCHE = \sqrt{13} \cdot 1 = \sqrt{13}$ . Now we can make the following adjustments to our 3d shape as shown in the diagram. All we need now is to solve for the height, or $XM$ . We can set up to following equation due to our knowledge of altitudes(of the hypotenuse)in right triangles. We can set up the following equations: \begin{align*} b(a+b) &= (MH_1)^2 \\ a(a+b) &= (MH_2)^2 \\ b\sqrt{13} &= 3^2 \\ a\sqrt{13} &= 2^2 \\ b &= \dfrac{9}{\sqrt{13}} \\ a &= \dfrac{4}{\sqrt{13}} \\ (MX)^2 &= ab \\ (MX)^2 &= \dfrac{9 \cdot 4}{13} \\ MX &= \dfrac {3 \cdot 2}{\sqrt{13}} \\ \end{align*} Thus $\triangle V_{BCHEM} = \dfrac{\text{(height)}\cdot \text{(base)}}{3} = \dfrac{MX \cdot BCHE}{3} = \dfrac {\sqrt{13} \cdot \dfrac{3 \cdot 2}{\sqrt{13}}}{3}$ $= \boxed{2}$


Answer: 2.0

Solution: If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is $\frac{1}{3}Bh$ , with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is $\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}$ . We can obtain the answer by subtracting twice this value from the diagonal half prism, or $(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})=$ $\boxed{2}$


Answer: 0.0

In the sequence $2001$ $2002$ $2003$ $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^\textrm{th}$ term in this sequence?
$\mathrm{(A) \ } -2004 \qquad \mathrm{(B) \ } -2 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 4003 \qquad \mathrm{(E) \ } 6007$

Solution:
We already know that $a_1=2001$ $a_2=2002$ $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ $a_6=2003+2000-2005=1998$ $a_7=2000+2005-1998=2007$ , and so on.
We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$

Solution:
Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$
Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots$
From the values given in the problem statement we see that $a_3=a_1+2$
From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$
From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$
Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}$


Answer: 0.0

Solution:
We already know that $a_1=2001$ $a_2=2002$ $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ $a_6=2003+2000-2005=1998$ $a_7=2000+2005-1998=2007$ , and so on.
We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$

Solution:
Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$
Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= \cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= \cdots$
From the values given in the problem statement we see that $a_3=a_1+2$
From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$
From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$
Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = \cdots = a_2 - 2002 = \boxed{0}$


Answer: 3.0

In the sequence \[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\] each term is the sum of the two terms to its left. Find $a$
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$

Solution:
We work backwards to find $a$
\[d+0=1\implies d=1\] \[c+1=0\implies c=-1\] \[b+(-1)=1\implies b=2\] \[a+2=-1\implies a=\boxed{3.}\]


Answer: 5.0

In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
$\textbf{(A)}\ 5 \qquad  \textbf{(B)}\ 4 \qquad  \textbf{(C)}\ 2 \qquad  \textbf{(D)}\ 1 \qquad  \textbf{(E)}\ -1$

Solution: First, write a few terms of the sequence: $1, 3, 2, -1, -3, -2, 1, 3, 2, \ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$ th term is $-2$ and the sum of the all those previous terms is $0$ . Then, write the 97th to the 100th terms down: $1, 3, 2,-1$ and add them up to get the sum of $\boxed{5}$


Answer: 29.0

In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$ $14$ $11$ , and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$ , what is the least number of points she could have scored in the tenth game?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$

Solution:
The sum of the scores for games $6$ through $9$ is $68$ .  The average in these four games is $\frac{68}{4} = 17$
The total points in all ten games is greater than $10\cdot 18 = 180$ .  Thus, it must be at least $181$
There are at least $181 - 68 = 113$ points in the other six games:  games $1-5$ and game $10$
Games $1-5$ must have an average of less than $17$ .  Thus we cannot put more than $16 + 17 + 17 + 17 + 17 = 84$ points in those five games.
Thus, the tenth game must have at least $113 - 84 = 29$ points, and the answer is $\boxed{29}$


Answer: 11.0

In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9$

Solution:
This problem asks to find largest $x$ that cannot be written as \[ 6 a + 10 b + 15 c = x, \hspace{1cm} (1) \] where $a, b, c \in \Bbb Z_+$
Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2.
Modulo 2 on Equation (1), we get
By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, we get $c \equiv r \pmod{2}$
Following from Chicken McNugget's theorem, we have that any number that is no less than $(3-1)(5-1) = 8$ can be expressed in the form of $3a + 5b$ with $a, b \in \Bbb Z_+$
Therefore, all even numbers that are at least equal to $2 \cdot 8 + 15 \cdot 0 = 16$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .
All odd numbers that are at least equal to $2 \cdot 8 + 15 \cdot 1 = 31$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$
The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$
Next, we need to prove that 29 cannot be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$
Because 29 is odd, we must have $c \equiv 1 \pmod{2}$ .
Because $a, b, c \in \Bbb Z_+$ , we must have $c = 1$ .
Plugging this into Equation (1), we get $3 a + 5 b = 7$ .
However, this equation does not have non-negative integer solutions.
All analysis above jointly imply that the largest $x$ that has no non-negative integer solution to Equation (1) is 29.
Therefore, the answer is $2 + 9 = \boxed{11}$

Solution:
Let the number of $6$ cent coins be $a$ , the number of $10$ cent coins be $b$ , and the number of $15$ cent coins be $c$ . We get the Diophantine equation
\[6a + 10b + 15c = k\]
and we wish to find the largest possible value of $k$
Construct the following $\mod 6$ table of $6$ $10$ , and $15$
\[\begin{array}{c|ccc}  &  &  & \\ \text{number of coins} & 6 & 10 & 15 \\  \hline  &  &  & \\  1 & 0 & 4 & 3\\  &  &  &  \\ 2 & 0 & 2 & 0 \\ \end{array}\]
There are only $6$ possible residues for $6$ , they are: $0$ $1$ $2$ $3$ $4$ , and $5$
Hence, the largest value in cents we cannot obtain using $6$ $10$ , and $15$ cent coins is $29$ $2 + 9 = \boxed{11}$

Solution:
We claim that the largest number that cannot be obtained using $6$ $10$ , and $15$ cent coins is $29$
Let's first focus on the combination of $6$ $10$ . As both of them are even numbers, we cannot obtain any odd numbers from these two but requires $15$ to sum up to an odd number. Notice that by Chicken McNugget Theorem, the largest even number cannot be obtained by $6$ $10$ is $2(3\cdot 5-3-5)=14$ . Add this with $29$ , we can easily verify that $29$ cannot be obtained by $6$ $10$ , and $15$ as it needs at least one odd number, with the remaining part cannot be represented by $6$ and $10$
Let's show that any number greater than $29$ can be obtained. First, any even numbers greater than $29$ can be obtained by $6$ and $10$ by the Chicken McNugget Theorem. Next, any odd number greater than $29$ can be obtained by adding one $15$ with some $6$ s and $10$ s, which is also shown by the Chicken McNugget Theorem. This completes the proof. So the answer is $2+9 = \boxed{11}$


Answer: 671.0

In the year $2001$ , the United States will host the International Mathematical Olympiad .  Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ .  What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$

Solution:
First, we need to recognize that a number is going to be lowest only if, of the $3$ factors , two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$ . It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$ . Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors.
Hence, we have $667 + 3 + 1 = \boxed{671}$

Solution:
The sum is the highest if two factors are the lowest.
So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{671}$

Solution: We see since $2 + 0 + 0 + 1$ is divisible by $3$ , we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{671}$ as our final answer.


Answer: 671.0

Solution:
First, we need to recognize that a number is going to be lowest only if, of the $3$ factors , two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$ . It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$ . Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors.
Hence, we have $667 + 3 + 1 = \boxed{671}$

Solution:
The sum is the highest if two factors are the lowest.
So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{671}$

Solution: We see since $2 + 0 + 0 + 1$ is divisible by $3$ , we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{671}$ as our final answer.


Answer: 17865.0

In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$
$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \  } 17865 \qquad \mathrm{(D) \  } 17874 \qquad \mathrm{(E) \  }19851$

Solution:
By the formula for the sum of a geometric series, \begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*} and similarly \[b = \frac{5\left(10^{1985}-1\right)}{9},\] so \begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}
We now compute the decimal expansion of this expression. Firstly, $10^{3971} = 100 \dotsb 0$ , with $1$ one and $3971$ zeroes, and $2 \cdot 10^{1986} = 200 \dotsb 0$ , with $1$ two and $1986$ zeroes. Subtracting therefore gives \[10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,\] where there are $3971-1986-1 = 1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99 \dotsb 9800 \dotsb 010$ , now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero.
Using long division, and noting that $80 = 8 \cdot 9 + 8$ and $81 = 9 \cdot 9$ , it follows that \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,\] with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9 \cdot 4 = 36$ $8 \cdot 4 = 32$ , and $8 \cdot 4 + 3 = 35$ , we obtain \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,\] where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{17865}


Answer: 3.0

In this addition problem, each letter stands for a different digit.
$\setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$
If T = 7 and the letter O represents an even number, what is the only possible value for W?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Since both T's are 7, then O has to equal 4, because 7 + 7 = 14. Then, F has to equal 1. To get R, we do 4 + 4 (since O = 4) to get R = 8. The value for W then has to be a number less than 5, otherwise it will change the value of O, and can't be a number that has already been used, like 4 or 1. The only other possibilities are 2 and 3. 2 doesn't work because it makes U = 4, which is what O already equals. So, the only possible value of W is 3 $\boxed{3}$


Answer: 12.25

In trapezoid $ABCD$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$

Solution: If $AB=x$ and $CD=y$ , then $BC=x+y$ . By the Pythagorean theorem , we have $(x+y)^2=(y-x)^2+49.$ Solving the equation, we get $4xy=49 \implies xy = \boxed{12.25}$

Solution:
Simpler is just drawing the trapezoid and then using what is given to solve.
Draw a line parallel to $\overline{AD}$ that connects the longer side to the corner of the shorter side. Name the bottom part $x$ and top part $a$ .
By the Pythagorean theorem, it is obvious that $a^{2} + 49 = (2x+a)^{2}$ (the RHS is the fact the two sides added together equals that). Then, we get $a^2 + 49 = 4x^2 + 4ax + a^2$ , cancel out and factor and we get $49 = 4x(x+a)$ . Notice that $x(x+a)$ is what the question asks, so the answer is $\boxed{12.25}$

Solution: We know it is a trapezoid and that $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ . If they are perpendicular to $\overline{AD}$ that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know $\overline{AD}$ is $7$ . We can then set the length of $\overline{AB}$ to be $x$ and the length of $\overline{DC}$ to be $y$ $\overline{BC}$ would then be $x+y$ . Let's draw a straight line down from point $B$ which is perpendicular to $\overline{DC}$ and parallel to $\overline{AD}$ . Let's name this line $M$ . Then let's name the point at which line $M$ intersects $\overline{DC}$ point $E$ . Line $M$ partitions the trapezoid into ▭ $ADEB$ and $\triangle$ $BEC$ . We will use the triangle to solve for $xy$ using the Pythagorean theorem. The line segment $\overline{EC}$ would be $y-x$ because $\overline{DC}$ is $y$ and $\overline{DE}$ is $x$ $\overline{DE}$ is $x$ because it is parallel to $\overline{AB}$ and both are of equal length. Because of the Pythagorean theorem, we know that $(EC)^2+(BE)^2=(BC)^2$ . Substituting the values we have we get $(y-x)^2+(7)^2=(x+y)^2$ . Simplifying this we get $(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)$ . Now we get rid of the $x^2$ and $y^2$ terms from both sides to get $(-2xy)+(49)=(2xy)$ . Combining like terms we get $(49)=(4xy)$ . Then we divide by $4$ to get $(12.25)=(xy)$ . Now we know that $xy\ =\ 12.25$ which is answer choice $\boxed{12.25}$

Solution:
Choose any value for $BC$ , and then use Pythagorean theorem to get $CD - AB$ , and $AB = (BC - (CD - AB))/2$ ). Then multiply $AB \cdot CD$
For example:
$BC=25$ $CD - AB = \sqrt{25^2 - 7^2} = 24$ $AB = (25 - 24)/2=0.5$ $CD = 0.5 + 24 = 24.5$ $AB \cdot CD = (0.5)(24.5)= \boxed{12.25}$


Answer: 4.5

In trapezoid $ABCD$ $\overline{AD}$ is perpendicular to $\overline{DC}$ $AD = AB = 3$ , and $DC = 6$ . In addition, $E$ is on $\overline{DC}$ , and $\overline{BE}$ is parallel to $\overline{AD}$ . Find the area of $\triangle BEC$ [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S); [/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$

Solution:
Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$
The area of $\triangle BEC$ is \[\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{4.5}.\] ~Aplus95 (Solution)

Solution:
Clearly, $ABED$ is a square with side-length $3.$
Let the brackets denote areas. We apply area subtraction to find the area of $\triangle BEC:$ \begin{align*} [BEC]&=[ABCD]-[ABED] \\ &=\frac{AB+CD}{2}\cdot AD - AB^2 \\ &=\frac{3+6}{2}\cdot 3 - 3^2 \\ &=\boxed{4.5} ~MRENTHUSIASM


Answer: 110.0

In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$

Solution:
Let $x = BC$ be the height of the trapezoid, and let $y = CD$ . Since $AC \perp BD$ , it follows that $\triangle BAC \sim \triangle CBD$ , so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$ . Then $AE = x$ , and $ADE$ is a right triangle . By the Pythagorean Theorem
\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]
The positive solution to this quadratic equation is $x^2 = \boxed{110}$

Solution: Let $BC=x$ . Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\sqrt{11}+\sqrt{1001-x^2}$ . Therefore, we know that vector $\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle$ and vector $\vec{AC}=\langle-\sqrt{11},-x\rangle$ . Now we know that these vectors are perpendicular, so their dot product is 0. \[\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0\] \[(x^2-11)^2=11(1001-x^2)\] \[x^4-11x^2-11\cdot 990=0.\] As above, we can solve this quadratic to get the positve solution $BC^2=x^2=\boxed{110}$

Solution: Let $BC=x$ and $CD=y+\sqrt{11}$ . From Pythagoras with $AD$ , we obtain $x^2+y^2=1001$ . Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$ , so we have \[\left(y+\sqrt{11}\right)^2+11=x^2+1001.\] Substituting $x^2=1001-y^2$ and simplifying yields \[y^2+\sqrt{11}y-990=0,\] and the quadratic formula gives $y=9\sqrt{11}$ . Then from $x^2+y^2=1001$ , we plug in $y$ to find $x^2=\boxed{110}$

Solution:
Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have \begin{align*} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{align*} Followed by dropping the perpendicular like in solution 1, we obtain system of equation \[BC^2=CD^2-990\] \[BC^2+CD^2-2\sqrt{11}CD=990\] Rearrange the first equation yields \[BC^2-CD^2=990\] Equating it with the second equation we have \[BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD\] Which gives $CD^2=\frac{BC^2}{11}$ .
Substituting into equation 1 obtains the quadratic in terms of $BC^2$ \[(BC^2)^2-11BC^2-11\cdot990=0\] Solving the quadratic to obtain $BC^2=\boxed{110}$


Answer: 34.0

Solution: There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$ . The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$ , because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$ $5$ $16$ , and $5$ . Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{34}$ units.


Answer: 5.0

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$

Solution:
Since the heights of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$
$\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$
$\frac{AB+EF}{2}=DC+EF$ , so
$AB+EF=2DC+2EF$
$EF$ is exactly halfway between $AB$ and $DC$ , so $EF=\frac{AB+DC}{2}$
$AB+\frac{AB+DC}{2}=2DC+AB+DC$ , so
$\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$ , and
$\frac{1}{2}AB=\frac{5}{2}DC$
$\frac{AB}{DC} = \boxed{5}$

Solution:
Mark $DC=z$ $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ $FECD$ are the same. Mark the height to be $h$
Then, we have that $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$
From this, we get that $x=2z+y$
We also get that $\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)$
Simplifying, we get that $2x=z+3y$
Notice that we want $\tfrac{AB}{DC}=\tfrac{x}{z}$
Dividing the first equation by $z$ , we get that $\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})$
Dividing the second equation by $z$ , we get that $2(\tfrac{x}{z})=1+3(\tfrac{y}{z})$
Now, when we subtract the top equation from the bottom, we get that $\tfrac{x}{z}=\boxed{5}$

Solution:
Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid $x$ . The area of the smaller trapezoid is $A$ $h \cdot \frac {b_1 + b_2}{2}$ . The area of the larger trapezoid is $A$ $h \cdot \frac {b_1 + x}{2}$ . Since this problem asks for proportions, assume that $b_1 = 1$ and $b_2 = 2$
The smaller trapezoid has area $h$ while the larger trapezoid must have area $5h$ . We have the equation $\frac {x}{2} = 5$ $x$ = 10, and our answer is $\boxed{5}$


Answer: 504.0

In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$

Solution:
Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$ . Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$
As $\angle AED = 90^{\circ}$ , note that the midpoint of $\overline{AD}$ $N$ , is the center of the circumcircle of $\triangle AED$ . We can do the same with the circumcircle about $\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$ ). It follows that
\[NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.\]
Thus $MN = NE - ME = \boxed{504}$

Solution:
Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$ , respectively. Let $x = NH$ , so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$ . Also, let $h = BF = CG = HM$
By AA~, we have that $\triangle AFB \sim \triangle CGD$ , and so \[\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.\]
By the Pythagorean Theorem on $\triangle MHN$ \[MN^{2} = x^2 + h^2 = 504^2,\] so $MN = \boxed{504}$

Solution:
Plot the trapezoid such that $B=\left(1000\cos 37^\circ, 0\right)$ $C=\left(0, 1000\sin 37^\circ\right)$ $A=\left(2008\cos 37^\circ, 0\right)$ , and $D=\left(0, 2008\sin 37^\circ\right)$
The midpoints of the requested sides are $\left(500\cos 37^\circ, 500\sin 37^\circ\right)$ and $\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)$
To find the distance from $M$ to $N$ , we simply apply the distance formula and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to get $MN=\boxed{504}$

Solution: Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $\boxed{504}$ . (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)

Solution: Let the height be h. Note that if $\overline{NH} = x$ then if we draw perpendiculars like in solution 1, $\overline{FN} = 500 - x, \overline{AF} = 504 + x, \overline{HG} = 500, \overline{GD} = 504 - x.$ Note that we wish to find $\overline{MN} = \sqrt{x^2 + h^2}.$ Let's find $\tan(53)$ in two ways. Finding $\tan(53)$ from $\triangle BAF$ yields $\tan(53) = \frac{504+x}{h}.$ Finding it from $\triangle CDG$ yields $\frac{h}{504-x}.$ Setting these equal yields \[\frac{504+x}{h}=\frac{h}{504-x} \rightarrow h^2 = 504^2-x^2 \rightarrow \sqrt{x^2+h^2} = \sqrt{504^2} = \boxed{504}\]

Solution:
Rotate trapezoid $MNCD$ 180 degrees around point $N$ so that $AN$ coincides with $ND$ . Let the image of trapezoid $MNCD$ be $ANM'C'$ . Since angles are preserved during rotations, $\angle BAC' = 37^{\circ} + 53 ^{\circ} =90 ^{\circ}$ . Since $BM=CM=C'M'$ and $BM || C'M'$ $BMM'C'$ is a parallelogram. Thus, $MM'=BC'$ .
Let the point where $BC'$ intersects $AD$ be $E$ . Since $BMNE$ is a parallelogram, $AE=AN-BM=1004-500-504$ . Since $BE=EC$ and $\angle BAC= 90^{\circ}$ $AE$ is a median to the hypotenuse of $BAC'$ . Therefore, $BC'=2 AE= 1008$ , and $BE=MN=\boxed{504}.$


Answer: 210.0

Solution:
It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$
Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$ , with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$ . Thus \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*} So the sides of $\triangle CDE$ are $15,36,39$ , which we recognize to be a $5 - 12 - 13$ right triangle . Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared), \[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{210}\]

Solution:
Draw altitudes from points $C$ and $D$
Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$ . We get the following triangle:
The length of $A'B$ in this triangle is equal to the length of the original $AB$ , minus the length of $CD$ .
Thus $A'B = 52 - 39 = 13$
Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$ , and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$
Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{210}$

Solution:
Draw altitudes from points $C$ and $D$
Call the length of $AD'$ to be $y$ , the length of $BC'$ to be $z$ , and the height of the trapezoid to be $x$ .
By the Pythagorean Theorem, we have: \[z^2 + x^2 = 144\] \[y^2 + x^2 = 25\]
Subtracting these two equation yields: \[z^2-y^2=119 \implies (z+y)(z-y)=119\]
We also have: $z+y=52-39=13$
We can substitute the value of $z+y$ into the equation we just obtained, so we now have:
\[(13) (z-y)=119 \implies z-y=\frac{119}{13}\]
We can add the $z+y$ and the $z-y$ equation to find the value of $z$ , which simplifies down to be $\frac{144}{13}$ . Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.
\[\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2}  \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}\]
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.
The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$ , and multiplying this and the height of the trapezoid gets us:
\[\frac{60 \cdot 91}{13 \cdot 2} = \boxed{210}\]

Solution:
We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$
By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{210}.$

Solution:
From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$ .
Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$ .
The numbers multiply nicely to get $180+30=\boxed{210}$ -harsha12345

Solution:
First note how the answer choices are all integers.
The area of the trapezoid is $\frac{39+52}{2} \cdot h = \frac{91}{2} h$ . So h divides 2. Let $x$ be $2h$ . The area is now $91x$ .
Trying $x=1$ and $x=2$ can easily be seen to not work. Those make the only integers possible so now you know x is a fraction.
Since the area is an integer the denominator of x must divide either 13 or 7 since $91 = 13\cdot7$ .
Seeing how $39 = 3\cdot13$ and $52 = 4\cdot13$ assume that the denominator divides 13. Letting $y = \frac{x}{13}$ the area is now $7y$ .
Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean $h$ is $4$ which we ruled out.
So the answer is $\boxed{210}$ . - megateleportingrobots


Answer: 150.0

In triangle $ABC$ $AB = 10$ $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$ , where $a$ $b$ , and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$

Solution:
First, by the Law of Cosines , we have \[\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},\] so $\angle BAC = 60^\circ$
Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$ , respectively.  We first compute \[\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.\] Because $\angle BDI_B$ and $\angle I_BBD$ are half of $\angle BDA$ and $\angle ABD$ , respectively, the above expression can be simplified to \[\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.\] Similarly, $\angle CO_2D = \angle ACD + \angle CDA$ .  As a result
\begin{align*}\angle CPB &= \angle CPD + \angle BPD \\&= \frac {1}{2} \cdot \angle CO_2D + \frac {1}{2} \cdot \angle BO_1D \\&= \frac {1}{2}(\angle ABD + \angle BDA + \angle ACD + \angle CDA) \\&= \frac {1}{2} (2 \cdot 180^\circ - \angle BAC) \\&= \frac {1}{2} \cdot 300^\circ = 150^\circ.\end{align*}
Therefore $\angle CPB$ is constant ( $150^\circ$ ). Also, $P$ is $B$ or $C$ when $D$ is $B$ or $C$ . Let point $L$ be on the same side of $\overline{BC}$ as $A$ with $LC = LB = BC = 14$ $P$ is on the circle with $L$ as the center and $\overline{LC}$ as the radius, which is $14$ . The shortest (and only) distance from $L$ to $\overline{BC}$ is $7\sqrt {3}$
When the area of $\triangle BPC$ is the maximum, the distance from $P$ to $\overline{BC}$ has to be the greatest. In this case, it's $14 - 7\sqrt {3}$ . The maximum area of $\triangle BPC$ is \[\frac {1}{2} \cdot 14 \cdot (14 - 7\sqrt {3}) = 98 - 49 \sqrt {3}\] and the requested answer is $98 + 49 + 3 = \boxed{150}$

Solution: From Law of Cosines on $\triangle{ABC}$ \[\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.\] Now, \[\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.\] Since $CI_CDP$ and $BI_BDP$ are cyclic quadrilaterals, it follows that \[\angle{BPC}=\angle{CPD}+\angle{DPB}=(180^\circ-\angle{CI_CD})+(180^\circ-\angle{BI_BD})=360^\circ-210^\circ=150^\circ.\] Next, applying Law of Cosines on $\triangle{CPB}$ \begin{align*} & BC^2=14^2=PC^2+PB^2+2\cdot PB\cdot PC\cdot\frac{\sqrt{3}}{2} \\ & \implies \frac{PC^2+PB^2-196}{PC\cdot PB}=-\sqrt{3} \\ & \implies \frac{PC}{PB}+\frac{PB}{PC}-\frac{196}{PC\cdot PB}=-\sqrt{3} \\ & \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right). \end{align*} By AM-GM, $\frac{PC}{PB}+\frac{PB}{PC}\geq{2}$ , so \[PB\cdot PC\leq 196\left(\frac{1}{2+\sqrt{3}}\right)=196(2-\sqrt{3}).\] Finally, \[[\triangle{BPC}]=\frac12 \cdot PB\cdot PC\cdot\sin{150^\circ}=\frac14 \cdot PB\cdot PC,\] and the maximum area would be $49(2-\sqrt{3})=98-49\sqrt{3},$ so the answer is $\boxed{150}$

Solution: Proceed as in Solution 2 until you find $\angle CPB = 150$ . The locus of points $P$ that give $\angle CPB = 150$ is a fixed arc from $B$ to $C$ $P$ will move along this arc as $D$ moves along $BC$ ) and we want to maximise the area of [ $\triangle BPC$ ]. This means we want $P$ to be farthest distance away from $BC$ as possible, so we put $P$ in the middle of the arc (making $\triangle BPC$ isosceles). We know that $BC=14$ and $\angle CPB = 150$ , so $\angle PBC = \angle PCB = 15$ . Let $O$ be the foot of the perpendicular from $P$ to line $BC$ . Then the area of [ $\triangle BPC$ ] is the same as $7OP$ because base $BC$ has length $14$ . We can split $\triangle BPC$ into two $15-75-90$ triangles $BOP$ and $COP$ , with $BO=CO=7$ and $OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3$ . Then, the area of [ $\triangle BPC$ ] is equal to $7 \cdot OP=98-49\sqrt{3}$ , and so the answer is $98+49+3=\boxed{150}$


Answer: 21.0

In triangle $ABC$ $AB = 13$ $BC = 14$ $AC = 15$ . Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$ . Which of the following is closest to the area of the triangle $ADE$
$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4$

Solution:
Let's find the area of $\Delta ABC$ by Heron,
$s=\frac{a+b+c}{2}\\\\s=\frac{14+15+13}{2}\to\boxed{21}$


Answer: 254.0

In triangle $ABC$ $AB = AC = 100$ , and $BC = 56$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$ , where $m$ $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$

Solution:
Refer to the above diagram. Let the larger circle have center $O_1$ , the smaller have center $O_2$ , and the incenter be $I$ . We can easily calculate that the area of $\triangle ABC = 2688$ , and $s = 128$ and $R = 21$ , where $R$ is the inradius.
Now, Line $\overline{AI}$ is the perpendicular bisector of $\overline{BC}$ , as $\triangle ABC$ is isosceles. Letting the point of intersection be $X$ , we get that $BX = 28$ and $IX = 21$ , and $B, O_2, I$ are collinear as $O_2$ is equidistant from $\overline{AB}$ and $\overline{BC}$ . By Pythagoras, $BI = 35$ , and we notice that $\triangle BIX$ is a 3-4-5 right triangle.
Letting $r$ be the desired radius and letting $Y$ be the projection of $O_2$ onto $\overline{BC}$ , we find that $BY = \frac{4r}{3}$ . Similarly, we find that the distance between the projection from $O_1$ onto $\overline{BC}$ $W$ , and $C$ , is $\frac{64}{3}$ . From there, we let the projection of $O_2$ onto $\overline{O_1W}$ be $Z$ , and we have $O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}$ $O_1Z = 16 - r$ , and $O_1O_2 = 16 + r$ . We finish with Pythagoras on $\triangle O_1O_2Z$ , whence we get the desired answer of $\boxed{254}$ . - Spacesam

Solution:
Let the incenter be O and the altitude from A to $\overline{BC}$ be T. Note that by AA, $\triangle BQY \sim \triangle OBT$ and $\triangle PXC \sim \triangle OTC.$ Note that from $A = rs$ , the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\overline{AT}$ is an altitude), we then have that $TB = TC = 28.$ From similar triangles, we can now find $\overline{BY}.$ We have \[\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r\] Now, note that as in solution 1, drawing the perpendicular from Q to $\overline{PX}$ (call it Z) yields $\overline{PZ} = 16 - r, \overline{ZX} = r.$ Then, from this, \[\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}\] Using similar similarity as was done to find $\overline{BY}$ we have $\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{3}$ .
Now adding all these up and equating them to $\overline{BC}$ yields \[\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}\]


Answer: 0.0

In triangle $ABC$ $AB=12$ $AC=7$ , and $BC=10$ . If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The median is unchanged} \qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0}$

Solution: If you double sides $AB$ and $AC$ , they become $24$ and $14$ respectively. If $BC$ remains $10$ , then this triangle has area $0$ because ${14} + {10} = {24}$ , so two sides overlap the third side. Therefore the answer is $\boxed{0}$


Answer: 56.0

In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$

Solution:
Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$ , respectively.  Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$ $\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$ , and $M$ is the midpoint of ${CP}$ . For the same reason, $AQ=AC=117$ , and $N$ is the midpoint of ${CQ}$ .
Hence $MN=\tfrac 12 PQ$ . Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$

Solution: Let $I$ be the incenter of $ABC$ . Since $I$ lies on $BM$ and $AN$ $IM \perp MC$ and $IN \perp NC$ , so $\angle IMC + \angle INC = 180^\circ$ . This means that $CMIN$ is a cyclic quadrilateral. By the Law of Sines, $\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI$ , where $R$ is the radius of the circumcircle of $CMIN$ . Since $\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI$ , we have that $MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI$ . Letting $H$ be the point of contact of the incircle of $ABC$ with side $BC$ , we have $MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH$ . Thus, $MN = s - AB = \frac{117+120-125}{2}=\boxed{056}$

Solution: Project $I$ onto $AC$ and $BC$ as $D$ and $E$ $ID$ and $IE$ are both in-radii of $\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$ . Since $IC$ is the hypotenuse for the 4 triangles ( $\triangle INC, \triangle IMC, \triangle IDC,$ and $\triangle IEC$ ), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\omega$ which is also the circumcircle of $\triangle CMN$ and $\triangle CDE$ . To find $MN$ , we can use the Law of Cosines on $\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})$ where $O$ is the center of $\omega$ . Now, the circumradius $R$ can be found with Pythagorean Theorem with $\triangle CDI$ or $\triangle CEI$ $r^2 + 56^2 = (2R)^2$ . To find $r$ , we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}$ . To find $\angle MCN$ , we can find $\angle MIN$ since $\angle MCN = 180 - \angle MIN$ $\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}$ . Thus, $\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}$ and since $\angle A + \angle B + \angle C = 180$ , we have $\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}$ . Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})$ . To find $\cos{\angle C}$ , we use LoC on $\triangle ABC \implies \cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}$ . Our formula now becomes $MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}$ . After simplifying, we get $MN^2 = 3136 \implies MN = \boxed{056}$

Solution:
Because $\angle CMI = \angle CNI = 90$ $CMIN$ is cyclic.
Applying Ptolemy's theorem on CMIN:
$CN \cdot MI+CM \cdot IN=CI \cdot MN$
$CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI \cdot MN$
$MN = CI \sin \angle MCN$ by sine angle addition formula.
$\angle MCN = 180 - \angle MIN = 90 - \angle BCI$
Let $H$ be where the incircle touches $BC$ , then $CI \cos \angle BCI = CH = \frac{a+b-c}{2}$ $a=120, b=117, c=125$ , for a final answer of $\boxed{056}$


Answer: 21.0

In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$

Solution:
Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\tfrac{12}{15} = \tfrac{4}{5}$ , and the ratio of the shorter leg to the hypotenuse is $\tfrac{9}{15} = \tfrac{3}{5}$ . It follows that $AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)$
Let $x=DF$ . By Ptolemy's Theorem , we have \[13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.\] Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $\boxed{21}$

Solution: From solution 1, we know that $AD = 12$ and $DC = 9$ . Since $\triangle ADC \sim \triangle DEC$ , we can figure out that $DE = \frac{36}{5}$ . We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°. Therefore, $\angle{B} = 180 - \angle{DFA} = \angle{EFA}$ , and triangles $\triangle AEF \sim \triangle ADB$ . Solving the resulting proportion gives $EF = 4$ . Therefore, $DF = ED - EF = \frac{16}{5}$ $m + n = 16 + 5 = 21$ and our answer is $\boxed{21}$

Solution: If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Using the Pythagorean theorem, we now get \[BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}\] and $AF$ can be found out noting that $AE$ is just $\tfrac{48}5$ through base times height (since $12\cdot 9 = 15 \cdot \tfrac{36}5$ , similar triangles gives $AE = \tfrac{48}5$ ), and that $EF$ is just $\tfrac{36}5 - x$ .  From there, \[AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.\] Now, $BF^2 + AF^2 = 169$ , and squaring and adding both sides and subtracting a 169 from both sides gives $2x^2 - \tfrac{32}5x = 0$ , so $x = \tfrac{16}5$ . Thus, the answer is $\boxed{21}$

Solution:
First, we find $BD = 5$ $DC = 9$ , and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$ $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$ . Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{48}{5}$
Points $A$ $B$ $D$ , and $F$ all lie on a circle whose diameter is $AB$ . Let the point where the circle intersects $AC$ be $G$ . Using power of a point, we can write the following equation to solve for $AG$ \[DC\cdot BC = CG\cdot AC\] \[9\cdot 14 = CG\cdot 15\] \[CG = 126/15\] Using that, we can find $AG = \frac{99}{15}$ , and using $AG$ , we can find that $GE = 3$
We can use power of a point again to solve for $DF$ \[FE\cdot DE = GE\cdot AE\] \[(\frac{36}{5} - DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}\] \[\frac{36}{5} - DF = 4\] \[DF = \frac{16}{5} = \frac{m}{n}\] Thus, $m+n = 16+5 = 21$ $\boxed{21}$


Answer: 21.0

Solution:
Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\tfrac{12}{15} = \tfrac{4}{5}$ , and the ratio of the shorter leg to the hypotenuse is $\tfrac{9}{15} = \tfrac{3}{5}$ . It follows that $AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)$
Let $x=DF$ . By Ptolemy's Theorem , we have \[13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.\] Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $\boxed{21}$

Solution: From solution 1, we know that $AD = 12$ and $DC = 9$ . Since $\triangle ADC \sim \triangle DEC$ , we can figure out that $DE = \frac{36}{5}$ . We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°. Therefore, $\angle{B} = 180 - \angle{DFA} = \angle{EFA}$ , and triangles $\triangle AEF \sim \triangle ADB$ . Solving the resulting proportion gives $EF = 4$ . Therefore, $DF = ED - EF = \frac{16}{5}$ $m + n = 16 + 5 = 21$ and our answer is $\boxed{21}$

Solution: If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Using the Pythagorean theorem, we now get \[BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}\] and $AF$ can be found out noting that $AE$ is just $\tfrac{48}5$ through base times height (since $12\cdot 9 = 15 \cdot \tfrac{36}5$ , similar triangles gives $AE = \tfrac{48}5$ ), and that $EF$ is just $\tfrac{36}5 - x$ .  From there, \[AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.\] Now, $BF^2 + AF^2 = 169$ , and squaring and adding both sides and subtracting a 169 from both sides gives $2x^2 - \tfrac{32}5x = 0$ , so $x = \tfrac{16}5$ . Thus, the answer is $\boxed{21}$

Solution:
First, we find $BD = 5$ $DC = 9$ , and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$ $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$ . Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{48}{5}$
Points $A$ $B$ $D$ , and $F$ all lie on a circle whose diameter is $AB$ . Let the point where the circle intersects $AC$ be $G$ . Using power of a point, we can write the following equation to solve for $AG$ \[DC\cdot BC = CG\cdot AC\] \[9\cdot 14 = CG\cdot 15\] \[CG = 126/15\] Using that, we can find $AG = \frac{99}{15}$ , and using $AG$ , we can find that $GE = 3$
We can use power of a point again to solve for $DF$ \[FE\cdot DE = GE\cdot AE\] \[(\frac{36}{5} - DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}\] \[\frac{36}{5} - DF = 4\] \[DF = \frac{16}{5} = \frac{m}{n}\] Thus, $m+n = 16+5 = 21$ $\boxed{21}$


Answer: 61.0

In triangle $ABC$ $AB=13$ $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
We let $[\ldots]$ denote area; then the desired value is
Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$ , we find that
and similarly that $\frac{[BDE]}{[ABC]} = q(1-p)$ and $\frac{[CEF]}{[ABC]} = r(1-q)$ . Thus, we wish to find \begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]}  \\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*} We know that $p + q + r = \frac 23$ , and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}$ . Substituting, the answer is $\frac 1{45} - \frac 23 + 1 = \frac{16}{45}$ , and $m+n = \boxed{061}$

Solution: By the barycentric area formula, our desired ratio is equal to \begin{align*} \begin{vmatrix} 1-p & p & 0 \\  0 & 1-q & q  \\  r & 0 & 1-r \notag \end{vmatrix} &=1-p-q-r+pq+qr+pr\\ &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\ &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\ &=\frac{16}{45} \end{align*}, so the answer is $\boxed{061}$

Solution: Since the only conditions are that $p + q + r = \frac{2}{3}$ and $p^2 + q^2 + r^2 = \frac{2}{5}$ , we can simply let one of the variables be equal to 0. In this case, let $p = 0$ . Then, $q + r = \frac{2}{3}$ and $q^2 + r^2$ $\frac{2}{5}$ . Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$ . Solving this system of equations, we get $q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}$ , and $r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}$ . Plugging back into $(1-q)(1-r)$ , we get $\frac{16}{45}$ , so the answer is $\boxed{061}$