Answer: 147.0

Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon 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. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: First note the useful fact that if $R$ is the circumradius of a dodecagon ( $12$ -gon) the area of the figure is $3R^2.$ If we connect the vertices of the $3$ squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is $3\sqrt{2}.$ Thus the area of the dodecagon is $3 \cdot (3\sqrt{2})^2 = 3 \cdot 18 = 54.$ But, the problem asked for the area of the combined figure which was made from the rotated squares. This area is the area of the dodecagon, which was found, subtracting the $12$ isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this $x.$ Then, we can use the Law of Cosines on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that $x = 3\sqrt{3} -3.$ Realize that the $12$ isosceles are congruent with an angle measure of $120^{\circ},$ this means that we can create $4$ congruent equilateral triangles with side length of $3\sqrt3 - 3.$ The area of the equilateral triangle is $\frac{\sqrt{3}}{4} \cdot (3\sqrt{3} -3)^2 =  \frac{\sqrt{3}}{4} \cdot (36 - 18\sqrt{3}) = \frac{36\sqrt{3} - 54}{4}.$ Thus, the area of all the twelve small equilateral traingles are $4 \cdot \frac{36\sqrt{3} - 54}{4} = 36\sqrt{3} - 54$ . Thus, the requested area is $54 - (36\sqrt{3} - 54) = 108 - 36\sqrt{3}.$ Thus, $a+b+c = 108 + 36 + 3 = 147,$ so the answer is $\boxed{147}.$

Solution:
To make things simpler, let's take only the original sheet and the 30 degree rotated sheet. Then the diagram is this;

The area of this diagram is the original square plus the area of the four triangles that 'jut' out of the square. Because the square is rotated $30^{\circ}$ , each triangle is a 30-60-90 triangle. Similarly, the triangles that are bounded on the inside of the original square outside of the rotated square are also congruent 30-60-90 triangles. Noting this, we can do some labelling:

Since the side lengths of the squares must be the same, and they are both 6, we have a system of equations; \[y+x+y\sqrt{3} = 6\] \[\frac{x\sqrt{3}}{2} + 2y + \frac{x}{2} = 6\]
We solve this to get $x = 6-2\sqrt{3}$ and $y = 3-\sqrt{3}$
The area of each triangle is $\frac{x}{2} \cdot \frac{x\sqrt{3}}{2} \cdot \frac{1}{2}  = 6\sqrt{3} - 9$ by plugging in $x$
The rotated 60 degree square is the same thing as rotating it 30 degrees counterclockwise, so it's triangles that jut out of the square will be congruent to the triangles we have found, and therefore they will have the same area.
Unfortunately, when drawing all three squares, we see the two triangles overlap; take the very top for example.

The area of this shape is twice the area of each of the triangles that we have already found minus the area of the small triangle that is overlapped by the two by PIE. Now we only need to find the area of $\bigtriangleup BCF$
$\angle GBD \cong \angle ECA \cong 30^{\circ}$ and by symmetry $\bigtriangleup BCF$ is isosceles, so it is a 30-30-120 triangle. If we draw a perpendicular, we split it into two 30-60-90 triangles;

By symmetry, the distance from A to the edge of the square is equal to the distance from D to the edge of the square is equal to $y$ . AC = BD = $x$ , and the side length of the square is 6, so we use PIE to obtain \[x+x-BC = 6-y-y \implies BC = 12 - 6\sqrt{3}\] To find the height of $\bigtriangleup BFC$ , we see that $HC = \frac{BC}{2} = 6-3\sqrt{3}$ . Then by 30-60-90 triangles, $HF = \frac{HC}{\sqrt{3}} = 2\sqrt{3} - 3$ . Finally, the area of $\bigtriangleup BFC = \frac{BC \cdot HF}{2} = 21\sqrt{3}-36$
Putting it all together, the area of the entire diagram is the area of the square plus four of these triangle-triangle intersections. The area of these intersections by PIE is $2 \cdot [ACE] - [BFC] = 12\sqrt{3}-18-(21\sqrt{3}-36) = 18-9\sqrt{3}$ .
Therefore the total area is $36 + 4 \cdot(18-9\sqrt{3}) = 36 + 72 - 36 \sqrt{3} = 108 - 36\sqrt{3}$
Thus $a + b + c = 108+36+3 = 147 = \boxed{147}$

Solution:
As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} &  \ [ABCDEFGHIJKL] - 12[MAB] \\ & = 12 [OAB] - 12 [MAB] \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36\sqrt{3} \end{align*}
Therefore, the answer is $\boxed{147}$

Solution:
Let $O$ be the center of the polygon, $A$ be the bottom right corner of the first square, $C$ be the next vertex to the left of $A$ , and $M$ be the midpoint between $A$ and $B$ , where $B$ is the bottom left corner of the first square. Note that because there are three $90^{\circ}$ squares separated by $\frac{90^{\circ}}{3} = 30^{\circ}$ , each side of the 24-sided polygon is equal in length, meaning to calculate the area of the whole polygon, we find the area of $\bigtriangleup OAC$ and multiply by 24.
To find $[\bigtriangleup OAC]$ , we already know the height $\overline{OM}$ is the sidelength of the square over $2$ , or $\frac{6}{2}=3$ , so we just need the length of the base $\overline{AC}$ . Notice that $\bigtriangleup OCM$ is a $30-60-90$ triangle since $\angle COM = \frac{360^{\circ}}{12} = 30^{\circ}$ , so $\overline{CM} = \frac{\overline{OM}}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ . Then $\overline{AC} = \overline{AM} - \overline{CM} = \frac{6}{2} - \sqrt{3} = 3 - \sqrt{3}$ , so \begin{align*} & [\bigtriangleup AOC] = \frac{1}{2} \cdot \overline{OM} \cdot \overline{AC} \\ & = \frac{1}{2} (3)(3 - \sqrt{3}) \\ & = \frac{9 - 3\sqrt{3}}{2} \end{align*}
Then the whole area of the polygon is $\frac{9 - 3\sqrt{3}}{2} \cdot 24 = 108 - 36\sqrt{3}$ . The desired solution is then $108 + 36 + 3 = 147$ , so the answer is $\boxed{147}$


Answer: 23.0

Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$ , one additional hour; and $R$ $x$ additional hours. The value of $x$ is:
$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

Solution:
Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.
We also know that all three working together take $x$ hours.
Now the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is, \[x=\frac1{\frac1{x+6}+\frac1{x+1}+\frac1{2x}}\] \[\frac{x}{x+6}+\frac{x}{x+1}+\frac{x}{2x}=1\] \[\frac{x}{x+6}+\frac{x}{x+1}=\frac12\] \[2x(x+1)+2x(x+6)=(x+1)(x+6)\] \[2x^2+2x+2x^2+12x=x^2+7x+6\] \[3x^2+7x-6=0\] \[(3x-2)(x+3)=0\] \[x\in\{\frac23, -3\}\] Obviously, the number of hours is positive, so the answer is $\boxed{23}$


Answer: 21.0

Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$

Solution:
We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\circ}$
Let the number of sides in our polygons be $3\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$
The integral angle of a regular $k$ -gon is $180 \frac{k-2}k$ . Therefore we are looking for integer solutions to:
\[360 = 180\left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Which can be simplified to:
\[2 = \left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Furthermore, we know that two of the polygons are congruent, thus WLOG $a=c$ . Our equation now becomes
\[2 = \left( 2\cdot\frac{a-2}a + \frac{b-2}b \right)\]
Multiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$
Using the standard technique for Diophantine equations , we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$
Remembering that $a,b\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$ $(2,4)$ $(4,2)$ , and $(8,1)$
These correspond to the following pairs $(a,b)$ $(5,10)$ $(6,6)$ $(8,4)$ , and $(12,3)$
The perimeters of the resulting polygon for these four cases are $14$ $12$ $14$ , and $\boxed{21}$

Solution: We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\boxed{21}$ sides.


Answer: 12.0

Three numbers $a,b,c$ , none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

Solution:
Let $d$ be the common difference of the arithmetic sequence , so $a = b-d$ and $c = b+d$
Since increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence \[\frac{b}{b-d+1} = \frac{b+d}{b}\] \[\frac{b}{b-d} = \frac{b+d+2}{b}\] Cross-multiply in both equations to get a system of equations \[b^2 = b^2 - d^2 + b + d\] \[b^2 = b^2 - d^2 + 2b - 2d\] Rearranging terms results in \[d^2 = b+d\] \[d^2 = 2b-2d\] Substitute and solve for $d$ \[b+d = 2b-2d\] \[d = \frac{b}{3}\] Finally, substitute $d$ and solve for $b$ .  Since $b \ne 0$ , dividing by $b$ is allowed. \[(\frac{b}{3})^2 = b + \frac{b}{3}\] \[\frac{b^2}{9} = \frac{4b}{3}\] \[\frac{b}{9} = \frac{4}{3}\] \[b = 12\] The answer is $\boxed{12}$


Answer: 12.0

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution: Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$ . WLOG assume $a$ is the largest. Then, $x+y>a$ , meaning the solution is $\boxed{12}$ , as shown in the graph below.

Solution: WLOG, let the largest of the three numbers drawn be $a>0$ .  Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ .  The probability that their sum is greater than $a$ is $\boxed{12}$

Solution:
This problem is going to require some geometric probability, so let's dive right in.
Take three integers $x$ $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true
\[x + y > z\] \[y + z > x\] \[x + z > y\]
Now, it would be really easy if these equations only had two variables instead of $3$ , because then we could graph it in a $2$ -dimensional plane instead of a $3$ -dimensional cube. So, we assume $z$ is a constant. We will deal with it later.
Now, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.
\[y > -x + z\] \[y > x - z\] \[y < x + z\]
Now, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$ -plane because $x$ $y \in [0, 1]$
I recommend drawing the following figure to get an idea of what is going on.
The first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.
Note: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$
Using the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to
Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.
Except, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$ ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!
To find the area of the triangles, our plan of attack will be
Find the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$
The area of the triangles is
\[\frac{z^2}{2} + (1-z)^2 = \frac{1}{2}\left(-3x^2 + 4x\right)\]
The integral from $0$ to $1$ is
\[\frac{1}{2}\int_0^1\left(-3x^2 + 4x\right)dx = \frac{1}{2}\]
The total area of all the possible unit squares is quite obviously
\[\int_0^11dx = 1\]
Thus, the area not enclosed by the triangles is $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and the total area of the square is $1$ . Thus, the desired probability is
\[\frac{\frac{1}{2}}{1} = \boxed{12}\]

Solution: Consider a stick of length $1$ . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\boxed{12}$ probability that the second cut is on the opposite side.


Answer: 5.0

Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution:
There is a total of $P(1000,6)$ possible ordered $6$ -tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$
There are $C(1000,6)$ possible sets $\{a_1,a_2,a_3,b_1,b_2,b_3\}.$ We have five valid cases for the increasing order of these six elements:
Note that the $a$ 's are different from each other, as there are $3!=6$ ways to permute them as $a_1,a_2,$ and $a_3.$ Similarly, the $b$ 's are different from each other, as there are $3!=6$ ways to permute them as $b_1,b_2,$ and $b_3.$
So, there is a total of $C(1000,6)\cdot5\cdot6^2$ valid ordered $6$ -tuples. The requested probability is \[p=\frac{C(1000,6)\cdot5\cdot6^2}{P(1000,6)}=\frac{C(1000,6)\cdot5\cdot6^2}{C(1000,6)\cdot6!}=\frac14,\] from which the answer is $1+4=\boxed{005}.$

Solution:
Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$ . Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick.
The total number of arrangements is ${6\choose3} = 20$ ; therefore, $p = \frac{3 + 2}{20} = \frac{1}{4}$ , and the answer is $1 + 4 = \boxed{005}$

Solution: As in Solutions 2 and 3, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected.  It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter.  We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain.  The formula for the $n$ th Catalan number (where $n$ is the number of pairs of rising and falling sections) is \[\frac{\binom{2n}{n}}{n+1}\] Thus, there are $\frac{\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled.  There are $\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\frac{\binom{6}{3}/4}{\binom{6}{3}}=\frac14\Longrightarrow \boxed{005}$


Answer: 105.0

Three planets orbit a star circularly in the same plane.  Each moves in the same direction and moves at constant speed.  Their periods are 60, 84, and 140 years.  The three planets and the star are currently collinear .  What is the fewest number of years from now that they will all be collinear again?

Solution:
Denote the planets $A, B, C$ respectively.  Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years.  These are given by $a(t) = \frac{t \pi}{30}$ $b(t) = \frac{t \pi}{42}$ $c(t) = \frac{t \pi}{70}$
In order for the planets and the central star to be collinear, $a(t)$ $b(t)$ , and $c(t)$ must differ by a multiple of $\pi$ .  Note that $a(t) - b(t) = \frac{t \pi}{105}$ and $b(t) - c(t) = \frac{t \pi}{105}$ , so $a(t) - c(t) = \frac{ 2 t \pi}{105}$ .  These are simultaneously multiples of $\pi$ exactly when $t$ is a multiple of $105$ , so the planets and the star will next be collinear in $\boxed{105}$ years.


Answer: 160.0

Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$

Solution: The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{160}$


Answer: 6.0

Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution:
Let the smallest number be $x.$ It follows that the largest number is $4x.$
Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{6} ~MRENTHUSIASM

Solution: Let the common difference of the arithmetic sequence be $d$ . Consequently, the smallest number is $15-d$ and the largest number is $15+d$ . As the largest number is $4$ times the smallest number, $15+d=60-4d\implies d=9$ . Finally, we find that the smallest number is $15-9=\boxed{6}$

Solution: Let the smallest number be $x$ . Since the integers are equally spaced, and there are three of them, the middle number ( $15$ ) is the arithmetic mean of the other two numbers ( $x$ and $4x$ ). Thus, we set up the equation $(4x + x)/3 = 15$ , and, solving for $x$ , get $x = 6$ . Since $6$ is the smallest number out of the list $6, 15, 24$ $24$ because it equals $4x$ ), the answer is $\boxed{6}$

Solution: Let the smallest number be $x$ . Because $x$ and $4x$ are equally spaced from $15$ $15$ must be the average. By adding $x$ and $4x$ and dividing by $2$ , we get that the mean is also $2.5x$ . We get that $2.5x=15$ , and solving gets $x=\boxed{6}$


Answer: 2.0

Three primes $p,q$ , and $r$ satisfy $p+q = r$ and $1 < p < q$ . Then $p$ equals
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$

Solution: We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$ , and $1$ is not prime). Thus, with one of either $p$ or $q$ being even, either $p$ or $q$ must be $2$ , and as $p < q$ , we deduce $p = 2$ (as $2$ is the smallest prime). This means the answer is $\boxed{2}$


Answer: 16.0

Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?
$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$

Solution:
There are $\frac{6!}{3!\cdot2!\cdot1!}=60$ total orderings.
Suppose we order the red and white beads first. If these two colors are ordered first, we must make sure that no neighboring beads are the same color, or only one pair of neighboring beads are the same color. There are five possible orderings:
$R\ W R\ W R$
$R\ R\ W R\ W$
$W R\ R\ W R$
$R\ W R\ R\ W$
$W R\ W R\ R$
For the first case, there are $6$ possible places we can put the blue bead. For the other 4 cases, there is only one place we can put the blue bead, which is between the pair of red beads. The desired probability is $\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{16}$


Answer: 2500.0

Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
$\textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000$

Solution:
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.
\[4.8x-4.4x=500 \Rightarrow x=1250\]
Because $4.4(1250)=5500$ is a multiple of 500, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\boxed{2,500}$ seconds.

Solution:
Working backwards from the answers starting with the smallest answer, if they had run $1000$ seconds, they would have run $4400, 4800, 5000$ meters, respectively.  The first two runners have a difference of $400$ meters, which is not a multiple of $500$ (one lap), so they are not in the same place.
If they had run $1250$ seconds, the runners would have run $5500, 6000, 6250$ meters, respectively.  The last two runners have a difference of $250$ meters, which is not a multiple of $500$
If they had run $2500$ seconds, the runners would have run $11000, 12000, 12500$ meters, respectively.  The distance separating each pair of runners is a multiple of $500$ , so the answer is $\boxed{2,500}$ seconds.

Solution: Let $t$ be the time run in seconds, then the difference in meters run between the three runners is $0.2t, 0.4t, 0.6t$ . For them to be at the same location all of them need to be multiples of 500. It is now easy to see that $0.2t=500, 0.4t=1000, 0.6t=1500$ , so $t=\boxed{2,500}$


Answer: 756.0

Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$

Solution:
This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$ . We take a cross-section that contains $A$ and $B$ , which contains these two spheres but not the third, as shown below:  Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \sqrt{560}$ . Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$ . Then we have $x^2 = 576-560 \implies x = 4$
We have $AO_A = BD$ because of the rectangle, so $\sqrt{11^2-r^2} = \sqrt{13^2-r^2}-4$ .
Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \cdot \sqrt{169-r^2}$ .
Subtracting, we get $8 \cdot \sqrt{169-r^2} = 64 \implies \sqrt{169-r^2} = 8 \implies 169-r^2 = 64 \implies r^2 = 105$ .
We also notice that since we had $\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$ $AO_A = 4$
We take a cross-section that contains $A$ and $C$ , which contains these two spheres but not the third, as shown below:  We have $CO_C = \sqrt{19^2-r^2} = \sqrt{361 - 105} = \sqrt{256} = 16$ . Since $AO_A = 4$ , we have $EO_C = 16-4 = 12$ . Using Pythagorean theorem, $O_AE = \sqrt{30^2 - 12^2} = \sqrt{900-144} = \sqrt{756}$ . Therefore, $O_AE^2 = AC^2  = \boxed{756}$

Solution:
Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately.
According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$
The desired value is $(11+19)^2-(16-4)^2=\boxed{756}.$

Solution:
Denote by $r$ the radius of three congruent circles formed by the cutting plane.
Denote by $O_A$ $O_B$ $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$ $B$ $C$ , respectively.
Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$ $O_A O_C = 11 + 19 = 30$
We have $O_A A^2 = 11^2 - r^2$ $O_B B^2 = 13^2 - r^2$ $O_C C^2 = 19^2 - r^2$
Because $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $\angle O_A A B = \angle O_B BA = 90^\circ$
Hence, \begin{align*} O_B B - O_A A & = \sqrt{O_A O_B^2 - AB^2} \\ & = 4 . \hspace{1cm} (1) \end{align*}
Recall that \begin{align*} O_B B^2 - O_A A^2 & = \left( 13^2 - r^2 \right) - \left( 11^2 - r^2 \right) \\ & = 48 . \hspace{1cm} (2) \end{align*}
Hence, taking $\frac{(2)}{(1)}$ , we get \[ O_B B + O_A A = 12 . \hspace{1cm} (3) \]
Solving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$
Thus, $r^2 = 11^2 - O_A A^2 = 105$
Thus, $O_C C = \sqrt{19^2 - r^2} = 16$
Because $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\angle O_A A C = \angle O_C CA = 90^\circ$
Therefore, \begin{align*} AC^2 & = O_A O_C^2 - \left( O_C C - O_A A \right)^2 \\ & = \boxed{756}


Answer: 42.0

Three times Dick's age plus Tom's age equals twice Harry's age.
Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age.
Their respective ages are relatively prime to each other. The sum of the squares of their ages is
$\textbf{(A) }42\qquad \textbf{(B) }46\qquad \textbf{(C) }122\qquad \textbf{(D) }290\qquad  \textbf{(E) }326$

Solution:
\[t=2h-3d\] \[3d^3+t^3=2h^3\]
First, substitute in t into the second equation and get $3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3$ . That turns into $h^3-6h^2d+9hd^2-4d^3=0$ which is factored into $(h-4d)(h-d)^2 =0.$ WLOG, $d=1$ and consequently $h=4$ . Then $t=8-3=5$ . Everything appears to be relatively prime already. The answer is thus $1+16+25=\boxed{42}.$ ~lopkiloinm


Answer: 15.0

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

Solution:
$AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\triangle ABC \sim \triangle BED$ . The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$
\[\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC\]
\[\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}\]
Since $AC=2BC$ $BC=\frac{1}{\sqrt{5}}$ $\triangle ABC$ is a right triangle so the area is just $\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{15}$

Solution:
Let $\text{E}$ be the origin. Then, $\text{D}=(1, 0)$ $\text{A}=(0, 2)$ $\text{B}=(1, 2)$ $\text{F}=(2, 1)$
${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$
Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$ .
Thus, $x=\frac{4}{5}$ .
Plugging this into the first equation gives us $y=\frac{8}{5}$
Since $\text{C} (0.8, 1.6)$ $G$ is $(0.8, 2)$
${AB}=1$ and ${CG}=0.4$
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}$ . The answer is $\boxed{15}$

Solution:
Triangle $EAB$ is similar to triangle $EHI$ ; line $HI = 1/2$
Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$
Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}$ .
The answer is $\boxed{15}$

Solution:
Let $L$ be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let $K$ be the top right corner of the top right unit square, where segment $ABL$ is 2 units in length.
Because of the Pythagorean Theorem, since $AC = 2$ and $LK$ = 1, the diagonal of triangle $ALK$ is $\sqrt{5}$
Triangle $ALK$ is clearly a similar triangle to triangle $ABC$ . Segment $AB$ is the hypotenuse of triangle $ABC$ . So, we can write down:
\[AK/AB = LK/BC\] , which is equal to: \[\frac{\sqrt{5}}{1} = \frac{1}{BC}\] Solving this equation yields:
\[BC = \frac{1}{\sqrt{5}}\]
By Pythagorean theorem, we can now find segment $AC$ \[(\frac{1}{\sqrt{5}})^2 + AC^2 = 1\] Solving this yields:
\[AC^2 = \frac{4}{5}\] , so $AC = \frac{2}{\sqrt{5}}$
So then we can use \[A = \frac{1}{2} * b * a.\] So \[A = \frac{1}{2} * \frac{1}{\sqrt{5}} *  \frac{2}{\sqrt{5}}\]
\[= \boxed{15}\]


Answer: 9.0

Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite.
The sum of the coordinates of vertex $S$ is:
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$

Solution: Graph the three points on the coordinate grid.  Noting that the opposite sides of a parallelogram are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$ .  The sum of the x-coordinates and y-coordinates is $9$ , so the answer is $\boxed{9}$


Answer: 96.0

Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution:
We can begin to put this into cases. Let's call the pairs $a$ $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.
Case $1$ :
Second Row: a b c
Third Row: b c a
Case $2$ :
Second Row: a c b
Third Row: c b a
Case $3$ :
Second Row: a b c
Third Row: c a b
Case $4$ :
Second Row: a c b
Third Row: b a c
For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has $2 \cdot 2 \cdot 2 = 8$ possibilities. Since there are four cases, when pair $a$ has someone in the leftmost seat of the second row, there are $32$ ways to rearrange it. However, someone from either pair $a$ $b$ , or $c$ could be sitting in the leftmost seat of the second row. So, we have to multiply it by $3$ to get our answer of $32 \cdot 3 = 96$ . So, the correct answer is $\boxed{96}$

Solution:
Lets call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$ . We can split our problem into two cases:
There is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.
Starting off with the first case, we see that there are $3!=6$ ways to arrange the A,B,C. Then, we have to choose which sibling sits. There are $2$ choices for each set of siblings meaning we have $2^3=8$ ways to arrange that. So, there are $48$ ways to arrange the siblings in the first row. The second row is a bit easier. We see that there are $2$ ways to place the A sibling and each placement yields only $1$ possibility. So, our first case has $48\cdot2=96$ possibilities.
In our second case, there are $3$ ways to choose which set of siblings will be in the same row and $4$ ways to choose the person in between. So, there $3*4 = 12$ ways to arrange the first row. In the second row, however, we see that it is impossible to make everything work out. So, there are $0$ possibilities for this case.
Thus, there are $96+0 = \boxed{96}$ possibilities for this trip.

Solution:
Call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$
There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that $A_1$ takes it ( $\circ$ denotes an empty seat):
\[A_1 \circ \circ\] \[\circ \ \circ \ \circ\]
Then there are 4 choices for the second seat ( $B_1$ $B_2$ $C_1$ , or $C_2$ ). Like before, it doesn't matter who takes the seat, so WLOG suppose it is $B_1$
\[A_1 B_1 \circ \\\] \[\circ \ \circ \ \circ\]
The last seat in the first row cannot be $A_2$ because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be $C_1$ or $C_2$ . Let's say WLOG that it is $C_1$ . There are two ways to create a second row:
\[A_1 B_1 C_1 \\\] \[B_2 C_2 A_2\]
and
\[A_1 B_1 C_1 \\\] \[C_2 A_2 B_2\]
Therefore, there are $6 \cdot 4 \cdot 2 \cdot 2= \boxed{96}$ possible seating arrangements.

Solution:
WLOG, define the three pairs of siblings to be: $A$ $B$ , and $C$ . Now, notice that you can only form a correct grouping either like this:
\[A B C\]
\[B C A\]
or this:
\[C B A\]
\[A C B\]
However, we need to consider the different orders. There are $3!$ ways to order each pair (eg. the same letters) and $2^3$ ways to order the people each of the three pairs. Now, we can just multiply everything together, yielding:
\[2\cdot3!\cdot2^3\]
Which is $\boxed{96}$

Solution: Let the families be $A$ $B$ $C$ . In any given possible arrangement, there are $3! = 6$ ways to arrange the families and $2 \cdot 2 \cdot 2 = 8$ ways to arrange the siblings. This means the answer has to be divisble by $6 \cdot 8 = 48$ . The only answer choice that satisfies this is $\boxed{96}$

Solution:
If a pair of siblings are in the same row, the other $2$ pairs of siblings cannot fit in the remaining $4$ seats to meet the requirements. Therefore, the siblings must be in different rows.
Let the first pair of siblings be $a_1$ $a_2$ , the second pair of siblings be $b_1$ $b_2$ , and the third pair of siblings be $c_1$ $c_2$ . For each row there must be $1$ child from each family. WLOG, let $a_1$ $b_1$ $c_1$ be in the first row. There are $3!$ arrangements. $a_2$ $b_2$ $c_2$ are in the second row, and they cannot be in the same column as their sibling. Now the problem becomes a $3$ element Derangement problem.
An $n$ element derangement problem is to find the number of permutations of $n$ elements where each element has a specified location, and no element is in it's specified location. For example, there are $3$ elements $1$ $2$ $3$ $1$ cannot be in the first location, $2$ cannot be in the second location, and $3$ cannot be in the third location. The derangements are: $(2,3,1)$ $(3,1,2)$ $D_3=2$
Each pair of siblings can swap their position. There are $2^3$ swaps. The total arrangements are: $3! \cdot D_3 \cdot 2^3=6 \cdot 2 \cdot 8 =\boxed{96}$


Answer: 6.0

Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square? \[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\]
[asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0)); draw((3,0)--(2,0)--(2,1)--(3,1)--cycle,linewidth(2));  label("$1$",(0,2),SE); label("$2$",(0,1),SE); [/asy]
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution:
The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ .
The only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with
a $\boxed{6}$


Answer: 2.0

To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$

Solution: The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$ . Therefore the number in the lower right-hand square is $\boxed{2}$

Solution: Note how the first and second row already contain a $2$ . Since the third row, last column already has a $4$ , the only possible place a $2$ could be in is the bottom right square. Thus our answer is $\boxed{2}$


Answer: 8.0

To control her blood pressure, Jill's grandmother takes one half of a pill every other day.  If one supply of medicine contains $60$ pills, then the supply of medicine would last approximately
$\text{(A)}\ 1\text{ month} \qquad \text{(B)}\ 4\text{ months} \qquad \text{(C)}\ 6\text{ months} \qquad \text{(D)}\ 8\text{ months} \qquad \text{(E)}\ 1\text{ year}$

Solution: If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 \times 4=240$ days which is about $\boxed{8}$


Answer: 189.0

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

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


Answer: 39.0

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

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


Answer: 21.0

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

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

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

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


Answer: 5.0

Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$

Solution: Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{5} Note that actual values were not found.


Answer: 5.0

Solution: Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{5} Note that actual values were not found.


Answer: 20.0

Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly.  During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175.  In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars.  What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution:
The total amount paid is $105 + 125 + 175 = 405$ .  To get how much each should have paid, we do $405/3 = 135$
Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars.  This means that $t-d = 30 - 10 = \boxed{20}$

Solution: The difference in the money that Tom paid and Dorothy paid is $20$ .  In order for them both to have paid the same amount, Tom must pay $20$ more than Dorothy.  The answer is $\boxed{20}$

Solution:
The meaning of sharing costs equally is meaning that, after the vacation, they are equally dividing the money in a way such that, each person would have the same amount left. As each person spends an amount of money, greater than 100, let it be that they all had $200$ dollars to spend. This means that after the vacation we want the amount of money, they currently have. After the trip, Tom would've $95$ dollars, Dorothy would've $75$ dollars, and Sammy had $25$ dollars. This gives us a total of $95+75+25=195$ dollars.
We want to equally split this money, as that is what happens after splitting the cost equally. This means that we want Dorothy, Tom, and Sammy to each have $65$ dollars. We know that Tom gave Sammy $t$ dollars meaning that we want to split this money first. As Tom gives money to no one else, we want him to reach $65$ dollars in this trade, meaning that as Tom has $95$ dollars and Sammy has $25$ dollars, we can do a trade of $30$ so $t=30$ . After this trade, we get that Tom has $65$ dollars, Sammy has $55$ dollars, and Dorothy has $75$ dollars.
Next trade is where Dorothy gives $d$ dollars to Sammy. Dorothy has $75$ dollars and Sammy has $55$ dollars. As both of these don't have $65$ dollars and this is the last trade, we need to make sure both have $65$ dollars at the end. This is possible if $d=10$
We want to find $t-d=30-10=\boxed{20}$


Answer: 20.0

Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ $105$ , Dorothy paid $ $125$ , and Sammy paid $ $175$ . In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution: Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: \[105 + t = 125 + d.\] Rearranging, we have \[t-d = \boxed{20}.\]

Solution:
Add up the amounts that Tom, Dorothy, and Sammy paid to get $ $405$ , and divide by 3 to get $ $135$ , the amount that each should have paid.
Tom, having paid $ $105$ , owes Sammy $ $30$ , and Dorothy, having paid $ $125$ , owes Sammy $ $10$
Thus, $t - d = 30 - 10 = 20$ , which is $\boxed{20}$


Answer: 13.0

Tony works $2$ hours a day and is paid $ $0.50$ per hour for each full year of his age. During a six month period Tony worked $50$ days and earned $ $630$ . How old was Tony at the end of the six month period?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$

Solution: Tony works $2$ hours a day and is paid $0.50$ dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is $12$ years old, he gets $12$ dollars a day. We also know that he worked $50$ days and earned $630$ dollars. If he was $12$ years old at the beginning of his working period, he would have earned $12 * 50 = 600$ dollars. If he was $13$ years old at the beginning of his working period, he would have earned $13 * 50 = 650$ dollars. Because he earned $630$ dollars, we know that he was $13$ for some period of time, but not the whole time, because then the money earned would be greater than or equal to $650$ . This is why he was $12$ when he began, but turned $13$ sometime during the six month period and earned $630$ dollars in total. So the answer is $13$ .The answer is $\boxed{13}$ . We could find out for how long he was $12$ and $13$ $12 \cdot x + 13 \cdot (50-x) = 630$ . Then $x$ is $20$ and we know that he was $12$ for $20$ days, and $13$ for $30$ days. Thus, the answer is $13$

Solution:
Let $x$ equal Tony's age at the end of the period. We know that his age changed during the time period (since $630$ does not evenly divide $50$ ). Thus, his age at the beginning of the time period is $x - 1$
Let $d$ be the number of days Tony worked while his age was $x$ . We know that his earnings every day equal his age (since $2 \cdot 0.50 = 1$ ). Thus, \[x \cdot d + (x - 1)(50 - d) = 630\] \[x\cdot d + 50x - x\cdot d - 50 + d = 630\] \[50x + d = 680\] \[x = \dfrac{680 - d}{50}\] Since $0 < d <50$ $d = 30$ . Then we know that $50x = 650$ and $x = \boxed{13}$

Solution:
Since Tony worked for $50$ days, he has worked for $100$ hours. Let $k$ be his hourly wage. Then, we know that $100k = 630$
Dividing both sides by $50$ , we get that $2k$ (Daily wage, which is equal to his age) $\approx$ $12.6$ . Since we now know that his age was either 12 or 13 during the 6 month span, and we are asked to find his age at the end of this time, the answer is $\boxed{13}$


Answer: 430.0

Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is
[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y[0]); for(int i = 1;i<y.length;++i) { pair p = (a,y[i]); pen pen = linewidth(.7); if(y[i]-prev.y!=1){  pen+=dotted; } draw((xscl*prev.x,prev.y)--(xscl*p.x,p.y),pen); prev = p; } }for(int a:y) { pair prev = (x[0],a); for(int i = 1;i<x.length;++i) { pair p = (x[i],a); pen pen = linewidth(.7); if(x[i]-prev.x!=1){  pen+=dotted; } draw((xscl*prev.x,prev.y)--(p.x*xscl,p.y),pen); prev = p; } } path lblx = (0,-.7)--(5*xscl,-.7); draw(lblx); label("$10$",lblx); path lbly = (5*xscl+.7,0)--(5*xscl+.7,5); draw(lbly); label("$20$",lbly);[/asy]
$\textrm{(A)}\ 30\qquad\textrm{(B)}\ 200\qquad\textrm{(C)}\ 410\qquad\textrm{(D)}\ 420\qquad\textrm{(E)}\ 430$

Solution: There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and $21\cdot10+11\cdot20=\boxed{430}$


Answer: 5.0

Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered 70% of the arithmetic, 40% of the algebra, and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% of the problems right. How many more problems would she have needed to answer correctly to earn a 60% passing grade?
$\text{(A)}\ 1 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11$

Solution:
First, calculate how many of each type of problem she got right:
Arithmetic: $70\% \cdot 10 = 0.70 \cdot 10 = 7$
Algebra: $40\% \cdot 30 = 0.40 \cdot 30 = 12$
Geometry: $60\% \cdot 35 = 0.60 \cdot 35 =  21$
Altogether, Tori answered $7 + 12 + 21 = 40$ questions correct.
To get a $60\%$ on her test overall, she needed to get $60\% \cdot 75 = 0.60 \cdot 75 = 45$ questions right.
Therefore, she needed to answer $45 - 40 = 5$ more questions to pass, so the correct answer is $\boxed{5}$


Answer: 127.0

Torus $T$ is the surface produced by revolving a circle with radius $3$ around an axis in the plane of the circle that is a distance $6$ from the center of the circle (so like a donut). Let $S$ be a sphere with a radius $11$ . When $T$ rests on the inside of $S$ , it is internally tangent to $S$ along a circle with radius $r_i$ , and when $T$ rests on the outside of $S$ , it is externally tangent to $S$ along a circle with radius $r_o$ . The difference $r_i-r_o$ can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] unitsize(0.3 inch); draw(ellipse((0,0), 3, 1.75)); draw((-1.2,0.1)..(-0.8,-0.03)..(-0.4,-0.11)..(0,-0.15)..(0.4,-0.11)..(0.8,-0.03)..(1.2,0.1)); draw((-1,0.04)..(-0.5,0.12)..(0,0.16)..(0.5,0.12)..(1,0.04)); draw((0,2.4)--(0,-0.15)); draw((0,-0.15)--(0,-1.75), dashed); draw((0,-1.75)--(0,-2.25)); draw(ellipse((2,0), 1, 0.9)); draw((2.03,-0.02)--(2.9,-0.4)); [/asy]

Solution:
First, let's consider a section $\mathcal{P}$ of the solids, along the axis.
By some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\mathcal{P}$ we took crosses one of the equator of the sphere.
Here I drew two graphs, the first one is the case when $T$ is internally tangent to $S$

and the second one is when $T$ is externally tangent to $S$

For both graphs, point $O$ is the center of sphere $S$ , and points $A$ and $B$ are the intersections of the sphere and the axis. Point $E$ (ignoring the subscripts) is one of the circle centers of the intersection of torus $T$ with section $\mathcal{P}$ . Point $G$ (again, ignoring the subscripts) is one of the tangents between the torus $T$ and sphere $S$ on section $\mathcal{P}$ $EF\bot CD$ $HG\bot CD$
And then, we can start our calculation.
In both cases, we know $\Delta OEF\sim \Delta OGH\Longrightarrow \frac{EF}{OE} =\frac{GH}{OG}$
Hence, in the case of internal tangent, $\frac{E_iF_i}{OE_i} =\frac{G_iH_i}{OG_i}\Longrightarrow \frac{6}{11-3} =\frac{r_i}{11}\Longrightarrow r_i=\frac{33}{4}$
In the case of external tangent, $\frac{E_oF_o}{OE_o} =\frac{G_oH_o}{OG_o}\Longrightarrow \frac{6}{11+3} =\frac{r_o}{11}\Longrightarrow r_o=\frac{33}{7}$
Thereby, $r_i-r_o=\frac{33}{4}-\frac{33}{7}=\frac{99}{28}$ . And there goes the answer, $99+28=\boxed{127}$


Answer: 45.0

Trapezoid $ABCD$ has $AD||BC$ $BD = 1$ $\angle DBA = 23^{\circ}$ , and $\angle BDC = 46^{\circ}$ .  The ratio $BC: AD$ is $9: 5$ .  What is $CD$
$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$

Solution:
Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$ .  Then
\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}
Thus $\triangle BDE$ is isosceles with $DE = BD$ .  Because $\overline {AD} \parallel \overline {BC}$ , it follows that the triangles $BCE$ and $ADE$ are similar.  Therefore \[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\] so $CD = \boxed{45}.$


Answer: 194.0

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$

Solution:
Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$ , therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$
Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$ , therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$
Since $\triangle ADB$ is right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{194}$

Solution:
Since $\triangle BCD$ is isosceles with base $\overline{BD},$ it follows that median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$
Since $\angle AOD=\angle COP$ by vertical angles, we conclude that $\triangle AOD\sim\triangle COP$ by AA, from which $\frac{AD}{CP}=\frac{OD}{OP},$ or \[AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$ ). Doubling both sides produces \begin{align*} 2[AOD]&=2[BOC] \\ OD\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$
Applying the Pythagorean Theorem to right $\triangle CPB,$ we have \[h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.\] Finally, we get \[AD=h\cdot\frac{x}{11}=4\sqrt{190},\] so the answer is $4+190=\boxed{194}.$

Solution:
Let $CP = y$ $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$
(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$
(2) Applying Pythagorean Theorem on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$
(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$
Thus, $\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $4+190=\boxed{194}.$

Solution:
Observe that $\triangle BPC$ is congruent to $\triangle DPC$ ; both are similar to $\triangle BDA$ . Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$ . Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ .
Let $M$ be the midpoint of $\overline{DE}$ . Note that $\triangle CME$ is congruent to $\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\overline{BE}.$
Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$ . This means that $O$ is the centroid of $\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ $DP = \frac{BD}{2}$ . The question tells us that $OP = 11$ $DP-DO=11$ ; we can write this in terms of $DB$ $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$
We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$ . Now, by Pythagorean theorem on $\triangle ABD$ $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$
The answer is $4+190 = \boxed{194}$

Solution:
Since $P$ is the midpoint of isosceles triangle $BCD$ , it would be pretty easy to see that $CP\perp BD$ . Since $AD\perp BD$ as well, $AD\parallel CP$ . Connecting $AP$ , it’s obvious that $[ADC]=[ADP]$ . Since $DP=BP$ $[APB]=[ADC]$
Since $P$ is the midpoint of $BD$ , the height of $\triangle APB$ on side $AB$ is half that of $\triangle ADC$ on $CD$ . Since $[APB]=[ADC]$ $AB=2CD$
As a basic property of a trapezoid, $\triangle AOB \sim \triangle COD$ , so $\frac{OB}{OD}=\frac{AB}{CD}=2$ , or $OB=2OD$ . Letting $OD=x$ , then $PB=DP=11+x$ , and $OB=22+x$ . Hence $22+x=2x$ and $x=22$
Since $\triangle AOD \sim \triangle COP$ $\frac{AD}{PC}=\frac{OD}{OP}=2$ . Since $PD=11+22=33$ $PC=\sqrt{43^2-33^2}=\sqrt{760}$
So, $AD=2\sqrt{760}=4\sqrt{190}$ . The correct answer is $\boxed{194}$

Solution:
Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$ -axis, and $A$ lie on the negative $y$ -axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\frac{b}{a}.$ Since $AB \parallel CD$ the slope of $CD$ is the same. Note that as $\triangle DCB$ is isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \frac{b}{a}x$ $D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \frac{3b}{a}x - 2b$ and intersects $BD$ $x$ -axis) at $O =\left(\frac{2}{3}a, 0\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\triangle DPC$ $\triangle DBC$ is isosceles), we have $b = \sqrt{43^2 - 33^2} = 2\sqrt{190},$ so $2b=4\sqrt{190}.$
Finally, the answer is $4+190=\boxed{194}.$


Answer: 194.0

Solution:
Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$ , therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$
Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$ , therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$
Since $\triangle ADB$ is right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{194}$

Solution:
Since $\triangle BCD$ is isosceles with base $\overline{BD},$ it follows that median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$
Since $\angle AOD=\angle COP$ by vertical angles, we conclude that $\triangle AOD\sim\triangle COP$ by AA, from which $\frac{AD}{CP}=\frac{OD}{OP},$ or \[AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$ ). Doubling both sides produces \begin{align*} 2[AOD]&=2[BOC] \\ OD\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$
Applying the Pythagorean Theorem to right $\triangle CPB,$ we have \[h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.\] Finally, we get \[AD=h\cdot\frac{x}{11}=4\sqrt{190},\] so the answer is $4+190=\boxed{194}.$

Solution:
Let $CP = y$ $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$
(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$
(2) Applying Pythagorean Theorem on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$
(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$
Thus, $\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $4+190=\boxed{194}.$

Solution:
Observe that $\triangle BPC$ is congruent to $\triangle DPC$ ; both are similar to $\triangle BDA$ . Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$ . Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ .
Let $M$ be the midpoint of $\overline{DE}$ . Note that $\triangle CME$ is congruent to $\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\overline{BE}.$
Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$ . This means that $O$ is the centroid of $\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ $DP = \frac{BD}{2}$ . The question tells us that $OP = 11$ $DP-DO=11$ ; we can write this in terms of $DB$ $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$
We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$ . Now, by Pythagorean theorem on $\triangle ABD$ $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$
The answer is $4+190 = \boxed{194}$

Solution:
Since $P$ is the midpoint of isosceles triangle $BCD$ , it would be pretty easy to see that $CP\perp BD$ . Since $AD\perp BD$ as well, $AD\parallel CP$ . Connecting $AP$ , it’s obvious that $[ADC]=[ADP]$ . Since $DP=BP$ $[APB]=[ADC]$
Since $P$ is the midpoint of $BD$ , the height of $\triangle APB$ on side $AB$ is half that of $\triangle ADC$ on $CD$ . Since $[APB]=[ADC]$ $AB=2CD$
As a basic property of a trapezoid, $\triangle AOB \sim \triangle COD$ , so $\frac{OB}{OD}=\frac{AB}{CD}=2$ , or $OB=2OD$ . Letting $OD=x$ , then $PB=DP=11+x$ , and $OB=22+x$ . Hence $22+x=2x$ and $x=22$
Since $\triangle AOD \sim \triangle COP$ $\frac{AD}{PC}=\frac{OD}{OP}=2$ . Since $PD=11+22=33$ $PC=\sqrt{43^2-33^2}=\sqrt{760}$
So, $AD=2\sqrt{760}=4\sqrt{190}$ . The correct answer is $\boxed{194}$

Solution:
Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$ -axis, and $A$ lie on the negative $y$ -axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\frac{b}{a}.$ Since $AB \parallel CD$ the slope of $CD$ is the same. Note that as $\triangle DCB$ is isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \frac{b}{a}x$ $D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \frac{3b}{a}x - 2b$ and intersects $BD$ $x$ -axis) at $O =\left(\frac{2}{3}a, 0\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\triangle DPC$ $\triangle DBC$ is isosceles), we have $b = \sqrt{43^2 - 33^2} = 2\sqrt{190},$ so $2b=4\sqrt{190}.$
Finally, the answer is $4+190=\boxed{194}.$


Answer: 25.0

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$ . The other two sides are of lengths $10$ and $14$ . The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$

Solution:
In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$ .
Denote $\overline{AE} = x$ and $\overline{DE} = h$ . In right triangle $AED$ , we have from the Pythagorean theorem: $x^2+h^2=100$ . Note that since $EF = DC$ , we have $BF = 33-DC-x = 12-x$ . Using the Pythagorean theorem in right triangle $BFC$ , we have $(12-x)^2 + h^2 = 196$
We isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$
Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$ . Now, we can determine that $h^2 = 100-4 \implies h = 4\sqrt{6}$

The two diagonals are $\overline{AC}$ and $\overline{BD}$ . Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$ , we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$ . Since $\sqrt{96+529}<\sqrt{96+961}$ $25$ is the shorter length*, so the answer is $\boxed{25}$

Solution: The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$ . This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$ . One may now proceed as in Solution $1$ to obtain an answer of $\boxed{25}$


Answer: 164.0

Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$

Solution:
Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$
Then $XD=xy-70, XC=y(92-x)-50,$ thus \[\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},\] which we can rearrange, expand and cancel to get $120x=70\cdot 92,$ hence $AP=x=\frac{161}{3}$ . This gives us a final answer of $161+3=\boxed{164}$

Solution: From $(1)$ above, $x = \frac{70r}{h}$ and $92-x = \frac{50r}{h}$ . Adding these equations yields $92 = \frac{120r}{h}$ . Thus, $x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}$ , and $m+n = \boxed{164}$

Solution:
The area of the trapezoid is $\frac{(19+92)h}{2}$ , where $h$ is the height of the trapezoid.
Draw lines $CP$ and $BP$ . We can now find the area of the trapezoid as the sum of the areas of the three triangles $BPC$ $CPD$ , and $PBA$
$[BPC] = \frac{1}{2} \cdot 50 \cdot r$ (where $r$ is the radius of the tangent circle.)
$[CPD] = \frac{1}{2} \cdot 19 \cdot h$
$[PBA] = \frac{1}{2} \cdot 70 \cdot r$
$[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}$
$60r = 46h$
$r = \frac{23h}{30}$
From Solution 1 above, $\frac{h}{70} = \frac{r}{x}$
Substituting $r = \frac{23h}{30}$ , we find $x = \frac{161}{3}$ , hence the answer is $\boxed{164}$


Answer: 5979.0

Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$ , then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers  ( $4$ $5$ $6$ ), then Tadd must say the next four numbers ( $7$ $8$ $9$ $10$ ), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$ th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

Solution:
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||}   \hline  Round & Tadd & Todd & Tucker \\ [0.5ex]   \hline\hline  1 & 1 & 2-3 & 4-6 \\   \hline  2 & 7-10 & 11-15 & 16-21 \\  \hline  3 & 22-28 & 29-36 & 37-45 \\  \hline  4 & 46-55 & 56-66 & 67-78 \\ [1ex]   \hline \end{tabular}$
Tadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round n. At the end of round n, the number of numbers Tadd has said so far is $1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n-1)}{2}$ , by the sum of arithmetic series formula.
We find that $\dfrac{37(110)}{2}=2035$ , so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37. At the end of Tadd's turn in round 37, the children have, in total, completed $36+36+37=109$ turns. In general, at the end of turn $n$ , the nth triangular number is said, or $\dfrac{n(n+1)}{2}$ . So at the end of turn 109 (or the end of Tadd's turn in round 37), Tadd says the number $\dfrac{109(110)}{2}=5995$ . Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was $5995-16=5979$
Thus, the answer is $\boxed{5979}$

Solution:
Firstly, as in Solution 1, we list how many numbers Tadd says, Todd says, and Tucker says in each round.
Tadd: $1, 4, 7, 10, 13 \cdots$
Todd: $2, 5, 8, 11, 14 \cdots$
Tucker: $3, 6, 9, 12, 15 \cdots$
We can find a general formula for the number of numbers each of the kids say after the $n$ th round. For Tadd, we can either use the arithmetic series sum formula (like in Solution 1) or standard summation results to get $\sum_{i=1}^n 3n-2=-2n+3\sum_{i=1}^n n=-2n+\frac{3n(n+1)}{2}=\frac{3n^2-n}{2}$
Now, to find the number of rotations Tadd and his siblings go through before Tadd says his $2019$ th number, we know the inequality $\frac{3n^2-n}{2}<2019$ must be satisfied, and testing numbers gives the maximum integer value of $n$ as $36$
The next main insight, in order to simplify the computation process, is to notice that the $2019$ th number Tadd says is simply the number of numbers Todd and Tucker say plus the $2019$ Tadd says, which will be the answer since Tadd goes first.
Carrying out the calculation thus becomes quite simple:
\[\left(\sum_{i=1}^{36} 3n+\sum_{i=1}^{36} 3n-1\right)+2019=\left(\sum_{i=1}^{36} 6n-1\right)+2019=(5+11+17...+215)+2019=\frac{36(220)}{2}+2019\]
At this point, we can note that the last digit of the answer is $9$ , which gives $\boxed{5979}$ . (Completing the calculation will confirm the answer, if you have time.)

Solution:
Think of a turn as adding a block or an interval of numbers to each person. Here are the first couple "blocks" that Tadd has (a block will be denoted in interval notation): [ $1$ ], [ $7$ $10$ ], [ $22$ $28$ ], [ $46$ $55$ ]. From simple inspection, we can notice two things.
$1$ ) The start points of each interval is increasing arithmetically by $9$
$2$ ) The length of each "block" is increasing by $3$
Since each block length is increasing by 3, then we want $1$ $4$ $7$ +.. $3n-2$ $2019$ or $n(3n-1)/2 = 2019$ , where $n$ is the amount of blocks we have. In a short amount of time, you will see that $n$ is not an integer from that equation, but thats perfectly okay since we can just find an $n$ that gets it close to 2019, then we can work from there. From inspection, you will see that a good n value that does that exact job is $37$
We can then deduce that the start point of the $37$ th block will be $1+(6+15+24..+321) = 5887$ if we refer to fact $1$ . We can easily find the length of 37th block by $36*3+1 = 109$ if we refer to fact $2$ . Remember that the length increases by $3$ for each block. So we then can deduce that our endpoint for the 37th block is $5887+109-1 = 5995$ . The endpoint of the $37$ th block will be the $2035$ th number said because we can plug $37$ into $n(3n-1)/2$ , which tells us how many numbers Tadd has said where $n$ is the number of blocks Tadd has. From there we know that the $2019$ th number is in $37$ th block, so that means we can take our endpoint and subtract $2035-2019 = 16$ from the endpoint of the $37$ th block, which is $5995$ , so we get $5995-16 = 5979$ or $\boxed{5979}$ .
~~triggod


Answer: 100.0

Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$

Solution: Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds $=2\cdot60 + 45 = 165$ seconds. We then set up a ratio: \[\frac{3}{5}=\frac{x}{165}\] \[3(165)=5x\] \[x=3(33)=99\approx\boxed{100}.\]

Solution:
$2$ minutes and $45$ seconds is equal to $120+45=165\text{ seconds}$
Since $6$ cars pass at around $10$ seconds, there are about $\left \lfloor{\dfrac{165}{10}}\right \rfloor =16$ groups of $6$ cars. There are about $16\cdot6=96\text{ cars}$ , so the closest answer choice is $\boxed{100}$


Answer: 95.0

Triangle $ABC$ , with sides of length $5$ $6$ , and $7$ , has one vertex on the positive $x$ -axis, one on the positive $y$ -axis, and one on the positive $z$ -axis. Let $O$ be the origin . What is the volume of tetrahedron $OABC$
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}$

Solution: Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7.  Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively.  Then by the Pythagorean Theorem \begin{align*} a^2+b^2 &=5^2 , \\  b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*} so $a^2 = (5^2+7^2-6^2)/2 = 19$ ; similarly, $b^2 = 6$ and $c^2 = 30$ . Since $OA$ $OB$ , and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice $\boxed{95}$


Answer: 5.0

Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution:
The product of two lengths with a common point brings to mind the Power of a Point Theorem
Since $PA = PB$ , we can make a circle with radius $PA$ that is centered on $P$ , and both $A$ and $B$ will be on that circle.  Since $\angle APB = \widehat {AB} = 2 \angle ACB$ , we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the measure of inscribed angle $\angle ACB$ , which is true for all inscribed angles.

Since $PDB$ is a line, we have $PD + DB = PB$ , which gives $3 = DB + 2$ , or $DB = 1$
We now extend radius $PB = 3$ to diameter $EB = 6$ .  Since $EDB$ is a line, we have $ED + DB = EB$ , which gives $ED + 1 = 6$ , or $ED = 5$
Finally, we apply the power of a point theorem to point $D$ .  This states that $AD \cdot DC = DB \cdot DE$ .  Since $DB = 1$ and $DE = 5$ , the desired product is $5$ , which is $\boxed{5}$

Solution: Construct the angle bisector of $\angle APD,$ and let it intersect $AD$ at $E.$ From the angle bisector theorem, we have $AE=3a$ and $DE=2a$ for some $a.$ Then, note that $\angle EPD = \angle BCD = x,$ so $EPCB$ is cyclic. Then, $\frac{PD}{ED} = \frac{CD}{BD}$ or $\frac{2}{2x} = \frac{CD}{1}.$ Thus, $AD \cdot DC = 5x \cdot DC = 5,$ or $\boxed{5}.$


Answer: 18.0

Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$
$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$

Solution: Let $D$ be the foot of the altitude to $\overline BC$ .  Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$ .  Thus $BC = BD + BC = 5 + 9 = 14$ or  assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ .  The sum of the two possible values is $14 + 4 = \boxed{18}$


Answer: 90.0

Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ$

Solution:
Let $\angle BAE = \angle ACD = x$
\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\  \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\  \angle EAC &= 60^\circ - x\\  \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}
Since $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$ , triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90}$


Answer: 90.0

Solution:
Let $\angle BAE = \angle ACD = x$
\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\  \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\  \angle EAC &= 60^\circ - x\\  \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}
Since $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$ , triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90}$


Answer: 923.0

Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $I$ be the incenter of $\triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles , and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$ . Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$ , which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$ . Thus, $m + n = \boxed{923}$

Solution:
The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$ . By Heron's formula , the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$ . Using the formula $A = rs$ , we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$ . Since $\triangle ADE \sim \triangle ABC$ , the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$ . From $A=\frac{1}{2}bh$ , we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$ . Thus, we have
Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$

Solution:
Let $P$ be the incenter ; then it is be the intersection of all three angle bisectors . Draw the bisector $AP$ to where it intersects $BC$ , and name the intersection $F$
Using the angle bisector theorem , we know the ratio $BF:CF$ is $21:22$ , thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$ , giving $F$ a weight of $43$ . In the same manner, using another bisector, we find that $A$ has a weight of $20$ . So, now we know $P$ has a weight of $63$ , and the ratio of $FP:PA$ is $20:43$ . Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$ . So, $DE$ is $43/63$ the size of $BC$ . Multiplying this ratio by the length of $BC$ , we find $DE$ is $860/63 = m/n$ . Therefore, $m+n=\boxed{923}$

Solution: More directly than Solution 2, we have \[DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.\]

Solution:
Diagram borrowed from Solution 3.
Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$
Applying the Angle Bisector Theorem on $\angle{A}$ we have \[\frac{AB}{BF}=\frac{AC}{CF}\] \[BF=BC\cdot(\frac{AB}{AB+AC})\] \[BF=\frac{420}{43}\] Since $BP$ is the angle bisector of $\angle{B}$ , we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives \[\frac{BA}{AP}=\frac{BF}{FP}\] \[\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have \[\frac{DE}{BC}=\frac{AP}{AF}\] \[DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})\] Solving gets $DE=\frac{860}{63}$ . Thus $m+n=860+63=\boxed{923}$

Solution:
Label $P$ the point the angle bisector of $A$ intersects ${BC}$ . First we find ${BP}$ and ${PC}$ . By the Angle Bisector Theorem, $\frac{BP}{PC} = \frac{21}{22}$ and solving for each using the fact that ${BC} = 20$ , we see that ${BP} = \frac{420}{43}$ and $PC = \frac{440}{43}$
\[{AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}\] \[{AP} = 21*22 - \frac{440\cdot420}{43^2}\]
Now we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}$
Now because $\triangle{APE} ~ \triangle{ABC}$ , we can find ${DE}$ by $\frac{AO}{AP} \cdot 20$ . Dividing and simplifying, we see that $\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}$ . So the answer is $\boxed{923}$


Answer: 15.0

Triangle $ABC$ has $AB=27$ $AC=26$ , and $BC=25$ .  Let $I$ be the intersection of the internal angle bisectors of $\triangle ABC$ .  What is $BI$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3}$

Solution:
Inscribe circle $C$ of radius $r$ inside triangle $ABC$ so that it meets $AB$ at $Q$ $BC$ at $R$ , and $AC$ at $S$ . Note that angle bisectors of triangle $ABC$ are concurrent at the center $O$ (also $I$ ) of circle $C$ . Let $x=QB$ $y=RC$ and $z=AS$ . Note that $BR=x$ $SC=y$ and $AQ=z$ . Hence $x+z=27$ $x+y=25$ , and $z+y=26$ . Subtracting the last 2 equations we have $x-z=-1$ and adding this to the first equation we have $x=13$
By Heron's formula for the area of a triangle we have that the area of triangle $ABC$ is $\sqrt{39(14)(13)(12)}$ . On the other hand the area is given by $(1/2)25r+(1/2)26r+(1/2)27r$ .  Then $39r=\sqrt{39(14)(13)(12)}$ so that $r^2=56$
Since the radius of circle $O$ is perpendicular to $BC$ at $R$ , we have by the pythagorean theorem $BO^2=BI^2=r^2+x^2=56+169=225$ so that $BI=\boxed{15}$

Solution: We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label $A$ with a mass of $25$ $B$ with $26$ , and $C$ with $27$ . We also label where the angle bisectors intersect the opposite side $A'$ $B'$ , and $C'$ correspondingly. It follows then that point $B'$ has mass $52$ . Which means that $\overline{BB'}$ is split into a $2:1$ ratio. We can then use Stewart's to find $\overline{BB'}$ . So we have $25^2\frac{27}{2} + 27^2\frac{25}{2} = \frac{25 \cdot 26 \cdot 27}{4} + 26\overline{BB'}^2$ . Solving we get $\overline{BB'} = \frac{45}{2}$ . Plugging it in we get $\overline{BI} = 15$ . Therefore the answer is $\boxed{15}$

Solution:
We can use POP(Power of a point) to solve this problem. First, notice that the area of $\triangle ABC$ is $\sqrt{39(39 - 27)(39 - 26)(39 - 25)} = 78\sqrt{14}$ . Therefore, using the formula that $sr = A$ , where $s$ is the semi-perimeter and $r$ is the length of the inradius, we find that $r = 2\sqrt{14}$
Draw radii to the three tangents, and let the tangent hitting $BC$ be $T_1$ , the tangent hitting $AB$ be $T_2$ , and the tangent hitting $AC$ be $T_3$ . Let $BI = x$ . By the pythagorean theorem, we know that $BT_1 = \sqrt{x^2 - 56}$ . By POP, we also know that $BT_2$ is also $\sqrt{x^2 - 56}$ . Because we know that $BC = 25$ , we find that $CT_1 = 25 - \sqrt{x^2 - 56}$ . We can rinse and repeat and find that $AT_2 = 26 - (25 - \sqrt{x^2 - 56}) = 1 + \sqrt{x^2 - 56}$ . We can find $AT_2$ by essentially coming in from the other way. Since $AB = 27$ , we also know that $AT_3 = 27 - \sqrt{x^2 - 56}$ . By POP, we know that $AT_2 = AT_3$ , so $1 + \sqrt{x^2 - 56} = 27 - \sqrt{x^2 - 56}$
Let $\sqrt{x^2 - 56} = A$ , for simplicity. We can change the equation into $1 + A = 27 - A$ , which we find $A$ to be $13$ . Therefore, $\sqrt{x^2 - 56} = 13$ , which further implies that $x^2 - 56 = 169$ . After simplifying, we find $x^2 = 225$ , so $x = \boxed{15}$


Answer: 820.0

Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have?

Solution:
First, consider the triangle in a coordinate system with vertices at $(0,0)$ $(9,0)$ , and $(a,b)$ . Applying the distance formula , we see that $\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}$
We want to maximize $b$ , the height, with $9$ being the base.
Simplifying gives $-a^2 -\frac{3200}{9}a +1600 = b^2$
To maximize $b$ , we want to maximize $b^2$ . So if we can write: $b^2=-(a+n)^2+m$ , then $m$ is the maximum value of $b^2$ (this follows directly from the trivial inequality , because if ${x^2 \ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \ge 0}$ ).
$b^2=-a^2 -\frac{3200}{9}a +1600=-\left(a +\frac{1600}{9}\right)^2 +1600+\left(\frac{1600}{9}\right)^2$
$\Rightarrow b\le\sqrt{1600+\left(\frac{1600}{9}\right)^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}$
Then the area is $9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}$

Solution:
Let $A, B$ be the endpoints of the side with length $9$ . Let $\Gamma$ be the Apollonian Circle of $AB$ with ratio $40:41$ ; let this intersect $AB$ at $P$ and $Q$ , where $P$ is inside $AB$ and $Q$ is outside. Then because $(A, B; P, Q)$ describes a harmonic set, $AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360$ . Finally, this means that the radius of $\Gamma$ is $\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}$
Since the area is maximized when the altitude to $AB$ is maximized, clearly we want the last vertex to be the highest point of $\Gamma$ , which just makes the altitude have length $180+\dfrac{20}{9}$ . Thus, the area of the triangle is $\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}$

Solution:
We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$ . Heron's gives
$\sqrt{\left(\frac{81x+9}{2} \right) \left(\frac{81x-9}{2} \right) \left(\frac{x+9}{2} \right) \left(\frac{-x+9}{2} \right)}$
This can be simplified to
$\frac{9}{4} \cdot \sqrt{(81x^2-1)(81-x^2)}$
We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.
We have that $-324x^3+13124x=0$ , so $x=\frac{\sqrt{3281}}{9}$
Plugging this into the expression, we have that the area is $\boxed{820}$

Solution:
We can start how we did above in solution 4 to get $\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}$ .
Then, we can notice the inside is a quadratic in terms of $x^2$ , which is $-81(x^2)^2+6562x^2-81$ . This is maximized when $x^2 = \frac{3281}{81}$ .If we plug it into the equation, we get $\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}$


Answer: 72.0

Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$

Solution:
Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$ , quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$
Thus,
\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]
Now let's apply the angle bisector theorem.
\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]
\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]
\[\frac {180}{LP}=\frac {5}{2}\]
\[LP=\boxed{072}\]

Solution:
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \[\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$ . Since $K$ is the midpoint of $A$ and $C$ , the weight of $A$ is equal to the weight of $C$ , which equals $2$ .
Also, since the weight of $L$ is $5$ and $C$ is $2$ , we can weight $P$ as $7$
By the definition of mass points, \[\frac{LP}{CP}=\frac{2}{5}\implies LP=\frac{2}{5}CP\] By vertical angles, angle $MKA =$ angle $PKC$ .
Also, it is given that $AK=CK$ and $PK=MK$
By the SAS congruence, $\triangle MKA$ $\triangle PKC$ . So, $MA$ $CP$ $180$ .
Since $LP=\frac{2}{5}CP$ $LP = \frac{2}{5}(180) = \boxed{072}$

Solution:
Using the diagram from solution $1$ , we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$
Applying the angle bisector theorem to $\triangle CKB$ , we get that $\frac{KP}{PB} = \frac{225}{300} = \frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$
Now, apply law of cosines on $\triangle CKP$ and $\triangle CPB.$
If we let $\angle KCP = \angle PCB = \alpha$ , then the law of cosines gives the following system of equations:
\[9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha\] \[16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos \alpha.\]
Bashing those out, we get that $x = 15 \sqrt{13}$ and $\cos \alpha = \frac{7}{10}.$
Since $\cos \alpha = \frac{7}{10}$ , we can use the double angle formula to calculate that $\cos \left(2 \alpha \right) = -\frac{1}{50}.$
Now, apply Law of Cosines on $\triangle ABC$ to find $AB$
We get: \[AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).\]
Bashing gives $AB = 30 \sqrt{331}.$
From the angle bisector theorem on $\triangle ABC$ , we know that $\frac{AL}{BL} = \frac{450}{300} = \frac{3}{2}.$ So, $AL = 18 \sqrt{331}$ and $BL = 12 \sqrt{331}.$
Now, we apply Law of Cosines on $\triangle ALC$ and $\triangle BLC$ in order to solve for the length of $LC$
We get the following system:
\[(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \cdot LC \cdot \frac{7}{10}\] \[(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}\]
The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$
The only value that satisfies both equations is $LC = 252$ , and since $LP = LC - PC$ , we have \[LC = 252 - 180 = \boxed{072}.\]


Answer: 38.0

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

Solution:
Let $E$ $F$ and $G$ be the points of tangency of the incircle with $BC$ $AC$ and $AB$ , respectively.  Without loss of generality, let $AC < AB$ , so that $E$ is between $D$ and $C$ .  Let the length of the median be $3m$ .  Then by two applications of the Power of a Point Theorem $DE^2 = 2m \cdot m = AF^2$ , so $DE = AF$ .  Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$ .  Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$
Now, by Stewart's Theorem in triangle $\triangle ABC$ with cevian $\overline{AD}$ , we have
\[(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.\]
Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$ , so we combine these two results to solve for $c$ and we get
\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.\]
Thus $c = 2$ or $= 10$ .  We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$ , so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$

Solution:
WLOG let E be be between C & D (as in solution 1). Assume $AD = 3m$ . We use power of a point to get that $AG = DE = \sqrt{2}m$ and $AB = AG + GB = AG + BE = 10+2\sqrt{2} m$
Since now we have $AC = 10$ $BC = 20, AB = 10+2\sqrt{2} m$ in triangle $\triangle ABC$ and cevian $AD = 3m$ .  Now, we can apply Stewart's Theorem
\[2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000\] \[1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}\] \[100 m^2 = 400\sqrt{2}m\]
$m = 4\sqrt{2}$ or $m = 0$ if $m = 0$ , we get a degenerate triangle, so $m = 4\sqrt{2}$ , and thus $AB = 26$ . You can now use Heron's Formula to finish. The answer is $24 \sqrt{14}$ , or $\boxed{038}$

Solution:
Let $E, F$ , and $G$ be the point of tangency (as stated in Solution 1). We can now let $AD$ be $3m$ . By using Power of a Point Theorem on A to the incircle, you get that $AG^2 = 2m^2$ . We can use it again on point D to the incircle to get the equation $(10 - CE)^2 = 2m^2$ . Setting the two equations equal to each other gives $(10 - CE)^2 = AG^2$ , and it can be further simplified to be $10 - CE = AG$
Let lengths $AC$ and $AB$ be called $b$ and $c$ , respectively. We can write $AG$ as $\frac{b + c - 20}{2}$ and $CE$ as $\frac{b + 20 - c}{2}$ . Plugging these into the equation, you get:
\[10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}\] \[b + c - 20 + b + 20 - c = 20 \rightarrow b = 10\]
Additionally, by Median of a triangle formula, you get that $3m = \frac{\sqrt{2c^2 - 200}}{2}$
Refer back to the fact that $AG^2 = 2m^2$ . We can now plug in our variables.
\[\left(\frac{b + c - 20}{2}\right)^2 = 2m^2 \rightarrow (c - 10)^2 = 8m^2\] \[c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}\] \[9c^2 - 180c + 900 = 4c^2 - 400\] \[5c^2 - 180c + 1300 = 0\] \[c^2 - 36c + 260 = 0\]
Solving, you get that $c = 26$ or $10$ , but the latter will result in a degenerate triangle, so $c = 26$ .
Finally, you can use Heron's Formula to get that the area is $24\sqrt{14}$ , giving an answer of $\boxed{038}$