Answer: 139.0

The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$

Solution:
Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$ . We can now use systems of equations to solve this problem.
Write two equations:
$2x + 4y = 522$
$x + y = 200$
Now multiply the latter equation by $2$
$2x + 4y = 522$
$2x + 2y = 400$
By subtracting the second equation from the first equation, we find that $2y = 122 \implies y = 61$ . Since there were $200$ heads, meaning that there were $200$ animals, there were $200 - 61 =  \boxed{139}$ two-legged birds.

Solution: Let's add 2 legs for each bird and assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be $4*200=800$ . Actually, there were only $522$ legs. The difference between these two numbers exactly gives us the number of all the legs we added for all birds: $800 - 522 = 278$ . Because each bird was added by 2 legs, so the total number of birds would be $278/2 = \boxed{139}$ two-legged birds. ---LarryFlora


Answer: 11.0

The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer (1,000 meters)?
$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$

Solution:
This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that $2^{10}=1024$
However, because the first term is $2^0=1$ and not $2^1=2$ , the solution to the problem is $10-0+1=\boxed{11}$


Answer: 96.0

The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?
$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192$

Solution: Within each division, there are $\binom {6}{2} = 15$ pairings, and each of these games happens twice. The same goes for the other division so that there are $4(15)=60$ games within their own divisions. The number of games between the two divisions is $(6)(6)=36$ . Together there are $60+36=\boxed{96}$ conference games.


Answer: 96.0

Solution: Each team plays 10 games in its own division and 6 games against teams in the other division. So each of the 12 teams plays 16 conference games. Because each game involves two teams, there are $\frac{12\times 16}{2}=\boxed{96}$ games scheduled. ~aopsav (Credit to AoPS Alcumus)


Answer: 10.0

Solution:
Draw the following four lines as shown:
We see these lines split the figure into five squares with side length $\sqrt2$ . Thus, the area is $5\cdot\left(\sqrt2\right)^2=5\cdot 2 = \boxed{10}$

Solution: There are $5$ lattice points in the interior of the logo and $12$ lattice points on the boundary of the logo. Because of Pick's Theorem, the area of the logo is $5+\frac{12}{2}-1=\boxed{10}$

Solution: Notice that the area of the figure is equal to the area of the $4 \times 4$ square subtracted by the $12$ triangles that are half the area of each square, which is $1$ . The total area of the triangles not in the figure is $12 \cdot \frac{1}{2} = 6$ , so the answer is $16-6 = \boxed{10}$

Solution:
Draw the following four lines as shown:

The area of the big square is $4$ , and the area of each triangle is $0.5$ . There are $12$ of these triangles, so the total area of all the triangles is $0.5\cdot12=6$ . Therefore, the area of the entire figure is $4+6=\boxed{10}$

Solution: The coordinates are $(1,2), (2,1), (3,2), (4,1), (5,2), (4,3), (5,4), (4,5), (3,4), (2,5), (1,4), (2,3)$ Use the Shoelace Theorem to get $\boxed{10}$

Solution: If the triangles are rearranged such that the gaps are filled, there would be a $4$ by $2$ rectangle, and two $1$ by $1$ squares are present. Thus, the answer is $\boxed{10}$


Answer: 10.0

The Pythagoras High School band has $100$ female and $80$ male members.  The Pythagoras High School orchestra has $80$ female and $100$ male members.  There are $60$ females who are members in both band and orchestra.  Altogether, there are $230$ students who are in either band or orchestra or both.  The number of males in the band who are NOT in the orchestra is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 70$

Solution: There are $100+80-60=120$ females in either band or orchestra, so there are $230-120=110$ males in either band or orchestra.  Suppose $x$ males are in both band and orchestra. \[80+100-x=110\Rightarrow x=70.\] Thus, the number of males in band but not orchestra is $80-70=10\rightarrow \boxed{10}$


Answer: 7.0

The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$

Solution:
Since the three angles of a triangle add up to $180^{\circ}$ and one of the angles is $90^{\circ}$ because it's a right triangle, $a^{\circ} + b^{\circ} = 90^{\circ}$
The greatest prime number less than $90$ is $89$ . If $a=89^{\circ}$ , then $b=90^{\circ}-89^{\circ}=1^{\circ}$ , which is not prime.
The next greatest prime number less than $90$ is $83$ . If $a=83^{\circ}$ , then $b=7^{\circ}$ , which IS prime, so we have our answer $\boxed{7}$ ~quacker88

Solution: Looking at the answer choices, only $7$ and $11$ are coprime to $90$ . Testing $7$ , the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\boxed{7}$

Solution:
It is clear that $\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \[\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,\] so $90$ and $b$ are relatively prime.
The least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\boxed{7}.$


Answer: 7.0

Solution:
Since the three angles of a triangle add up to $180^{\circ}$ and one of the angles is $90^{\circ}$ because it's a right triangle, $a^{\circ} + b^{\circ} = 90^{\circ}$
The greatest prime number less than $90$ is $89$ . If $a=89^{\circ}$ , then $b=90^{\circ}-89^{\circ}=1^{\circ}$ , which is not prime.
The next greatest prime number less than $90$ is $83$ . If $a=83^{\circ}$ , then $b=7^{\circ}$ , which IS prime, so we have our answer $\boxed{7}$ ~quacker88

Solution: Looking at the answer choices, only $7$ and $11$ are coprime to $90$ . Testing $7$ , the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\boxed{7}$

Solution:
It is clear that $\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \[\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,\] so $90$ and $b$ are relatively prime.
The least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\boxed{7}.$


Answer: 8.0

The addition below is incorrect. The display can be made correct by changing one digit $d$ , wherever it occurs, to another digit $e$ . Find the sum of $d$ and $e$
$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }$

Solution:
If we change $0$ , the units column would  be incorrect.
If we change $1$ , then the leading $1$ in the sum would be incorrect.
However, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.
Changing the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$ .  This does not match up to the bottom.
Changing the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$ .  This is the number on the bottom if the $2$ s were changed to $6$ s.
Thus $d=2$ and $e=6$ . so $d+e= 8 \boxed{8}$


Answer: 7.0

The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline  &\texttt{2 4 5 6}\end{tabular}$
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$

Solution:
Doing the addition as is, we get $641 + 852 + 973 = 2466$ .  This number is $10$ larger than the desired sum of $2456$ .  Therefore, we must make one of the three numbers $10$ smaller.
We may either change $641 \rightarrow 631$ $852 \rightarrow 842$ , or $973 \rightarrow 963$ .  Either change results in a valid sum.  The largest digit that could be changed is thus the $7$ in the number $973$ , and the answer is $\boxed{7}$


Answer: 44.0

Solution:
The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$
The four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$
Thus, the perimeter is $2(12+10) = 44$ , which is option $\boxed{44}$


Answer: 175.0

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$
[asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Solution:
Let $M$ be the midpoint of the chord $BC$ . From right triangle $OMB$ , we have $OM = \sqrt{OB^2 - BM^2} =4$ . This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$
Notice that the distance $OM$ equals $PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)$ , where $r$ is the radius of circle $P$
Hence \[\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5\] (where $R$ represents the radius, $5$ , of the large circle given in the question). Therefore, since $\angle AOM$ is clearly acute, we see that \[\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3\]
Next, notice that $\angle AOB = \angle AOM - \angle BOM$ . We can therefore apply the subtraction formula for $\tan$ to obtain \[\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}\] It follows that $\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}$ , such that the answer is $7 \cdot 25=\boxed{175}$

Solution:
This solution, while similar to Solution 2, is arguably more motivated and less contrived.
Firstly, we note the statement in the problem that " $AD$ is the only chord starting at $A$ and bisected by $BC$ " – what is its significance? What is the criterion for this statement to be true?
We consider the locus of midpoints of the chords from $A$ . It is well-known that this is the circle with diameter $AO$ , where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor $\frac{1}{2}$ and center $A$ . Thus, the locus is the result of the dilation with scale factor $\frac{1}{2}$ and centre $A$ of circle $O$ . Let the center of this circle be $P$
Now, $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$
The rest of this problem is straightforward.
Our goal is to find $\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}$ , where $M$ is the midpoint of $BC$ . We have $BM=3$ and $OM=4$ .
Let $R$ be the projection of $A$ onto $OM$ , and similarly let $Q$ be the projection of $P$ onto $OM$ . Then it remains to find $AR$ so that we can use the addition formula for $\sin$
As $PN$ is a radius of circle $P$ $PN=2.5$ , and similarly, $PO=2.5$ . Since $OM=4$ , we have $OQ=OM-QM=OM-PN=4-2.5=1.5$ . Thus $PQ=\sqrt{2.5^2-1.5^2}=2$
Further, we see that $\triangle OAR$ is a dilation of $\triangle OPQ$ about center $O$ with scale factor $2$ , so $AR=2PQ=4$
Lastly, we apply the formula: \[\sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}\] Thus the answer is $7\cdot25=\boxed{175}$

Solution:
Let the center of the circle be $O$ . Fix $B,C,$ and $A$ . Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$ , it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$
Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$ . Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$ . Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$ . Clearly, $KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\frac{5}{2}$ . The same applies for $FO$ , which also equals $\frac{5}{2}$ . By the Pythagorean theorem, we deduce that $FL = 2$ , so $EK = 2$ . This is very important information! Now we know that $BE = 1$ , so by Power of a Point, $AE = ED = \sqrt{5}$
We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$ , we deduce that $EO = 2\sqrt{5}$ $EC=OC=5$ , so $\sin (CEO) = \frac{2\sqrt{5}}{5}$ . Furthermore, since $\sin (CEO) = \cos(DEC)$ , we know that $\cos (DEC) = \frac{2\sqrt{5}}{5}$ . By the law of cosines, \[DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10\] Therefore, $DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}$ . Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$ . Then, by the Pythagorean theorem, $OZ = \frac{7\sqrt{2}}{2}$ . Thus, $\sin (BOZ) = \frac{\sqrt{2}}{10}$ and $\cos (BOZ) = \frac{7\sqrt{2}}{10}$ . As a result, $\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}$ $7 \cdot 25 = \boxed{175}$

Solution:
Let I be the intersection of AD and BC.
Lemma: $AI = ID$ if and only if $\angle AIO = 90$
Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$ , We can get $\triangle AIO \cong \triangle OID$
Let  be this the circle with diameter AO.
Thus, we get $\angle AIO = 90$ , implying I must lie on $\omega$ . I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.
Now, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C.
Let Z be (0,5).
Let Y be (-5,0).
Let X be the center of $\omega$ . Since $\omega$ 's radius is $\frac{5}{2}$ , the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \frac{3}{5}$ $sin(BOZ) =  \frac{3}{5}$ . If we let $sin(\theta) = \frac{3}{5}$ , we can find that what we are looking for is $sin(90 - 2\theta)$ , which we can evaluate and get $\frac{7}{25} \implies \boxed{175}$

Solution:
Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$ . Additionally, this circle must be tangent to $\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ $N$ be the foot of the perpendicular from $P$ to $\overline{BM}$ , and $K$ be the foot of the perpendicular from $B$ to $\overline{AP}$ . Let $x=BK$
From right triangle $BKO$ , we get $KO = \sqrt{25-x^2}$ . Thus, $KP = \sqrt{25-x^2}-\frac52$
Since $BO = 5$ $BM = 3$ , and $\angle BMO$ is right, $MO=4$ . From quadrilateral $MNPO$ , we get $MN = \sqrt{PO^2 - (MO - NP)^2} = \sqrt{(5/2)^2 - (4 - 5/2)^2} = \sqrt{(5/2)^2 - (3/2)^2} = 2$ . Thus, $BN = 1$
Since angles $BNP$ and $BKP$ are right, we get \[BK^2+KP^2 = BN^2 + NP^2 \implies x^2 + \left(\sqrt{25-x^2}-\frac52\right)^2 = \left(\frac52\right)^2 + 1\] \[25 - 5\sqrt{25-x^2} = 1\] \[5\sqrt{25-x^2} = 24\] \[25(25-x^2) = 24^2\] \[25x^2 = 25^2 - 24^2 = 49\] \[x = \frac75\] Thus, $\sin \angle AOB = \frac{x}{5} = \frac{7}{25}\implies \boxed{175}$


Answer: 22.0

The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24$ , while the other two multiply to $30$ . What is the sum of the ages of Jonie's four cousins?
$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25$

Solution:
First look at the two cousins' ages that multiply to $24$ . Since the ages must be single-digit, the ages must either be $3 \text{ and } 8$ or $4 \text{ and } 6.$
Next, look at the two cousins' ages that multiply to $30$ . Since the ages must be single-digit, the only ages that work are $5 \text{ and } 6.$ Remembering that all the ages must all be distinct, the only solution that works is when the ages are $3, 8$ and $5, 6$
We are required to find the sum of the ages, which is \[3 + 8 + 5 + 6 = \boxed{22}.\]


Answer: 90.0

The altitudes of a triangle are $12, 15,$ and $20.$ The largest angle in this triangle is
$\text{(A) }72^\circ \qquad \text{(B) }75^\circ \qquad \text{(C) }90^\circ \qquad \text{(D) }108^\circ \qquad \text{(E) }120^\circ$

Solution: Let $a, b,$ and $c$ denote the bases of altitudes $12, 15,$ and $20,$ respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so $\frac{12a}{2}=\frac{15b}{2}=\frac{20c}{2}.$ Multiplying by $2$ , we get $12a=15b=20c.$ Notice that a simple solution to the equation is if all of them equal $12 \cdot 15 \cdot 20.$ That means $a=15 \cdot 20, b=12 \cdot 20,$ and $c=12 \cdot 15.$ Simplifying our solution to check for Pythagorean triples we see that this is just a Pythagorean triple, namely a $3-4-5$ triangle. Since the other two angles of a right triangle must be acute, the right angle must be the greatest angle. Therefore, our answer is $\boxed{90}.$


Answer: 108.0

The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$

Solution:
If the angles are in an arithmetic progression, they can be expressed as $a$ $a+n$ $a+2n$ $a+3n$ , and $a+4n$ for some real numbers $a$ and $n$ .
Now we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$ .
Adding our expressions for the five angles together, we get $5a+10n=540$ .
We now divide by 5 to get $a+2n=108$ . It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\boxed{108}$ .
(In fact, for any arithmetic progression with an odd number of terms,
the middle term is equal to the average of all the terms.)


Answer: 173.0

The angles of quadrilateral $ABCD$ satisfy $\angle A=2 \angle B=3 \angle C=4 \angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?
$\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180$

Solution: The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{173}\]


Answer: 173.0

Solution: The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{173}\]


Answer: 882.0

The annual incomes of $1,000$ families range from $8200$ dollars to $98,000$ dollars.  In error, the largest income was entered on the computer as $980,000$ dollars.  The difference between the mean of the incorrect data and the mean of the actual data is
$\text{(A)}\ \text{882 dollars} \qquad \text{(B)}\ \text{980 dollars} \qquad \text{(C)}\ \text{1078 dollars} \qquad \text{(D)}\ \text{482,000 dollars} \qquad \text{(E)}\ \text{882,000 dollars}$

Solution: Let $S$ be the sum of all the incomes but the largest one.  For the actual data, the mean is $\frac{S+98000}{1000}$ , and for the incorrect data the mean is $\frac{S+980000}{1000}$ .  The difference is $882, or \rightarrow \boxed{882}$


Answer: 8.0

Solution: The base is $\overline{BC}=4$ . The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is $4\cdot 2=8\Rightarrow \boxed{8}$


Answer: 72.0

The area of a circle inscribed in an equilateral triangle is $48\pi$ . The perimeter of this triangle is:
$\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72$

Solution: We can see that the radius of the circle is $4\sqrt{3}$ . We know that the radius is $\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\sqrt{3}$ . Since the median line completes a $30$ $60$ $90$ triangle, we can conclude that one of the sides of the triangle is $24$ . Triple the side length and we get our answer, $\boxed{72}$


Answer: 144.0

The area of a circle whose circumference is $24\pi$ is $k\pi$ . What is the value of $k$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 36 \qquad \mathrm{(E)}\ 144$

Solution: If the circumference of a circle is $24\pi$ , the radius would be $12$ . Since the area of a circle is $\pi r^2$ , the area is $144\pi$ . The answer is $\boxed{144}$

Solution: By definition, $\pi$ is the ratio of the circumference to the diameter. Since the circumference is $24\pi$ , the diameter must be $24$ and the radius is $12$ . Therefore, by the area of circle formula $A=\pi r^{2}$ the area is $12^{2}\pi=144\pi$ and $k=144 \Longrightarrow \boxed{144}$


Answer: 78.0

The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$
$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$

Solution: The area of the larger pizza is $16\pi$ , while the area of the smaller pizza is $9\pi$ . Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$ , which is closest to $\boxed{78}$


Answer: 46.0

Solution:
Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.
If we continue segment $\overline{FE}$ until it reaches the right side at $G$ , we create two rectangles - one on the top and one on the bottom.
We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times5 = 30$ . For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.
Note that $GC+GB=9$ , and $GB=AF=5$ , so we must have \[GC+5=9\Rightarrow GC=4\]
The area of the bottom rectangle is then \[(DC)(GC)=4\times 4=16\]
Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$
$\boxed{46}$

Solution:
Let $\langle ABCDEF \rangle$ be the area of polygon $ABCDEF$ . Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.
Clearly, \[\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle\]
Since $AB=6$ and $BC=9$ $\langle ABCG \rangle =6\times 9=54$
To compute the area of $GFED$ , note that \[AB=GD+DC\] \[BC=GF+FA\]
We know that $AB=6$ $DC=4$ $BC=9$ , and $FA=5$ , so \[6=GD+4\Rightarrow GD=2\] \[9=GF+5\Rightarrow GF=4\]
Thus $\langle GFED \rangle = 4\times 2=8$
Finally, we have \begin{align*} \langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\ &= 54-8 \\ &= 46 \\ \end{align*}
This is answer choice $\boxed{46}$


Answer: 9.0

Solution:
Notice that $AF + DE = BC$ , so $DE=4$ . Let $O$ be the intersection of the extensions of $AF$ and $DC$ , which makes rectangle $ABCO$ . The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$
\[\text{Area} = 52 = 8 \cdot 9- EF \cdot 4\]
Solving for the unknown, $EF=5$ , therefore $DE+EF=4+5=\boxed{9}$


Answer: 27.0

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]
$\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

Solution:
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension.  Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions.  Labelling the right midpoint as $M$ , and the bottom midpoint as $N$ , we know that $DN = NC = 6$ , and $BM = MC = 3$
$[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$
$[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$
$[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$
$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$
$[\triangle AMN] = 72 - 18 - 9 - 18$
$[\triangle AMN] = 27$ , and the answer is $\boxed{27}$

Solution:
The above answer is fast, but satisfying, and assumes that the area of $\triangle AMN$ is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label $AB = CD = l$ and $BC = DA = h$
Labelling $m$ and $n$ as the right and lower midpoints respectively, and redoing all the work above, we get:
$[\triangle ABN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}$
$[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}$
$[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}$
$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$
$[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}$
$[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27$ , and the answer is $\boxed{27}$

Solution: Let's assume, for simplicity, that the sides of the rectangle are $9$ and $8.$ The area of the 3 triangles would then be $8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,$ $4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,$ $4\cdot 9\cdot\frac{1}{2} = 18.$ Adding these up, we get $45$ , and subtracting that from $72$ , we get $27$ , so the answer is $\boxed{27}$


Answer: 2.0

The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$

Solution: The area of a triangle is $\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\frac12 \cdot 2r \cdot r = \boxed{2}$


Answer: 54.0

The area of the region bounded by the graph of \[x^2+y^2 = 3|x-y| + 3|x+y|\] is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$

Solution:
In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$
Case 1: $|x-y|=x-y, |x+y|=x+y$
Substituting and simplifying, we have $x^2-6x+y^2=0$ , i.e. $(x-3)^2+y^2=3^2$ , which gives us a circle of radius $3$ centered at $(3,0)$
Case 2: $|x-y|=y-x, |x+y|=x+y$
Substituting and simplifying again, we have $x^2+y^2-6y=0$ , i.e. $x^2+(y-3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,3)$
Case 3: $|x-y|=x-y, |x+y|=-x-y$
Doing the same process as before, we have $x^2+y^2+6y=0$ , i.e. $x^2+(y+3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,-3)$
Case 4: $|x-y|=y-x, |x+y|=-x-y$
One last time: we have $x^2+y^2+6x=0$ , i.e. $(x+3)^2+y^2=3^2$ . This gives us a circle of radius $3$ centered at $(-3,0)$
After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:
 Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$ .
The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$ . The answer is $36+18$ which is $\boxed{54}$

Solution:
A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by $90^{\circ}$ about the origin. This allows us to quickly sketch the region after solving Case 1.
Upon simplifying Case 1, we obtain $(x-3)^2 + y^2 = 3^2$ which is a circle of radius 3 centered at $(3,0)$ . We remark that only the points on the semicircle where $x \ge 3$ work here, since Case 1 assumes $x-y \ge 0$ and $x+y \ge 0$ . Next, we observe that an ordered pair is a solution to the given equation if and only if any of its $90^{\circ}$ rotations about the origin is a solution. This follows as the value of $x^2+y^2-3(|x-y|+|x+y|)$ is invariant to $90^{\circ}$ rotations, since $x^2+y^2$ simply represents the square of the distance to the origin (which is unchanged upon rotation), and $|x-y|+|x+y|$ is the sum of the distances to the lines $y=x$ and $y=-x$ , multiplied by $\sqrt{2}$ (also unchanged upon $90^{\circ}$ rotation).
By the above observation, we can quickly sketch the remainder of the region, and the area is $\boxed{54}$ as above.

Solution:
Assume $y$ $0$ . We get that $x$ $6$ . That means that this figure must contain the points $(0,6), (6,0), (0, -6), (-6, 0)$ . Now, assume that $x$ $y$ . We get that $x$ $3 \sqrt 3$ . We get the points $(3,3), (3,-3), (-3, 3), (-3, -3)$
Since this contains $x^2 + y^2$ , assume that there are circles. Therefore, we can guess that there is a center square with area $6 \cdot 6$ $36$ and $4$ semicircles with radius $3$ . We get $4$ semicircles with area $4.5 \pi$ , and therefore the answer is $36+18$ $\boxed{54}$


Answer: 64.0

The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution:
Let $[ABC]$ denote the area of figure $ABC$
Clearly, $[BEDC]=[ABCD]-[ABE]$ .  Using basic area formulas,
Since $AE+ED=BC=10$ and $ED=6$ $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$
Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{64}$


Answer: 145.0

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution:
Lemma: If $x,y$ satisfy $px+qy=1$ , then the minimal value of $\sqrt{x^2+y^2}$ is $\frac{1}{\sqrt{p^2+q^2}}$
Proof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}$ . In particular, the distance between the origin and any point $(x,y)$ on the line $px+qy=1$ is at least $\frac{1}{\sqrt{p^2+q^2}}$
Let the vertices of the right triangle be $(0,0),(5,0),(0,2\sqrt{3}),$ and let $(a,0),(0,b)$ be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is $\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)$ . This point must lie on the hypotenuse $\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1$ , i.e. $a,b$ must satisfy \[\frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,\] which can be simplified to \[\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.\]
By the lemma, the minimal value of $\sqrt{a^2+b^2}$ is \[\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},\] so the minimal area of the equilateral triangle is \[\frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},\] and hence the answer is $75+3+67=\boxed{145}$

Solution:
Let $AB=2\sqrt{3}, BC=5$ $D$ lies on $BC$ $F$ lies on $AB$ and $E$ lies on $AC$
Set $D$ as the origin, $BD=a,BF=b$ $F$ can be expressed as $-a+bi$ in argand plane, the distance of $CD$ is $5-a$
We know that $(-a+bi)\cdot\cos(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)$ . We know that the slope of $AC$ is $-\frac{2\sqrt{3}}{5}$ , we have that $\frac{5-a-(-\frac{a+\sqrt{3}b}{2})}{(\sqrt{3}a+b)/2}=\frac{5}{2\sqrt{3}}$ , after computation, we have $11b+7\sqrt{3}a=20\sqrt{3}$
Now the rest is easy with C-S inequality, $(a^2+b^2)(147+121)\geq (7\sqrt{3}a+11a)^2, a^2+b^2\geq \frac{300}{67}$ so the smallest area is $\frac{\sqrt{3}}{4}\cdot \frac{300}{67}=\frac{75\sqrt{3}}{67}$ , and the answer is $\boxed{145}$

Solution:
Let $\triangle ABC$ be the right triangle with sides $AB = x$ $AC = y$ , and $BC = z$ and right angle at $A$
Let an equilateral triangle touch $AB$ $AC$ , and $BC$ at $D$ $E$ , and $F$ respectively, having side lengths of $c$
Now, call $AD$ as $a$ and $AE$ as $b$ . Thus, $DB = x-a$ and $EC = y-b$
By Law of Sines on triangles $\triangle DBF$ and $ECF$
$BF = \frac{z(a\sqrt{3}+b)} {2y}$ and $CF = \frac{z(a+b\sqrt{3})} {2x}$
Summing,
$BF+CF =  \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z$
Now substituting $AB = x = 2\sqrt{3}$ $AC = y = 5$ , and $BC = \sqrt{37}$ and solving, $\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1$
We seek to minimize $[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}$
This is equivalent to minimizing $a^2+b^2$
Using the lemma from solution 1, we conclude that $\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}$
Thus, $[DEF] = \frac{75\sqrt{3}}{67}$ and our final answer is $\boxed{145}$

Solution:
We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are $5$ and $2\sqrt{3}i$ , respectively. Now let the vertex of the equilateral triangle on the real axis be $a$ and let the vertex of the equilateral triangle on the imaginary axis be $bi$ . Then, the third vertex of the equilateral triangle is given by: \[(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{b}{2})i\]
For this to be on the hypotenuse of the right triangle, we also have the following: \[\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}\]
Note that the area of the equilateral triangle is given by $\frac{\sqrt{3}(a^2+b^2)}{4}$ , so we seek to minimize $a^2+b^2$ . This can be done by using the Cauchy Schwarz Inequality on the relation we derived above: \[1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}\]
Thus, the minimum we seek is simply $\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}$ , so the desired answer is $\boxed{145}$

Solution: We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be $a$ and the point on the imaginary axis be $bi$ . Then, we see that $(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).$ Now we switch back to Cartesian coordinates. The equation of the hypotenuse is $y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.$ This means that the point $\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)$ is on the line. Plugging the numbers in, we have $\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.$ Now, we note that the side length of the equilateral triangle is $a^2+b^2$ so it suffices to minimize that. By Cauchy-Schwarz, we have $(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.$ Thus, the area of the smallest triangle is $\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}$ so our desired answer is $\boxed{145}$

Solution:
Let $S$ be the triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37}$
We will think about this problem backwards, by constructing a triangle as large as possible (We will call it $T$ , for convenience) which is similar to $S$ with vertices outside of a unit equilateral triangle $\triangle ABC$ , such that each vertex of the equilateral triangle lies on a side of $T$ . After we find the side lengths of $T$ , we will use ratios to trace back towards the original problem.
First of all, let $\theta = 90^{\circ}$ $\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)$ , and $\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)$ (These three angles are simply the angles of triangle $S$ ; out of these three angles, $\alpha$ is the smallest angle, and $\theta$ is the largest angle). Then let us consider a point $P$ inside $\triangle ABC$ such that $\angle APB = 180^{\circ} - \theta$ $\angle BPC = 180^{\circ} - \alpha$ , and $\angle APC = 180^{\circ} - \beta$ . Construct the circumcircles $\omega_{AB}, ~\omega_{BC},$ and $\omega_{AC}$ of triangles $APB, ~BPC,$ and $APC$ respectively.
From here, we will prove the lemma that if we choose points $X$ $Y$ , and $Z$ on circumcircles $\omega_{AB}, ~\omega_{BC},$ and $\omega_{AC}$ respectively such that $X$ $B$ , and $Y$ are collinear and $Y$ $C$ , and $Z$ are collinear, then $Z$ $A$ , and $X$ must be collinear. First of all, if we let $\angle PAX = m$ , then $\angle PBX = 180^{\circ} - m$ (by the properties of cyclic quadrilaterals), $\angle PBY = m$ (by adjacent angles), $\angle PCY = 180^{\circ} - m$ (by cyclic quadrilaterals), $\angle PCZ = m$ (adjacent angles), and $\angle PAZ = 180^{\circ} - m$ (cyclic quadrilaterals). Since $\angle PAX$ and $\angle PAZ$ are supplementary, $Z$ $A$ , and $X$ are collinear as desired. Hence, $\triangle XYZ$ has an inscribed equilateral triangle $ABC$
In addition, now we know that all triangles $XYZ$ (as described above) must be similar to triangle $S$ , as $\angle AXB = \theta$ and $\angle BYC = \alpha$ , so we have developed $AA$ similarity between the two triangles. Thus, $\triangle XYZ$ is the triangle similar to $S$ which we were desiring. Our goal now is to maximize the length of $XY$ , in order to maximize the area of $XYZ$ , to achieve our original goal.
Note that, all triangles $PYX$ are similar to each other if $Y$ $B$ , and $X$ are collinear. This is because $\angle PYB$ is constant, and $\angle PXB$ is also a constant value. Then we have $AA$ similarity between this set of triangles. To maximize $XY$ , we can instead maximize $PY$ , which is simply the diameter of $\omega_{BC}$ . From there, we can determine that $\angle PBY = 90^{\circ}$ , and with similar logic, $PA$ $PB$ , and $PC$ are perpendicular to $ZX$ $XY$ , and $YZ$ respectively We have found our desired largest possible triangle $T$
All we have to do now is to calculate $YZ$ , and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within $S$ . First of all, we will prove that $\angle ZPY = \angle ACB + \angle AXB$ . By the properties of cyclic quadrilaterals, $\angle AXB = \angle PAB + \angle PBA$ , which means that $\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC$ . Now we will show that $\angle ZPY =  180^{\circ} - \angle PAC - \angle PBC$ . Note that, by cyclic quadrilaterals, $\angle YZP = \angle PAC$ and $\angle ZYP = \angle PBC$ . Hence, $\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC$ (since $\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}$ ), proving the aforementioned claim. Then, since $\angle ACB = 60^{\circ}$ and $\angle AXB = \theta = 90^{\circ}$ $\angle ZPY = 150^{\circ}$
Now we calculate $PY$ and $PZ$ , which are simply the diameters of circumcircles $\omega_{BC}$ and $\omega_{AC}$ , respectively. By the extended law of sines, $PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}$ and $PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}$
We can now solve for $ZY$ with the law of cosines:
\[(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)\]
\[(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}\]
\[(ZY)^2 = \frac{37 \cdot 67}{300}\]
\[ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}\]
Now we will apply this discovery towards our original triangle $S$ . Since the ratio between $ZY$ and the hypotenuse of $S$ is $\frac{\sqrt{67}}{10\sqrt{3}}$ , the side length of the equilateral triangle inscribed within $S$ must be $\frac{10\sqrt{3}}{\sqrt{67}}$ (as $S$ is simply as scaled version of $XYZ$ , and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within $S$ is $\frac{75\sqrt{3}}{67}$ , implying that the answer is $\boxed{145}$

Solution:
Let the right triangle's lower-left point be at $O(0,0)$ . Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the $x$ -axis ( $B$ ) and the $y$ -axis ( $A$ ) and label them $(b, 0)$ and $(0, a)$ respectively.  The third point ( $C$ ) will then be located on the hypotenuse.  We proceed to find the third point's coordinates in terms of $a$ and $b$
1. Find the slope of $AB$ and take the negative reciprocal of it to find the slope of the line containing $C$ .  Notice the line contains the midpoint of $AB$ so we can then have an equation of the line.
2. Let $AB=x.$ For $ABC$ to be an equilateral triangle, the altitude from $C$ to $AB$ must be $\frac{x\sqrt{3}}{2}.$
We then have two equations and two variables, so we can solve for $C$ 's coordinates.
We can find $C(\frac{a+b\sqrt{3}}{2}), (\frac{b+a\sqrt{3}}{2}).$ Also, note that $C$ must be on the hypotenuse of the triangle $\frac{x}{5}+\frac{y}{2\sqrt{3}}=1.$ We can plug in $x$ and $y$ as the coordinates of $C$ ,  which simplifies to
\[11b+7\sqrt{3}a=20\sqrt{3}.\]
We aim to minimize the side length of the triangle, which is $\sqrt{a^2+b^2}.$ Applying the Cauchy inequality gives us
\[(a^2+b^2)(7\sqrt{3}^2+11^2)\geq (11b+7\sqrt3a)^2 = 1200\]
From which we obtain $\sqrt{a^2+b^2} \geq \sqrt{\frac{300}{67}}.$ Thus, the area of the triangle = $\frac{75\sqrt{3}}{67}$ which leads to the answer $75+3+67=\boxed{145}.$


Answer: 64.0

The area of the smallest square that will contain a circle of radius 4 is
$\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution: Draw a square circumscribed around the circle.  (Alternately, the circle is inscribed in the square.)  If the circle has radius $4$ , it has diameter $8$ .  Two of the diameters of the circle will run parallel to the sides of the square.  Thus, the smallest square that contains it has side length $8$ , and area $8\times8=64$ $\boxed{64}$


Answer: 50.0

Solution:
Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$
There are $10$ sides of this length, so the perimeter is $10(5)=50\rightarrow \boxed{50}$


Answer: 10.0

Solution: Using the formula for the area of a trapezoid, we have $164=8(\frac{BC+AD}{2})$ . Thus $BC+AD=41$ . Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$ . From the Pythagorean Theorem, $AE=6$ and $DF=15$ thus $AD=BC+21$ . Substituting back into our original equation we have $BC+BC+21=41$ thus $BC=\boxed{10}$


Answer: 3.0

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]
$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

Solution: The shaded region is a right trapezoid. Assume WLOG that $YZ=8$ . Then because the area of $\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$ . Because the line $AB$ is a midsegment, the top base of the trapezoid is $\frac12 AB = \frac14 YZ = 2$ . Also, $AB$ divides $XC$ in two, so the height of the trapezoid is $\frac12 (2) = 1$ . The bottom base is $\frac12 YZ = 4$ . The area of the shaded region is $\frac12 (2+4)(1) = \boxed{3}$

Solution:
Since $A$ and $B$ are the midpoints of $XY$ and $XZ$ , respectively, $AY=AX=BX=BZ$ .
Draw segments $AC$ and $BC$ .
Drawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that $YC=CZ$ $AB$ is the line that connects the midpoints of two sides of a triangle together, which means that $AB$ is parallel to and half in length of $YZ$ . Then $AB=YC=CZ$ .
Since $AB$ is parallel to $YZ$ , and $XY$ is the transversal, $\angle XAB=\angle AYC.$ Similarly, $\angle XBA=\angle BZC.$ Then, by SAS, $\triangle XAB=\triangle AYC=\triangle BZC$ .
Since corresponding parts of congruent triangles are congruent, $AC=BC=XA$ .
Since ACY and BCZ are now isosceles triangles, $\angle AYC=\angle ACY$ and $\angle BZC=\angle BCZ$ Using the fact that $AB$ is parallel to $YZ$ $\angle ACY=\angle CAB$ and $\angle BCZ=\angle CBA$ .
Now $\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC$ .
Draw an altitude through each of them such that each triangle is split into two congruent right triangles. Now there are a total of 8 congruent small triangles, each with area 1. The shaded area has three of these triangles, so it has area 3.
Basically the proof is to show $\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC$ . If you just look at the diagram you can easily see that the triangles are congruent and you would solve this a lot faster. Anyways, since those triangles are congruent, you can split each in half to find eight congruent triangles with area 1, and since the shaded region has three of these triangles, its area is $\boxed{3}$

Solution:
We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\frac{bh}{2}$ or in this problem $\frac{XC\cdot YZ}{2}$ . By solving, we have \[\frac{XC\cdot YZ}{2} = 8,\] \[XC\cdot YZ = 16.\]
With SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY = \frac{8}{2} = 4$ . (Let's say point $D$ is the intersection between line segments $XC$ and $AB$ .) We can find the area of the trapezoid $ADCY$ by subtracting the area of triangle $XAD$ from $4$
We find the area of triangle $XAD$ by the $\frac{bh}{2}$ formula- $\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}$ $AD$ is $\frac{1}{4}$ of $YZ$ from solution 1. The area of $XAD$ is \[\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1\]
Therefore, the area of the shaded area- trapezoid $ADCY$ has area $4-1 = \boxed{3}$

Solution:
$\usepackage[dvipsnames]{xcolor} \textcolor{BlueViolet}{\text{Super fast after convincing yourself they are congruent}}$
We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates $90$ °. (This is not a random idea - the two original two triangles on the top inspires us to do so.)
Note that $\overline{AD}$ is perpendicular to $\overline{YC}$ $\overline{BE}$ is perpendicular to $\overline{CZ}$
$\overline{AB}$ is perpendicular to $\overline{YZ}$ , so because of corresponding angles theorem, $\angle YAD=\angle AXF$ . We are also told $\overline{YA}=\overline{AX}$ . Both $\bigtriangleup \text{XAF}$ and $\bigtriangleup \text{AYD}$ have a $90$ ° angle. Because of AAS Congruency (Angle-Angle-Side; $90$ °, $\angle YAD$ , and side $\overline{YA}$ ), $\bigtriangleup \text{XAF}$ is congruent to $\bigtriangleup \text{AYD}$ . Because of symmetry, the same goes for $\bigtriangleup \text{XFB}$ and $\bigtriangleup \text{BEZ}$
$\bigtriangleup \text{YAD}$ is congruent to $\bigtriangleup \text{CAD}$ , which is congruent to $\bigtriangleup \text{ACF}$ because of symmetry. So there are $4$ congruent triangles on the left side. There is also $4$ congruent triangles on the right side because of symmetry.
There are $8$ congruent triangles in all. We are told the area of the triangle $\bigtriangleup \text{XYZ}$ is 8. $8\div8=1$ , so each of the small congruent triangles is $1$ . The shaded area contains $3$ small triangles, so the area of the shaded section is $1\cdot3=\boxed{3}$


Answer: 37.5

The arithmetic mean (average) of a set of $50$ numbers is $38$ . If two numbers, namely, $45$ and $55$ , are discarded, the mean of the remaining set of numbers is: $\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52$

Solution:
Since the arithmetic mean of the $50$ numbers is $38$ , their sum must be $50*38 = 1900$ . After $45$ and $55$ are discarded, the sum decreases by $45 + 55 = 100$ , so it must become $1900 - 100 = 1800$ .
But this means that the new mean of the remaining $50 - 2 = 48$ numbers must be $\frac{1800}{48} = 37.5$ . Thusly, our answer is $\boxed{37.5}$ , and we are done.


Answer: 81.0

The arithmetic mean (average) of four numbers is $85$ .  If the largest of these numbers is $97$ , then the mean of the remaining three numbers is
$\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3$

Solution: Say that the four numbers are $a, b, c,$ $97$ . Then $\frac{a+b+c+97}{4} = 85$ . What we are trying to find is $\frac{a+b+c}{3}$ . Solving, \[\frac{a+b+c+97}{4} = 85\] \[a+b+c+97 = 340\] \[a+b+c = 243\] \[\frac{a+b+c}{3} = \boxed{81}\]


Answer: 1.0

The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$ , when $x \not = 0$ , is:
$\textbf{(A)}\ {2}\text{, if }{a \not = 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$

Solution: We have $\frac{1}{2}\cdot \left(\frac{x + a}{x} + \frac{x - a}{x}\right) = \frac{2}{2} = \boxed{1}$


Answer: 37.5

The arithmetic mean of a set of $50$ numbers is $38$ . If two numbers of the set, namely $45$ and $55$ , are discarded,
the arithmetic mean of the remaining set of numbers is:
$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$

Solution: If the arithmetic mean of a set of $50$ numbers is $38$ , then the sum of the $50$ numbers equals $1900$ .  Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$ .  Therefore, the new mean is $37.5$ , which is answer choice $\boxed{37.5}$


Answer: 0.0

The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$

Solution: We wish to find $\frac{9+99+\cdots +999999999}{9}$ , or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\boxed{0}$

Solution: Notice that the final number is guaranteed to have the digits $\{1, 3, 5, 7, 9\}$ and that each of these digits can be paired with an even number adding up to 9. $\boxed{0}$ can be taken out, with the other digits fulfilling divisibility by 9.

Solution:
The arithmetic mean is $\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789$ . So select $\boxed{0}$ .
~hastapasta


Answer: 0.0

Solution: We wish to find $\frac{9+99+\cdots +999999999}{9}$ , or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\boxed{0}$

Solution: Notice that the final number is guaranteed to have the digits $\{1, 3, 5, 7, 9\}$ and that each of these digits can be paired with an even number adding up to 9. $\boxed{0}$ can be taken out, with the other digits fulfilling divisibility by 9.

Solution:
The arithmetic mean is $\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789$ . So select $\boxed{0}$ .
~hastapasta


Answer: 55.0

The average (arithmetic mean) of $10$ different positive whole numbers is $10$ .  The largest possible value of any of these numbers is
$\text{(A)}\ 10 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 55 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 91$

Solution:
If the average of the numbers is $10$ , then their sum is $10\times 10=100$
To maximize the largest number of the ten, we minimize the other nine.  Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$ .  Their sum is $45$
The sum of nine of the numbers is $45$ , and the sum of all ten is $100$ so the last number must be $100-45=55\rightarrow \boxed{55}$


Answer: 24.0

The average (mean) of $20$ numbers is $30$ , and the average of $30$ other numbers is $20$ . What is the average of all $50$ numbers?
$\textbf{(A) } 23\qquad \textbf{(B) } 24\qquad \textbf{(C) } 25\qquad \textbf{(D) } 26\qquad \textbf{(E) } 27$

Solution:
Since the average of the first $20$ numbers is $30$ , their sum is $20\cdot30=600$
Since the average of $30$ other numbers is $20$ , their sum is $30\cdot20=600$
So the sum of all $50$ numbers is $600+600=1200$
Therefore, the average of all $50$ numbers is $\frac{1200}{50}=\boxed{24}$


Answer: 24.75

The average age of $33$ fifth-graders is $11$ . The average age of $55$ of their parents is $33$ . What is the average age of all of these parents and fifth-graders?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$

Solution: The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{24.75}$


Answer: 24.75

Solution: The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{24.75}$


Answer: 33.0

The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$

Solution:
Let $x$ be the average of the remaining $4$ people.
The equation we get is $\frac{4x + 18}{5} = 30$
Simplify,
$4x + 18 = 150$
$4x = 132$
$x = 33$
Therefore, the answer is $\boxed{33}$

Solution: Since an $18$ year old left from a group of people averaging $30$ , The remaining people must total $30 - 18 = 12$ years older than $30$ . Therefore, the average is $\dfrac{12}{4} = 3$ years over $30$ . Giving us $\boxed{33}$

Solution: The total ages would be $30*5=150$ . Then, if one $18$ year old leaves, we subtract $18$ from $150$ and get $132$ . Then, we divide $132$ by $4$ to get the new average, $\boxed{33}$


Answer: 34.0

The average age of the $6$ people in Room A is $40$ . The average age of the $4$ people in Room B is $25$ . If the two groups are combined, what is the average age of all the people?
$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$

Solution:
The total of all their ages over the number of people is
\[\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{34}.\]


Answer: 28.0

The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30$

Solution: First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\div5=\boxed{28}$


Answer: 80.0

The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance
call in the USA in $2005$ was $7$ cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.
$\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathrm{(C)}\ 34 \qquad\mathrm{(D)}\ 41 \qquad\mathrm{(E)}\ 80$

Solution:
The percent decrease is (the amount of decrease)/(original amount)
the amount of decrease is $41 - 7 = 34$
so the percent decrease is $\frac{34}{41}$ which is about $\boxed{80}$


Answer: 58.0

The average of the five numbers in a list is $54$ . The average of the first two
numbers is $48$ . What is the average of the last three numbers?
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$

Solution: Let the $5$ numbers be $a, b, c, d$ , and $e$ . Thus $\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$ . Since $\frac{a+b}{2}=48$ $a+b=96$ . Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$ . Dividing by $3$ gives the average of $\boxed{58}$


Answer: 0.0

The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$

Solution: Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{0}$ dimes.

Solution: If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\boxed{0}.$


Answer: 0.0

Solution: Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{0}$ dimes.

Solution: If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\boxed{0}.$


Answer: 138.0

The average weight of $6$ boys is $150$ pounds and the average weight of $4$ girls is $120$ pounds.  The average weight of the $10$ children is
$\text{(A)}\ 135\text{ pounds} \qquad \text{(B)}\ 137\text{ pounds} \qquad \text{(C)}\ 138\text{ pounds} \qquad \text{(D)}\ 140\text{ pounds} \qquad \text{(E)}\ 141\text{ pounds}$

Solution:
Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ .  We are given
\begin{align*} \frac{S_B}{6} &= 150 \\ \frac{S_G}{4} &= 120  \end{align*}
We want the average of the $10$ children, which is \[\frac{S_B+S_G}{10}\] From the first two equations , we can determine that $S_B=900$ and $S_G=480$ , so $S_B+S_G=1380$ .  Therefore, the average we desire is \[\frac{1380}{10}=138 \rightarrow \boxed{138}\]


Answer: 24.0

Solution: The total frequency is $50+60+40+60+40=250$ , with the blue frequency of $60$ . Therefore, the precentage that preferred blue is $\frac{60}{250}=\boxed{24}$


Answer: 13.0

The base of isosceles $\triangle ABC$ is $24$ and its area is $60$ . What is the length of one
of the congruent sides?
$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$

Solution:
The area of a triangle is shown by $\frac{1}{2}bh$ . We set the base equal to $24$ , and the area equal to $60$ , and we get the height, or altitude, of the triangle to be $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , we can solve for one of the legs of the triangle (it will be the the hypotenuse, $c$ ). $a = 12$ $b = 5$ $c = 13$ .
The answer is $\boxed{13}$


Answer: 19.0

The base of the decimal number system is ten, meaning, for example, that $123 = 1\cdot 10^2 + 2\cdot 10 + 3$ . In the binary system, which has base two, the first five positive integers are $1,\,10,\,11,\,100,\,101$ . The numeral $10011$ in the binary system would then be written in the decimal system as:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 10011\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 7$

Solution:
Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ $1111$ in base $2$ ) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$
Using this same logic, $10011_2$ would be $1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{19}$


Answer: 3.0

The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$

Solution: Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: \begin{align*} N&=27006000052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ &\equiv 2-7+6-5+2 &&\pmod{5} \\ &\equiv -2 &&\pmod{5} \\ &\equiv \boxed{3} ~Aidensharp ~Kante314 ~MRENTHUSIASM

Solution:
We need to first convert $N$ into a regular base- $10$ number: \[N = 27006000052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.\]
Now, consider how the last digit of $9$ changes with changes of the power of $9:$ \begin{align*} 9^0&=1, \\ 9^1&=9, \\ 9^2&=\ldots 1, \\ 9^3&=\ldots 9, \\ 9^4&=\ldots 1, \\ & \ \vdots \end{align*} Note that if $x$ is odd, then $9^x \equiv 4\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \equiv 1\pmod{5}.$
Therefore, we have \begin{align*} N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &&\pmod{5} \\ &\equiv 2+28+6+20+2 &&\pmod{5} \\ &\equiv 58 &&\pmod{5} \\ &\equiv \boxed{3}$ may simplify the process further, as given by Solution 1.


Answer: 3.0

Solution: Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: \begin{align*} N&=27006000052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ &\equiv 2-7+6-5+2 &&\pmod{5} \\ &\equiv -2 &&\pmod{5} \\ &\equiv \boxed{3} ~Aidensharp ~Kante314 ~MRENTHUSIASM

Solution:
We need to first convert $N$ into a regular base- $10$ number: \[N = 27006000052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.\]
Now, consider how the last digit of $9$ changes with changes of the power of $9:$ \begin{align*} 9^0&=1, \\ 9^1&=9, \\ 9^2&=\ldots 1, \\ 9^3&=\ldots 9, \\ 9^4&=\ldots 1, \\ & \ \vdots \end{align*} Note that if $x$ is odd, then $9^x \equiv 4\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \equiv 1\pmod{5}.$
Therefore, we have \begin{align*} N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &&\pmod{5} \\ &\equiv 2+28+6+20+2 &&\pmod{5} \\ &\equiv 58 &&\pmod{5} \\ &\equiv \boxed{3}$ may simplify the process further, as given by Solution 1.


Answer: 12.0

The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ $M$ , and $H$ denote digits that are not given. What is $T+M+H$
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$

Solution: We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$ . Their divisibility rules (see Solution 2) tell us that $T + M \equiv 3 \;(\bmod\; 9)$ and that $T - M \equiv 7 \;(\bmod\; 11)$ . By guess and checking, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = \boxed{12}$

Solution: We know that $H = 0$ , because $19!$ ends in three zeroes (see Solution 1). Furthermore, we know that $9$ and $11$ are both factors of $19!$ . We can simply use the divisibility rules for $9$ and $11$ for this problem to find $T$ and $M$ . For $19!$ to be divisible by $9$ , the sum of digits must simply be divisible by $9$ . Summing the digits, we get that $T + M + 33$ must be divisible by $9$ . This leaves either $\text{A}$ or $\text{C}$ as our answer choice. Now we test for divisibility by $11$ . For a number to be divisible by $11$ , the alternating sum must be divisible by $11$ (for example, with the number $2728$ $2-7+2-8 = -11$ , so $2728$ is divisible by $11$ ). Applying the alternating sum test to this problem, we see that $T - M - 7$ must be divisible by 11. By inspection, we can see that this holds if $T=4$ and $M=8$ . The sum is $8 + 4 + 0 = \boxed{12}$

Solution: Multiplying it out, we get $19! = 121,645,100,408,832,000$ . Evidently, $T = 4$ $M = 8$ , and $H = 0$ . The sum is $8 + 4 + 0 = \boxed{12}$

Solution:
7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.
The answer is $4 + 8 + 0 = \boxed{12}$ . ~ AliciaWu


Answer: 392.0

The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position.

Solution: Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly $131 - 1 = 130$ numbers must be in front of it. There are $\frac{130}{2} = 65$ cards from each of piles A, B in front of card 131. This suggests that $n = 131 + 65 = 196$ ; the total number of cards is $196 \cdot 2 = \boxed{392}$

Solution: If you index the final stack $1,2,\dots,2n$ , you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers $1,2,\dots, n$ . This all leads us to the following equation. \[131=2n-2(131)+1\implies2n=\boxed{392}\]


Answer: 10.0

Solution: The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\frac{3}{2}$ , so the volume of one pyramid is $\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\cdot5=\boxed{10}$


Answer: 10.0

Solution: The "base" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ of the rectangular prism's volume, meaning that the answer is $3 \cdot 4 \cdot 5 \cdot \dfrac{1}{6} = \boxed{10}$


Answer: 40.0

The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches.
The length of the common internal tangent is:
$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

Solution:
Let $A$ be the center of the circle with radius $5$ , and $B$ be the center of the circle with radius $4$ . Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$ . Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp\overline{AE}$ . Since $\overline{AC}\perp\overline{CD}$ and $\overline{BD}\perp\overline{CD}$ $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$
Since the centers of the two circles are $41$ inches apart, $AB=41$ . Also, $BE=4+5=9$ . Using the Pythagorean Theorem on right triangle $ABE$ $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$ . The length of the common internal tangent is $\boxed{40}$


Answer: 1.05

The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$ . The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at the same place will come to
$\textbf{(A)}\ $1.70 \qquad \textbf{(B)}\ $1.65 \qquad \textbf{(C)}\ $1.20 \qquad \textbf{(D)}\ $1.05 \qquad \textbf{(E)}\ $0.95$

Solution:
Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie.  With the information,
\[3s+7c+p=3.15\] \[4s+10c+p=4.20\]
Subtract the first equation from the second to get
\[s+3c=1.05\]
That means $s=1.05-3c$ .  Substituting it back in the second equation results in.
\[4.20-12c+10c+p=4.20\]
Solving for $p$ yields $p=2c$ .  With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is $(1.05-3c)+c+(2c) = \boxed{1.05}$


Answer: 23.0

The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$ , respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$ . By $\frac{MB}{BO}=\frac{BO}{BQ}$ $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$ . However, since $O$ is the circumcenter of triangle $ABC$ $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$ . Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle BNO$ , we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$ . Thus, $BP = \frac{18}{5}$ $m + n=\boxed{023}$

Solution: Minor arc $BC = 2A$ so $\angle{BOC}=2A$ . Since $\triangle{BOC}$ is isosceles ( $BO$ and $OC$ are radii), $\angle{CBO}=(180-2A)/2=90-A$ $\angle{CBO}=90-A$ , so $\angle{BQO}=A$ . From this we get that $\triangle{BPQ}\sim \triangle{BCA}$ . So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$ , plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$ , so $BP=\dfrac{18}{5}$ , and $m+n=\boxed{023}$

Solution: Let $r=BO$ . Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$ . From there, \[OM=\sqrt{r^2-4}\] Thus, \[OQ=\frac{\sqrt{4r^2+9}}{2}\] Using $\triangle{BOQ}$ , we get $r=3$ . Now let's find $NP$ . After some calculations with $\triangle{BON}$ $\triangle{OPN}$ ${NP=11/10}$ . Therefore, \[BP=\frac{5}{2}+\frac{11}{10}=18/5\] $18+5=\boxed{023}$

Solution:
Let $\angle{BQO}=\alpha$ . Extend $OB$ to touch the circumcircle at a point $K$ . Then, note that $\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$ . But since $BK$ is a diameter, $\angle{KAB}=90^\circ$ , implying $\angle{CAB}=\alpha$ . It follows that $APCQ$ is a cyclic quadrilateral.
Let $BP=x$ . By Power of a Point, \[5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.\] The answer is $18+5=\boxed{023}$

Solution:
$\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$
Denote the circumradius of $ABC$ to be $R$ , the circumcircle of $ABC$ to be $O$ , and the shortest distance from $Q$ to circle $O$ to be $x$
Using Power of a Point on $Q$ relative to circle $O$ , we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$ . Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$ . Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$ . Now, set $\cos{ABC} = y$ . Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$ . Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$ . Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$ (You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$ , we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$ . Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$ . For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$ . The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$ . By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$ . Solution by hyxue

Solution:
 Reflect $A$ $P$ across $OB$ to points $A'$ and $P'$ , respectively with $A'$ on the circle and $P, O, P'$ collinear. Now, $\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB$ by parallel lines. From here, $\angle P'PB = \angle PP'B = \angle A'P'Q$ as $P, P', Q$ collinear. From here, $A'P'QC$ is cyclic, and by power of a point we obtain $\frac{18}{5} \implies \boxed{023}$ .
~awang11's sol


Answer: 14.0

The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$

Solution:
By the Binomial Theorem , each term of the expansion is $\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n$
We want the exponent of $x$ to be $7$ , so \[2(8-n)-n=7\] \[16-3n=7\] \[n=3\]
If $n=3$ , then the corresponding term is \[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\] \[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\] \[-14x^7\]
The answer is $\boxed{14}$


Answer: 15.0

The complex number $z$ is equal to $9+bi$ , where $b$ is a positive real number and $i^{2}=-1$ .  Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to?

Solution:
Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$ . Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i$
Setting these two equal, we get that $18bi = 243bi - b^3i$ , so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$ . Since $b > 0$ , the solution is $\boxed{015}$


Answer: 289.0

The complex number $z$ satisfies $z + |z| = 2 + 8i$ . What is $|z|^{2}$ ? Note: if $z = a + bi$ , then $|z| = \sqrt{a^{2} + b^{2}}$
$\textbf{(A)}\ 68\qquad \textbf{(B)}\ 100\qquad \textbf{(C)}\ 169\qquad \textbf{(D)}\ 208\qquad \textbf{(E)}\ 289$

Solution:
Let the complex number $z$ equal $a+bi$ . Then the preceding equation can be expressed as \[a+bi+\sqrt{a^2+b^2} = 2+8i\] Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$ , giving $b = 8$ . Plugging this in back to our equation gives us $a+\sqrt{a^2+64} = 2$ .
Rearranging this into $2-a = \sqrt{a^2+64}$ , we can square each side of the equation resulting in \[4-4a+a^2 = a^2+64\] Further simplification will yield $60 = -4a$ meaning that $-15 = a$ . Knowing both $a$ and $b$ , we can plug them in into $a^2+b^2$ . Our final answer is $\boxed{289}$


Answer: 71.0

The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$

Solution: Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$ nd root of unity, and thus, by De Moivre's theorem, the imaginary part of $z$ will be of the form $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},$ where $k \in \{1, 2, \ldots, 70\}.$ Note that $71$ is prime and $k<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71.$ Thus, $n = \boxed{071}.$


Answer: 40.0

The complex numbers $z$ and $w$ satisfy the system \[z + \frac{20i}w = 5+i\] \[w+\frac{12i}z = -4+10i\] Find the smallest possible value of $\vert zw\vert^2$

Solution: Multiplying the two equations together gives us \[zw + 32i - \frac{240}{zw} = -30 + 46i\] and multiplying by $zw$ then gives us a quadratic in $zw$ \[(zw)^2 + (30-14i)zw - 240 =0.\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$


Answer: 11.0

Solution: The triangle has base $6,$ which means its height satisfies \[\dfrac{6h}{2}=3h=12.\] This means that $h=4,$ so the answer is $7+4=\boxed{11}$

Solution:
 Label point $D(3,7)$ as the point at which $CD\perp DA$ . We now have $[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$ , where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of $\triangle ACD$ are $y-7$ and $5-3=2$ . The two side lengths of $\triangle BCD$ are $y-7$ and $11-3 = 8.$ Now,
\[[\triangle ABC] = 12  = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2  = 3(y-7)\]
Dividing by $3$ gives $y -7 = 4,$ so $y = \boxed{11}.$

Solution:
By the Shoelace Theorem, $\triangle ABC$ has area \[\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|\] . From the problem, this is equal to $12$ . We now solve for y.
$\frac{1}{2}|6y - 42| = 12$
$|6y-42| = 24$
$6y - 42 = 24$ OR $6y - 42 = -24$
$6y = 66$ OR $6y = 18$
$y = 11$ OR $y = 3$
However, since, as stated in the problem, $y > 7$ , our only valid solution is $\boxed{11}$

Solution: As in the figure, the triangle is determined by the vectors $\begin{bmatrix}-2 \\ y-7\end{bmatrix}$ and $\begin{bmatrix}6\\0\end{bmatrix}$ . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that $\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24$ . Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$ . Solving each equation individually, we find that $y = 3$ or $y = 11$ . However, the problem states that $y > 7$ , so the only valid solution is $\boxed{11}$


Answer: 131.0

The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$ . Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$ , and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram , and $\overrightarrow{AB} = \overrightarrow{D'C}$ . Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.
Let the slope of the perpendicular be $m$ . Then the midpoint of $\overline{DD'}$ lies on the line $y=mx$ , so $\frac{b+7}{2} = m \cdot \frac{a+1}{2}$ . Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$ . Combining these two equations yields
\[a^2 + \left(7 - (a+1)m\right)^2 = 50\]
Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$ . We exclude the cases $(\pm 1, \pm 7)$ because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have
\[7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5\]
These yield $m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12$ . Therefore, the corresponding slopes of $\overline{AB}$ are $-1, -\frac 43, 1, \frac 34, -\frac 12, -3, \frac 13$ , and $2$ .  The sum of their absolute values is $\frac{119}{12}$ . The answer is $m+n= \boxed{131}$

Solution:
A very natural solution:
. Shift $A$ to the origin. Suppose point $B$ was $(x, kx)$ . Note $k$ is the slope we're looking for. Note that point $C$ must be of the form: $(x \pm 1, kx \pm 7)$ or $(x \pm 7, kx \pm 1)$ or $(x \pm 5, kx \pm 5)$ . Note that we want the slope of the line connecting $D$ and $C$ so also be $k$ , since $AB$ and $CD$ are parallel.
Instead of dealing with the 12 cases, we consider
point $C$ of the form $(x \pm Y, kx \pm Z)$ where
we plug in the necessary values for $Y$ and $Z$ after simplifying.
Since the slopes of $AB$ and $CD$ must both be $k$ $\frac{7 - kx \pm Z}{1 - x \pm Y} = k \implies k = \frac{7 \pm Z}{1 \pm Y}$ . Plugging in the possible values of $\pm 7, \pm 1, \pm 5$ in their respective pairs and ruling out degenerate cases, we find the sum is $\frac{119}{12} \implies m + n = \boxed{131}$ - whatRthose


Answer: 1.0

The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$ , first quartile $Q_1 = 33$ , and third quartile $Q_3 = 43$ . An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ( $Q_1$ ) or more than $1.5$ times the interquartile range above the third quartile ( $Q_3$ ), where the interquartile range is defined as $Q_3 - Q_1$ . How many outliers does this data set have?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: The interquartile range is defined as $Q3 - Q1$ , which is $43 - 33 = 10$ $1.5$ times this value is $15$ , so all values more than $15$ below $Q1$ $33 - 15 = 18$ is an outlier. The only one that fits this is $6$ . All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so there is only $\boxed{1}$ outlier in total.


Answer: 26.0

The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution:
We have
\[\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]
Now we do some estimation. Notice that $2^{20} = 1024^2$ , which means that $2^{20}$ is a little more than $1000^2=1,000,000$ . Multiplying it with $10^{20}$ , we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$ . Notice that when we divide $1$ by an $n$ digit number as long as n is not a power of 10, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$ , there are $\boxed{26}$ zeros in the initial string after the decimal point. -PCChess

Solution:
First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$ . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$
$\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fact that $\log{5} = 1 - \log{2}$ ), $\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14$ digits.
Our answer is $\boxed{26}$

Solution: Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of $\boxed{26}$ zeroes to be formed. -OreoChocolate

Solution:
We see that $\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}$ . We see that this has $27-1=26$ zeros after the decimal point before coming to $9$
Therefore, the answer is $\boxed{26}$

Solution: \begin{align*}|\lceil \log \dfrac{1}{20^{20}} \rceil| &= |\lceil \log 20^{-20} \rceil| \\ &= |\lceil -20 \log(20) \rceil| \\ &= |\lceil -20(\log 10 + \log 2) \rceil| \\ &= |\lceil -20(1 + 0.301) \rceil| \\ &= |\lceil -26.02 \rceil| \\ &= |-26| \\ &= \boxed{26}


Answer: 30.0

Solution:
Angle-chasing using the small triangles:
Use the line below and to the left of the $110^\circ$ angle to find that the rightmost angle in the small lower-left triangle is $180 - 110 = 70^\circ$
Then use the small lower-left triangle to find that the remaining angle in that triangle is $180 - 70 - 40 = 70^\circ$
Use congruent vertical angles to find that the lower angle in the smallest triangle containing $A$ is also $70^\circ$
Next, use line segment $AB$ to find that the other angle in the smallest triangle containing $A$ is $180 - 100 = 80^\circ$
The small triangle containing $A$ has a $70^\circ$ angle and an $80^\circ$ angle.  The remaining angle must be $180 - 70 - 80 = \boxed{30}$

Solution: The third angle of the triangle containing the $100^\circ$ angle and the $40^\circ$ angle is $180^\circ - 100^\circ - 40^\circ = 40^\circ$ . It follows that $A$ is the third angle of the triangle consisting of the found $40^\circ$ angle and the given $110^\circ$ angle. Thus, $A$ is a $180^\circ - 110^\circ - 40^\circ = 30^\circ$ angle, and so the answer is $\boxed{30}$


Answer: 143.0

The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle

Solution: Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\frac{360}{18}=20^\circ$ . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is $20$ . Since there are even number of exterior angles, the middle two must be $19^\circ$ and $21^\circ$ , and the difference between terms must be $2$ . Check to make sure the smallest exterior angle is greater than $0$ $19-2(8)=19-16=3^\circ$ . It is, so the greatest exterior angle is $21+2(8)=21+16=37^\circ$ and the smallest interior angle is $180-37=\boxed{143}$

Solution: The sum of the angles in a 18-gon is $(18-2) \cdot 180^\circ = 2880 ^\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18a+153d = 2880 ^\circ,$ or $2a+17d = 320^\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=\boxed{143}$

Solution:
Each individual angle in a $18$ -gon is $\frac {(18-2) \cdot 180^\circ}{18} = 160^\circ$ . Since no angle in a convex polygon can be larger than $180^\circ$ , the smallest angle possible is in the set $159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177$
Our smallest possible angle is $\boxed{143}$


Answer: 17.0

The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$

Solution: It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\boxed{17}$


Answer: 240.0

The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximum possible number of points in a growing path, and let $r$ be the number of growing paths consisting of exactly $m$ points. Find $mr$

Solution:
We label our points using coordinates $0 \le x,y \le 3$ , with the bottom-left point being $(0,0)$ . By the Pythagorean Theorem , the distance between two points is $\sqrt{d_x^2 + d_y^2}$ where $0 \le d_x, d_y \le 3$ ; these yield the possible distances (in decreasing order) \[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\ \sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2},\ \sqrt{1}\] As these define $9$ lengths, so the maximum value of $m$ is $10$ . For now, we assume that $m = 10$ is achievable. Because it is difficult to immediately impose restrictions on a path with increasing distances, we consider the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former.
The $\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$
The $\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$
From $(1,0)$ , there are two possible ways to move $\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$ . However, from $(0,3)$ , there is no way to move $\sqrt{9}$ away, so we discard it as a possibility.
From $(2,3)$ , the lengths of $\sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2}$ fortunately are all determined, with the endpoint sequence being $(2,3)-(2,0)-(0,2)-(2,1)-(0,1)-(1,2)$
From $(1,2)$ , there are $3$ possible lengths of $\sqrt{1}$ (to either $(1,1),(2,2),(1,3)$ ). Thus, the number of paths is $r = 4 \cdot 2 \cdot 3 = 24$ , and the answer is $mr = 10 \cdot 24 = \boxed{240}$