Answer: 12.0

A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$

Solution:
Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of the shortest 3 sides must always be longer than the longest side. Thus, the answer is $\frac{26}{2}-1=13-1=\boxed{12}$

Solution:
Say the chosen side is $a$ and the other sides are $b,c,d$
By the Generalised Polygon Inequality, $a<b+c+d$ . We also have $a+b+c+d=26\Rightarrow b+c+d=26-a$
Combining these two, we get $a<26-a\Rightarrow a<13$
The largest length that satisfies this is $a=\boxed{12}$

Solution: The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing the side length. By Brahmagupta's Formula, the area of the quadrilateral is defined by $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi-perimeter. If the perimeter of the quadrilateral is $26$ , then the semi-perimeter will be $13$ . The area of the quadrilateral must be positive so the difference between the semi-perimeter and a side length must be greater than $0$ as otherwise, the area will be $0$ or negative. Therefore, the longest a side can be in this quadrilateral is $\boxed{12}$

Solution: This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$ . Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$ . So the top side equals $8$ , and the bottom side length equals $4+4+4$ $=$ $\boxed{12}$ ~ kabbybear


Answer: 4.0

A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12  \qquad\textbf{(E)}\ 13$

Solution: Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$ .  Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles.  This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . Simplifying we get $(a-b)(a+b) = 8$ . Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$ . So the answer is $\boxed{4}$

Solution:
Solution by e_power_pi_times_i
By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{4}$


Answer: 500.0

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$

Solution:
Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}$

Solution:
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.

Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, \[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$ \[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\] \[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\] \[4(1-x^2)(2-x^2)=(2x^2-1)^2\] \[8-12x^2+4x^4=4x^4-4x^2+1\] \[8x^2=7\] \[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem \[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values, \[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\] \[1+AD=\frac{7}{2}\] \[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{500}$

Solution:
Let $s = 200$ .  Let $O$ be the center of the circle.  Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$ .  Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$ , hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{500.}\]

Solution:
Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$ . Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$ .
Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$ ,we seek to find $2r\sin 3\theta$ .
First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean. \[\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}\] \[2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{500}\]

Solution: For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$ , so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$ . By Brahmagupta's formula , the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ $DA=5$ . Scaling up by 100, we get $\boxed{500}$

Solution:
 Label the points as shown, and let $\angle{EOF} = \theta$ . Since $\overline{OB} = \overline{OC}$ , and $\triangle{OFE} \sim \triangle{OCB}$ , we get that $\angle{EFO} = 90-\frac{\theta}{2}$ . We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$ . Also, since $\triangle{OCB} \cong \triangle{ODC}$ $\angle{OCD} = \alpha$ . This means that $\angle{CDF} = 180-2\alpha = \theta$ , which leads to $\triangle{OCB} \sim \triangle{DCF}$ .
Since we know that $\overline{CD} = 200$ $\overline{DF} = 200$ , and by similar reasoning $\overline{AE} = 200$ .
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$ , which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$ , and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{500}$ - ColtsFan10

Solution: We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\lvert z \rvert = 1$ and $\lvert z-1 \rvert = 1/\sqrt{2}$ , and we wish to find $\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\sqrt{7}/4$ . Thus, $AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{500}$

Solution:
Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$ , we can use the formula \[R=\frac{abc}{4A}\] and \[A=\frac{absin(C)}{2}\] and plug in all the values we got to get \[200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin(2a)}{2})}\] . This boils down to \[\sqrt{2}=\frac{sin(a)}{sin{2a}}\] . This expression can further be simplified by the trig identity \[sin(2a)=2sin(a)cos(a)\] . This leads to the final simplified form \[2\sqrt{2}=\frac{1}{cos(a)}\] . Solving this expression gives us \[cos(a)=\frac{\sqrt{2}}{4}\] . However, as we want $sin(a)$ , we use the identity $sin^2+cos^2=1$ , and substitute  to get that $sin(a)=\frac{\sqrt{14}}{4}$ , and therefore BD is $100\sqrt{14}$
Then, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$ ) to get \[14 \cdot 100^2=200x+200^2\] . Finally, we solve to get $\boxed{500}$


Answer: 500.0

Solution:
Let quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$
By the Pythagorean Theorem, the length of $OH$ is \begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore, \begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$ , we get: \[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\] \[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\] \[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\] \[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\] \[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\] \[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\] \[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\] \[7x^2 - 5600x + 1120000 = 320000 - x^2\] \[8x^2 - 5600x + 800000 = 0\] \[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula: \begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$ , then $ABCD$ would be a square. Thus, the radius of the circle would be \[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\] Which is a contradiction. Therefore, our answer is $\boxed{500}.$

Solution: Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply the law of cosines on $\Delta BOC$ ; let $\theta = \angle BOC$ . We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\] Because $\angle AOB$ $\angle BOC$ , and $\angle COD$ are congruent, $\angle AOD = 3\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\Delta AOD$ . Now, to find $\cos 3\theta$ , we can derive a formula that only uses $\cos\theta$ \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ( $\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$ ).  Plugging in $\cos\theta=\frac{3}{4}$ , we get $\cos 3\theta= -\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\]

Solution:
Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From the Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$ , so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$ , so $\angle DCF = \theta$
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \boxed{500}$

Solution:
 Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$ . Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$ . Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$ . Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$ . Let $F$ the feet of the altitude from $O$ to $AD$ . Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$ , because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$ . Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$ . Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$ , but $0<\alpha<180 \implies cos(\alpha)=3/4$ . Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$


Answer: 500.0

Solution:
Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}$

Solution:
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.

Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, \[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$ \[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\] \[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\] \[4(1-x^2)(2-x^2)=(2x^2-1)^2\] \[8-12x^2+4x^4=4x^4-4x^2+1\] \[8x^2=7\] \[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem \[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values, \[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\] \[1+AD=\frac{7}{2}\] \[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{500}$

Solution:
Let $s = 200$ .  Let $O$ be the center of the circle.  Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$ .  Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$ , hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{500.}\]

Solution:
Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$ . Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$ .
Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$ ,we seek to find $2r\sin 3\theta$ .
First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean. \[\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}\] \[2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{500}\]

Solution: For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$ , so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$ . By Brahmagupta's formula , the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ $DA=5$ . Scaling up by 100, we get $\boxed{500}$

Solution:
 Label the points as shown, and let $\angle{EOF} = \theta$ . Since $\overline{OB} = \overline{OC}$ , and $\triangle{OFE} \sim \triangle{OCB}$ , we get that $\angle{EFO} = 90-\frac{\theta}{2}$ . We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$ . Also, since $\triangle{OCB} \cong \triangle{ODC}$ $\angle{OCD} = \alpha$ . This means that $\angle{CDF} = 180-2\alpha = \theta$ , which leads to $\triangle{OCB} \sim \triangle{DCF}$ .
Since we know that $\overline{CD} = 200$ $\overline{DF} = 200$ , and by similar reasoning $\overline{AE} = 200$ .
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$ , which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$ , and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{500}$ - ColtsFan10

Solution: We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\lvert z \rvert = 1$ and $\lvert z-1 \rvert = 1/\sqrt{2}$ , and we wish to find $\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\sqrt{7}/4$ . Thus, $AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{500}$

Solution:
Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$ , we can use the formula \[R=\frac{abc}{4A}\] and \[A=\frac{absin(C)}{2}\] and plug in all the values we got to get \[200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin(2a)}{2})}\] . This boils down to \[\sqrt{2}=\frac{sin(a)}{sin{2a}}\] . This expression can further be simplified by the trig identity \[sin(2a)=2sin(a)cos(a)\] . This leads to the final simplified form \[2\sqrt{2}=\frac{1}{cos(a)}\] . Solving this expression gives us \[cos(a)=\frac{\sqrt{2}}{4}\] . However, as we want $sin(a)$ , we use the identity $sin^2+cos^2=1$ , and substitute  to get that $sin(a)=\frac{\sqrt{14}}{4}$ , and therefore BD is $100\sqrt{14}$
Then, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$ ) to get \[14 \cdot 100^2=200x+200^2\] . Finally, we solve to get $\boxed{500}$


Answer: 500.0

Solution:
Let quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$
By the Pythagorean Theorem, the length of $OH$ is \begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore, \begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$ , we get: \[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\] \[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\] \[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\] \[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\] \[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\] \[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\] \[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\] \[7x^2 - 5600x + 1120000 = 320000 - x^2\] \[8x^2 - 5600x + 800000 = 0\] \[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula: \begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$ , then $ABCD$ would be a square. Thus, the radius of the circle would be \[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\] Which is a contradiction. Therefore, our answer is $\boxed{500}.$

Solution: Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply the law of cosines on $\Delta BOC$ ; let $\theta = \angle BOC$ . We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\] Because $\angle AOB$ $\angle BOC$ , and $\angle COD$ are congruent, $\angle AOD = 3\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\Delta AOD$ . Now, to find $\cos 3\theta$ , we can derive a formula that only uses $\cos\theta$ \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ( $\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$ ).  Plugging in $\cos\theta=\frac{3}{4}$ , we get $\cos 3\theta= -\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\]

Solution:
Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From the Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$ , so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$ , so $\angle DCF = \theta$
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \boxed{500}$

Solution:
 Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$ . Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$ . Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$ . Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$ . Let $F$ the feet of the altitude from $O$ to $AD$ . Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$ , because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$ . Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$ . Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$ , but $0<\alpha<180 \implies cos(\alpha)=3/4$ . Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$


Answer: 321.0

A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$

Solution:
First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.
$11_{12}=15_8$
$22_{12}=32_8$
$33_{12}=47_8$
$44_{12}=64_8$
$55_{12}=101_8$
We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number.
Compare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$ .
Evaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$ .
When the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$ . Since $d\neq0$ , the first value is correct. Compiling the necessary digits leaves us a final answer of $\boxed{321}$

Solution: The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$ .  Simplifying the first equation gives $a = (3/2)b$ .  Plugging this into the second equation gives $3b/32 = c/8 + d/64$ . Multiplying both sides by 64 gives $6b = 8c + d$ $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$ $(a,b) = (3,2)$ or $(6,4)$ .  Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$ , and $d = 4$ .  Therefore $abc = \boxed{321}$

Solution:
Converting to base $10$ we get
$4604a+72c+9d=6960b$
Since $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get
$46a=70b$
Writing out the first few values of $a$ and $b$ , the first possible tuple is
$a=3, b=2, c=1, d=4$
and the second possible tuple is
$a=6, b=4, c=3, d=0$
Note that $d$ can not be $0$ , therefore the answer is $\boxed{321}$

Solution:
In the problem, we are given that \[8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.\] We multiply by the LCM of the denominators, which is $576$ to get \[4608a+576b+72c+9d=6912b+576b+48b+4a.\] We then group like terms and factor to get \[4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.\]
Observe that the coefficients of $a$ $c$ , and $b$ are all divisible by $4$ . Therefore, we know that $d$ must also be divisible by $4$ to compensate. (To observe this, one could rearrange the terms to see $9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)$ . At this point, it is obvious that $9d$ must be divisible by $4$ , so $d$ must be divisible by $4$ .) Thus, we let $d=4y$ to see $4604a+72c+9\cdot4y=24\cdot290b$ . We divide by $4$ in the equation to get \[1151a+18c+9y=6\cdot290b.\]
Observe that the coefficients of $c$ $y$ , and $b$ are all divisible by $3$ . By similar reasoning, $a$ must be divisible by $3$ , so we let $a=3x$ . Substituting and dividing by $3$ , we get \[1151x+6c+3y=2\cdot290b.\] We observe that this can no longer be reduced by similar means.
We know that $0<a,b,c<8\implies0<3x,b,4y<8$ . We examine $0<4y<8$ ; this becomes $0<y<2$ . It is apparent that $y=1\implies d=4$ . However, the problem does not even ask for $d$ , so it may appear that this find is meaningless. However, we substitute our value of $y$ in to get \[1151x+6c+3=580b.\]
We know that $x$ is either $1$ or $2$ , since $0<3x<8$ . We take our equation modulo $6$ to find \[5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.\] If $x=2$ , then $1\equiv4b\pmod{6}$ - this is equivalent to saying that $1+6n=4b\implies 1=4b-6n$ . However, $4b-6n$ is always even, and $1$ is odd. Thus, this case is not possible, so $x=1\implies a=3$ . We know that $x=1$ ; thus, $3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}$ . Obviously, $b=2$ and $b=5$ work; however, if $b=5$ , then the RHS of our original equation ( $2\cdot290b$ ) is much too large to be equal to the LHS (the maximum possible value of the LHS is $1151\cdot1+6\cdot7+3\cdot1$ , which is less than $1200$ while the RHS would become $2\cdot290\cdot5=2900$ ), so we have $b=2$
We recall our equation of $1151x+6c+3=580b$ . Plugging in what we know, we have $1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1$
Therefore, $\overline{abc}=\boxed{321}$


Answer: 10.0

A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$ ) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$ ). If $\measuredangle CDA=8^\circ$ , what is the largest value $n$ can have?
[asy] unitsize(1.5cm); pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0); draw((-6.5,0)--D--C,black); draw(A--E--F--G--B,black); dotfactor=4; dot("$A$",A,S); dot("$C$",C,N); dot("$R_1$",E,N); dot("$R_2$",F,S); dot("$R_3$",G,N); dot("$B$",B,S); markscalefactor=0.015; draw(rightanglemark(G,B,D)); draw(anglemark(C,E,A,12)); draw(anglemark(F,E,G,12)); draw(anglemark(E,F,A)); draw(anglemark(E,F,A,12)); draw(anglemark(B,F,G)); draw(anglemark(B,F,G,12)); draw(anglemark(E,G,F)); draw(anglemark(E,G,F,12)); draw(anglemark(E,G,F,16)); draw(anglemark(B,G,D)); draw(anglemark(B,G,D,12)); draw(anglemark(B,G,D,16)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$

Solution:
Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$ , but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$ , the reflection would give us a triangle with angles $16, 90$ , and $74$ . Then, when we reflect again, we will have $180 - 74 - 74$ $32$ . Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle.
Notice that we keep decreasing by increments of $8$ . This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$ , we must decrease by $8$ every time. This is the most optimal path of the light beam.
The pattern of light will be $82-74-66-58-50-42-34-26-18-10$ . When we get to the angle of $2$ degrees, we have reached angle $A$ . Therefore, we don't count the $2$ , so our total number of reflections between $CD$ and $AD$ is $\boxed{10}$


Answer: 37.0

A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$ . The probability that the roots of the polynomial
$x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2$
are all real can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal 0 and thus, they are roots that we can factor out.
The polynomial becomes:
$(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)$
Since we know $1$ and $-2$ are real numbers, we only need to focus on the quadratic.
We should set the discriminant of the quadratic greater than or equal to 0.
$(2a - 1)^2 - 4 \geq 0$
This simplifies to:
$a \geq \dfrac{3}{2}$
or
$a \leq -\dfrac{1}{2}$
This means that the interval $\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)$ is the "bad" interval. The length of the interval where $a$ can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long.
$\dfrac{36}{38} = \dfrac{18}{19}$
$18 + 19 = \boxed{037}$


Answer: 2.0

A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b)  + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$

Solution: Substituting $a = b$ we get \[f(2a) + f(0) = 2f(a)^2\] Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\] This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\] Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$ , so answer choice $\boxed{2}$ is impossible.

Solution:
First, we set $a \leftarrow 0$ and $b \leftarrow 0$ .
Thus, the equation given in the problem becomes $[ f(0) + f(0) = 2 f(0) \times f(0) . ]$
Thus, $f(0) = 0$ or 1.
Case 1: $f(0) = 0$
We set $b \leftarrow 0$ .
Thus, the equation given in the problem becomes $[ 2 f(a) = 0 . ]$
Thus, $f(a) = 0$ for all $a$
Case 2: $f(0) = 1$
We set $b \leftarrow a$ .
Thus, the equation given in the problem becomes \[[ f(2a) + 1 = 2 \left( f(a) \right)^2. ]\]
Thus, for any $a$ \begin{align*} f(2a) & = -1 + 2 \left( f(a) \right)^2 \\ & \geq -1 . \end{align*}
Therefore, an infeasible value of $f(1)$ is $\boxed{2}.$

Solution:
The relationship looks suspiciously like a product-to-sum identity. In fact, \[\cos(\alpha)\cos(\beta) = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))\] which is basically the relation. So we know that $f(x) = \cos(x)$ is a valid solution to the function. However, if we define $x=ay,$ where $a$ is arbitrary, the above relation should still hold for $f(x) = \cos(ay) = \cos(a(1))$ so any value in $[-1,1]$ can be reached, so choices $A,B,$ and $C$ are incorrect.
In addition, using the similar formula for hyperbolic cosine, we know \[\cosh(\alpha)\cosh(\beta) = \frac{1}{2}(\cosh(\alpha-\beta)+\cosh(\alpha+\beta))\] The range of $\cosh(ay)$ is $[1,\infty)$ so choice $D$ is incorrect.
Therefore, the remaining answer is choice $\boxed{2}.$


Answer: 750.0

A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$

Solution:
We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
Let's say that $500\text{ sheets}=5\text{ cm}\Rightarrow \frac{500 \text{ sheets}}{5 \text{ cm}} = 1$ . So by multiplying $7.5 \text{ cm}$ by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
\begin{align*} \frac{7.5 \times 500}{5} &= 7.5 \times 100 \\ &= 750 \text{ sheets} \\ \end{align*}
$750$ is $\boxed{750}$

Solution:
We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness.
Our proportion is \[\frac{5}{500}=\frac{7.5}{x}\] where $x$ is the number we are looking for.
Next, we cross-multiply to get $5x=500 \times 7.5$ so $x=750$ which is $\boxed{750}$


Answer: 720.0

A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$
[asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]

Solution:
When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.

By Heron's Formula, the area of each isosceles triangle is $\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}$ . So the area of both is $144\sqrt{5}$ . From the rectangle, our original area is $36a$ . The area of the rectangle in the hexagon is $24a$ . So we have \[24a+144\sqrt{5}=36a\implies 12a=144\sqrt{5}\implies a=12\sqrt{5}\implies a^2=\boxed{720}.\]

Solution: Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$ . Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitude of the end triangles is $\sqrt{3^2-2^2}=\sqrt{5}$ given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area $\sqrt{5}$ $4$ . Then, the area of the hexagon is $4p+4\sqrt{5}$ , and the area of the rectangle is $6p$ . Equating them, $p=2\sqrt{5}$ . Multiply by scale factor of 6 and square it to get $36(20)= 720 \implies a^2=\boxed{720}$


Answer: 7.0

A rectangle intersects a circle as shown: $AB=4$ $BC=5$ , and $DE=3$ . Then $EF$ equals:
[asy]defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW);  label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

Solution: Draw $BE$ and $CF$ , forming a trapezoid . Since it's cyclic, this trapezoid must be isosceles . Also, drop altitudes from $E$ to $AC$ $B$ to $DF$ , and $C$ to $DF$ , and let the feet of these altitudes be $G$ $H$ , and $I$ respectively. $AGED$ is a rectangle since it has $4$ right angles . Therefore, $AG=DE=3$ , and $GB=4-3=1$ . By the same logic, $GBHE$ is also a rectangle, and $EH=GB=1$ $BH=CI$ since they're both altitudes to a trapezoid, and $BE=CF$ since the trapezoid is isosceles. Therefore, $\triangle BHE \cong \triangle CIF$ by HL congruence , so $IF=EH=1$ . Also, $BCIH$ is a rectangle from $4$ right angles, and $HI=BC=5$ . Therefore, $EF=EH+HI+IF=1+5+1=\boxed{7}$


Answer: 540.0

A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); [/asy]
$\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720$

Solution: The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\cdot4\cdot3\cdot3\cdot3=\boxed{540}$

Solution:
Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.
Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\cdot4\cdot3\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$
Case 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\cdot4\cdot3 = 60$
Therefore, we have $120 + 360 + 60 = \boxed{540}$


Answer: 540.0

Solution: The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\cdot4\cdot3\cdot3\cdot3=\boxed{540}$

Solution:
Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.
Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\cdot4\cdot3\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$
Case 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\cdot4\cdot3 = 60$
Therefore, we have $120 + 360 + 60 = \boxed{540}$


Answer: 448.0

A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $\sqrt{N}\,$ , for a positive integer $N\,$ . Find $N\,$

Solution:
Put the rectangle on the coordinate plane so its vertices are at $(\pm4,\pm3)$ , for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$
Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$ $B(-x,3)$ $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$ . Then this is a rectangle, so $OA=OB$ , or $16+y^2=9+x^2$ , so $x^2=y^2+7$
Reflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$ . Then $BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$ . The midpoint of $\overline{BD'}$ is $(-4-x,0)$ , and since $-4>-4-x$ and $-y\le0$ $C$ always lies below $\overline{BD'}$
If $y$ is positive, it can be decreased to $y'<y$ . This causes $x$ to decrease as well, to $x'$ , where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$ , which must lie within $\triangle BCD'$ . Then we must have $BC'+C'D'<BC+CD'$ , and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$ , so $x=\sqrt7$
By the distance formula, this minimum perimeter is \[2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)\] \[=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.\] Therefore $N$ would equal $\boxed{448}.$

Solution: Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let $a$ and $b$ be the sides of the rectangle. Then $ab = 3(8) = 24$ since both are twice the area of the same right triangle, and $a^2+b^2 = 64$ . So $(a+b)^2 = 64+2(24) = 112$ , so $2(a+b) = \sqrt{\boxed{448}$


Answer: 80.0

A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram.  What is the perimeter of one of the five congruent rectangles? [asy] defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,3)--foot((r,3),(0,1),(4,1))); draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]
$\mathrm{(A)\ } 35.2 \qquad \mathrm{(B) \ }76 \qquad \mathrm{(C) \  } 80 \qquad \mathrm{(D) \  } 84 \qquad \mathrm{(E) \  }86$

Solution:
Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$
From the large rectangle, we see that the top has length $3w$ , the right has length $l + w$ , the bottom has length $2l$ , and the left has length $l + 2$
Since the perimeter of the large rectangle is $176$ , we know that $172 = 3w + l + w + 2l + l + w$ , or $172 = 5w + 4l$
From the top and bottom of the large rectangle, we know that $3w = 2l$ , or $l = 1.5w$
Plugging that into the first equation, we get $176 = 5w + 4(1.5)w$
$176 = 11w$
$w = 16$
$l = 1.5w = 24$
$P = 2l + 2w = 2(16 + 24) = 80$ , and the answer is $\boxed{80}$


Answer: 102.0

A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ . Which of the following numbers cannot equal $A+P$
$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$

Solution:
Let the rectangle's length be $a$ and its width be $b$ . Its area is $ab$ and the perimeter is $2a+2b$
Then $A + P = ab + 2a + 2b$ . Factoring, we have $(a + 2)(b + 2) - 4$
The only one of the answer choices that cannot be expressed in this form is $102$ , as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$ , keeping $a$ and $b$ as positive integers.
Our answer is then $\boxed{102}$


Answer: 67.0

Solution:
We use Image:2021_AMC_10B_(Nov)_Problem_25,_sol.png to facilitate our analysis.
Denote $\angle AFE = \theta$ . Thus, $\angle FIB = \angle CEF = \angle EKG = \angle KLC = \theta$
Hence, \begin{align*} AB & = AF + FB \\ & = EF \cos \angle EFA + IF \sin \angle FIB \\ & = 3 \cos \theta + \sin \theta , \end{align*} and \begin{align*} AC & = AE + EK + KC \\ & = EF \sin \angle EFA + EG \cos \angle CEG + KG \cos \angle EKG + KL \sin \angle CLK \\ & = 3 \sin \theta + \cos \theta + \cos \theta + \sin \theta \\ & = 2 \cos \theta + 4 \sin \theta . \end{align*}
Because $AB = AC$ $3 \cos \theta + \sin \theta = 2 \cos \theta + 4 \sin \theta$ .
Hence, $\tan \theta = \frac{1}{3}$ .
Hence, $\sin \theta = \frac{1}{\sqrt{10}}$ and $\cos \theta = \frac{3}{\sqrt{10}}$
Hence, $AB = AC = BD = CD = \sqrt{10}$
Now, we put the graph to a coordinate plane by setting point $A$ as the origin, putting $AB$ in the $x$ -axis and $AC$ on the $y$ -axis.
Hence, $A = \left( 0 , 0 \right)$ $B = \left( \sqrt{10} , 0 \right)$ $C = \left( 0 , \sqrt{10} \right)$ $D = \left( \sqrt{10} , \sqrt{10} \right)$ $E = \left( 0 , \frac{3}{\sqrt{10}} \right)$ $F = \left( \frac{9}{\sqrt{10}} , 0 \right)$ $G = \left( \frac{1}{\sqrt{10}} , \frac{6}{\sqrt{10}} \right)$ $H = \left( \frac{4}{\sqrt{10}} , \frac{7}{\sqrt{10}} \right)$ $I = \left( \sqrt{10} , \frac{3}{\sqrt{10}} \right)$
Denote $P = \left( \frac{10 - u}{\sqrt{10}} , \sqrt{10} \right)$ $Q = \left( \sqrt{10} , \frac{10 - v}{\sqrt{10}} \right)$
Because $HPQJ$ is a rectangle, $HP \perp PQ$ . Hence, $m_{HP} m_{PQ} = -1$ .
We have $m_{HP} = \frac{3}{6 - u}$ and $m_{PQ} = - \frac{v}{u}$ .
Hence, \[ \frac{3}{6 - u} \cdot \left( - \frac{v}{u} \right) = - 1 . \hspace{1cm} (1) \]
Because $HPQJ$ is a rectangle, $x_J + x_P = x_H + x_Q$ and $y_J + y_P = y_H + y_Q$ .
Hence, $J = \left( \frac{4 + u}{\sqrt{10}} , \frac{7 - v}{\sqrt{10}} \right)$
The equation of line $GI$ is \begin{align*} y & = \frac{\frac{3}{\sqrt{10}} - \frac{6}{\sqrt{10}}}{\sqrt{10} - \frac{1}{\sqrt{10}}} \left( x - \frac{1}{\sqrt{10}} \right) + \frac{6}{\sqrt{10}} \\ & = - \frac{x}{3} + \frac{19}{3 \sqrt{10}} . \end{align*} Because point $J$ is on line $GI$ , plugging the coordinates of $J$ into the equation of line $GI$ , we get \[ \frac{7 - v}{\sqrt{10}} = - \frac{\frac{4 + u}{\sqrt{10}}}{3} + \frac{19}{3 \sqrt{10}} . \hspace{1cm} (2) \]
By solving Equations (1) and (2), we get $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$ or $\left( 3 , 3 \right)$
Case 1: $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$
We have $P = \left( \frac{8}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{22}{3 \sqrt{10}} \right)$ .
Thus, $HP = \frac{5}{\sqrt{10}}$ and $PQ = \frac{10}{3\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{5}{3}$
Case 2: $\left( u , v \right) = \left( 3 , 3 \right)$
We have $P = \left( \frac{7}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{7}{ \sqrt{10}} \right)$ .
Thus, $HP = \frac{3 \sqrt{2}}{\sqrt{10}}$ and $PQ = \frac{3 \sqrt{2}}{\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{9}{5}$
Putting these two cases together, the sum of all possible values of the area of $R$ is $\frac{5}{3} + \frac{9}{5} = \frac{52}{15}$
Therefore, the answer is $\boxed{67}$

Solution:
 We will scale every number up by a factor of $\sqrt{10}.$ This implies our final area will be $\frac{1}{10}$ of the answer we receive.
We have \[FAE\sim EOH \sim IOH\sim JDI\sim KCJ \sim NQK\sim GPF.\] Let $AE=a$ and $FA=b.$ We have \[FP=AE=OH=JC=\frac13ID=a\] and \[PG=AF=EO=OI=KC=\frac13 DJ=b\] As $ABCD$ is a square, we have $AD=DC$ or \[a+2b+3a=3b+a \Rightarrow 3a=b.\] Since $a^2+b^2=10,$ we have \[a=1,b=3.\] We have $\triangle GPL\cong \triangle MQN$ which implies \[MQ=GP=3.\] Denote $QK=x.$ As $\triangle NQK\sim \triangle JDI,$ we have $NQ=3x.$
We have \begin{align*}BM&=BC-(CK+QK+MQ)\\ &=4-x.\end{align*}
In addition, \begin{align*}LB&=AB-(AF+FP+PL)\\&=6-3x.\end{align*}
Since $\triangle LBM\sim \triangle MQN,$ we have \[\frac{LB}{BM}=\frac{MQ}{QN}\Rightarrow \frac{6-3x}{4-x}=\frac{3}{3x}=\frac{1}{x}.\] Simplifying we have \[3x^2-7x+4=0 \Rightarrow x=\frac43, 1.\] We have \begin{align*}[GLMN]&=MN\cdot LM\\ &= 3\sqrt{x^2+1}\cdot \sqrt{10x^2-44x+52}.\end{align*}
Plugging in $x=1,$ we have $[GLMN]=18.$
Plugging in $x=\frac43,$ we have $[GLMN]=\frac{50}{3}.$
The sum of the two possible $R$ s is \[\frac{1}{10}\cdot\frac{104}{3}=\frac{52}{15}.\] Hence, $52+15=\boxed{67}.$


Answer: 1.0

Solution: If we extend the lines, we have the following diagram:  Therefore, we see that the answer is $\boxed{1}.$

Solution: Note that the $y$ -intercepts of line $AB$ and line $CD$ are $0$ and $10$ . If the analytic expression for line $AB$ is $y=k_{1}x$ , and the analytic expression for line $CD$ is $y=k_{2}x+10$ , we have equations: $3k_{1} = 1$ and $2k_{2} + 10 = 9$ . Solving these equations, we can find out that $k_{1} = \frac{1}{3}$ and $k_{2} = -\frac{1}{2}$ . Therefore, we can determine that the expression for line $AB$ is $y=\frac{1}{3}x$ and the expression for line $CD$ is $y=-\frac{1}{2}x + 10$ . When $x=15$ , the coordinates that line $AB$ and line $CD$ pass through are $(15, 5)$ and $\left(15, \frac{5}{2}\right)$ , and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $\left(15, \frac{5}{2}\right)$ is out of the range $3 \leq y \leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$ ). Considering that the $y$ value of the line $CD$ will only decrease, and the $y$ value of the line $AB$ will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is $\boxed{1}.$


Answer: 120.0

Solution:
The numbers that are shaded are the triangular numbers, which are numbers in the form $\frac{(n)(n+1)}{2}$ for positive integers.  They can also be generated by starting with $1$ , and adding $1, 2, 3, 4...$ as in the description of the problem.
Squares that have the same remainder after being divided by $8$ will be in the same column.  Thus, we want to find when the last remainder, from $0$ to $7$ , is found.
So, instead of adding $1, 2, 3, 4, 5, 6, 7, 8, 9, 10...$ , we can effectively either add $1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...$ or subtract $7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...$ if we are only concerned about remainders when divided by $8$ .  We will pick the number that keeps the terms on the list between $1$ and $8$ .  We get:
$1$
$1 + 2 = 3$
$3 + 3 = 6$
$6 - 4 = 2$
$2 + 5 = 7$
$7 - 2 = 5$
$5 - 1 = 4$
$4 + 0 = 4$
$4 + 1 = 5$
$5 + 2 = 7$
$7 - 5 = 2$
$2 + 4 = 6$
$6 - 3 = 3$
$3 - 2 = 1$
$1 - 1 = 0$
Finally, a term with $0$ is found, and checking, all numbers $1$ through $7$ are also on the right side of the list.  This means the last term in our sequence is the first time that column $8$ is shaded.  There are $15$ terms in the sequence, leading to an answer of $\frac{15\cdot 16}{2} = 120$ , which is choice $\boxed{120}$

Solution:
Note that the triangular numbers up to $120$ are $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120$ .  When you divide each of those numbers by $8$ , all remainders must be present.  We first search for number(s) that are evenly divisible by $8$ ; if two such numbers exist, we search for numbers that leave a remainder of $1$ , etc.
Quickly scanning the list, only $6, 10, 28, 36, 66, 78$ and $120$ are even.  That smaller list doesn't have any multiples of $8$ until it hits $120$ .  So $\boxed{120,}$ must be the answer.

Solution: The numbers shaded are triangular numbers of the form $\frac{(n)(n+1)}{2}$ .  For this number to be divisible by $8$ , the numerator must be divisible by $16$ .  Since only one of $n$ and $n+1$ can be even, only one of them can have factors of $2$ .  Therefore, the first time the whole expression is divisible by $8$ is when either $n+1=16$ or when $n=16$ .  This gives $n=15$ as the first time $\frac{n(n+1)}{2}$ is divisible by $8$ , which gives $120$ .  No other triangular number lower than that is divisible by $8$ , and thus the $8^{th}$ column on the checkerboard won't be filled until then.  That gives $\boxed{120}$ as the right answer.


Answer: 10.0

A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$

Solution:
Box P has dimensions $l$ $w$ , and $h$ .
Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]
The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\] so the diameter is \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{10}.\]

Solution:
As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$ , and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$
Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$ . (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$ , and we are solving for $\sqrt{l^2 + w^2}$ . Because \[(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,\] this means that \[l^2 + w^2 = 28^2 - 384 = 400,\] so the space diagonal is $\sqrt{400} = 20$ . Since the diameter of the sphere is $20$ , the radius is $\boxed{10}$ . ~ emerald_block


Answer: 96.0

A rectangular box has integer side lengths in the ratio $1: 3: 4$ . Which of the following could be the volume of the box?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144$

Solution: Let the smallest side length be $x$ . Then the volume is $x \cdot 3x \cdot 4x =12x^3$ . If $x=2$ , then $12x^3 = 96 \implies \boxed{96.}$


Answer: 41.0

A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$

Solution:
Let the height of the box be $x$
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{\left(\frac{x}{2}\right)^2 + 64}$ , and $\sqrt{\left(\frac{x}{2}\right)^2 + 36}$ . Since the area of the triangle is $30$ , the altitude of the triangle from the base with length $10$ is $6$
Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of $10$
We find: \[10 = \sqrt{\left(28+x^2/4\right)}+x/2\]
Solving for $x$ gives us $x=\frac{36}{5}$ . Since this fraction is simplified: \[m+n=\boxed{041}\]

Solution:
We may use vectors. Let the height of the box be $2h$ . Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$ . Let the center point of the bottom face be $P_1$ , the center of the left face be $P_2$ and the center of the front face be $P_3$
We are given that the area of the triangle $\triangle P_1 P_2 P_3$ is $30$ . Thus, by a well known formula, we note that $\frac{1}{2}|\vec{P_1P_2} \text{x} \vec{P_1P_3}|=30$ We quickly attain that $\vec{P_1P_2}=<-6,0,h>$ and $\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry)
Computing the cross product, we find: \[\vec{P_1P_2} x \vec{P_1P_3}=-<6h,8h,48>\]
Thus: \[\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60\] \[h=3.6\] \[2h=7.2\]
\[2h=36/5\]
\[m+n=\boxed{041}\]

Solution:
Let the height of the box be $x$
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{(x/2)^2 + 64}$ , and $\sqrt{(x/2)^2 + 36}$ . Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
The semiperimeter is $\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2$ . Therefore, when we square Heron's formula, we find
\begin{align*}900 &= \frac{1}{2}\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2\right)\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - 10\right)\\&\qquad\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - \sqrt{(x/2)^2 + 64}\right)\\&\qquad\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - \sqrt{(x/2)^2 + 36}\right).\end{align*}
Solving, we get $\boxed{041}$

Solution:
Let half the height be $a$ (we want to find $2a$ ), then we see that the three sides of the triangle are (by Pyth Theorem) $10, \sqrt{a^2+36}, \sqrt{a^2+64}$ . Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is $\frac{a^2}{\sqrt{(a^2+36)(a^2+64)}}$ by the Law of Cosines, meaning that its sine is $\frac{\sqrt{100a^2+2304}}{\sqrt{(a^2+36)(a^2+64)}}$ . Finally, multiply the two square-rooted sides by this sine and one-half, and equate to 30.
You get $\sqrt{25a^2+576} = 30$ , giving $a=\frac{18}{5}$ , so our answer is $\boxed{041}$

Solution:
Let $x$ be $\frac12$ the height of the box. We will solve for $x$ and then multiply by $2$ at the end. The three side lengths of the triangle, by the Pythagorean Theorem, are $\sqrt{x^2+6^2},\sqrt{x^2+8^2},$ and $10$
Heron's formula states that the area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter. Applying difference of squares to make the formula less computational, we get $\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}=\frac{\sqrt{(a^2+b^2+2ab-c^2)(c^2-b^2-a^2+2ab)}}4=\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}4$
Now, we plug in $\sqrt{x^2+6^2}$ for $a$ $\sqrt{x^2+8^2}$ for $b$ , and $10$ for $c$ . This gives us \[\frac{\sqrt{4(x^2+6^2)(x^2+8^2)-(x^2+6^2+x^2+8^2-10^2)^2}}4=30\] \[\sqrt{4x^4+400x^2+4\cdot48^2-4x^4}=120\] \[400x^2+4\cdot48^2=120^2\] \[25x^2+24^2=30^2\] \[25x^2=6^2(5^2-4^2)\] \[25x^2=18^2\] \[5x=18\] \[x=\frac{18}5\]
Now multiplying by $2$ , we get that the height is $\frac{36}5$ , and $m+n=36+5=\boxed{041}$


Answer: 10.0

A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution:
We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.

Solution:
The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\geqslant3$
Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b}  \geq \frac{1}{c}$ .
Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$ , so $a \leq 6$
So we have $a=3, 4, 5$ or $6$
Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$
When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.
When $a=4$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.
When $a=5$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$
When $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.


Answer: 10.0

Solution:
We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.

Solution:
The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\geqslant3$
Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b}  \geq \frac{1}{c}$ .
Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$ , so $a \leq 6$
So we have $a=3, 4, 5$ or $6$
Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$
When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.
When $a=4$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.
When $a=5$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$
When $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.


Answer: 2.0

A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$ . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the whole floor. How many possibilities are there for the ordered pair $(a,b)$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution:
Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$ . With this information we can make the equation:
\begin{eqnarray*} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4a - 4b + 8 \\ ab - 4a - 4b + 8 &=& 0  \end{eqnarray*} Applying Simon's Favorite Factoring Trick , we get
\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}
Since $b > a$ , then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$ , or $(a-4) = 2$ and $(b-4) = 4$ . This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\boxed{2}$


Answer: 26.0

A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$

Solution: The number of tiles the bug visits is equal to $1$ plus the number of times it crosses a horizontal or vertical line.  As it must cross $16$ horizontal lines and $9$ vertical lines, it must be that the bug visits a total of $16+9+1 = \boxed{26}$ squares.

Solution:
We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ feet sections.

Though it may appear that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count the number of squares the line passes through using the diagram, we get $13$ squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of $2 \cdot 13 = \boxed{26}$


Answer: 400.0

A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?
$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

Solution: We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$ . To get the greatest area while keeping a perimeter of $160$ , the sides should all be $40$ . that means an area of $1600$ . Right now, the area is $20 \times 60$ which is $1200$ $1600-1200=400$ which is $\boxed{400}$


Answer: 12.0

A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side.  Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall.  What is the fewest number of posts required to fence an area $36$ m by $60$ m?
[asy] unitsize(12); draw((0,0)--(16,12)); draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4)); label("WALL",(7,4),SE); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$ m by $60$ m rectangle, so the $60$ m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$ m. We can start by counting $60\div12+1$ on the $60$ m fence (since we also count the $0$ m post). Next, we have the two $36$ m fences. There are a total of $36\div12+1-1$ fences on that side since the $0$ m and $60$ m fence posts are also part of the $36$ m fences. So we have $6+2\cdot3=12$ minimum fence posts needed to box a $36$ m by $60$ m grazing area, or answer $\boxed{12}$


Answer: 62.0

A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution: Let the sides of the rectangular parking lot be $a$ and $b$ . Then $a^2 + b^2 = 625$ and $ab = 168$ . Add the two equations together, then factor. \begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \boxed{62}$

Solution: We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy $a^2+b^2=25^2$ . In other words, $(a,b,25)$ is a set of Pythagorean triples. Guessing and checking, we have $(7,24,25)$ as the triplet, as the area is $7 \cdot 24 = 168$ as requested. Therefore, the perimeter is $2(7+24)=\boxed{62}$


Answer: 88.0

A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures $8$ inches high and $10$ inches wide. What is the area of the border, in square inches?
$\textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88$

Solution: In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$ , and the width of the whole frame, $10+2+2 = 14$ . Therefore, the area of the whole figure would be $12 \times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $168-80 = \boxed{88}$


Answer: 15.0

A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds?
$\mathrm{(A)}\frac {1}{8}\qquad \mathrm{(B)}\frac {1}{6}\qquad \mathrm{(C)}\frac {1}{5}\qquad \mathrm{(D)}\frac {1}{4}\qquad \mathrm{(E)}\frac {1}{3}$

Solution: Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ square meters.  Thus the flower beds have a total area of $25$ square meters.  The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters.  The fraction of the yard occupied by the flower beds is $\frac {25}{125} = \boxed{15}$


Answer: 15.0

Solution: Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ square meters.  Thus the flower beds have a total area of $25$ square meters.  The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters.  The fraction of the yard occupied by the flower beds is $\frac {25}{125} = \boxed{15}$


Answer: 79.0

A region $S$ in the complex plane is defined by \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] A complex number $z = x + iy$ is chosen uniformly at random from $S$ . What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78$

Solution: We multiply $z$ and $(\frac{3}{4}+\frac{3}{4}i)$ to get \[(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).\] Since we want to find the probability that this number is in $S$ , we need the real and complex coefficients of this number to be less than or equal to $1$ or greater than or equal to $-1.$ This gives us the equations \[-1\le \frac{3}{4}x-\frac{3}{4}y \le 1\] and \[-1\le \frac{3}{4}x+\frac{3}{4}y\le 1.\] Now, we see that we can solve this by graphing. We can graph our barriers $-1\le x\le 1$ and $-1\le y\le 1$ to form a $2$ by $2$ square centered at the origin. Graphing our two equations gives us the four lines \[x-y=\frac{4}{3},\] \[x-y=-\frac{4}{3},\] \[x+y=\frac{4}{3},\] \[x+y=-\frac{4}{3}.\] The square that is formed is the region that satisfies these four equations. Now, the barriers and this square gives us an octagon as the desired region. The area of this octagon is the total area of the square minus the 4 small triangles on each corner, each with $\frac{2}{9}$ area. Therefore, the octagon has area of $\frac{28}{9}.$ Finally, to find the probability of it working, we find the area of the octagon divided by the area of the entire square which is $\frac{\frac{28}{9}}{4}=\frac{7}{9}$ or $\boxed{79}.$


Answer: 4.0

A region is bounded by semicircular arcs constructed on the side of a square whose sides measure $\frac{2}{\pi}$ , as shown. What is the perimeter of this region?
[asy] size(90); defaultpen(linewidth(0.7)); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle,gray(0.5)); filldraw(arc((1,0),1,180,0, CCW)--cycle,gray(0.7)); filldraw(arc((0,1),1,90,270)--cycle,gray(0.7)); filldraw(arc((1,2),1,0,180)--cycle,gray(0.7)); filldraw(arc((2,1),1,270,90, CCW)--cycle,gray(0.7)); [/asy]
$\textbf{(A) } \frac{4}{\pi}\qquad \textbf{(B) } 2\qquad \textbf{(C) } \frac{8}{\pi}\qquad \textbf{(D) } 4\qquad \textbf{(E) } \frac{16}{\pi}$

Solution:
Since the side of the square is the diameter of the semicircle, the radius of the semicircle is $\frac{1}{2}\cdot\frac{2}{\pi}=\frac{1}{\pi}$
Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is $\frac{1}{2}\cdot2\cdot\pi\cdot\frac{1}{\pi}=1$
Since the desired perimeter is made up of four of these arcs, the perimeter is $4\cdot1=\boxed{4}$


Answer: 39.0

A regular 15-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$
$\textbf{(A)}\; 24 \qquad\textbf{(B)}\; 27 \qquad\textbf{(C)}\; 32 \qquad\textbf{(D)}\; 39 \qquad\textbf{(E)}\; 54$

Solution: From consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence $L=15$ , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjacent side, hence $R = 360^\circ / 15 = 24^\circ$ , the number of degrees between adjacent sides. Therefore the answer is $L + R = 15 + 24 = \boxed{39}$


Answer: 6.0

A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?
$\mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6$

Solution:
$A_{\triangle} = \frac{s_t^2\sqrt{3}}{4}$
$A_{hex} = \frac{6s_h^2\sqrt{3}}{4}$ since a regular hexagon is just six equilateral triangles.
Setting the areas equal, we get:
$s_t^2 = 6s_h^2$
$\left(\frac{s_t}{s_h}\right)^2 = 6$
$\frac{s_t}{s_h} = \sqrt{6}$ , and the answer is $\boxed{6}$


Answer: 5.0

A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$ .) If $m=10$ , what is the value of $n$ [asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 26$

Solution:
To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\frac{(n-2)*180}{n}$ , the measure of the decagon's interior angle is $\frac{8*180}{10} = 144$ degrees.
The regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\boxed{5}$


Answer: 12.0

A regular polygon of $n$ sides is inscribed in a circle of radius $R$ . The area of the polygon is $3R^2$ . Then $n$ equals:
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$

Solution:
Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius $R$ .  Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.
That means each of these angles between the two lines from the center to the two vertices of a side equals $\frac{360}{n}$ degrees.  Thus, the area of the polygon is \[n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2\] Dividing both sides by $R^2$ yields \[\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3\] Multiply both sides by $\frac{2}{n}$ to get \[\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}\] At this point, use trial-and-error for each of the answer choices.  When checking $n = 12$ , the equation results in $\sin(30^{\circ}) = \frac{1}{2}$ , which is correct.  Thus, the answer is $\boxed{12}$


Answer: 8.0

A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8$

Solution:
Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\times m\times3=6m^2$ . Since the customer wants to eat a different dinner in all $365$ days of $2003$ , we must have
\begin{align*} 6m^2 &\geq 365\\ m^2 &\geq 60.83\ldots.\end{align*} Also, year 2003 is not a leap year, because 2003 divided by 4 does not equal an integer.
The smallest integer value that satisfies this is $\boxed{8}$

Solution: Let $m$ denote the number of main courses needed to meet the requirement. Then the number of dinners available is $3\cdot m \cdot 2m = 6m^2$ . Thus $m^2$ must be at least $365/6 \approx 61$ . Since $7^2 = 49<61<64 = 8^2$ $\boxed{8}$

Solution:
Let $x$ be the number of main courses and let $2x$ be the number of appetizers.
Since there are 3 desserts, the number of possible dinner choices would be $2x \cdot x \cdot 3 = 6x^2$ for any number $x$ . Since a year has $365$ days, we can assume that: \begin{align*} 6x^2 \ge 365 \end{align*} \begin{align*} x^2 \ge 61 \end{align*} \begin{align*} x \ge 7.8 \end{align*} The least option that is greater than $7.8$ is $8$ , so the answer is $\boxed{8}$


Answer: 8.0

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution:
Note Euler's formula where $\text{Vertices}+\text{Faces}-\text{Edges}=2$ . There are $12$ faces and the number of edges is $24$ because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $14$ vertices on the figure. Now note that the sum of the degrees of all the points is twice the number of edges. Let $x=$ the amount of vertices with $3$ edges. Now we know $\frac{3x+4(14-x)}{2}=24$ . Solving this system of equations gives $x = 8$ so the answer is $\boxed{8}$ .
~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)

Solution: Let $x$ be the number of vertices with three edges, and $y$ be the number of vertices with four edges. Since there are $\frac{4*12}{2}=24$ edges on the polyhedron, we can see that $\frac{3x+4y}{2}=24$ . Then, $3x+4y=48$ . Notice that by testing the answer choices, (D) is the only one that yields an integer solution for $y$ . Thus, the answer is $\boxed{8}$

Solution:
With $12$ rhombi, there are $4\cdot12=48$ total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have $\dfrac{48}2=24$ total edges.
Let $A$ be the number of vertices with $3$ edges (this is what the problem asks for) and $B$ be the number of vertices with $4$ edges. We have $3A + 4B = 48$
Euler's formula states that, for all convex polyhedra, $V-E+F=2$ . In our case, $V-24+12=2\implies V=14.$ We know that $A+B$ is the total number of vertices as we are given that all vertices are connected to either $3$ or $4$ edges. Therefore, $A+B=14.$
We now have a system of two equations. There are many ways to solve for $A$ ; choosing one yields $A=\boxed{8}$

Solution:
Note that Euler's formula is $V+F-E=2$ . We know $F=12$ from the question. We also know $E = \frac{12 \cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces.  We can solve for the vertices based on this information.
Using the formula we can find: \[V + 12 - 24 = 2\] \[V = 14\] Let $t$ be the number of vertices with $3$ edges and $f$ be the number of vertices with $4$ edges. We know $t+f = 14$ from the question and $3t + 4f = 48$ . The second equation is because the total number of points is $48$ because there are 12 rhombuses of $4$ vertices.
Now, we just have to solve a system of equations. \[3t + 4f = 48\] \[3t + 3f = 42\] \[f = 6\] \[t = 8\] Our answer is simply just $t$ , which is $\boxed{8}$ ~musicalpenguin

Solution:
Each of the twelve rhombi has two pairs of angles across from each other that must be congruent. If both pairs of angles occur at $4$ -point intersections, we have a grid of squares. If both occur at $3$ -point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a $3$ -point intersection and two at a $4$ -point intersection.
Since each $3$ -point intersection has $3$ adjacent rhombuses, we know the number of $3$ -point intersections must equal the number of $3$ -point intersections per rhombus times the number of rhombuses over $3$ . Since there are $12$ rhombuses and two $3$ -point intersections per rhombus, this works out to be:
$\frac{2\cdot12}{3}$
Hence: $\boxed{8}$ ~hollph27
~Minor edits by FutureSphinx

Solution: Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here ), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is $\boxed{8}$

Solution: Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has $8$ triangular faces, which correspond to $\boxed{8}$ vertices on a rhombic dodecahedron that have $3$ edges.


Answer: 8.0

Solution: Let $m$ be the number of $4$ -edge vertices, and $n$ be the number of $3$ -edge vertices. The total number of vertices is $m+n$ . Now, we know that there are $4 \cdot 12 = 48$ vertices, but we have overcounted. We have overcounted $m$ vertices $3$ times and overcounted $n$ vertices $2$ times. Therefore, we subtract $3m$ and $2n$ from $48$ and set it equal to our original number of vertices. \[48 - 3m - 2n = m+n\] \[4m + 3n = 48\] From here, we reduce both sides modulo $4$ . The $4m$ disappears, and the left hand side becomes $3n$ . The right hand side is $0$ , meaning that $3n$ must be divisible by $4$ . Looking at the answer choices, this is only possible for $n = \boxed{8}$


Answer: 14.0

A right circular cone has base radius $r$ and height $h$ .  The cone lies on its side on a flat table.  As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered  at the point where the vertex touches the table.  The cone first returns to its original position on the table after making $17$ complete rotations.  The value of $h/r$ can be written in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime.  Find $m + n$

Solution:
The path is a circle with radius equal to the slant height of the cone, which is $\sqrt {r^{2} + h^{2}}$ . Thus, the length of the path is $2\pi\sqrt {r^{2} + h^{2}}$
Also, the length of the path is 17 times the circumference of the base, which is $34r\pi$ . Setting these equal gives $\sqrt {r^{2} + h^{2}} = 17r$ , or $h^{2} = 288r^{2}$ . Thus, $\dfrac{h^{2}}{r^{2}} = 288$ , and $\dfrac{h}{r} = 12\sqrt {2}$ , giving an answer of $12 + 2 = \boxed{014}$


Answer: 52.0

A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution:
We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.
Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \sqrt{3}$ with 6 cases or $\sqrt{3}, \sqrt{3}, \sqrt{3}$ with 2 cases. This can be repeated on the other base for $16$ cases.
Case 2: The vertices span two bases. WLOG call the only vertex on one of the bases $X$ . Call the closest vertex on the other base $B$ , and label clockwise $C, D, E, F, G$ . We will multiply the following scenarios by $12$ , because the top vertex can have $6$ positions and the top vertex can be on the other base. We can have $XCG, XDF$ , but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is $2$ and the height is $2$ , so $XBE$ is also correct! Those are the only three cases, so there are $12*3=36$ cases for this case.
In total there's $\boxed{052}$ cases.

Solution: If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \cdot 2\cdot 6 = 28$ ways. Then, if one length was $\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\sqrt{3}$ side, and two ways to put the other point. $2 \cdot 12 = 24$ for four ways to put the third point. Adding the number up, the final answer is $24+28 = \boxed{052}.$

Solution:
To start, let's find the distances from any vertex (call it A, doesn't matter since the prism is symmetrical) to all the other 11 vertices. Using Pythagorean relations, we find that the distances are $1$ $\sqrt{3}$ $2$ $\sqrt{3}$ $1$ $\sqrt{7}$ $\sqrt{5}$ $2$ $\sqrt{5}$ $\sqrt{7}$ , and $\sqrt{8}$
We can clearly form an isosceles triangle using any vertex and any two neighboring edges that are equal. There are 12 total vertices, all of which are symmetrical to the vertex A. Thus, for each vertex, we can form 5 isosceles triangles (with neighboring edges $1$ $1$ $2$ $2$ $\sqrt{3}$ $\sqrt{3}$ $\sqrt{5}$ $\sqrt{5}$ , and $\sqrt{7}$ $\sqrt{7}$ ) , so have so far $12 \times 5 = 60$ isosceles triangles.
To check for overcounting, note that isosceles triangles can be split into two major categories: equilateral isosceles, and non-equilateral isosceles.
We only counted non-equilateral isosceles triangles once since there is only one vertex whose two neighboring edges are equal.
But equilateral triangles were counted three times (namely once for each vertex).
By observation, there are only 4 equilateral triangles (1-1-1 side lengths), two on each hexagonal face. Since we counted each two more times than we should of (three instead of once), we will subtract ( $4$ eq. triangles) $\times$ $2$ overcounts per eq. triangle) = $8$ overcounts.
Finally, we have $60 - 8$ overcounts = $\boxed{052}$ isoceles triangles.

Solution:
Start by drawing the hexagonal prism. Start from one of the points, let's call it $A$ . From simple inspection, we can see that each point we choose creates $4$ isosceles triangles; $A$ and the two adjacent points, $A$ and the two points adjacent to those, $A$ and the adjacent point on the other face, and $A$ and the two points adjacent to those. Here, we have $12\cdot4\Rightarrow48$ isosceles triangles.
Next, notice that you can create an isosceles triangle with the bases of the hexagonal prism. Let's say that our base, starting from the very left-most point of the hexagon, has points $A, B, C, D, E, F$ , rotating counter-clockwise. We can create a triangle with either points $\bigtriangleup{BCE}$ or $\bigtriangleup{BCF}$ . WLOG, assuming we use $\bigtriangleup{BCE}$ , we can use the Law of Cosines and find that $\overline{CE}$ $\sqrt{3}$ . Thus, this would mean $\overline{CF}$ is $2$ , which is the exactly the same length as the height of the hexagonal prism. There are $4$ possibilities of this case, so our final solution is $48+4=\boxed{052}$