Answer: 230.0

Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$ . A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$ , which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$ . The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$ , where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$

Solution:
When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane.
In each reflection, the distance traveled by the light is $\sqrt{ (12^2) + (5^2) + (7^2) }$ $\sqrt{218}$ . This happens 12 times, so the total distance is $12\sqrt{218}$ $m=12$ and $n=218$ , so therefore, the answer is $m+n=\boxed{230}$

Solution:
We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided into cubes identical to the one we have. Now let's follow two photons of light that start in $A$ at the same time: one of them will reflect as given in the problem statement, the second will simply fly straight through all cubes. It can easily be seen that at any moment in time the photons are in exactly the same position relative to the cubes they are inside at the moment. In other words, we can take the cube with the first photon, translate it and flip if necessary, to get the cube with the other photon.
It follows that both photons will hit a vertex at the same time, and at this moment they will have travelled the same distance.
Now, the path of the second photon is simply a half-line given by the vector $(12,7,5)$ . That is, the points visited by the photon are of the form $(12t,7t,5t)$ for $t\geq 0$ . We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube).
Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$ . At this moment the second photon will be at the coordinates $(12\cdot 12,7\cdot 12,5\cdot 12)$
Then the distance it travelled is $\sqrt{ (12\cdot 12)^2 + (7\cdot 12)^2 + (5\cdot 12)^2 } = 12\sqrt{12^2 + 7^2 + 5^2}=12\sqrt{218}$ .
And as the factorization of $218$ is $218=2\cdot 109$ , we have $m=12$ and $n=218$ , hence $m+n=\boxed{230}$


Answer: 112.0

Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$

Solution:
For reference, $2\sqrt{65} \approx 16$ , so $\overline{AD}$ is the longest of the four sides of $ABCD$ . Let $h_1$ be the length of the altitude from $B$ to $\overline{AC}$ , and let $h_2$ be the length of the altitude from $D$ to $\overline{AC}$ . Then, the triangle area equation becomes
\[\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP\]
What an important finding! Note that the opposite sides $\overline{AB}$ and $\overline{CD}$ have equal length, and note that diagonal $\overline{DB}$ bisects diagonal $\overline{AC}$ . This is very similar to what happens if $ABCD$ were a parallelogram with $AB = CD = 10$ , so let's extend $\overline{DB}$ to point $E$ , such that $AECD$ is a parallelogram. In other words, \[AE = CD = 10\] and \[EC = DA = 2\sqrt{65}\] Now, let's examine $\triangle ABE$ . Since $AB = AE = 10$ , the triangle is isosceles, and $\angle ABE \cong \angle AEB$ . Note that in parallelogram $AECD$ $\angle AED$ and $\angle CDE$ are congruent, so $\angle ABE \cong \angle CDE$ and thus \[\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB\] Define $\alpha := \text{m}\angle CDB$ , so $180^\circ - \alpha = \text{m}\angle ABD$
We use the Law of Cosines on $\triangle DAB$ and $\triangle CDB$
\[\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha\]
\[14^2 = 10^2 + BD^2 - 20BD\cos\alpha\]
Subtracting the second equation from the first yields
\[260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}\]
This means that dropping an altitude from $B$ to some foot $Q$ on $\overline{CD}$ gives $DQ = \frac{8}{5}$ and therefore $CQ = \frac{42}{5}$ . Seeing that $CQ = \frac{3}{5}\cdot BC$ , we conclude that $\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \frac{56}{5}$ . Then, the area of $\triangle BCD$ is $\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56$ . Since $AP = CP$ , points $A$ and $C$ are equidistant from $\overline{BD}$ , so \[\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56\] and hence \[\left[ABCD\right] = 56 + 56 = \boxed{112}\] -kgator

Solution:
Denote $\angle APB$ by $\alpha$ . Then $\sin(\angle APB)=\sin \alpha = \sin(\angle APD)$ .
Using the formula for the area of a triangle, we get \[\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha ,\] so \[(AP-CP)(BP-DP)=0\] Hence $AP=CP$ (note that $BP=DP$ makes no difference here).
Now, assume that $AP=CP=x$ $BP=y$ , and $DP=z$ . Using the cosine rule for $\triangle APB$ and $\triangle BPC$ , it is clear that \[x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)\] or \begin{align}x^2+y^2=148\end{align}. Likewise, using the cosine rule for triangles $APD$ and $CPD$ \begin{align}\tag{2}x^2+z^2=180\end{align}. It follows that \begin{align}\tag{3}z^2-y^2=32\end{align}. Since $\sin\alpha=\sqrt{1-\cos^2\alpha}$ \[\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}\] which simplifies to \[\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.\] Plugging this back to equations $(1)$ $(2)$ , and $(3)$ , it can be solved that $x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}$ . Then, the area of the quadrilateral is \[x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}\] --Solution by MicGu

Solution:
As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$ , but it's an AIME problem, we can take $AP=CP$ , and assume the other choice will lead to the same result (which is true).
From $AP=CP$ , we have $[DAP]=[DCP]$ , and $[BAP]=[BCP] \implies [ABD] = [CBD]$ , therefore, \begin{align} \nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\ \Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A  \end{align} By Law of Cosines, \begin{align} \nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\ \Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &=  \sqrt{65}\cos A \tag{2} \end{align} Square $(1)$ and $(2)$ , and add them, to get \[\left(\frac 85\right)^2  + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65\] Solve, $\cos C = 3/5  \implies \sin C = 4/5$ \[[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}\] -Mathdummy

Solution: Either $PA=PC$ or $PD=PB$ . Let $PD=PB=s$ . Applying Stewart's Theorem on $\triangle ABD$ and $\triangle BCD$ , dividing by $2s$ and rearranging, \[\tag{1}CP^2+s^2=148\] \[\tag{2}AP^2+s^2=180\] Applying Stewart on $\triangle CAB$ and $\triangle CAD$ \[\tag{3} 5CP^2=3AP^2\] Substituting equations 1 and 2 into 3 and rearranging, $s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}$ . By Law of Cosines on $\triangle APB$ $\cos(\angle APB)=\frac{4\sqrt{65}}{65}$ so $\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}$ . Using $[\triangle ABC]=\frac{ab\sin(\angle C)}{2}$ to find unknown areas, $[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}$

Solution: Now we prove P is the midpoint of $BD$ . Denote the height from $B$ to $AC$ as $h_1$ , height from $D$ to $AC$ as $h_2$ .According to the problem, $AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2$ implies $h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2$ . Then according to basic congruent triangles we get $BP=DP$ Firstly, denote that $CP=a,BP=b,CP=c,AP=d$ . Applying Stewart theorem, getting that $100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c$ , denote $b=5x,c=3x$ Applying Stewart Theorem, getting $260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)$ solve for a, getting $a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}$ Now everything is clear, we can find $cos\angle{BPA}=\frac{4}{\sqrt{65}}$ using LOC, $sin\angle{BPA}=\frac{7\sqrt{65}}{65}$ , the whole area is $\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}$

Solution:
$BP = PD$ as in another solutions.
Let $D'$ be the reflection of $D$ across $C$ .
Let points $E, E',$ and $H$ be the foot of perpendiculars on $AC$ from $D,D'$ , and $B$ respectively. \begin{align*} &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ \Rightarrow &\qquad \triangle ABH =  \triangle CDE =  \triangle CD'E' \\ \Rightarrow &\qquad \angle BAC = \angle ACD' \\ \Rightarrow &\qquad \triangle ABC =  \triangle AD'C \\ \Rightarrow &\qquad BC = AD'. \end{align*} The area of quadrilateral $ABCD$ is equal to the area of triangle $ADD'$ with sides $AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20$ .
The semiperimeter is $s = 17 + \sqrt{65},$ the area \[[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.\]

Solution:
Use your favorite method to get that $P$ is the midpoint of one of the two diagonals (suppose it's the midpoint of $\overline{AC}$ ). From here, let $x=AP=PC, y=BP, z=PD, a=\cos\theta$ where $\theta$ is the angle that the diagonals make. Then we have a system of four equations:
\begin{align*} x^2+y^2+2xya &= 100 \\ z^2+x^2+2xza &= 100 \\ x^2+y^2-2xya &= 196 \\ x^2+z^2-2xza &= 260 \\ \end{align*}
From these equations we get that \begin{align*} xya &= -24 \\ xza &= -40 \\ x^2+y^2-48 &= 10 \\ x^2+z^2-80 &= 10 \end{align*}
From here we can see that $\frac{z}{y}={5}{3}, z^2-y^2=32,$ so $z=5\sqrt{2}, y=3\sqrt{2}.$ Furthermore, this implies $x=\sqrt{130}$ and $xa=-4\sqrt{2},$ which implies $a=\cos\theta=\frac{4}{\sqrt{65}}.$ Then note that the area of the quadrilateral is \[\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.\]

Solution:
Note that $\angle APB = \angle CPD = 180-\angle APD = 180-\angle BPC.$ (All angles are in degrees)
Since $\sin(\theta)=\sin(180-\theta),$ we can use sine area formula to get the following(after some simplifying steps): \begin{align*} BP \times AP + CP \times DP = BP \times PC + AP \times PD. \end{align*} For convenience, let $AP=a, BP=b, CP=c,DP=d.$ The above equation simplifies to: \begin{align*} ab + cd = bc + ad \\ab-ad+cd-bc=0 \\a(b-d)-c(b-d)=0 \\(a-c)(b-d)=0 \end{align*} From here, we see that $a=c$ or $b=d$ . Without loss of generality, let $a=c$ . Since triangles $ABP$ and $CDP$ are obviously not congruent, we see that one triangle is obtuse and the other one is acute.(Refer to the diagram) However, if we drop perpendiculars from $B$ to $AC$ and $D$ to $AC$ , we do get congruent triangles. If the foot of the perpendicular from $B$ is $M$ , and the foot of the perpendicular from $D$ is $N$ , then right triangle $BMP$ is congruent to right triangle $DNP$ .
From here, we see that the altitudes of triangles $ABC$ and $ADC$ to $AC$ are equal. Since they share base $AC$ , their areas are equal. We can use Heron's formula. To not have any fractions, let $AC=2x.$ \begin{align*} \sqrt{(12+x)(12-x)(x+2)(x-2)}=\sqrt{5+\sqrt{65}+x)(5+\sqrt{65}-x)(5-\sqrt{65}+x)(\sqrt{65}-5+x)} \end{align*} Even though this looks bad at first, it actually isn't too complicated to simplify. Expanding the differences of squares and simplifying completely, we get $x^2=32.$ Plugging this $x$ back into the Heron's formula, we get that the area of $ABC$ (or $ADC$ ) is $56$ . Since these triangles have equal area, the area of the quadrilateral is $2 \times 56 = \boxed{112}$ , and we are done. $\blacksquare$

Solution:
Use any method to derive that $P$ is the midpoint of $A$ and $C$ . Now, either construct a parallelogram with as shown in the diagram to the right, or use law of sines. We will take the parallelogram route here, but the same thing can be done with law of sines on triangles $\triangle \textnormal{ABP}$ and $\triangle \textnormal{CPD}$ . Reflect $D$ across $P$ to get $D'$ . Since $CD = AD' = AB = 10$ $\triangle \textnormal{ABD'}$ is isosceles. Thus, $\angle AD'B = \angle ABD'$ , and because $ADCD'$ is a parallelogram (since $AP = PC$ and $DP = PD'$ ), $\angle AD'B = \angle BDC = \angle ABD'$ . So, $\angle ABD = 180 - \angle ABD' = 180 - \angle BDC$ . Now, apply law of cosines on $\triangle \textnormal{ABD}$ and $\triangle \textnormal{CDB}$ . We get:
\begin{align}
100 + BD^2 - 20BD \cos{\angle ABD} &= 100 + BD^2 - 20 BD \cos {(180 - \angle BDC)} = \\
100 + BD^2 + 20 BD \cos{\angle BDC} &= 260 \\
&\textnormal{and} \\
100 + BD^2 - 20 BD \cos{\angle BDC} &= 196 \\
\textnormal{summing }&\textnormal{and simplifying,} \\
BD &= 8\sqrt{2}
\end{align}
Then, applying law of cosines on $\triangle \textnormal{BCD}$ again, we obtain \[100 + 196 - 280 \cos{\angle BCD} = BD^2 = 128 \implies \cos{\angle BCD} = \frac{3}{5} \implies \sin{\angle BCD} = \frac{4}{5}\] Since $AP = PD$ $[ABD] = [BCD] \implies [ABCD] = [ABD] + [BCD] = 2[BCD]$ . Thus, $[ABCD] = 2[BCD] = 2 \cdot \frac{1}{2} \cdot 10 \cdot 14 \sin{\angle BCD} = 140 \cdot \frac{4}{5} = \boxed{112}$


Answer: 180.0

Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution:
According to the problem, we have $AB=AB'=2$ $DC=DC'=3$ $MB=MB'$ $MC=MC'$ , and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$ , we have $BM=MC$ , so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$
This is when we found that points $M$ $C$ $D$ , and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$ . This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$
Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6$
Point $H$ is the midpoint of $B'C'$ , and $MH \perp AD$ $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$
The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}\] \[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\] \[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\] \[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]
Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$

Solution:
Denote by $M$ the midpoint of segment $BC$ .
Let points $P$ and $Q$ be on segment $AD$ , such that $AP = AB$ and $DQ = DC$
Denote $\angle DAM = \alpha$ $\angle BAD = \beta$ $\angle BMA = \theta$ $\angle CMD = \phi$
Denote $BM = x$ . Because $M$ is the midpoint of $BC$ $CM = x$
Because $AM$ is the angle bisector of $\angle BAD$ and $AB = AP$ $\triangle BAM \cong \triangle PAM$ .
Hence, $MP = MB$ and $\angle AMP = \theta$ .
Hence, $\angle MPD = \angle MAP + \angle PMA = \alpha + \theta$
Because $DM$ is the angle bisector of $\angle CDA$ and $DC = DQ$ $\triangle CDM \cong \triangle QDM$ .
Hence, $MQ = MC$ and $\angle DMQ = \phi$ .
Hence, $\angle MQA = \angle MDQ + \angle QMD = \beta + \phi$
Because $M$ is the midpoint of segment $BC$ $MB = MC$ .
Because $MP = MB$ and $MQ = MC$ $MP = MQ$
Thus, $\angle MPD = \angle MQA$
Thus, \[ \alpha + \theta = \beta + \phi . \hspace{1cm} (1) \]
In $\triangle AMD$ $\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta$ .
In addition, $\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi$ .
Thus, \[ \alpha + \beta = \theta + \phi . \hspace{1cm} (2) \]
Taking $(1) + (2)$ , we get $\alpha = \phi$ .
Taking $(1) - (2)$ , we get $\beta = \theta$
Therefore, $\triangle ADM \sim \triangle AMB \sim \triangle MDC$
Hence, $\frac{AD}{AM} = \frac{AM}{AB}$ and $\frac{AD}{DM} = \frac{DM}{CD}$ .
Thus, $AM = \sqrt{AD \cdot AD} = \sqrt{14}$ and $DM = \sqrt{AD \cdot CD} = \sqrt{21}$
In $\triangle ADM$ , by applying the law of cosines, $\cos \angle AMD  = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}$ .
Hence, $\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}$ .
Hence, ${\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}$
Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}
Therefore, the square of ${\rm Area} \ ABCD$ is $\left( 6 \sqrt{5} \right)^2 = \boxed{180}$

Solution:
Claim
In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$
Proof
Let $A = 2\alpha.$ Then $\angle DBC = \angle DCB = \alpha.$
Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the circle.
Then the sum of arcs $\overset{\Large\frown} {BE} + \overset{\Large\frown}{AC} + \overset{\Large\frown}{CD} = 180^\circ.$ \[\overset{\Large\frown} {BE} = 180^\circ – 2\alpha – \overset{\Large\frown}{AC}.\]
Let $E'$ be the point of intersection of the line $CM$ with the circle. $CM$ is perpendicular to $AD, \angle AMC = 90^\circ – \alpha,$ the sum of arcs $\overset{\Large\frown}{A}C +  \overset{\Large\frown}{BE'} = 180^\circ – 2\alpha \implies E'$ coincides with $E.$
The inscribed angles $\angle DEM = \angle DEB, M$ is symmetric to $B$ with respect to $DE, DM = DB.$
Solution
Let $AB' = AB, DC' = DC, B'$ and $C'$ on $AD.$
Then $AB' = 2, DC' = 3, B'C' = 2 = AB'.$
Quadrilateral $ABMC'$ is cyclic.
Let $\angle A = 2\alpha.$ Then $\angle MBC' = \angle MC'B = \alpha.$
Circle $BB'C'C$ centered at $M, BC$ is its diameter, $\angle BC'C = 90^\circ.$ $\angle DMC' = \angle MC'B,$ since they both complete $\angle MC'C$ to $90^\circ.$
$\angle MB'A = \angle MC'D,$ since they are the exterior angles of an isosceles $\triangle MB'C'.$ $\triangle AMB' \sim \triangle MDC'$ by two angles. $\frac {AB'}{MC'} = \frac {MB'}{DC'}, MC' =\sqrt{AB' \cdot C'D} = \sqrt{6}.$
The height dropped from $M$ to $AD$ is $\sqrt{MB'^2 - (\frac{B'C'}{2})^2} =\sqrt{6 - 1} = \sqrt{5}.$
The areas of triangles $\triangle AMB'$ and $\triangle MC'B'$ are equal to $\sqrt{5},$ area of $\triangle MC'D$ is $\frac{3}{2} \sqrt{5}.$
\[\triangle AMB' = \triangle AMB, \triangle MC'D = \triangle MCD \implies\] The area of $ABCD$ is $(1 + 2 + 3) \sqrt{5} = 6\sqrt{5} \implies 6^2 \cdot 5 = \boxed{180}.$

Solution:
Extend $AB$ and $CD$ so they intersect at a point $X$ . Then note that $M$ is the incenter of $\triangle{XAD}$ , implying that $M$ is on the angle bisector of $X$ . Now because $XM$ is both an angle bisector and a median of $\triangle{XBC}$ $\triangle{XBC}$ is isosceles. Then we can start angle chasing:
Let $\angle{BAM}=a, \angle{CDM}=b,$ and $\angle{XBC}=c$ . Then $\angle{AMD}=\pi-(a+b), \angle{ABM}=\pi-c, \angle{DCM}=\pi-c$ , implying that $\angle{BMA}+\angle{CMD}=a+b$ , implying that $2c-(a+b)=(a+b)$ , or that $c=a+b$ . Substituting this into the rest of the diagram, we find that $\triangle{BMA} \sim \triangle{CDM} \sim \triangle{MDA}$
Then $\frac{AB}{BM}=\frac{MC}{CD}$ , or $BM=CM=\sqrt{6}$ . Moreover, $\frac{AB}{AM}=\frac{AM}{AD}$ , or $AM=\sqrt{14}$ . Similarly, $\frac{CD}{MD}=\frac{MD}{AD}$ , or $DM=\sqrt{21}$ . Then using Law of Cosines on $\triangle{AMD}$ , to get that $cos\angle{AMD}=-\frac{\sqrt{6}}{6}$ , or $sin\angle{AMD}=\frac{\sqrt{30}}{6}$
We finish by using the formula $K=\frac{1}{2}absinC$ , as follows:
$[ABCD]=[ABM]+[CDM]+[ADM]=\frac{\frac{\sqrt{30}}{6}(2\sqrt{6}+3\sqrt{6}+7\sqrt{6})}{2}=6\sqrt{5}$
$(6\sqrt{5})^2=\boxed{180}$

Solution:
As shown in paragraph one of solution 4, extending $AB$ and $CD$ to $X$ , we realize that $\triangle{XBC}$ is isosceles, thus $XM \perp BC$ . Let $XB = XC = x$ . And, midpoint $M$ is the incenter of $\triangle{XAD}$ . Construct perpendiculars $ME, MF, MG$ to sides $AD, AX, DX$ respectively (constructing the radii of the incircle). Let $EM = FM = GM = r$ . The semiperimeter $s = \frac{2x + 2 + 3 + 7}{2} = x+6$ . Since $FX$ is the tangent off the incircle, $FX = s - AD = x-1$ . So, $BF = BX - FX = 1$ . Because $\triangle{BFM} \sim \triangle{MFX}$
\[\frac{BF}{FM} = \frac{FM}{FX} \implies {FM}^2 = BF \cdot FX = x - 1 \implies r^2 = x - 1\]
By Heron's formula and the inradius area formula,
\[(x+6)r = \sqrt{(x+6)\cdot 4 \cdot 3 \cdot (x-1)} \implies (x+6)r^2 = 12(x-1) \implies (x+6)(x-1) = 12(x-1) \implies x=6\]
Then, $r^2 = x - 1 = 5 \implies r = \sqrt{5}$ . Finally,
\[[ABCD] = [ABM] + [CDM] + [AMD] = \frac{AB \cdot FM}{2} + \frac{CD \cdot GM}{2} + \frac{AD \cdot EM}{2} = \frac{2r}{2} + \frac{3r}{2} + \frac{7r}{2} = 6r = 6\sqrt{5}\]
Thus, our answer is $(6\sqrt{5})^{2} = \boxed{180}$

Solution:
Let the midpoint of $BC$ be $M$ . Angle-chase and observe that $\Delta AMD~\Delta ABM~\Delta MCD$ . Let $BM=CM=a$ and $AM=x$ and $DM=y$ . As a result of this similarity, we write
\[\dfrac2a=\dfrac a3,\]
which gives $a=\sqrt 6$ . Similarly, we write
\[\dfrac2x=\dfrac x7\]
and
\[\dfrac3y=\dfrac y7\]
to get $x=\sqrt{14}$ and $y=\sqrt{21}$
We now have all required side lengths; we can find the area of $\Delta AMD$ with Heron's formula. Doing so yields $\dfrac72\sqrt5$ . We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between $\Delta AMD$ and $\Delta ABM$ is $\dfrac{\sqrt{14}}7=\sqrt{\dfrac27}$ , and between $\Delta AMD$ and $\Delta MCD$ is $\dfrac{\sqrt{21}}7=\sqrt{\dfrac37}$ . Thus, the area ratios are $\dfrac27$ and $\dfrac37$ , respectively, so adding together we have $\dfrac27+\dfrac37=\dfrac57$ . Multiplying this by our $\dfrac72\sqrt5$ , we have $\dfrac52\sqrt5$ as their total area. Adding this to our original area, we have $\dfrac52\sqrt5+\dfrac72\sqrt5=\sqrt5\left(\dfrac52+\dfrac72\right)=\sqrt5\left(\dfrac{12}2\right)=6\sqrt5$
The square of this is $\boxed{180}$

Solution:
As in solution 4, let $X=AB\cap CD$ , so $M$ is the incenter of $ADX$ and $XB=XC$ . Let $XB=XC=x$ . Then the normalized barycentric coordinates of $B$ $C$ , and $M$ with respect to $ADX$ are $\left[\frac{x}{x+2}:0:\frac{2}{x+2}\right]$ $\left[0:\frac{x}{x+3}:\frac{3}{x+3}\right]$ , and $\left[\frac{x+3}{2x+12}:\frac{x+2}{2x+12}:\frac{7}{2x+12}\right]$ . So we have $\frac{1}{2}\frac{x}{x+2}=\frac{x+3}{2x+12}$ giving $x=6$ . The sidelengths of $ADX$ are thus $AD=7$ $AX=8$ , and $DX=9$ giving $[ADX]=12\sqrt 5$ . Also, we have $[BCX]=\frac{6}{8}\cdot\frac{6}{9}[ADX]=6\sqrt 5$ so that $[ABCD]=[ADX]-[BCX]=6\sqrt 5$ . The area squared is thus $\boxed{180}$ .
~~ golue3120


Answer: 301.0

Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively: \begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos\angle BEC&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\angle CED&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos\angle DEA&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ \end{alignat*} We subtract $(1\star)+(3\star)$ from $(2\star)+(4\star):$ \begin{align*} 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ 2\cdot\cos\theta\cdot59&=22 \\ \cos\theta&=\frac{11}{59}. \end{align*} Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$

Solution:
Let the brackets denote areas.
We find $[ABCD]$ in two different ways:
Equating the expressions for $[ABCD],$ we have \[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.\] Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$

Solution: We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09]* WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \sqrt{59} - x.$ By Pythagorean theorem, we have \[49 - x^2 = 16 - \left(\sqrt{59} - x\right)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.\] Now let $y$ be $C_1D.$ (thus making $C_1B = \sqrt{59} - y$ ). Similarly, we have \[36 - y^2 = 25 - \left(\sqrt{59} - y\right)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.\] We see that $A_1C_1$ , the scaled down diagonal is just $x - y = \frac{11\sqrt{59}}{59},$ which is $\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\frac{11}{59} \cdot 22 = \frac{242}{59},$ making our answer $242+59 = \boxed{301}.$

Solution:
Solution
In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to $\cos \theta,$ where $\theta$ is the acute angle between the diagonals. \begin{align*}  s &= A'B' + B'C' + C'D' + D'A' \\    &= (AB + BC + CD + DA)\cos \theta \\    &= (a + b + c + d)\cos \theta \\    &= 22\cos \theta. \end{align*} In accordance with Claim 2, \begin{align*} 2(ac + bd)\cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*} \[2 \cdot 59 \cos \theta = |13 + 9|.\] \[s = 22\cos \theta = \frac{22 \cdot 11}{59} = \frac{242}{59}.\] Therefore, the answer is $242+59=\boxed{301}.$


Answer: 308.0

Let $ABCD$ be a parallelogram .  Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$

Solution: We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$ . We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$ . Equating the two results gives $\frac{112}{RC} =  \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$


Answer: 1.0

Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$ $\overrightarrow{BB^\prime}$ $\overrightarrow{CC^\prime}$ , and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$ . If $AA^{\prime} = 10$ $BB^{\prime}= 8$ $CC^\prime = 18$ , and $DD^\prime = 22$ and $M$ and $N$ are the midpoints of $A^{\prime} C^{\prime}$ and $B^{\prime}D^{\prime}$ , respectively, then $MN =$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Let $ABCD$ be a unit square with $A(0,0,0)$ $B(0,1,0)$ $C(1,1,0)$ , and $D(1,0,0)$ .  Assume that the rays go in the +z direction.  In this case, $A^\prime(0,0,10)$ $B^\prime(0,1,8)$ $C^\prime(1,1,18)$ , and $D^\prime(1,0,22)$ .  Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}, \frac{1}{2}, 14)$ and $N(\frac{1}{2}, \frac{1}{2}, 15)$ .  The distance $MN$ is $15 - 14 = 1$ , and the answer is $\boxed{1}$


Answer: 5.0

Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$ . Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$ , respectively, with $AE=BF=AG=2$ . Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$ . The area of quadrilateral $EFHG$ is
$\text{(A) } 4\quad \text{(B) } 4.5\quad \text{(C) } 5\quad \text{(D) } 5.5\quad \text{(E) } 6$

Solution:
We note that $ABFG$ is a parallelogram because $AG = BF = 2$ and $AG \parallel BF$ . Using the same reasoning, $GFCD$ is also a parallelogram.
Assume that the height of parallelogram $ABFG$ with respect to base $AB$ is $x$ . Then, the area of parallelogram $ABFG$ is $AB * x$ . The area of triangle $EFG$ is $\frac{AB * x}{2}$ , which is half of the area of parallelogram $ABFG$
Likewise, the area of triangle $FGH$ is half the area of parallelogram $GFCD$
Thus, $[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}$


Answer: 81.0

Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

Solution:
Let $X$ denote the intersection point of the diagonals $AC$ and $BD$ . Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$ , so $XP = XQ = 3$ and $XR = XS = 4$ . Now note that since $\angle APB = \angle ARB = 90^\circ$ , quadrilateral $ARPB$ is cyclic, and so \[XR\cdot XA = XP\cdot XB,\] which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$
Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$ . Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$ , and so \[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15\] Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$ , and so \[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\] The requested answer is $32 + 8 + 41 = \boxed{81}$

Solution: Let $X$ denote the intersection point of the diagonals $AC$ and $BD,$ and let $\theta = \angle{COB}$ . Then, by the given conditions, $XR = 4,$ $XQ = 3,$ $[XCB] = \frac{15}{4}$ . So, \[XC = \frac{3}{\cos \theta}\] \[XB \cos \theta = 4\] \[\frac{1}{2} XB XC \sin \theta = \frac{15}{4}\] Combining the above 3 equations, we get \[\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.\] Since we want to find $d^2 = 4XB^2 =  \frac{64}{\cos^2 \theta},$ we let $x = \frac{1}{\cos^2 \theta}.$ Then \[\frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.\] Solving this, we get $x = \frac{4 + \sqrt{41}}{8},$ so $d^2 = 64x = 32 + 8\sqrt{41} \rightarrow 32+8+41=\boxed{81}$

Solution: Let $X$ be the intersection of diagonals $AC$ and $BD$ . By symmetry $[\triangle XCB] = \frac{15}{4}$ $XQ = 3$ and $XR = 4$ , so now we have reduced all of the conditions one quadrant. Let $CQ = x$ $XC = \sqrt{x^2+9}$ $RB = \frac{4x}{3}$ by similar triangles and using the area condition we get $\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}$ . Note that it suffices to find $XB = \frac{4}{3}\sqrt{x^2+9}$ because we can double and square it to get $d^2$ . Solving for $a = x^2$ in the above equation, and then using $d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow 8+41+32=\boxed{81}$

Solution: Again, Let $X$ be the intersection of diagonals $AC$ and $BD$ . Note that triangles $\triangle QXC$ and $\triangle BXR$ are similar because they are right triangles and share $\angle CXQ$ . First, call the length of $XB = \frac{d}{2}$ . By the definition of an area of a parallelogram, $CQ \cdot 2XB = 15$ , so $CQ = \frac{15}{d}$ . Using similar triangles on $\triangle QXC$ and $\triangle BXR$ $\frac{CQ}{XQ} = \frac{BR}{XR}$ . Therefore, finding $BR$ $BR = \frac{XR}{XQ} \cdot CQ = \frac{4}{3} \cdot \frac{15}{d} = \frac{20}{d}$ . Now, applying the Pythagorean theorem once, we find $(\frac{20}{d}) ^2$ $(4)^2$ $(\frac{d}{2}) ^2$ . Solving this equation for $d^2$ , we find $d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \rightarrow 32+8+41= \boxed{81}$

Solution: Let $BQ = PD = x.$ We know that the area of the parallelogram is $15,$ so it follows that $[\triangle{BCD}] = [\triangle{BAD}] = \tfrac{15}{2}$ and the height of each triangle, which are also the lengths of $QC$ and $AP,$ is $\tfrac{15}{2(x+3)}.$ Suppose that $E = RS \cap BD.$ Because $\angle{BRE} = \angle{CQE}$ and $\angle{BER} = \angle{CQD},$ we have $\triangle{BRE} \sim \triangle{CQE}.$ The length of $CE,$ by the Pythagorean Theorem is $\sqrt{3^2+(\tfrac{15}{2(x+3)})^2}$ and the length of $BR,$ by the Pythagorean Theorem on $\triangle{BRE},$ is $\sqrt{(x+3)^2 - 4}.$ Note that \[\sin{\angle QEC} = \frac{CQ}{CE} = \frac{BR}{BE}\] Substituting in our values, \[\frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}\] To rid unnecessary computation, we let $(x+3)^2 = a.$ The equation simplifies, after cross multiplying, to \[\sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2}\] \[36a^2 - 576a - 15^2\cdot 16 = 0\] \[a^2-16a-100 =0\] By the quadratic formula, $a \in \{\tfrac{16 - \sqrt{656}}{2}, \tfrac{16 + \sqrt{656}}{2}\},$ so we discard the negative solution. The value of $BD^2$ is \[BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}\] and the desired answer is $32+8+41 = \boxed{81}$ ~skyscraper

Solution:
Let the intersection of $RS$ and $BD$ be $X$
$\because$ $\angle APX = \angle DSX$ and $\angle AXP = \angle DXS$ $\triangle APX \sim \triangle DSX$ by $AA$
$\therefore$ $\frac{PA}{DS} = \frac68 = \frac34$ $DS = \frac43 \cdot PA$
By the Pythagorean theorem and the property of projection, $BD^2 = (DS+BR)^2 + RS^2 = 4DS^2 + 64 = 4(\frac43 \cdot PA)^2 + 64 = \frac{64}{9} \cdot PA^2 + 64$ $\frac{64}{9} \cdot PA^2 = BD^2 - 64$
$\because [ABCD] = PA \cdot BD = 15$ $\therefore PA = \frac{15}{BD}$
\[\frac{64}{9} (\frac{15}{BD})^2 = BD^2 - 64\]
\[\frac{1600}{BD^2} = BD^2 - 64\]
\[BD^4 - 64 BD^2 - 1600 = 0\]
\[BD^2 = \frac{64 + \sqrt{64^2 - 4 (-1600)}}{2} = 32 + 8 \sqrt{41}\]
Therefore, the answer is $32 + 8 + 41 = \boxed{81}$


Answer: 7.0

Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$ . Let $N$ be a point on $\overline{AB}$ , and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$ , respectively. Which of the following is closest to the minimum possible value of $PQ$
$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75  \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$

Solution: Let the intersection of $\overline{AC}$ and $\overline{BD}$ be $E$ . Since $ABCD$ is a rhombus, we have $\overline{AC} \perp \overline{BD}$ and $AE = CE = \dfrac{AC}{2} = 8$ . Since $\overline{NQ} \perp \overline{BD}$ , we have $\overline{NQ} \parallel \overline{AC}$ , so $\triangle{BNQ} \sim \triangle{BAE} \sim \triangle{NAP}$ . Therefore, \[\dfrac{NP}{AP} = \dfrac{NP}{8 - NQ} = \dfrac{BE}{AE} = \dfrac{15}{8} \Rightarrow NP = 15 - \dfrac{15}{8} NQ.\] By Pythagorean Theorem, \[PQ^2 = NQ^2 + NP^2 = NQ^2 + \left(15 - \dfrac{15}{8} NQ \right)^2 = \dfrac{289}{64} NQ^2 - \dfrac{225}{4} NQ + 225.\] The minimum value of $PQ^2$ would give the minimum value of $PQ$ , so we take the derivative (or use vertex form) to find that the minimum occurs when $NQ = \dfrac{225 \cdot 8}{289}$ which gives $PQ^2 = \dfrac{225 \cdot 64}{289}$ . Hence, the minimum value of $PQ$ is $\sqrt{\dfrac{225 \cdot 64}{289}} = \dfrac{120}{17}$ , which is closest to $7 \Rightarrow \boxed{7}$

Solution: Let the intersection $\overline{AC}$ and $\overline{BD}$ be $E.$ Let $\overline{PE}=y \implies \overline{AP}=8-y$ and $\overline{EQ}=x \implies \overline {QB}=15-x.$ Since $\triangle NQB \sim \triangle APN,$ we have $\frac{8-y}{x}=\frac{y}{15-x} \implies 120=15y+8x.$ We want to minimize $\overline{PQ}=\sqrt{x^2+y^2}.$ By Cauchy, $17^2(x^2+y^2)=(x^2+y^2)(8^2+15^2) \ge (8x+15y)^2=120^2 \implies \sqrt{x^2+y^2} \ge \frac{120}{17} \approx 7.$ So, choice $\boxed{7}$ is closest to the minimum.


Answer: 113.0

Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

Solution:
Without loss of generality, we assume the length of each side of $ABCD$ is $2$ .
Because $E$ is the midpoint of $CD$ $CE = 1$
Because $ABCD$ is a rhombus, $\angle BCE = 180^\circ - \angle D$
In $\triangle BCE$ , following from the law of sines, \[ \frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} . \]
We have $\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$
Hence, \[ \frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} . \]
By solving this equation, we get $\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}$
Because $AF \perp BF$ \begin{align*} BF & = AB \cos \angle ABF \\ & = 2 \cos \left( 46^\circ - \angle FBC \right) . \end{align*}
In $\triangle BFC$ , following from the law of sines, \[ \frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} . \]
Because $\angle BCF = 180^\circ - \angle BFC - \angle FBC$ , the equation above can be converted as \[ \frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} . \]
Therefore, \begin{align*} \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\ & = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\ & = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\ & = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\ & = - \tan \frac{134^\circ}{2} \\ & = - \tan 67^\circ \\ & = \tan \left( 180^\circ - 67^\circ \right) \\ & = \tan 113^\circ . \end{align*}
Therefore, $\angle BFC = \boxed{113}$

Solution:
Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$
Because $\overline{AB} \parallel \overline{ED}$ , we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$ , so $\triangle ABG \sim \triangle DEG$ by AA.
Because $ABCD$ is a rhombus, $AB = CD = 2DE$ , so $AG = 2GD$ , meaning that $D$ is a midpoint of segment $\overline{AG}$
Now, $\overline{AF} \perp \overline{BE}$ , so $\triangle GFA$ is right and median $FD = AD$
So now, because $ABCD$ is a rhombus, $FD = AD = CD$ . This means that there exists a circle from $D$ with radius $AD$ that passes through $F$ $A$ , and $C$
AG is a diameter of this circle because $\angle AFG=90^\circ$ . This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$ , so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$ , which means that $\angle BFC = \boxed{113}$

Solution: Let $\overline{AC}$ meet $\overline{BD}$ at $O$ , then $AOFB$ is cyclic and $\angle FBO = \angle FAO$ . Also, $AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE$ , so $\frac{AF}{BO} = \frac{AC}{BE}$ , thus $\triangle AFC \sim \triangle BOE$ by SAS, and $\angle OEB = \angle ACF$ , then $\angle CFE = \angle EOC = \angle DAC = 67^\circ$ , and $\angle BFC = \boxed{113}$

Solution:
Observe that all answer choices are close to $112.5 = 90+\frac{45}{2}$ . A quick solve shows that having $\angle D = 90^\circ$ yields $\angle BFC = 135^\circ = 90 + \frac{90}{2}$ , meaning that $\angle BFC$ increases with $\angle D$ .
Substituting, $\angle BFC = 90 + \frac{46}{2} = \boxed{113}$

Solution:
This solution refers to the Diagram section.
We extend $AD$ and $BE$ to point $G$ , as shown below:  We know that $AB=AD=2$ and $CE=DE=1$
By AA Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$ . This implies that $2AD=AG$ and $AD \cong DG$ , so $AG=2AD=2\cdot2=4$ . That is, $D$ is the midpoint of $AG$
Now, let's redraw our previous diagram, but construct a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$ , which is on the circle, as shown below:  Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$
Let's call $\angle CFE = \theta$
Note that $\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$ , So $\angle CDG = 2\angle CFE$
Let $\angle CDG = 2\theta$ . Notice how $\angle CDG$ and $\angle ADC$ are supplementary to each other. We conclude that \begin{align*} 2\theta &= 180-\angle ADC \\ 2\theta &= 180-46 \\ 2\theta &= 134 \\ \theta &= 67.  \end{align*} Since $\angle BFC=180-\theta$ , we have $\angle BFC=180-67=\boxed{113}$

Solution:
If angle $ADC$ was a right angle, it would be much easier. Thus, first pretend that $ADC$ is a right angle. $ABCD$ is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line $AE$ , which is $\sqrt{5}/2$ . We want the measure of angle $BFC$ , so to work closer to it, we should try finding the length of line $BF$ . Angle $FAB$ and angle $ABF$ are complementary. Angle $ABF$ and angle $FBC$ are also complementary. Thus, $\sin FAB=\cos ABF=\sin FBC$ $\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}$ . Since $\sin FAB=1\sqrt{5}$ ,and $AB=1$ $FB=\sin FAB$ . It follows now that $FE=3*\sqrt{5}/10$
Now, zoom in on triangle $BEC$ . To use the Law of Cosines on triangle $FBC$ , we need the length of $FC$ . Use the Law of Cosines on triangle $EFC$ . Cos $E=1/\sqrt{5}$ . Thus, after using the Law of Cosines, $FC=\sqrt{2/5}$
Since we now have SSS on $BEC$ , we can get use the Law of Cosines. $\cos BFC=1/-\sqrt{2}$ $\arccos 1/-\sqrt{2}$ is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. $180-45=135$ . Angle $BFC$ is $135^\circ$
Realize that, around point F, there will always be 3 right angles, regardless of what angle $ADC$ is. There are only two angles that change when $ADC$ changes. Break up angle $BFC$ into angle $BFB'$ , which is always 90 degrees, and angle $B'FC$ , which we have discovered to to be half of $ADC$ . Thus, when angle $ADC$ is 46 degrees, then $B'FC$ will be 23. $23+90=113$ . Angle $BFC$ is $\boxed{113}$ degrees.


Answer: 18.0

Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$

Solution:
It's important to note that $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$
We define $a$ as the length of the side of larger inner square, which is also $EB$ $b$ as the length of the side of the smaller inner square which is also $AE$ , and $s$ as the side length of $ABCD$ . Since we are given that the sum of the areas of the two squares is $\frac{9}{10}$ of the the area of ABCD, we can represent that as $a^2 + b^2 = \frac{9s^2}{10}$ . The sum of the two nonsquare rectangles can then be represented as $2ab  = \frac{s^2}{10}$
Looking back at what we need to find, we can represent $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ as $\dfrac{a^2 + b^2}{ab}$ . We have the numerator, and dividing $\frac{s^2}{10}$ by two gives us the denominator $\frac{s^2}{20}$ . Dividing $\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}$ gives us an answer of $\boxed{018}$

Solution:
Let the side of the square be $1$ .  Therefore the area of the square is also $1$ .
We label $AE$ as $a$ and $EB$ as $b$ .  Notice that what we need to find is equivalent to: $\frac{a^2+b^2}{ab}$ .
Since the sum of the two squares ( $a^2+b^2$ ) is $\frac{9}{10}$ (as stated in the problem) the area of the whole square, it is clear that the
sum of the two rectangles is $1-\frac{9}{10} \implies \frac{1}{10}$ .  Since these two rectangles are congruent, they
each have area: $\frac{1}{20}$ .  Also note that the area of this is $ab$ .  Plugging this into our equation we get:
$\frac{\frac{9}{10}}{\frac{1}{20}} \implies \boxed{018}$

Solution: Let $AE$ be $x$ , and $EB$ be $1$ . Then we are looking for the value $x+\frac{1}{x}$ . The areas of the smaller squares add up to $9/10$ of the area of the large square, $(x+1)^2$ . Cross multiplying and simplifying we get $x^2-18x+1=0$ . Rearranging, we get $x+\frac{1}{x}=\boxed{018}$

Solution: As before, $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ . Let $x$ represent the value of $AE=CF$ . Since $EB=FB=1-x,$ the area of the two rectangles is $2x(1-x)=-2x^2+2x=\frac1{10}$ . Adding $2x^2-2x$ to both sides and dividing by $2$ gives $x^2-x+\frac1{20}=0.$ Note that the two possible values of $x$ in the quadratic both sum to $1,$ like how $AE$ and $EB$ does. Therefore, $EB$ must be the other root of the quadratic that $AE$ isn't. Applying Vietas and manipulating the numerator, we get $\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{\frac{1}{20}}=\frac{1^2-\frac1{10}}{\frac1{20}}=\frac{\frac9{10}}{\frac{1}{20}}=\boxed{018}$

Solution: Let $AE = x$ and $BE = y$ . From this, we get $AB = x + y$ . The problem is asking for $\frac{x}{y} + \frac{y}{x}$ , which can be rearranged to give $\frac{x^2 + y^2}{xy}$ . The problem tells us that $x^2 + y^2 = \frac{9(x+y)^2}{10}$ . We simplify to get $x^2 + y^2 = 18xy$ . Finally, we divide both sides by $xy$ to get $\frac{x^2 + y^2}{xy} = \boxed{018}$ . - Spacesam

Solution: After we get the polynomial $x^2 - 18x + 1,$ we want to find $x + \frac 1 {x}.$ Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \frac 1 {x}.$ Hence $x + \frac 1 {x}$ is just $18$ by Vieta's formula, or $\boxed{018}$

Solution:
We have the equation $x^2 + y^2$ $\frac {9}{10} \cdot (x+y)^2$ . We get $x^2 + y^2 = 18xy$ . We rearrange to get $x^2 + y^2 - 18xy = 0$ . Since the problem only asks us for a ratio, we assume $x$ $1$ . We have $y^2 - 18y + 1$ $0$ . Solving the quadratic yields $9 + 4 \sqrt 5$ and $9 - 4 \sqrt 5$ . It doesn't really matter which one it is, since both of them are positive. We will use $9 + 4 \sqrt 5$
We have $9 + 4 \sqrt 5 + \frac {1}{9+4 \sqrt 5}$ . Rationalizing the denominator gives us $9 + 4 \sqrt 5 + \frac {9 - 4 \sqrt 5}{81-80} = (9 + 4 \sqrt 5) + (9 - 4 \sqrt 5) = 18$ . Our answer is $\boxed{018}$


Answer: 104.0

Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$

Solution:
Notice that \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal
\begin{equation*}
\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.
\end{equation*}
Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is
\begin{equation*}
h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.
\end{equation*}
Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for cones,
\begin{align*}
\frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\
&=\frac13Sr\cdot4.
\end{align*}
Hence, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).

Solution:
Inscribe tetrahedron $ABCD$ in an rectangular prism as shown above.
By the Pythagorean theorem, we note
\[OA^2 + OB^2 = AB^2 = 41,\] \[OA^2 + OC^2 = AC^2 = 80, \text{and}\] \[OB^2 + OC^2 = BC^2 = 89.\]
Solving yields $OA = 4, OB = 5,$ and $OC = 8.$
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of $ABCD.$ We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of $ABCD.$
We know that the distance from all $4$ faces must be the same, so we only need to find the distance from the center to plane $ABC$
Let $O = (0,0,0), A = (4,0,0), B = (0,5,0),$ and $C = (0,0,8).$ We obtain that the plane of $ABC$ can be marked as $\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,$ or $10x + 8y + 5z - 40 = 0,$ and the center of the prism is $(2,\frac{5}{2},4).$
Using the Point-to-Plane distance formula, our distance is
\[d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.\]
Our answer is $20 + 21 + 63 = \boxed{104}.$

Solution:
We use the formula for the volume of iscoceles tetrahedron. $V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}$
Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find \[\cos{\angle ACB} = \frac{80 + 89 - 41}{2\sqrt{80\cdot 89}}= \frac{16}{9\sqrt{5}}.\]
From this, we find \[\sin{\angle ACB} = \sqrt{1-\cos^2{\angle ACB}} = \sqrt{1 - \frac{256}{405}} = \sqrt{\frac{149}{405}}\] and can find the area of $\triangle ABC$ as \[A = \frac{1}{2} \sqrt{89\cdot 80}\cdot \sin{\angle ACB} = 6\sqrt{21}.\]
Let $R$ be the distance we want to find. By taking the sum of (equal) volumes \[[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,\] We have \[V = \frac{4AR}{3}.\] Plugging in and simplifying, we get $R = \frac{20\sqrt{21}}{63}$ for an answer of $20 + 21 + 63 = \boxed{104}$

Solution:
Let $AH$ be perpendicular to $BCD$ that meets this plane at point $H$ .
Let $AP$ $AQ$ , and $AR$ be heights to lines $BC$ $CD$ , and $BD$ with feet $P$ $Q$ , and $R$ , respectively.
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as $A$ , is $A = 6 \sqrt{21}$
Hence, by using this area, we can compute $AP$ $AQ$ and $AR$ .
We have $AP = \frac{2 A}{BC} = \frac{2A}{\sqrt{89}}$ $AQ = \frac{2 A}{CD} = \frac{2A}{\sqrt{41}}$ , and $AR = \frac{2 A}{BC} = \frac{2A}{\sqrt{80}}$
Because $AH \perp BCD$ , we have $AH \perp BC$ . Recall that $AP \perp BC$ .
Hence, $BC \perp APH$ . Hence, $BC \perp HP$
Analogously, $CD \perp HQ$ and $BD \perp HR$
We introduce a function $\epsilon \left( l \right)$ for $\triangle BCD$ that is equal to 1 (resp. -1) if point $H$ and the opposite vertex of side $l$ are on the same side (resp. opposite sides) of side $l$
The area of $\triangle BCD$ is
\begin{align*}
A & = \epsilon_{BC} {\rm Area} \ \triangle HBC
+ \epsilon_{CD} {\rm Area} \ \triangle HCD
+ \epsilon_{BD} {\rm Area} \ \triangle HBD \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot HP
+ \frac{1}{2} \epsilon_{CD} CD \cdot HQ +
\frac{1}{2} \epsilon_{BD} BD \cdot HR \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2}
+ \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\
& \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1)
\end{align*}
Denote $B = 2A$ .
The above equation can be organized as
\begin{align*}
B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
+ \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\
& \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
This can be further reorganized as
\begin{align*}
B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
& = \epsilon_{CD} \sqrt{B^2 - 41 AH^2}
+ \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
Taking squares on both sides and reorganizing terms, we get
\begin{align*}
& 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\
& = \epsilon_{CD} \epsilon_{BD}
\sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} .
\end{align*}
Taking squares on both sides and reorganizing terms, we get \[ - \epsilon_{BC} 2 B \sqrt{B^2 - 89 AH^2} = - 2 B^2 + 189 AH^2 . \]
Taking squares on both sides, we finally get
\begin{align*}
AH & = \frac{20B}{189} \\
& = \frac{40A}{189}.
\end{align*}
Now, we plug this solution to Equation (1). We can see that $\epsilon_{BC} = -1$ $\epsilon_{CD} = \epsilon_{BD} = 1$ .
This indicates that $H$ is out of $\triangle BCD$ . To be specific, $H$ and $D$ are on opposite sides of $BC$ $H$ and $C$ are on the same side of $BD$ , and $H$ and $B$ are on the same side of $CD$
Now, we compute the volume of the tetrahedron $ABCD$ , denoted as $V$ . We have $V = \frac{1}{3} A \cdot AH = \frac{40 A^2}{3 \cdot 189}$
Denote by $r$ the inradius of the inscribed sphere in $ABCD$ .
Denote by $I$ the incenter.
Thus, the volume of $ABCD$ can be alternatively calculated as
\begin{align*}
V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\
& = \frac{1}{3} r \cdot 4A .
\end{align*}
From our two methods to compute the volume of $ABCD$ and equating them, we get
\begin{align*}
r & = \frac{10A}{189} \\
& = \frac{20 \sqrt{21}}{63} .
\end{align*}
Therefore, the answer is $20 + 21 + 63 = \boxed{104}$

Solution:
Consider the following construction of the tetrahedron. Place $AB$ on the floor. Construct an isosceles vertical triangle with $AB$ as its base and $M$ as the top vertex. Place $CD$ on the top vertex parallel to the ground with midpoint $M.$ Observe that $CD$ can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project $AB$ onto the plane of $CD$ , let the minor angle $\theta$ be this discrepancy.
By Median formula or Stewart's theorem, $AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.$ Consequently the area of $\triangle AMB$ is $\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.$ Note the altitude $8$ is also the distance between the parallel planes containing $AB$ and $CD.$
By Distance Formula, \begin{align*} (\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \\ (\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \\ \implies CD \cos{\theta} \sqrt{41} &= 9 \\ \sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}. \end{align*} Then the volume of the tetrahedron is given by $\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.$
The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from $I$ w.r.t each of the faces. If $r$ is the inradius, i.e the distance to the faces, then $\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])$ must the volume. Each face has the same area by SSS congruence, and by Heron's it is $\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.$
Therefore the answer is, $\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.$


Answer: 134.0

Let $ABCD$ be an isosceles trapezoid , whose dimensions are $AB = 6, BC=5=DA,$ and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, n,$ and $p$ are positive integers $n$ is not divisible by the square of any prime , and $k$ and $p$ are relatively prime . Find $k+m+n+p.$

Solution:
Let the radius of the center circle be $r$ and its center be denoted as $O$
Clearly line $AO$ passes through the point of tangency of circle $A$ and circle $O$ . Let $y$ be the height from the base of the trapezoid to $O$ . From the Pythagorean Theorem \[3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.\]
We use a similar argument with the line $DO$ , and find the height from the top of the trapezoid to $O$ $z$ , to be $z = \sqrt {r^2 + 4r}$
Now $y + z$ is simply the height of the trapezoid. Let $D'$ be the foot of the perpendicular from $D$ to $AB$ ; then $AD' = 3 - 2 = 1$ . By the Pythagorean Theorem, $(AD')^2 + (DD')^2 = (AD)^2 \Longrightarrow DD' = \sqrt{24}$ so we need to solve the equation $\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt {24}$ . We can solve this by moving one radical to the other side, and squaring the equation twice to end with a quadratic equation
Solving this, we get $r = \frac { - 60 + 48\sqrt {3}}{23}$ , and the answer is $k + m + n + p = 60 + 48 + 3 + 23 = \boxed{134}$


Answer: 567.0

Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

Solution:
Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below.  From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.
Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\] Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$
Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$
By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\] Solving this system of equations ( $(1)$ and $(2)$ ), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$

Solution:
First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$ , and the length of the altitude from $B$ to $AC$ is also $10$ . Let the foot of this altitude be $F$ , and let the foot of the altitude from $A$ to $BC$ be denoted as $E$ . Then, $\Delta BCF \sim \Delta ACE$ . So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$ . Now, notice that $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$ , where $[ABC]$ denotes the area of triangle $ABC$ . Letting $AB = x$ , this equality becomes $AC = \frac{9x}{5}$ . Also, from $\frac{BC}{AC} = \frac{2}{3}$ , we have $BC = \frac{6x}{5}$ . Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$ , we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Notice that $AC = AF + CF$ , so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$ , and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$ . Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$ . Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$
Now, let the foot of the perpendicular from $A$ to $CD$ be $G$ . Then let $DG = y$ . Let the foot of the perpendicular from $B$ to $CD$ be $H$ . Then, $CH$ is also equal to $y$ . Notice that $ABHG$ is a rectangle, so $GH = x$ . Now, we have $CG = GH + CH = x + y$ . By the Pythagorean theorem applied to $\Delta AGC$ , we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$ . We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$ , so we can plug this into this equation. Solving for $x+y$ , we get $x+y=\frac{63}{2\sqrt{2}}$
Finally, to find $[ABCD]$ , we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$ . The problem asks us for $K \cdot \sqrt{2}$ , which comes out to be $\boxed{567}$

Solution:
Make $AE$ perpendicular to $BC$ $AG$ perpendicular to $BD$ $AF$ perpendicular $DC$
It's obvious that $\triangle{AEB} \sim \triangle{AFD}$ . Let $EB=5x; AB=5y; DF=6x; AD=6y$ . Then make $BQ$ perpendicular to $DC$ , it's easy to get $BQ=18$
Since $AB$ parallel to $DC$ $\angle{ABG}=\angle{BDQ}$ , so $\triangle{ABG} \sim \triangle{BDQ}$ . After drawing the altitude, it's obvious that $FQ=AB=5y$ , so $DQ=5y+6x$ . According to the property of similar triangles, $AG/BQ=BG/DQ$ . So, $\frac{5}{9}=\frac{GB}{(6x+5y)}$ , or $GB=\frac{(30x+25y)}{9}$
Now, we see the $\triangle AEB$ , pretty easy to find that $15^2+(5x)^2=(5y)^2$ , then we get $x^2+9=y^2$ , then express $y$ into $x$ form that $y=\sqrt{x^2+9}$ we put the length of $BG$ back to $\triangle AGB$ $BG^2+100=AB^2$ . So, \[\frac{[30x+25\sqrt{(x^2+9)}]^2}{81}+100=(5\sqrt{x^2+9})^2.\] After calculating, we can have a final equation of $x^2+9=\sqrt{x^2+9}\cdot3x$ . It's easy to find $x=\frac{3\sqrt{2}}{4}$ then $y=\frac{9\sqrt{2}}{4}$ . So, \[\sqrt{2}\cdot K = \sqrt{2}\cdot(5y+5y+6x+6x)\cdot9=\boxed{567}.\] ~bluesoul

Solution:
Let the foot of the altitude from $A$ to $BC$ be $P$ , to $CD$ be $Q$ , and to $BD$ be $R$
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$ . As $\angle QAB = 90^\circ$ , we have that $\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ$ , with the last equality coming from cyclic quadrilateral $APBR$ . Thus, $\triangle QAR \sim \triangle QPA$ and we have that $\frac{AQ}{AR} = \frac{PQ}{PA}$ or that $\frac{18}{10} = \frac{QP}{15}$ , which we can see gives us that $QP = 27$ . Further ratios using the same similar triangles gives that $QR = \frac{25}{3}$ and $RP = \frac{56}{3}$
We also see that quadrilaterals $APBR$ and $ARQD$ are both cyclic, with diameters of the circumcircles being $AB$ and $AD$ respectively. The intersection of the circumcircles are the points $A$ and $R$ , and we know $DRB$ and $QRP$ are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center $A$ taking $\triangle APQ$ to $\triangle APD$ . Because we know a lot about $\triangle APQ$ but very little about $\triangle APD$ and we would like to know more, we wish to find the ratio of similitude between the two triangles.
To do this, we use the one number we have for $\triangle APD$ : we know that the altitude from $A$ to $BD$ has length $10$ . As the two triangles are similar, if we can find the height from $A$ to $PQ$ , we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$ . Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$ . Then, let $QH = x$ and thus $HP=27-x$ . We then have by the Pythagorean Theorem on $\triangle AQH$ and $\triangle APH$ \begin{align*} 15^2 - x^2 &= 18^2 - (27-x)^2 \\ 225 - x^2 &= 324 - (x^2-54x+729) \\ 54x &= 630 \\ x &= \frac{35}{3}. \end{align*} Then, $RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}$ . This gives us then from right triangle $\triangle ARH$ that $AH = \frac{20\sqrt{2}}{3}$ and thus the ratio of $\triangle APQ$ to $\triangle ABD$ is $\frac{3\sqrt{2}}{4}$ . From this, we see then that \[AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}\] and \[AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.\] The Pythagorean Theorem on $\triangle AQD$ then gives that \[QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.\] Then, we have the height of trapezoid $ABCD$ is $AQ = 18$ , the top base is $AB = \frac{45\sqrt{2}}{4}$ , and the bottom base is $CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}$ . From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}$ , so the answer is $K\sqrt{2} = \boxed{567}$

Solution:
Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$ , respectively.
Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$
Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: \begin{align*} \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square \end{align*} Let $AD=a$ . We obtain from the similarities $AB = \frac{5a}{6}$ and $AC=BD=\frac{3a}{2}$
By Ptolemy, $\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD$ , so $\frac{5a^2}{4} = \frac{5a}{6} \cdot CD$
We obtain $CD=\frac{3a}{2}$ , so $DF=\frac{CD-AB}{2}=\frac{a}{3}$
Applying the Pythagorean theorem on $\triangle ADF$ , we get $324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}$
Thus, $a=\frac{27}{\sqrt{2}}$ , and $[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}$ , yielding $\sqrt2\cdot[ABCD]=\boxed{567}$

Solution:
Let $AD=BC=a$ . Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$ $F$ be the foot of the perpendicular from $A$ to line $BC$ , and $H$ be the foot of the perpendicular from $A$ to $DC$
Note that $\triangle CBG\sim\triangle CAF$ , and we get that $\frac{10}{15}=\frac{a}{AC}$ . Therefore, $AC=\frac32 a$ . It then follows that $\triangle ABF\sim\triangle ADH$ . Using similar triangles, we can then find that $AB=\frac{5}{6}a$ . Using the Law of Cosines on $\triangle ABC$ , We can find that the $\cos\angle ABC=-\frac{1}{3}$ . Since $\angle ABF=\angle ADH$ , and each is supplementary to $\angle ABC$ , we know that the $\cos\angle ADH=\frac{1}{3}$ . It then follows that $a=\frac{27\sqrt{2}}{2}$ . Then it can be found that the area $K$ is $\frac{567\sqrt{2}}{2}$ . Multiplying this by $\sqrt{2}$ , the answer is $\boxed{567}$

Solution: Draw the distances in terms of $B$ , as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$ . As a result, let $AB=u$ , then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$ . The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$ . By angle subtraction, $\cos(180-\theta)=-\cos\theta$ . Therefore, $AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}$ and $AD=BC=\frac{27}{\sqrt{2}}$ . By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}$ which $\sqrt{2}\cdot k=\boxed{567}$

Solution:
 Let the points formed by dropping altitudes from $A$ to the lines $BC$ $CD$ , and $BD$ be $E$ $F$ , and $G$ , respectively.
We have \[\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB\] and \[BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.\] For convenience, let $AB = 5x$ . By Heron's formula on $\triangle ABD$ , we have sides $5x,6x,9x$ and semiperimeter $10x$ , so \[\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},\] so $AB = 5x = \frac{45}{2\sqrt{2}}$
Then, \[BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}\] and \[\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.\] Finally, recalling that $ABCD$ is isosceles, \[K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},\] so $\sqrt{2}\cdot K = \boxed{567}$

Solution:
Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$ .
Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$ . Denote $\theta = \angle{CBH}.$ It is clear that \[BG' = AG, CH = AF, \triangle CBH \ =\triangle ADF,\] the area of $ABCD$ is equal to the area of the rectangle $AFCH.$
The problem is reduced to finding $AH$
In triangle $ABC$ all altitudes are known: \[AB : BC : AC =  \frac{1}{CH}\ : \frac{1}{AE}\ : \frac{1}{BG'}\ =\] \[= \frac{1}{AF}\ :  \frac{1}{AE}\ :  \frac{1}{AG}\ = 5 : 6 : 9.\] We apply the Law of Cosines to $\triangle ABC$ and get $:$ \begin{align*} 2\cdot AB\cdot BC \cdot \cos\theta = AC^2 – AB^2 – BC^2, \end{align*} \begin{align*} 2\cdot 5\cdot 6\cdot \cos\theta  = 60 \cos\theta  = 9^2 – 5^2 – 6^2 = 20, \cos\theta  =\frac{1}{3}. \end{align*} \begin{align*} BH = BC \cos\theta = \frac{BC}{3}.\end{align*} We apply the Pythagorean Law to $\triangle HBC$ and get $:$ \begin{align*} HC^2 = 18^2 = BC^2 – BH^2 = 9\cdot BH^2 – BH^2 = 8 BH^2.\end{align*} \begin{align*} BH =  \frac{9}{\sqrt2}, AH = (\frac{5}{2} + 1)\cdot BH =  \frac{63}{2\cdot \sqrt2}. \end{align*} Required area is \begin{align*}  K =  \frac{63}{2\cdot \sqrt{2}} \cdot 18 =  \frac{567}{\sqrt{2}} \implies \sqrt{2} K=\boxed{567}

Solution:
Let $F$ be on $DC$ such that $AF \| DC$ . Let $G$ be on $BD$ such that $AG \| BD$
Let $m$ be the length of $AB$ . Let $n$ be the length of $AD$
The area of $\triangle ABD$ can be expressed in three ways: $\frac{1}{2}(15)(BC) = \frac{1}{2}(15)(n)$ $\frac{1}{2}(18)(m)$ , and $\frac{1}{2}(10)(BD)$
\[\frac{1}{2}(15)(n) = \frac{1}{2}(18)(m)\] \[15n = 18m\] \[5n = 6m\] \[n = \frac{6}{5}m\]
Now, $BD = BG + GD = \sqrt{m^2-100} + \sqrt{n^2-100}$ . We can substitute in $n = \frac{6}{5}m$ to get
$BD = \sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}$
We have \[\frac{1}{2}(10)\left(\sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}\right) = \frac{1}{2}(18)(m)\] After a fairly straightforward algebraic bash, we get $m = \frac{45\sqrt{2}}{4}$ , and $n = (\frac{6}{5})(\frac{45\sqrt{2}}{4}) = \frac{27\sqrt{2}}{2}$ . By the Pythagorean Theorem on $\triangle ADF$ $DF^2 = n^2 - 18^2 = \frac{729}{2} - 324 = \frac{81}{2}$ , and $DF = \frac{9\sqrt{2}}{2}$
Thus, $DC = 2DF + AB = 9\sqrt{2}+\frac{45\sqrt{2}}{4} = \frac{81\sqrt{2}}{4}$ . Therefore, $K = \frac{1}{2}(\frac{45\sqrt{2}}{4}+\frac{81\sqrt{2}}{4}) \cdot 18 = \frac{63\sqrt{2}}{2} \cdot 18 = \frac{567\sqrt{2}}{2}$ . The requested answer is $K \cdot \sqrt{2} = \boxed{567}$


Answer: 32.0

Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$ . The diagonals have length $10\sqrt {21}$ , and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$ , respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$ . The distance $EF$ can be expressed 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:
Key observation. $AD = 20\sqrt{7}$
Proof 1. By the triangle inequality , we can immediately see that $AD \geq 20\sqrt{7}$ . However, notice that $10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}$ , so by the law of sines, when $AD = 20\sqrt{7}$ $\angle ACD$ is right and the circle centered at $A$ with radius $10\sqrt{21}$ , which we will call $\omega$ , is tangent to $\overline{CD}$ . Thus, if $AD$ were increased, $\overline{CD}$ would have to be moved even farther outwards from $A$ to maintain the angle of $\frac{\pi}{3}$ and $\omega$ could not touch it, a contradiction.
Proof 2. Again, use the triangle inequality to obtain $AD \geq 20\sqrt{7}$ . Let $x = AD$ and $y = CD$ . By the law of cosines on $\triangle ADC$ $2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0$ . Viewing this as a quadratic in $y$ , the discriminant $\Delta$ must satisfy $\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}$ . Combining these two inequalities yields the desired conclusion.
This observation tells us that $E$ $A$ , and $D$ are collinear, in that order.
Then, $\triangle ADC$ and $\triangle ACF$ are $30-60-90$ triangles. Hence $AF = 15\sqrt {7}$ , and
Finally, the answer is $25+7=\boxed{032}$

Solution:
Extend $\overline {AB}$ through $B$ , to meet $\overline {DC}$ (extended through $C$ ) at $G$ $ADG$ is an equilateral triangle because of the angle conditions on the base.
If $\overline {GC} = x$ then $\overline {CD} = 40\sqrt{7}-x$ , because $\overline{AD}$ and therefore $\overline{GD}$ $= 40\sqrt{7}$
By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2}$ ,
and $\overline{CF} = \frac{40\sqrt{21} - \sqrt{3}x}{2}$
Similarly $CAF$ is a 30-60-90 triangle and thus $\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}$
Equating and solving for $x$ $x = 30\sqrt{7}$ and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5\sqrt{7}$
$\overline{ED}-\overline{FD} = \overline{EF}$
$30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7}$ and $25 + 7 = \boxed{032}$


Answer: 484.0

Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution:
Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$ . Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram , and so $\triangle ABC \cong \triangle CFA$ . Also, as $AC \parallel DE$ , it follows that $\triangle ABC \sim \triangle EFD$
By the Law of Cosines $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^{\circ} = 49 \Longrightarrow AC = 7$ . Thus the length similarity ratio between $\triangle ABC$ and $\triangle EFD$ is $\frac{AC}{ED} = \frac{7}{15}$
Let $h_{ABC}$ and $h_{BDE}$ be the lengths of the altitudes in $\triangle ABC, \triangle BDE$ to $AC, DE$ respectively. Then, the ratio of the areas $\frac{[ABC]}{[BDE]} = \frac{\frac 12 \cdot h_{ABC} \cdot AC}{\frac 12 \cdot h_{BDE} \cdot DE} = \frac{7}{15} \cdot \frac{h_{ABC}}{h_{BDE}}$
However, $h_{BDE} = h_{ABC} + h_{CAF} + h_{EFD}$ , with all three heights oriented in the same direction. Since $\triangle ABC \cong \triangle CFA$ , it follows that $h_{ABC} = h_{CAF}$ , and from the similarity ratio, $h_{EFD} = \frac{15}{7}h_{ABC}$ . Hence $\frac{h_{ABC}}{h_{BDE}} = \frac{h_{ABC}}{2h_{ABC} + \frac {15}7h_{ABC}} = \frac{7}{29}$ , and the ratio of the areas is $\frac{7}{15} \cdot \frac 7{29} = \frac{49}{435}$ . The answer is $m+n = \boxed{484}$


Answer: 0.0

Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$ . The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\  \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$

Solution:
The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$ , and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure $180^\circ - 108^\circ = 72^\circ$ . The base angles are equal, so the triangles must be isosceles.
Let one of the sides of the pentagon have length $x_1$ (and the others $x_2, x_3, x_4, x_5$ ). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length $\frac{x_1}{2} \sec 72^\circ$ , and so the two sides together have length $x_1 \sec 72^\circ$ . To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get $(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ$ (because the perimeter of the pentagon is $1$ ). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is $\boxed{0}$

Solution: The extreme case, that results in the minimum and/or maximum, would probably be a pentagon that approaches a degenerate pentagon. However, due to the way the problem is phrased, we know there exists a minimum and maximum; therefore, we can reasonably assume that the star's perimeter is constant, and answer with $\boxed{0}$


Answer: 11.0

Let $ABCDEF$ be a regular hexagon . Let $G$ $H$ $I$ $J$ $K$ , and $L$ be the midpoints of sides $AB$ $BC$ $CD$ $DE$ $EF$ , and $AF$ , respectively. The segments $\overline{AH}$ $\overline{BI}$ $\overline{CJ}$ $\overline{DK}$ $\overline{EL}$ , and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Without loss of generality, let $BC=2.$
Note that $\angle BMH$ is the vertical angle to an angle of the regular hexagon, so it has a measure of $120^\circ$
Because $\triangle ABH$ and $\triangle BCI$ are rotational images of one another, we get that $\angle{MBH}=\angle{HAB}$ and hence $\triangle ABH \sim \triangle BMH \sim \triangle BCI$
Using a similar argument, $NI=MH$ , and
\[MN=BI-NI-BM=BI-(BM+MH).\]
Applying the Law of cosines on $\triangle BCI$ $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}$
\begin{align*}\frac{BC+CI}{BI}&=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH} \\ BM+MH&=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}} \\ MN&=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}} \\ \frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}&=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}
Thus, the answer is $4 + 7 = \boxed{011}.$

Solution:
We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$
then $B$ is at $(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$C$ is at $(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
$D$ is at $(\cos(180^\circ),\sin(180^\circ))=(-1,0)$
\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\ &I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}
Line $AH$ has the slope of $-\frac{\sqrt{3}}{2}$ and the equation of $y=-\frac{\sqrt{3}}{2}(x-1)$
Line $BI$ has the slope of $\frac{\sqrt{3}}{5}$ and the equation $y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)$
Let's solve the system of equation to find $M$
\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\ -5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\ x&=\frac{1}{7} \\ y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}
Finally,
\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\ &\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}
Thus, the answer is $\boxed{011}$


Answer: 55.0

Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$
$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

Solution:
Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively.
The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles.
We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively.
By equilateral triangles and segment addition, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{55}.$

Solution: Let the length $AB=x, BC=y.$ Then, we have \begin{align*} (y+2x)^2\cdot\frac{\sqrt 3}{4}&=324\sqrt3, \\ (x+2y)^2\cdot\frac{\sqrt 3}{4}&=192\sqrt3. \end{align*} We get \begin{align*} y+2x&=36, \\ x+2y&=16\sqrt3. \end{align*} We want $3x+3y,$ and it follows that \[3x+3y=(y+2x)+(x+2y)=36+16\sqrt3.\] Finally, the answer is $36+16+3=\boxed{55}.$


Answer: 17.0

Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$

Solution: We can iteratively calculate out the first few $A_i$ and $B_i$ . We know that $A_0 = (0,0)$ and the line through $A_0$ and $B_1$ needs to make a $60^{\circ}$ angle with the x-axis (because the triangle is equilateral). The equation of a line that makes an angle $\theta$ with the x-axis and passes through the origin has equation $\tan(\theta)x$ , so the line through $A_0$ and $B_1$ has equation $\sqrt{3}x$ . We set this equal to $\sqrt{x}$ (to find where the two curves intersect) and, when solving, find that $x = \frac{1}{3}$ . Therefore, $B_1 = (\frac{1}{3}, \frac{\sqrt{3}}{3})$ . By equilateral triangle properties, we must then have that $A_1 = (\frac{2}{3}, 0)$ . To find $B_2$ , we find the equation of the line through $A_1$ that makes a $60^{\circ}$ angle with the $x$ -axis. This is the same line as the one through the origin (which we already found) shifted $\frac{2}{3}$ to the right, so it has equation $y = \sqrt{3}(x - \frac{2}{3})$ . Setting this equal to $\sqrt{x}$ and solving for $x$ , we get $x = \frac{4}{3}$ , so $B_2 = (\frac{4}{3}, \frac{2\sqrt{3}}{3})$ . By equilateral triangle properties, we have that $A_2 = (2, 0)$ . Repeating this process, we find that $B_3 = (3, \sqrt{3})$ and $A_3 = (4,0)$ . At this point, we notice that the $y$ -coordinate of each $B_i$ is $\frac{\sqrt{3}}{3}$ more than that of $B_{i-1}$ , so $B_4 = (\frac{16}{3}, \frac{4\sqrt{3}}{3})$ and $A_4 = (\frac{20}{3}, 0)$ . It may be helpful to continue calculating out the $A_i$ and $B_i$ if the pattern in the $x$ -coordinates of the $A_i$ isn't visible yet. At this point, it can be noticed that the differences between the x coordinates of consecutive $A_i$ form an arithmetic sequence ( $x_{A_1} - x_{A_0} = \frac{2}{3}$ $x_{A_2} - x_{A_1} = \frac{4}{3}$ , etc.). Additionally, the numerator of each of the fractions is a consecutive even number, and the sum of the first $n$ even numbers is $n(n+1)$ . Thus, we must find the least integral solution to $\frac{n(n+1)}{3} \geq 100$ . This least solution is $\boxed{17}$


Answer: 17.0

Solution:
Let $a_n=|A_{n-1}A_n|$ . We need to rewrite the recursion into something manageable. The two strange conditions, $B$ 's lie on the graph of $y=\sqrt{x}$ and $A_{n-1}B_nA_n$ is an equilateral triangle, can be compacted as follows: \[\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1\] which uses $y^2=x$ , where $x$ is the height of the equilateral triangle and therefore $\frac{\sqrt{3}}{2}$ times its base.
The relation above holds for $n=k$ and for $n=k-1$ $(k>1)$ , so \[\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=\] \[=\left(\frac{a_k}{2}+a_{k-1}+a_{k-2}+\cdots+a_1\right)-\left(\frac{a_{k-1}}{2}+a_{k-2}+a_{k-3}+\cdots+a_1\right)\] Or, \[a_k-a_{k-1}=\frac23\] This implies that each segment of a successive triangle is $\frac23$ more than the last triangle. To find $a_{1}$ , we merely have to plug in $k=1$ into the aforementioned recursion and we have $a_{1} - a_{0} = \frac23$ . Knowing that $a_{0}$ is $0$ , we can deduce that $a_{1} = 2/3$ .Thus, $a_n=\frac{2n}{3}$ , so $A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}$ . We want to find $n$ so that $n^2<300<(n+1)^2$ $n=\boxed{17}$ is our answer.

Solution:
Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the $x$ axis and denote the resulting heights $h_n$ and $h_{n+1}$ . From 30-60-90 rules, the distance between the points where these altitudes meet the x-axis is \[\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} = \frac{h_{n+1}+h_n}{\sqrt{3}}\]
But the square root curve means that this distance is also expressible as $h_{n+1}^2-h_n^2$ (the $x$ coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by $h_{n+1}+h_n$ leaves $h_{n+1}-h_n=\frac{1}{\sqrt{3}}$ . So the difference in height of successive triangles is $\frac{1}{\sqrt{3}}$ , meaning their bases are wider by $2/3$ units each time. From here, one can proceed as in Solution 1 to arrive at $n=\boxed{17}$

Solution:
Note that $A_{1}$ is of the form $(2x,0)$ for some $x$ , and thus $B_{1}$ is of the form $(x, x \sqrt{3}).$ Then, we are told that $B_{1}$ lies on the graph of $y = \sqrt{x}$ , so \[(x \sqrt{3})^{2} = x.\] Solving for x, we get that $x = \frac{1}{3},$ and so $A_{1} = (2/3,0)$
Now, similarly to before, let $|A_{1}A_{2}|=2y.$ Then, $B_{2} = (y+\frac{2}{3},y \sqrt{3})$ , and so \[(y \sqrt{3})^{2} = (y+\frac{2}{3}).\] Solving using the quadratic formula gives \[y = \frac{-(-1) \pm \sqrt{1^{2}-4(3)(-\frac{2}{3})}}{6} = \frac{1 \pm \sqrt{1^{2}+8}}{6} = \frac{2}{3}.\] Then, $2y = \frac{4}{3},$ so $A_{2} = (2,0).$
In general, if $A_{n} = (a_{n},0)$ for all integers $n$ , and $|A_{n}A_{n+1}| = 2x$ for some real number $x$ , we have the following equation for $x$ \[(x \sqrt{3})^{2} = x+a_{n},\] which give us when plugged into the quadratic formula gives \[x = \frac{1 \pm \sqrt{1+12a_{n}}}{6},\] but since $x$ must be positive, we have that \[x = \frac{1 + \sqrt{1+12a_{n}}}{6},\] and so \[a_{n+1} = a_{n}+2x = a_{n}+\frac{1 + \sqrt{1+12a_{n}}}{3}.\] Computing a few terms of $a_{n}$ using this method gives $a_{3} = 4,$ $a_{4} = \frac{20}{3},$ and $a_{5} = 10$
Notice how all our $a_{n}$ terms so far are rational, even though there is an abundance of radicals in the recurrence. This motivates us to look at our discriminants in the quadratic formula that is solved for $x$
The discriminant of \[(x \sqrt{3})^{2} = x+a_{2}\] is \[1+12 \cdot \frac{2}{3} = 9.\] Similarly, the discriminant of \[(x \sqrt{3})^{2} = x+a_{3}\] is $1+12 \cdot a_{3} = 25,$ and $1+12 \cdot a_{4} = 49$
Note how our results keep coming out as the squares of the odd integers. Moreover, it seems that \[1+12a_{n} = (2n+1)^{2}.\] We will prove this with induction. The base case, $n=2$ , we have already verified.
Now, for the Inductive step, assume that \[1+12a_{n-1} = (2n-1)^{2}.\] for some integer $n-1$ . We will prove that this is true for $n$ as well.
Plugging this into our recurrence formula gives us \[a_{n} = a_{n}+2x = a_{n-1}+\frac{1 + \sqrt{1+12a_{n-1}}}{3}\] \[a_{n} = \frac{(2n-1)^{2}-1}{12} + \frac{1 + (2n-1)}{3}\] \[a_{n} = \frac{4n^{2}-4n+1-1}{12} + \frac{8n}{12}\] \[a_{n} = \frac{4n^{2}+4n+1-1}{12}\] \[a_{n} = \frac{(2n+1)^{2}-1}{12}.\] Therefore, we have proved our claim. Now, we have that $|A_{0}A_{n}|=a_{n},$ so we just need the least integer $n$ so that \[a_{n} > 100,\] or \[(2n+1)^{2} > 1201.\] Then, we see that $35^{2} = 1225$ is the smallest odd square larger than $1201.$ Therefore, we have $2n+1=35$ , so $n = \boxed{17}.$


Answer: 37.0

Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon.  Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively.  For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ .  Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively.  If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers.  Find $m + n$

Solution:
We use coordinates. Let the octagon have side length $2$ and center $(0, 0)$ . Then all of its vertices have the form $(\pm 1, \pm\left(1+\sqrt{2}\right))$ or $(\pm\left(1+\sqrt{2}\right), \pm 1)$
By symmetry, $B_{1}B_{3}B_{5}B_{7}$ is a square. Thus lines $\overleftrightarrow{B_{1}B_{3}}$ and $\overleftrightarrow{B_{5}B_{7}}$ are parallel, and its side length is the distance between these two lines. However, this is given to be the side length of the octagon, or $2$
Suppose the common slope of the lines is $m$ and let $m=\tan\theta$ . Then, we want to find \[\cos 2\left(90-\theta\right)=2\cos^{2}\left(90-\theta\right)-1=2\sin^{2}\theta-1.\]
It can easily be seen that the equations of the lines are \begin{align*} B_{1}B_{3}: y-mx+m\left(1+\sqrt{2}\right)=0 \\ B_{5}B_{7}: y-mx-m\left(1+\sqrt{2}\right)=0.\end{align*} By the distance between parallel lines formula , a corollary of the point to line distance formula , the distance between these two lines is \[\frac{|c_{2}-c_{1}|}{\sqrt{a^{2}+b^{2}}}=\frac{2m\left(1+\sqrt{2}\right)}{\sqrt{m^{2}+1}}.\] Since we want this to equal $2$ , we have \begin{align*}\frac{2m\left(1+\sqrt{2}\right)}{\sqrt{m^{2}+1}}&=2 \\ 4m^{2}\left(3+2\sqrt{2}\right)&=4m^{2}+4 \\ \left(12+8\sqrt{2}\right)m^{2}&=4m^{2}+4 \\ \left(8+8\sqrt{2}\right)m^{2}&=4 \\ m^{2}&=\frac{4}{8+8\sqrt{2}} \\ \Rightarrow m^{2}=\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}&=\frac{1}{2+2\sqrt{2}}.\end{align*} Since $\sin^{2}\theta+\cos^{2}\theta=1,$ we have $\sin^{2}\theta=\frac{1}{3+2\sqrt{2}}$ . Thus \[2\sin^{2}\theta-1=\frac{2}{3+2\sqrt{2}}-1=\frac{-1-2\sqrt{2}}{3+2\sqrt{2}}=\frac{\left(-1-2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}=\frac{5-4\sqrt{2}}{1}=5-\sqrt{32}.\] The answer is $\boxed{037}$

Solution:
Let $\theta=\angle M_1 M_3 B_1$ . Thus we have that $\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta$
Since $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ is a regular octagon and $B_1 B_3 = A_1 A_2$ , let $k=A_1 A_2 = A_2 A_3 = B_1 B_3$
Extend $\overline{A_1 A_2}$ and $\overline{A_3 A_4}$ until they intersect. Denote their intersection as $I_1$ . Through similar triangles & the $45-45-90$ triangles formed, we find that $M_1 M_3=\frac{k}{2}(2+\sqrt2)$
We also have that $\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3$ through ASA congruence ( $\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3$ $M_7 M_1 = M_1 M_3$ $\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1$ ). Therefore, we may let $n=M_1 B_7 = M_3 B_1$
Thus, we have that $\sin\theta=\frac{n-k}{\frac{k}{2}(2+\sqrt2)}$ and that $\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}$ . Therefore $\cos\theta-\sin\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2$
Squaring gives that $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2$ and consequently that $-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta$ through the identities $\sin^2\theta + \cos^2\theta = 1$ and $\sin2\theta = 2\sin\theta\cos\theta$
Thus we have that $\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}$ . Therefore $m+n=5+32=\boxed{037}$

Solution: Let $A_1A_2 = 2$ .  Then $B_1$ and $B_3$ are the projections of $M_1$ and $M_5$ onto the line $B_1B_3$ , so $2=B_1B_3=-M_1M_5\cos x$ , where $x = \angle A_3M_3B_1$ .  Then since $M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}$ $\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}$ , and $m+n=\boxed{037}$

Solution:
Notice that $R_3$ and $R_7$ are parallel ( $B_1B_3B_5B_7$ is a square by symmetry and since the rays are perpendicular) and $B_1B_3=B_3B_5=s=$ the distance between the parallel rays. If the regular hexagon as a side length of $s$ , then $M_3M_7$ has a length of $s+s\sqrt{2}$ . Let $X$ be on $R_3$ such that $M_7X$ is perpendicular to $M_3X$ , and $\phi=\angle M_7M_3X$ . The distance between $R_3$ and $R_7$ is $s=M_7X$ , so $\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}$
Since we are considering a regular hexagon, $M_3$ is directly opposite to $M_7$ and $\angle A_3M_3B_1=90 ^\circ +\phi$ . All that's left is to calculate $\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi$ . By drawing a right triangle or using the Pythagorean identity, $\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}$ and $\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}$ , so $m+n=\boxed{037}$

Solution: Assume that $A_1A_2=1.$ Denote the center $O$ , and the midpoint of $B_1$ and $B_3$ as $B_2$ . Then we have that \[\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{1/2}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.\] Thus, by the cosine double-angle theorem, \[\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},\] so $m+n=\boxed{037}$


Answer: 183.0

Let $A_1,A_2,A_3,\cdots,A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\cdots,A_{12}\} ?$

Solution: Proceed as above to initially get 198 squares (with overcounting). Then note that any square with all four vertices on the dodecagon has to have three sides "between" each vertex, giving us a total of three squares. However, we counted these squares with all four of their sides plus both of their diagonals, meaning we counted them 6 times. Therefore, our answer is $198-3(6-1)=198-15=\boxed{183}.$