Answer: 80.0

Triangle $ABC$ has $\angle BAC = 60^{\circ}$ $\angle CBA \leq 90^{\circ}$ $BC=1$ , and $AC \geq AB$ . Let $H$ $I$ , and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$ , respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$
$\textbf{(A)}\ 60^{\circ} \qquad \textbf{(B)}\ 72^{\circ} \qquad \textbf{(C)}\ 75^{\circ} \qquad \textbf{(D)}\ 80^{\circ} \qquad \textbf{(E)}\ 90^{\circ}$

Solution:
Let $\angle CAB=A$ $\angle ABC=B$ $\angle BCA=C$ for convenience.
It's well-known that $\angle BOC=2A$ $\angle BIC=90+\frac{A}{2}$ , and $\angle BHC=180-A$ (verifiable by angle chasing). Then, as $A=60$ , it follows that $\angle BOC=\angle BIC=\angle BHC=120$ and consequently pentagon $BCOIH$ is cyclic. Observe that $BC=1$ is fixed, hence the circumcircle of cyclic pentagon $BCOIH$ is also fixed. Similarly, as $OB=OC$ (both are radii), it follows that $O$ and also $[BCO]$ is fixed. Since $[BCOIH]=[BCO]+[BOIH]$ is maximal, it suffices to maximize $[BOIH]$
Verify that $\angle IBC=\frac{B}{2}$ $\angle HBC=90-C$ by angle chasing; it follows that $\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}$ since $A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90$ by Triangle Angle Sum. Similarly, $\angle OBC=(180-120)/2=30$ (isosceles base angles are equal), hence \[\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}\] Since $\angle IBH=\angle IBO$ $IH=IO$ by Inscribed Angles.
There are two ways to proceed.
Letting $O'$ and $R$ be the circumcenter and circumradius, respectively, of cyclic pentagon $BCOIH$ , the most straightforward is to write $[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]$ , whence \[[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))\] and, using the fact that $R$ is fixed, maximize $2\sin(60-C)+\sin(2C-60)$ with Jensen's Inequality.
A more elegant way is shown below.
Lemma: $[BOIH]$ is maximized only if $HB=HI$
Proof by contradiction: Suppose $[BOIH]$ is maximized when $HB\neq HI$ . Let $H'$ be the midpoint of minor arc $BI$ be and $I'$ the midpoint of minor arc $H'O$ . Then $[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]$ since the altitude from $H'$ to $BI$ is greater than that from $H$ to $BI$ ; similarly $[BH'I'O]>[BOIH']>[BOIH]$ . Taking $H'$ $I'$ to be the new orthocenter, incenter, respectively, this contradicts the maximality of $[BOIH]$ , so our claim follows. $\blacksquare$
With our lemma( $HB=HI$ ) and $IH=IO$ from above, along with the fact that inscribed angles that intersect the same length chords are equal, \[\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{80}\]


Answer: 108.0

Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$

Solution:
Let $D$ be the midpoint of $\overline{BC}$ . Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$ , so $\angle ADB = \angle ADC = 90^o$
Now let $BD=y$ $AB=x$ , and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$
Then $\mathrm{cos}{(\theta)} = \dfrac{y}{8}$
and $\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}$
Cross-multiplying yields $32y = x(y^2-32)$
Since $x,y>0$ $y^2-32$ must be positive, so $y > 5.5$
Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$ $BD=y < 8$
Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$ $6.5$ $7$ , and $7.5$
However, only one of these values, $y=6$ , yields an integral value for $AB=x$ , so we conclude that $y=6$ and $x=\dfrac{32(6)}{(6)^2-32}=48$
Thus the perimeter of $\triangle ABC$ must be $2(x+y) = \boxed{108}$

Solution:
Angle bisectors motivate trig bash.
Define angle $IBC = x$ . Foot of perpendicular from $I$ to $BC$ is point $P$ $\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N$ , where $N$ is an integer. Thus, $\cos(x) = \frac{N}{16}$ . Via double angle, we calculate $\overline{AB}$ to be $\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}$ . This is to be an integer.  We can bound $N$ now, as $N > 11$ to avoid negative values and $N < 16$ due to triangle inequality. Testing, $N = 12$ works, giving $\overline{AB} = 48, \overline{BC} = 12$ .
Our answer is $2 * 48 + 12 = \boxed{108}$ .
- whatRthose

Solution:
Let $M$ be midpoint $BC, BM = x, AB = y, \angle IBM = \alpha.$
$BI$ is the bisector of $\angle ABM$ in $\triangle ABM.$ $BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.$ \[y = \frac {32 x} {x^2 - 32}.\] $BC = 2x$ is integer, $5.5^2 < 32 \implies x \ge 6.$ $BM < BI \implies x =\{ 6, 6.5, 7, 7.5 \}.$
If $x > 6$ then $y$ is not integer. \[x = 6 \implies y = 48 \implies 2(x+y) =  \boxed{108}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 33.0

Triangle $ABC$ has right angle at $B$ , and contains a point $P$ for which $PA = 10$ $PB = 6$ , and $\angle APB = \angle BPC = \angle CPA$ .  Find $PC$
[asy] unitsize(0.2 cm);  pair A, B, C, P;  A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180));  draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P);  label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$P$", P, NE); [/asy]

Solution:
Let $PC = x$ . Since $\angle APB = \angle BPC = \angle CPA$ , each of them is equal to $120^\circ$ .  By the Law of Cosines applied to triangles $\triangle APB$ $\triangle BPC$ and $\triangle CPA$ at their respective angles $P$ , remembering that $\cos 120^\circ = -\frac12$ , we have
\[AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x\]
Then by the Pythagorean Theorem $AB^2 + BC^2 = CA^2$ , so
\[x^2 + 10x + 100 = x^2 + 6x + 36 + 196\]
and
\[4x = 132 \Longrightarrow x = \boxed{033}.\]


Answer: 30.0

Triangle $ABC$ has side lengths $AB = 11, BC=24$ , and $CA = 20$ . The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$ , and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$ . The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$ . What is $CF$
$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

Solution:
By the Law of Cosine $\cos A = \frac{AC^2 + AB^2 - BC^2}{ 2 \cdot AC \cdot AB} = \frac{20^2 + 11^2 - 24^2}{2\cdot20\cdot11} = -\frac18$
As $ABEC$ is a cyclic quadrilateral, $\angle CEA = \angle CBA$ . As $BDEF$ is a cyclic quadrilateral, $\angle CBA = \angle FEA$
$\because \quad \angle CEA = \angle FEA \quad \text{and} \quad \angle CAE = \angle FAE$
$\therefore \quad \triangle AFE \cong \triangle ACE$ by $ASA$
Hence, $AF = AC = 20$
By the Law of Cosine $CF = \sqrt{20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)} = \sqrt{900} = \boxed{30}$

Solution: Construct the $E$ -antipode, $E^{\prime}\in(ABC)$ . Notice $\triangle CE^{\prime}A\stackrel{+}{\sim}\triangle CBF$ by spiral similarity at $C$ , thus $CF=\dfrac{CB\cdot CA}{CE^{\prime}}=\frac{480}{CE^{\prime}}$ . Let $CE^{\prime}=x$ ; by symmetry $BE^{\prime}=x$ as well and $\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{11^{2}+20^{2}-24^{2}}{2\cdot 11\cdot 20}=-\tfrac{1}{8}$ from Law of Cosines in $\triangle ABC$ , so by Law of Cosines in $\triangle BE^{\prime}C$ we have \[x^{2}+x^{2}+\left(2x^{2}\right)\left(-\dfrac{1}{8}\right)=24^{2}\] from which $x=16$ . Now, $CF=\dfrac{480}{16}=\boxed{30}$

Solution: Applying Stewart's theorem on $\triangle ABC$ with cevian $\overline{CF}$ using the directed lengths $AF = AC = 20$ and $FB = 11-20 = -9$ , we obtain \begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\ 11CF^{2} - 1980 &= 11520 - 3600\end{align*} so $CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqrt{\frac{9900}{11}}=\sqrt{900}=\boxed{30}$

Solution:
Note that $\angle CAF = \angle CAB$ so we may plug into Law of Cosines to find the angle's cosine: \[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.\]
So, we observe that we can use Law of Cosines again to find $CF$ \[CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = 900 \to CF=\boxed{30}\] both ways.

Solution:
This solution is based on this figure: 2021 AMC 12B (Nov) Problem 24, sol.png
Denote by $O$ the circumcenter of $\triangle BED$ .
Denote by $R$ the circumradius of $\triangle BED$
In $\triangle BCF$ , following from the law of cosines, we have \begin{align*} CF^2 & = BC^2 + BF^2 - 2 BC \cdot BF \cos \angle CBF \\ & = BC^2 + BF^2 + 2 BC \cdot BF \cos \angle ABC .  \hspace{1cm} (1) \end{align*} For $BF$ , we have \begin{align*} BF & = 2 R \cos \angle FBO \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \angle CBO \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - \angle BOD}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BED}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BCA}{2} \right) \\ & = 2 R \cos \left( 90^\circ - \angle ABC + \angle BCA \right) \\ & = 2 R \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BED} \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BCA} \sin \left( \angle ABC - \angle BCA \right) \\ & = BD \left( \sin \angle ABC \cot \angle BCA - \cos \angle ABC \right) . \hspace{1cm} (2) \end{align*} The fourth equality follows from the property that $B$ $D$ $E$ are concyclic.
The fifth and the ninth equalities follow from the property that $A$ $B$ $C$ $E$ are concyclic.
Because $AD$ bisects $\angle BAC$ , following from the angle bisector theorem, we have \[ \frac{BD}{CD} = \frac{AB}{AC} . \] Hence, $BD = \frac{24 \cdot 11}{31}$
In $\triangle ABC$ , following from the law of cosines, we have \begin{align*} \cos \angle ABC & = \frac{AB^2 + BC^2 - AC^2}{2 AB \cdot BC} \\ & = \frac{9}{16} \end{align*} and \begin{align*} \cos \angle BCA & = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} \\ & = \frac{57}{64} . \end{align*} Hence, $\sin \angle ABC = \frac{5 \sqrt{7}}{16}$ and $\sin \angle BCA = \frac{11 \sqrt{7}}{64}$ .
Hence, $\cot \angle BCA = \frac{57}{11 \sqrt{7}}$
Now, we are ready to compute $BF$ whose expression is given in Equation (2).
We get $BF = 9$
Now, we can compute $CF$ whose expression is given in Equation (1).
We have $CF = 30$
Therefore, the answer is $\boxed{30}$


Answer: 161.0

Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
If $\omega = 25$ ,  the area of rectangle $PQRS$ is $0$ , so
\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]
and $\alpha = 25\beta$ .  If $\omega = \frac{25}{2}$ , we can reflect $APQ$ over $PQ$ $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the area of the triangle.  Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$
\[[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90\]
so
\[45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}\]
and
\[\beta = \frac{180}{625} = \frac{36}{125}\]
so the answer is $m + n = 36 + 125 = \boxed{161}$

Solution:
Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$
After finding the area, solve for the altitude to $BC$ . Let $E$ be the intersection of the altitude from $A$ and side $BC$ . Then $AE = \frac{36}{5}$ .
Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$ . We then know that $CE = \frac{77}{5}$
Now consider the rectangle $PQRS$ . Since $SR$ is collinear with $BC$ and parallel to $PQ$ $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$
Let $F$ be the intersection between $AE$ and $PQ$ . By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$ . Since $PF+FQ=PQ=\omega$ . We can solve for $PF$ and $FQ$ in terms of $\omega$ . We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$
Let's work with $PF$ . We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$ . We can set up the proportion:
$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$ . Solving for $AF$ $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$
We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$
Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$
This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$

Solution: Using the diagram from Solution 2 above, label $AF$ to be $h$ . Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$ , so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$ . Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$ , you find $\frac{w}{25}=\frac{h}{36/5}$ . Thus, $h=\frac{36w}{125}$ . To find the height of the rectangle, subtract $h$ from $\frac{36}{5}$ to get $\left(\frac{36}{5}-\frac{36w}{125}\right)$ , and multiply this by the other given side $w$ to get $\frac{36w}{5}-\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\boxed{161}$

Solution:
Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$ . Hence, $h=\frac{36}{5}$
Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$
Hence, $PS=\frac{3}{5}(12-L)$
Now, we see that $[PQRS]=\frac{3}{5}(12-L)(w)$
Now, it is obvious that we want to find $L$ in terms of $W$
Looking at the diagram, we see that because $PQRS$ is a rectangle, $\triangle{APQ}\sim{\triangle{ABC}}$
Hence.. we can now set up similar triangles.
We have that $\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}$
Plugging back in..
$[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}$
Simplifying, we get $\frac{36W}{5}-\frac{36W^2}{125}$
Hence, $125+36=\boxed{161}$

Solution: Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$ , so $\beta = \frac{36}{125}$ , yielding $\boxed{161}$


Answer: 20.0

Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length.

Solution: For each $k$ between $2$ and $2450$ , the area of the trapezoid with $\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\triangle{ABC}$ . Let $d_k$ be the length of segment $\overline{P_kQ_k}$ . The area of the trapezoid with bases $\overline{P_{k-1}Q_{k-1}}$ and $P_kQ_k$ is $\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}$ times the area of $\triangle{ABC}$ . (This logic also applies to the topmost triangle if we notice that $d_0 = 0$ .) However, we also know that the area of each shape is $\frac{1}{2450}$ times the area of $\triangle{ABC}$ . We then have $\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}$ . Simplifying, $d_k^2-d_{k-1}^2 = \frac{3}{98}$ . However, we know that $d_0^2 = 0$ , so $d_1^2 = \frac{3}{98}$ , and in general, $d_k^2 = \frac{3k}{98}$ and $d_k = \frac{\sqrt{\frac{3k}{2}}}{7}$ . The smallest $k$ that gives a rational $d_k$ is $6$ , so $d_k$ is rational if and only if $k = 6n^2$ for some integer $n$ .The largest $n$ such that $6n^2$ is less than $2450$ is $20$ , so $k$ has $\boxed{020}$ possible values.

Solution: We have that there are $2449$ trapezoids and $1$ triangle of equal area, with that one triangle being $AP_1Q_1$ . Notice, if we "stack" the trapezoids on top of $\bigtriangleup AP_1Q_1$ the way they already are, we'd create a similar triangle, all of which are similar to $\bigtriangleup ABC$ , and since the trapezoids and $\bigtriangleup AP_1Q_1$ have equal area, each of these similar triangles $AP_kQ_k$ have area $\frac{k}{2450}\left[ ABC\right]$ , and so $\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}$ . We want the ratio of the side lengths $P_kQ_k:BC$ . Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or \[\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}\] \[\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}\] \[\implies k=6n^2<2450\] \[\implies 0<n\leq 20\] so there are $\boxed{020}$ solutions.

Solution: Let $T_1$ stand for $AP_1Q_1$ , and $T_k = AP_kQ_k$ . All triangles $T$ are similar by AA. Let the area of $T_1$ be $x$ . The next trapezoid will also have an area of $x$ , as given. Therefore, $T_k$ has an area of $kx$ . The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, $P_k Q_k=P_1 Q_1\cdot \sqrt{k}$ , and the same if $Q$ is substituted for $P$ throughout. We want the side $P_k Q_k$ to be rational. Setting up proportions: \[5\sqrt{3} : \sqrt{2450}=35\sqrt{2}\] \[\sqrt{6} : 14\] which shows that $P_1 Q_1=\frac{\sqrt{6}}{14}$ . In order for $\sqrt{k} P_1 Q_1$ to be rational, $\sqrt{k}$ must be some rational multiple of $\sqrt{6}$ . This is achieved at $\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}$ . We end there as $21\sqrt{6}=\sqrt{2646}$ . There are 20 numbers from 1 to 20, so there are $\boxed{020}$ solutions.


Answer: 715.0

Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$ , and $15$ , respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$ , and $\ell_C$

Solution:
Let the points of intersection of $\ell_A, \ell_B,\ell_C$ with $\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$ . Furthermore, let the desired triangle be $\triangle XYZ$ , with $X$ closest to side $BC$ $Y$ closest to side $AC$ , and $Z$ closest to side $AB$ . Hence, the desired perimeter is $XE+EF+FY+YG+GH+HZ+ZI+ID+DX=(DX+XE)+(FY+YG)+(HZ+ZI)+115$ since $HG=55$ $EF=15$ , and $ID=45$
Note that $\triangle AHG\sim \triangle BID\sim \triangle EFC\sim \triangle ABC$ , so using similar triangle ratios, we find that $BI=HA=30$ $BD=HG=55$ $FC=\frac{45}{2}$ , and $EC=\frac{55}{2}$
We also notice that $\triangle EFC\sim \triangle YFG\sim \triangle EXD$ and $\triangle BID\sim \triangle HIZ$ . Using similar triangles, we get that \[FY+YG=\frac{GF}{FC}\cdot \left(EF+EC\right)=\frac{225}{45}\cdot \left(15+\frac{55}{2}\right)=\frac{425}{2}\] \[DX+XE=\frac{DE}{EC}\cdot \left(EF+FC\right)=\frac{275}{55}\cdot \left(15+\frac{45}{2}\right)=\frac{375}{2}\] \[HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200\] Hence, the desired perimeter is $200+\frac{425+375}{2}+115=600+115=\boxed{715}$ -ktong

Solution:
Let the diagram be set up like that in Solution 1.
By similar triangles we have \[\frac{AH}{AB}=\frac{GH}{BC}\Longrightarrow AH=30\] \[\frac{IB}{AB}=\frac{DI}{AC}\Longrightarrow IB=30\] Thus \[HI=AB-AH-IB=60\]
Since $\bigtriangleup IHZ\sim\bigtriangleup ABC$ and $\frac{HI}{AB}=\frac{1}{2}$ , the altitude of $\bigtriangleup IHZ$ from $Z$ is half the altitude of $\bigtriangleup ABC$ from $C$ , say $\frac{h}{2}$ . Also since $\frac{EF}{AB}=\frac{1}{8}$ , the distance from $\ell_C$ to $AB$ is $\frac{7}{8}h$ . Therefore the altitude of $\bigtriangleup XYZ$ from $Z$ is \[\frac{1}{2}h+\frac{7}{8}h=\frac{11}{8}h\]
By triangle scaling, the perimeter of $\bigtriangleup XYZ$ is $\frac{11}{8}$ of that of $\bigtriangleup ABC$ , or \[\frac{11}{8}(220+180+120)=\boxed{715}\]

Solution: Notation shown on diagram. By similar triangles we have \[k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},\] \[k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},\] \[k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.\] So, \[\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 - k_1 - k_2,\] \[\frac{FY}{BC} = \frac{FE'}{AC} = \frac{AC - AF - CE'}{AC} = 1 - k_1 - k_3.\] \[k = \frac{ZY}{BC} = \frac{ZE + EF + FY}{BC} = (1 - k_1 - k_2) + k_1 + (1 - k_1 - k_3)\] \[k = 2 -  k_1 - k_2 - k_3 = 2 - \frac{1}{4} - \frac{1}{4}  - \frac{1}{8} = \frac{11}{8}.\] \[\frac{ZY+YX +XZ}{BC +AB + AC} = k \implies ZY + YX + XZ =\frac{11}{8} (220 + 120 + 180) = \boxed{715}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 32.0

Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$

Solution: Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\] By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\] Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\] Let $X=\overline{AB}\cap\overline{CF}$ $a=XB$ , and $b=XD$ . Then, \[XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.\] However, since $\triangle XFD\sim\triangle XAC$ $XF=\tfrac{4+a}3$ , but since $\triangle XFE\sim\triangle XBC$ \[\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,\] and the requested sum is $5+21+2+4=\boxed{032}$

Solution:
Define $\omega_1$ to be the circumcircle of $\triangle ACD$ and $\omega_2$ to be the circumcircle of $\triangle EBC$
Because of exterior angles,
$\angle ACB = \angle CBE - \angle CAD$
But $\angle CBE = \angle CFE$ because $CBFE$ is cyclic. In addition, $\angle CAD = \angle CFD$ because $CAFD$ is cyclic. Therefore, $\angle ACB = \angle CFE - \angle CFD$ . But $\angle CFE - \angle CFD = \angle DFE$ , so $\angle ACB = \angle DFE$ . Using Law of Cosines on $\triangle ABC$ , we can figure out that $\cos(\angle ACB) = \frac{3}{4}$ . Since $\angle ACB = \angle DFE$ $\cos(\angle DFE) = \frac{3}{4}$ . We are given that $DF = 2$ and $FE = 7$ , so we can use Law of Cosines on $\triangle DEF$ to find that $DE = 4\sqrt{2}$
Let $G$ be the intersection of segment $\overline{AE}$ and $\overline{CF}$ . Using Power of a Point with respect to $G$ within $\omega_1$ , we find that $AG \cdot GD = CG \cdot GF$ . We can also apply Power of a Point with respect to $G$ within $\omega_2$ to find that $CG \cdot GF = BG \cdot GE$ . Therefore, $AG \cdot GD = BG \cdot GE$
$AG \cdot GD = BG \cdot GE$
$(AB + BG) \cdot GD = BG \cdot (GD + DE)$
$AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$
$AB \cdot GD = BG \cdot DE$
$4 \cdot GD = BG \cdot 4\sqrt{2}$
$GD = BG \cdot \sqrt{2}$
Note that $\triangle GAC$ is similar to $\triangle GFD$ $GF = \frac{BG + 4}{3}$ . Also note that $\triangle GBC$ is similar to $\triangle GFE$ , which gives us $GF = \frac{7 \cdot BG}{5}$ . Solving this system of linear equations, we get $BG = \frac{5}{4}$ . Now, we can solve for $BE$ , which is equal to $BG(\sqrt{2} + 1) + 4\sqrt{2}$ . This simplifies to $\frac{5 + 21\sqrt{2}}{4}$ , which means our answer is $\boxed{032}$

Solution:
Construct $FC$ and let $FC\cap AE=K$ . Let $FK=x$ . Using $\triangle FKE\sim \triangle BKC$ \[BK=\frac{5}{7}x\] Using $\triangle FDK\sim ACK$ , it can be found that \[3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}\] This also means that $BK=\frac{21}{4}-4=\frac{5}{4}$ . It suffices to find $KE$ . It is easy to see the following: \[180-\angle ABC=\angle KBC=\angle KFE\] Using reverse Law of Cosines on $\triangle ABC$ $\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}$ . Using Law of Cosines on $\triangle EFK$ gives $KE=\frac{21\sqrt 2}{4}$ , so $BE=\frac{5+21\sqrt 2}{4}\to \boxed{032}$ .
-franchester

Solution: Let $P=AE\cap CF$ . Let $CP=5x$ and $BP=5y$ ; from $\triangle{CBP}\sim\triangle{EFP}$ we have $EP=7x$ and $FP=7y$ . From $\triangle{CAP}\sim\triangle{DFP}$ we have $\frac{6}{4+5y}=\frac{2}{7y}$ giving $y=\frac{1}{4}$ . So $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$ . These similar triangles also gives us $DP=\frac{5}{3}x$ so $DE=\frac{16}{3}x$ . Now, Stewart's Theorem on $\triangle{FEP}$ and cevian $FD$ tells us that \[\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,\] so $x=\frac{3\sqrt{2}}{4}$ . Then $BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}$ so the answer is $\boxed{032}$ as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)

Solution:
Connect $CF$ meeting $AE$ at $J$ . We can observe that $\triangle{ACJ}\sim \triangle{FJD}$ Getting that $\frac{AJ}{FJ}=\frac{AC}{FD}=3$ . We can also observe that $\triangle{CBJ}\sim \triangle{EFJ}$ , getting that $\frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7}$
Assume that $BJ=5x;FJ=7x$ , since $\frac{AJ}{FJ}=3$ , we can get that $\frac{AJ}{FJ}=\frac{AB+BJ}{FJ}=\frac{4+5x}{7x}=3$ , getting that $x=\frac{1}{4}; BJ=\frac{5}{4}; FJ=\frac{7}{4}$
Using Power of Point, we can get that $BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ$ Assume that $DJ=5k,CJ=15k$ , getting that $JE=21k, DE=16k$
Now applying Law of Cosine on two triangles, $\triangle{ACJ};\triangle{FJE}$ separately, we can get two equations
$(1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36$
$(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49$
Since $\angle{CJA}=\angle{FJE}$ , we can use $15(2)-7(1)$ to eliminate the $cos$ term
Then we can get that $5040k^2=630$ , getting $k=\frac{\sqrt{2}}{4}$
$BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4}$ , so the desired answer is $\frac{21\sqrt{2}+5}{4}$ , which leads to the answer $\boxed{032}$

Solution:
Nice problem!
First, let $AE$ and $CF$ intersect at $X$ . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that \[\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC\] By the so-called "Reverse Law of Cosines" on $\triangle ABC$ we have \[\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}\] Applying on $\triangle DFE$ gives \[DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)\] \[= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}\] \[=32\] So $DE = 4 \sqrt{2}$ , now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to $BX$ and $XD$ , which are crucial lengths in the problem. Suppose $BX = r, XD = s$ for simplicity. We have:
$\bullet~~~~\triangle AXC \sim \triangle FXD$ $\bullet~~~~\triangle BXC \sim \triangle FXE$
So \[\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3\] \[\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}\] \[\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}\] \[\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}\] So $BE = r + s + 4 \sqrt{2} = \frac{5 + 21 \sqrt{2}}{4}$ . The requested sum is $5 + 21 + 2 + 4 = \boxed{032}$


Answer: 11.0

Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
 -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$ , respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$ $^{(*)}$ , so $\triangle AKB\sim \triangle CKA$ . Using similarity ratios, we can easily find \[AK^2=BK*KC\] However, since $AB=7$ and $CA=9$ , we can use similarity ratios to get \[BK=\frac{7}{9}AK, CK=\frac{9}{7}AK\]
Giving us \[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\] \[\implies \frac{196}{81}AK^2=49\] \[AK=\frac{9}{2}\] so our answer is $9+2=\boxed{011}$

Solution:
Consider an inversion with center $A$ and radius $r=AK$ . Then, we have $AB\cdot AB^*=AK^2$ , or $AB^*=\frac{AK^2}{7}$ . Similarly, $AC^*=\frac{AK^2}{9}$ . Notice that $AB^*KC^*$ is a parallelogram, since $\omega_1$ and $\omega_2$ are tangent to $AC$ and $AB$ , respectively. Thus, $AC^*=B^*K$ . Now, we get that \[\cos(\angle AB^*K)=\cos(180-\angle BAC)=-\frac{11}{21}\] so by Law of Cosines on $\triangle AB^*K$ we have \[(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)\] \[\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}\] \[\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{22AK^2}{63\cdot21}\] \[\Rightarrow AK=\frac{9}{2}\] Then, our answer is $9+2=\boxed{11}$ .
-brianzjk

Solution:
Let the centers of the circles be $O_{1}$ and $O_{2}$ where the $O_{1}$ has the side length $7$ contained in the circle. Now let $\angle BAC =x.$ This implies \[\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x\] by the angle by by tangent. Then we also know that \[\angle AO_{1}B = \angle AO_{2}C = 2x\] Now we first find $\cos x.$ We use law of cosines on $\bigtriangleup ABC$ to obtain \[64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}\] \[\implies \cos{x} =\frac{11}{21}\] \[\implies \sin{x} =\frac{8\sqrt{5}}{21}\] Then applying law of sines on $\bigtriangleup AO_{1}B$ we obtain \[\frac{7}{\sin{2x}} =\frac{O_{1}B}{\sin{90^{\circ}-x}}\] \[\implies\frac{7}{2\sin{x}\cos{x}} =\frac{O_{1}B}{\cos{x}}\] \[\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}\] Using similar logic we obtain $O_{2}A =\frac{189}{16\sqrt{5}}.$
Now we know that $\angle O_{1}AO_{2}=180^{\circ}-x.$ Thus using law of cosines on $\bigtriangleup O_{1}AO_{2}$ yields \[O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}\] While this does look daunting we can write the above expression as \[\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}\] Then factoring yields \[\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}\] The area \[[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] Now $AK$ is twice the length of the altitude of $\bigtriangleup O_{1}AO_{2}$ so we let the altitude be $h$ and we have \[\frac{1}{2} \cdot h \cdot\frac{147}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] \[\implies h =\frac{9}{4}\] Thus our desired length is $\frac{9}{2} \implies m+n = \boxed{11}.$

Solution: By the definition of $K$ , it is the spiral center mapping $BA\to AC$ , which means that it is the midpoint of the $A$ -symmedian chord. In particular, if $M$ is the midpoint of $BC$ and $M'$ is the reflection of $A$ across $K$ , we have $\triangle ABM'\sim\triangle AMC$ . By Stewart's Theorem, it then follows that \[AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.\]

Solution:
We start by assigning coordinates to point $A$ , labeling it $(0,0)$ and point $B$ at $(7,0)$ , and letting point $C$ be above the $x$ -axis. Through an application of the Pythagorean Theorem and dropping an altitude to side $AB$ , it is easy to see that $C$ has coordinates $(33/7, 24\sqrt{5}/7)$
Let $O1$ be the center of circle $\omega_1$ and $O2$ be the center of circle $\omega_2$ . Since circle $\omega_1$ contains both points $A$ and $B$ $O1$ must lie on the perpendicular bisector of line $AB$ , and similarly $O2$ must lie on the perpendicular bisector of line $AC$ . Through some calculations, we find that the perpendicular bisector of $AB$ has equation $x = 3.5$ , and the perpendicular bisector of $AC$ has equation $y = {-11\sqrt{5}/40 \cdot x} + 189\sqrt{5}/80$
Since circle $\omega_1$ is tangent to line $AC$ at $A$ , its radius must be perpendicular to $AC$ at $A$ .
Therefore, the radius has equation $y = {{-11\cdot\sqrt{5}/40} \cdot x}$ . Substituting the $x$ -coordinate of $O1$ into this, we find the y-coordinate of $O1 = {{-11 \cdot \sqrt{5}/40} \cdot 7/2} = {-77 \cdot \sqrt{5}/80}$
Similarly, since circle $\omega_2$ is tangent to line $AB$ at $A$ , its radius must be perpendicular to $AB$ at $A$ . Therefore, the radius has equation $x = 0$ and combining with the previous result for $O2$ we get that the coordinates of $O2$ are $(0, 189\sqrt{5}/80)$
We now find the slope of $O1O2$ , the line joining the centers of circles $\omega_1$ and $\omega_2$ , which turns out to be ${({266 \cdot \sqrt{5} / 80}) \cdot -2/7} = {-19 \cdot \sqrt{5}/20}$ . Since the $y$ -intercept of that line is at $O2(0,189\sqrt{5}/80)$ , the equation is $y = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}$ .  Since circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $K$ , line $AK$ is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles, $AK$ has slope ${4 \cdot \sqrt{5}/19}$ . Since point $A$ is $(0,0)$ , this line has a $y$ -intercept of $0$ , so it has equation $y$ ${{4 \cdot \sqrt{5}/19} \cdot x}$
We set ${{4 \cdot \sqrt{5}/19} \cdot x} = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}$ in order to find the intersection $I$ of the radical axis $AK$ and $O1O2$ . Through some moderate bashing, we find that the intersection point is $I(57/28, {3 \cdot \sqrt{5}/7})$ . We know that either intersection point of two circles is the same distance from the intersection of radical axis and line joining the centers of those circles, so reflecting $A$ over $I$ yields $K$ and $AK$ $2AI$ = (This is the most tedious part of the bash) ${2 \cdot \sqrt{(57/28)^2 + ({3 \cdot \sqrt{5}/7)^2)}}} = {2 \cdot \sqrt{3969/784}} = {2 \cdot 63/28} = {2 \cdot 9/4} = 9/2$ . Therefore the answer is $9 + 2 = \boxed{011}.$


Answer: 30.0

Triangle $ABC$ has side-lengths $AB = 12, BC = 24,$ and $AC = 18.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\  33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$

Solution:
Let $O$ be the incenter of $\triangle{ABC}$ . Because $\overline{MO} \parallel \overline{BC}$ and $\overline{BO}$ is the angle bisector of $\angle{ABC}$ , we have
\[\angle{MBO} = \angle{CBO} = \angle{MOB} = \frac{1}{2}\angle{MBC}\]
It then follows due to alternate interior angles and base angles of isosceles triangles that $MO = MB$ . Similarly, $NO = NC$ . The perimeter of $\triangle{AMN}$ then becomes \begin{align*} AM + MN + NA &= AM + MO + NO + NA \\ &= AM + MB + NC + NA \\ &= AB + AC \\ &= 30 \rightarrow \boxed{30}

Solution:
Let $O$ be the incenter. $AO$ is the angle bisector of $\angle MAN$ . Let the angle bisector of $\angle BAC$ meets $BC$ at $P$ and the angle bisector of $\angle ABC$ meets $AC$ at $Q$ . By applying both angle bisector theorem and Menelaus' theorem,
$\frac{AO}{OP} \times \frac{BP}{BC} \times \frac{CQ}{QA} = 1$
$\frac{AO}{OP} \times \frac{12}{30} \times \frac{24}{12} = 1$
$\frac{AO}{OP}=\frac{5}{4}$
$\frac{AO}{AP}=\frac{5}{9}$
Perimeter of $\triangle AMN = \frac{12+24+18}{9} \times 5 = 30  \rightarrow \boxed{30}$

Solution: Like in other solutions, let $O$ be the incenter of $\triangle ABC$ . Let $AO$ intersect $BC$ at $D$ . By the angle bisector theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{18} = \frac{2}{3}$ . Since $BD+DC = 24$ , we have $\frac{BD}{24-BD} = \frac{2}{3}$ , so $3BD = 48 - 2BD$ , so $BD = \frac{48}{5} = 9.6$ . By the angle bisector theorem on $\triangle ABD$ , we have $\frac{DO}{OA} = \frac{BD}{BA} = \frac{4}{5} = 0.8$ , so $\frac{DA}{OA} = 1 + \frac{DO}{OA} = \frac{9}{5} = 1.8$ , so $\frac{AO}{AD} = \frac{5}{9}$ . Because $\triangle AMN \sim \triangle ABC$ , the perimeter of $\triangle AMN$ must be $\frac{5}{9} (12 + 18 + 24) = 30$ , so our answer is $\boxed{30}$

Solution: We know that the ratio of the perimeter of $\triangle AMN$ and $\triangle ABC$ is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from $A$ to $BC$ is $\frac{9\sqrt{15}}{4}$ from Herons and then $A=\frac{bh}{2}$ and also that the height from $A$ to $MN$ is simply the height from $A$ to $BC$ minus the inradius. We know the area and the semiperimeter so $r=\frac{A}{s}$ which gives us $r=\sqrt{15}$ . Now we know that the altitude from $A$ to $MN$ is $\frac{5\sqrt{15}}{4}$ so the ratios of the heights from $A$ for $\triangle AMN$ and $\triangle ABC$ is $\frac{5}{9}$ . Thus the perimeter of $\triangle AMN$ is $\frac{5}{9} \times 54 = 30$ so our answer is $\boxed{30}$


Answer: 6.0

Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ .  What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$

Solution: The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$ .  Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$ , which is $\frac 4{\sqrt 2} = 2\sqrt 2$ . Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$


Answer: 6.0

Solution: Because the line $x + y = 7$ is parallel to $\overline {AB}$ , the area of $\triangle ABC$ is independent of the location of $C$ on that line.  Therefore it may be assumed that $C$ is $(7,0)$ .  In that case the triangle has base $AC = 4$ and altitude $3$ , so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed{6}$


Answer: 578.0

Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ 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:
We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$ $MN=BN-BM$ , and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get $CN=\frac{168} {25}.$
By the Pythagorean Theorem in $\Delta BCN,$ we have
$BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$
Thus, $MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$
In $\Delta ADM$ , we use the Pythagorean Theorem to get $DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.$
Thus, $[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$
Hence, the answer is $527+11+40=\boxed{578}.$

Solution:
Let $E$ be the intersection of lines $BC$ and $DM$ .  Since triangles $\Delta CME$ and $\Delta CMD$ share a side and height, the area of $\Delta CDM$ is equal to $\frac{DM}{EM}$ times the area of $\Delta CME$ .
By AA similarity, $\Delta EMB$ is similar to $\Delta ACB$ $\frac{EM}{AC}=\frac{MB}{CB}$ . Solving yields $EM=\frac{175}{48}$ . Using the same method but for $EB$ yields $EB=\frac{625}{48}$ . As in previous solutions, by the Pythagorean Theorem $DM=\frac{5\sqrt{11}}{2}$ . So, $\frac{DM}{EM}=\frac{24\sqrt{11}}{35}$ .
Now, since we know both $CB$ and $EB$ , we can find that $CE=\frac{527}{48}$ . The height of $\Delta CME$ is the length from point $M$ to $CB$ . Since $M$ is the midpoint of $AB$ , the height is just $\frac{1}{2}\cdot7=\frac{7}{2}$ . Using this, we can find that the area of $\Delta CMD$ is $\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$ , giving our answer of $\boxed{578}$


Answer: 12.0

Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$
$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Solution: Observe that $\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ since $m\angle MEI = m\angle MAI = 45^\circ$ ), so $MI=2,MC=\frac{3}{\sqrt{2}}$ . With $\angle{MCI}=45^\circ$ , we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\boxed{12}$

Solution:
We first claim that $\triangle{EMI}$ is isosceles and right.
Proof: Construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ . Since $\overline{AM}$ bisects $\angle{BAC}$ , one can deduce that $MF=MG$ . Then by AAS it is clear that $MI=ME$ and therefore $\triangle{EMI}$ is isosceles. Since quadrilateral $AIME$ is cyclic, one can deduce that $\angle{EMI}=90^\circ$ . Q.E.D.
Since the area of $\triangle{EMI}$ is 2, we can find that $MI=ME=2$ $EI=2\sqrt{2}$
Since $M$ is the mid-point of $\overline{BC}$ , it is clear that $AM=\frac{3\sqrt{2}}{2}$
Now let $AE=a$ and $AI=b$ . By Ptolemy's Theorem, in cyclic quadrilateral $AIME$ , we have $2a+2b=6$ . By Pythagorean Theorem, we have $a^2+b^2=8$ . One can solve the simultaneous system and find $b=\frac{3+\sqrt{7}}{2}$ . Then by deducting the length of $\overline{AI}$ from 3 we get $CI=\frac{3-\sqrt{7}}{2}$ , giving the answer of $\boxed{12}$ . (Surefire2019)

Solution: Like above, notice that $\triangle{EMI}$ is isosceles and right, which means that $\dfrac{ME \cdot MI}{2} = 2$ , so $MI^2=4$ and $MI = 2$ . Then construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ as well as $\overline{MI}$ . It's clear that $MG^2+GI^2 = MI^2$ by Pythagorean, so knowing that $MG = \dfrac{AB}{2} = \dfrac{3}{2}$ allows one to solve to get $GI = \dfrac{\sqrt{7}}{2}$ . By just looking at the diagram, $CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}$ . The answer is thus $3+7+2=\boxed{12}$

Solution: Let $A$ lie on $(0,0)$ $E$ on $(0,y)$ $I$ on $(x,0)$ , and $M$ on $\left(\frac{3}{2},\frac{3}{2}\right)$ . Since ${AIME}$ is cyclic, $\angle EMI$ (which is opposite of another right angle) must be a right angle; therefore, $\overrightarrow{ME} \cdot \overrightarrow{MI} = \left<\frac{-3}{2}, y-\frac{3}{2}\right> \cdot \left<x-\frac{3}{2}, -\frac{3}{2}\right> = 0$ . Compute the dot product to arrive at the relation $y=3-x$ . We can set up another equation involving the area of $\triangle EMI$ using the Shoelace Theorem . This is \[2=\frac{1}{2}\left[\frac{3}{2}\left(y-\frac{3}{2}\right)-xy+\frac{3}{2}\left(x+\frac{3}{2}\right)\right].\] Multiplying, substituting $3-x$ for $y$ , and simplifying, we get $x^2 -3x + \frac{1}{2}=0$ . Thus, $(x,y)=\left(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2}\right)$ . But $AI>AE$ , meaning $x=AI=\frac{3 + \sqrt{7}}{2}$ and $CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}$ , and the final answer is $3+7+2=\boxed{12}$

Solution:
From $AIME$ cyclic we get $\angle{MEI} = \angle{MAI} = 45^\circ$ and $\angle{MIE} = \angle{MAE} = 45^\circ$ , so $\triangle{EMI}$ is an isosceles right triangle.
From $[EMI]=2$ we get $EM=MI=2$
Notice $\triangle{AEM} \cong \triangle{CIM}$ , because $\angle{AEM}=180-\angle{AIM}=\angle{CIM}$ $EM=IM$ , and $\angle{EAM}=\angle{ICM}=45^\circ$
Let $CI=AE=x$ , so $AI=3-x$
By Pythagoras on $\triangle{EAI}$ we have $x^2+(3-x)^2=EI^2=8$ , and solve this to get $x=CI=\dfrac{3-\sqrt{7}}{2}$ for a final answer of $3+7+2=\boxed{12}$

Solution: Let $CI=a$ $BE=b$ . Because opposite angles in a cyclic quadrilateral are supplementary, we have $\angle EMI=90^{\circ}$ . By the law of cosines, we have $MI^2=a^2+\frac{9}{4}-\frac{3}{2}a$ , and $ME^2=b^2+\frac{9}{4}-\frac{3}{2}b$ . Notice that $EI=2MO$ , where $O$ is the origin of the circle mentioned in the problem. Thus $\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}$ . By the Pythagorean Theorem, we have $ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8$ . By the Pythagorean Theorem, we have $AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8$ . Thus we have $18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}$ $=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a$ . We know that \begin{align*} (3-a)^2+(3-b)^2&=8 \\ (3-a)^2+a^2&=8 \\ 2a^2-6a+9&=8 \\ 2a^2-6a+1&=0 \\ a&=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}. \end{align*} We take the smaller solution because we have $AI>AE\implies 3-AI<3-AE\implies CI<CE$ , and we want $CI$ , not $CE$ , thus $CI=\frac{3-\sqrt{7}}{2}$ . Thus our final answer is $3+7+2=\boxed{12}$


Answer: 5.5

Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$ . Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$ , and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$ . What is the length of $\overline{BD}$
[asy] size(100); draw((0,0)--(2,4)--(4,0)--(6,4)--cycle--(4,0),linewidth(1)); label("$A$", (0,0), SW); label("$B$", (2,4), N); label("$C$", (4,0), SE); label("$D$", shift(0.2,0.1)*intersectionpoint((0,0)--(6,4),(2,4)--(4,0)), N); label("$E$", (6,4), NE);[/asy]
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6$

Solution: Since triangle $ABD$ is congruent to triangle $ECD$ and $\overline{CE} =11$ $\overline{AB}=11$ . Since $\overline{AB}=\overline{BC}$ $\overline{BC}=11$ . Because point $D$ is the midpoint of $\overline{BC}$ $\overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{5.5}$


Answer: 12.0

Triangle $ABC$ is equilateral with side length $6$ . Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$ $O$ , and $C$
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution:
Construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$
Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^\circ$
Consider another point $M$ on Circle $X$ opposite to point $O$
As $AOCM$ is an inscribed quadrilateral of Circle $X$ $\angle AMC=180^\circ-120^\circ=60^\circ$
Afterward, deduce that $\angle AXC=2·\angle AMC=120^\circ$
By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$
$2r^2(1-\cos(120^\circ))=6^2$
$r^2=12$
The area is therefore $\pi r^2 = \boxed{12}$

Solution:
We have $\angle AOC = 120^\circ$
Denote by $R$ the circumradius of $\triangle AOC$ .
In $\triangle AOC$ , the law of sines implies \[ 2 R = \frac{AC}{\sin \angle AOC} = 4 \sqrt{3} . \]
Hence, the area of the circumcircle of $\triangle AOC$ is \[ \pi R^2 = 12 \pi . \]
Therefore, the answer is $\boxed{12}$

Solution:
As in the previous solution, construct the circle that passes through $A$ $O$ , and $C$ , centered at $X$ . Let $Y$ be the intersection of $\overline{OX}$ and $\overline{AB}$
Note that since $\overline{OA}$ is the angle bisector of $\angle BAC$ that $\angle OAC=30^\circ$ . Also by symmetry, $\overline{OX}$ $\perp$ $\overline{AB}$ and $AY = 3$ . Thus $\tan(30^\circ) = \frac{OY}{3}$ so $OY = \sqrt{3}$
Let $r$ be the radius of circle $X$ , and note that $AX = OX = r$ . So $\triangle AYX$ is a right triangle with legs of length $3$ and $r - \sqrt{3}$ and hypotenuse $r$ . By Pythagoras, $3^2 + (r - \sqrt{3})^2 = r^2$ . So $r = 2\sqrt{3}$
Thus the area is $\pi r^2 = \boxed{12}$

Solution:
The semiperimeter is $\frac{6+6+6}{2}=9$ units.
The area of the triangle is $9\sqrt{3}$ units squared.  By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius $r=\sqrt{3}$ . As $\angle{AOC}=120^\circ$ , we can form an altitude from point $O$ to side $AC$ at point $M$ , forming two 30-60-90 triangles. As $CM=MA=3$ , we can solve for $OC=2\sqrt{3}$ . Now, the area of the circle is just $\pi*(2*\sqrt{3})^2 = 12\pi$ . Select $\boxed{12}$


Answer: 12.0

Solution: Call the diameter of the circle $d$ . If we extend points $A$ and $C$ to meet at a point on the circle and call it $E$ , then $\bigtriangleup OAE=\bigtriangleup OCE$ . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, $CE=AE=\sqrt{d^2-12}$ . We know this since $OC=OA=OB$ and $OC$ is the hypotenuse of a $30-60-90$ right triangle, with the longer leg being $\frac{6}{2}=3$ so $OC=2\sqrt{3}$ . Applying Ptolemy's Theorem on cyclic quadrilateral $OCEA$ , we get $2({\sqrt{d^2-12}})\cdot{2\sqrt{3}}=6d$ . Squaring and solving we get $d^2=48 \Longrightarrow (2r)^2=48$ so $r^2=12$ . Therefore, the area of the circle is $\boxed{12}$


Answer: 9.0

Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$ . If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$ [asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$

Solution: We solve for $\angle A$ as follows: \[4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.\] That means that minor arc $\widehat{BC}$ has measure $40^\circ$ . We can fit a maximum of $\frac{360}{40}=\boxed{9}$ of these arcs in the circle.


Answer: 61.0

Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$ $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$ . Then one interior angle of the triangle is:
$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$

Solution: Because the triangle is inscribed, the sum of the minor arcs equals $360^\circ$ .  Thus, \[x+75+2x+25+3x-22=360\] \[6x+78=360\] Solving this yields $x = 47$ , so the inscribed angles are $122^\circ$ $99^\circ$ , and $119^\circ$ .  Noting that an angle of $\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\triangle ABC$ are $59.5^\circ$ $49.5^\circ$ , and $\boxed{61}$


Answer: 43.0

Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
 Let $\angle ACP=\alpha$ $\angle PCQ=\beta$ , and $\angle QCB=\gamma$ . Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$ , so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\] Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$
Now Law of Sines on $\triangle ACS$ $\triangle SCT$ , and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\] Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\] so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\] Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$

Solution: Projecting through $C$ we have \[\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}\] which easily gives $ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{043}$

Solution:
By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[5\cdot 7+13\cdot ST=AT\cdot BS.\] Therefore, in order to find $ST$ , it suffices to find $AT\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.
As $\triangle APS\sim \triangle CPB$ , we find \[\frac{4}{PC}=\frac{7}{BC}.\] Therefore, $\frac{BC}{PC}=\frac{7}{4}$
As $\triangle BQT\sim\triangle CQA$ , we find \[\frac{6}{CQ}=\frac{5}{AC}.\] Therefore, $\frac{AC}{CQ}=\frac{5}{6}$
As $\triangle ATQ\sim\triangle CBQ$ , we find \[\frac{AT}{BC}=\frac{7}{CQ}.\] Therefore, $AT=\frac{7\cdot BC}{CQ}$
As $\triangle BPS\sim \triangle CPA$ , we find \[\frac{9}{PC}=\frac{BS}{AC}.\] Therefore, $BS=\frac{9\cdot AC}{PC}$ . Thus we find \[AT\cdot BS=\left(\frac{7\cdot BC}{CQ}\right)\left(\frac{9\cdot AC}{PC}\right).\] But now we can substitute in our previously found values for $\frac{BC}{PC}$ and $\frac{AC}{CQ}$ , finding \[AT\cdot BS=63\cdot \frac{7}{4}\cdot \frac{5}{6}=\frac{21\cdot 35}{8}.\] Substituting this into our original expression from Ptolemy's Theorem, we find \begin{align*}35+13ST&=\frac{21\cdot 35}{8}\\13ST&=\frac{13\cdot 35}{8}\\ST&=\frac{35}{8}.\end{align*} Thus the answer is $\boxed{43}$

Solution:
Extend $\overline{AB}$ past $B$ to point $X$ so that $CPTX$ is cyclic. Then, by Power of a Point on $CPTX$ $(CQ)(QT) =  (PQ)(QX)$ . By Power of a Point on $CATB$ $(CQ)(QT) = (AQ)(QB) = 42$ . Thus, $(PQ)(QX) = 42$ , so $BX = 8$
By the Inscribed Angle Theorem on $CPTX$ $\angle SCT = \angle BXT$ . By the Inscribed Angle Theorem on $ASTC$ $\angle SCT = \angle SAT$ , so $\angle BXT = \angle SAT$ . Since $ASTB$ is cyclic, $\angle AST = \angle TBX$ . Thus, $\triangle AST \sim \triangle XBT$ , so $AS/XB = ST/BT$ . Solving for $ST$ yields $ST = \frac{35}{8}$ , for a final answer of $35+8 = \boxed{043}$

Solution: By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[AS\cdot BT+AB\cdot ST=AT\cdot BS.\] Projecting through $C$ we have \[\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}.\] Therefore \[AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies\] \[\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies\] \[ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}\] \[ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed{43}.\]

Solution:
Connect $AT$ and $\angle{SCT}=\angle{SAT}, \angle{ACS}=\angle{ATS}, \frac{ST}{\sin \angle{SAT}}=\frac{AS}{\sin \angle{ATS}}$
So we need to get the ratio of $\frac{\sin \angle{ACS}}{\sin \angle{SCT}}$
By clear observation $\triangle{CAQ}\sim \triangle{BTQ}$ , we have $\frac{CQ}{AC}=\frac{6}{5}$ , LOS tells $\frac{AC}{\sin \angle{CPA}}=\frac{4}{\sin \angle{ACS}}; \frac{CQ}{\sin \angle{CPQ}}=\frac{3}{\sin \angle{PCQ}}$ so we get $\frac{\sin \angle{PCQ}}{\sin \angle{ACS}}=\frac{5}{8}$ , the desired answer is $7\cdot \frac{\sin \angle{SAT}}{\sin \angle{ATS}}=\frac{35}{8}$ leads to $\boxed{043}$


Answer: 919.0

Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Take a force-overlaid inversion about $A$ and note $D$ and $E$ map to each other. As $DE$ was originally the diameter of $\gamma$ $DE$ is still the diameter of $\gamma$ . Thus $\gamma$ is preserved. Note that the midpoint $M$ of $BC$ lies on $\gamma$ , and $BC$ and $\omega$ are swapped. Thus points $F$ and $M$ map to each other, and are isogonal. It follows that $AF$ is a symmedian of $\triangle{ABC}$ , or that $ABFC$ is harmonic. Then $(AB)(FC)=(BF)(CA)$ , and thus we can let $BF=5x, CF=3x$ for some $x$ . By the LoC, it is easy to see $\angle{BAC}=120^\circ$ so $(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49$ . Solving gives $x^2=\frac{49}{19}$ , from which by Ptolemy's we see $AF=\frac{30}{\sqrt{19}}$ . We conclude the answer is $900+19=\boxed{919}$

Solution:
Use the angle bisector theorem to find $CD=\tfrac{21}{8}$ $BD=\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\tfrac{49}{8}$ , and so $AE=8$ . Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$ , hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$  In triangle $AEF$ , let $X$ be the foot of the altitude from $A$ ; then $EF=EX+XF$ , where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$ , we get \begin{align}\tag{1}     EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{align} Note $\angle AFE = \angle ACE$ , and the Law of Cosines in $\triangle ACE$ gives $\cos \angle ACE = -\tfrac 17$ .
Also, $\angle AEF = \angle DEF$ , and $\angle DFE = \tfrac{\pi}{2}$ $DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} =  \tfrac{8}{49}\cdot EF$
Plugging in all our values into equation $(1)$ , we get: \[EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.\] The Law of Cosines in $\triangle AEF$ , with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives \[8^2 = AF^2 + \tfrac{49}{225}  AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2\] Thus $AF^2 = \frac{900}{19}$ . The answer is $\boxed{919}$

Solution:
Let $a = BC$ $b = CA$ $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$ . We claim that $\angle MAD=\angle DAF$
$\textit{Proof}$ . Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$ . Therefore, $M\in \gamma$ . Now let $X = FD\cap \omega$ . Since $\angle EFX=90^\circ$ $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$ ; hence $E$ $M$ $X$ are collinear. From $\angle DAG = \angle DMX = 90^\circ$ , quadrilateral $ADMX$ is cyclic. Therefore, $\angle MAD = \angle MXD$ . But $\angle MXD$ and $\angle EAF$ are both subtended by arc $EF$ in $\omega$ , so they are equal. Thus $\angle MAD=\angle DAF$ , as claimed.  As a result, $\angle CAM = \angle FAB$ . Combined with $\angle BFA=\angle MCA$ , we get $\triangle ABF\sim\triangle AMC$ and therefore \[\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}\] By Stewart's Theorem on $\triangle ABC$ (with cevian $AM$ ), we get \[AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},\] so $AF^2 = \tfrac{900}{19}$ , so the answer is $900+19=\boxed{919}$

Solution:
Use the angle bisector theorem to find $CD=\tfrac{21}{8}$ $BD=\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\tfrac{49}{8}$ , and so $AE=8$ . Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}$  Since $DE$ is the diameter of circle $\gamma$ $\angle DFE$ is $90^\circ$ . Extending $DF$ to intersect circle $\omega$ at $X$ , we find that $XE$ is the diameter of $\omega$ (since $\angle DFE$ is $90^\circ$ ). Therefore, $XE=\tfrac{14\sqrt{3}}{3}$
Let $EF=x$ $XD=u$ , and $DF=v$ . Then $XE^2-XF^2=EF^2=DE^2-DF^2$ , so we get \[(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}\] which simplifies to \[u^2+2uv = \frac{5341}{192}.\] By Power of Point $D$ $uv=BD \cdot DC=735/64$ . Combining with above, we get \[XD^2=u^2=\frac{931}{192}.\] Note that $\triangle XDE\sim \triangle ADF$ and the ratio of similarity is $\rho = AD : XD = \tfrac{15}{8}:u$ . Then $AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R$ and \[AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.\] The answer is $900+19=\boxed{919}$

Solution:
Denote $AB = c, BC = a, AC = b, \angle A = 2 \alpha.$ Let M be midpoint BC. Let $\theta$ be the circle centered at $A$ with radius $\sqrt{AB \cdot AC} =\sqrt{bc}.$
We calculate the length of some segments.
The median $AM = \sqrt{\frac {b^2}{2} + \frac {c^2}{2} -  \frac {a^2}{4}}.$ The bisector $AD = \frac {2 b c \cos \alpha}{b+c}.$ One can use Stewart's Theorem in both cases.
$AD$ is bisector of $\angle A \implies BD = \frac {a c}{b + c}, CD = \frac {a b}{b + c} \implies$ \[BD \cdot CD = \frac {a^2 bc }{(b+c)^2}.\] We use Power of Point $D$ and get $AD \cdot DE = BD \cdot CD.$ \[AE = AD + DE = AD + \frac {BD \cdot CD}{AD},\] \[AE =\frac {2 b c \cos \alpha}{b+c} + \frac {a^2 bc \cdot (b+c) }{(b+c)^2 \cdot 2 b c \cos \alpha} =\] \[=  \frac {b c \cos^2 \alpha + a^2}{2(b+c)\cos \alpha} =\frac {4bc \cos^2 \alpha  + b^2 +c^2 -2 b c \cos 2\alpha}{2(b+c) \cos \alpha} = \frac {b+c}{2} \implies AD \cdot AE = 2 bc \cos \alpha.\] We consider the inversion with respect $\theta.$
$B$ swap $B' \implies AB' = AC, B'  \in AB \implies B'$ is symmetric to $C$ with respect to $AE.$
$C$ swap $C' \implies AC' = AB, C'$ lies on line $AC \implies C'$ is symmetric to $B$ with respect to $AE.$
$BC^2 = AB^2 + AC^2 + AB \cdot BC \implies \alpha = 60^\circ \implies AD \cdot AE = bc \implies D$ swap $E.$
Points $D$ and $E$ lies on $\Gamma \implies \Gamma$ swap $\Gamma.$
$DE$ is diameter $\Gamma, \angle DME = 90^\circ \implies  M \in \Gamma.$ Therefore $M$ is crosspoint of $BC$ and $\Gamma.$
Let $\Omega$ be circumcircle $AB'C'.  \Omega$ is image of line $BC.$ Point $M$ maps into $M' \implies M' = \Gamma \cap \Omega.$
Points $A, B',$ and $C'$ are symmetric to $A, C,$ and $B,$ respectively.
Point $M'$ lies on $\Gamma$ which is symmetric with respect to $AE$ and on $\Omega$ which is symmetric to $\omega$ with respect to $AE  \implies$
$M'$ is symmetric $F$ with respect to $AE \implies AM' = AF.$
We use Power of Point $A$ and get \[AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}} = \frac {30}{\sqrt{19}} \implies \boxed{919}.\]


Answer: 616.0

Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers.  Find $a+b.$
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Solution: Let $x=\angle CAM$ , so $3x=\angle CDM$ . Then, $\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11$ . Expanding $\tan 3x$ using the angle sum identity gives \[\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.\] Thus, $\frac{3-\tan^2x}{1-3\tan^2x}=11$ . Solving, we get $\tan x= \frac 12$ . Hence, $CM=\frac{11}2$ and $AC= \frac{11\sqrt{5}}2$ by the Pythagorean Theorem . The total perimeter is $2(AC + CM) = \sqrt{605}+11$ . The answer is thus $a+b=\boxed{616}$

Solution: In a similar fashion, we encode the angles as complex numbers, so if $BM=x$ , then $\angle BAD=\text{Arg}(11+xi)$ and $\angle BDM=\text{Arg}(1+xi)$ . So we need only find $x$ such that $\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)$ . This will happen when $\frac{363x-x^3}{1331-33x^2}=x$ ,  which simplifies to $121x-4x^3=0$ . Therefore, $x=\frac{11}{2}$ . By the Pythagorean Theorem, $AB=\frac{11\sqrt{5}}{2}$ , so the perimeter is $11+11\sqrt{5}=11+\sqrt{605}$ , giving us our answer, $\boxed{616}$

Solution: Let $\angle BAD=\alpha$ , so $\angle BDM=3\alpha$ $\angle BDA=180-3\alpha$ , and thus $\angle ABD=2\alpha.$ We can then draw the angle bisector of $\angle ABD$ , and let it intersect $\overline{AM}$ at $E.$ Since $\angle BAE=\angle ABE$ $AE=BE.$ Let $AE=x$ . Then we see by the Pythagorean Theorem, $BM=\sqrt{BE^2-ME^2}=\sqrt{x^2-(11-x)^2}=\sqrt{22x-121}$ $BD=\sqrt{BM^2+1}=\sqrt{22x-120}$ $BA=\sqrt{BM^2+121}=\sqrt{22x}$ , and $DE=10-x.$ By the angle bisector theorem, $BA/BD=EA/ED.$ Substituting in what we know for the lengths of those segments, we see that \[\frac{\sqrt{22x}}{\sqrt{22x-120}}=\frac{x}{10-x}.\] multiplying by both denominators and squaring both sides yields \[22x(10-x)^2=x^2(22x-120)\] which simplifies to $x=\frac{55}{8}.$ Substituting this in for x in the equations for $BA$ and $BM$ yields $BA=\frac{\sqrt{605}}{2}$ and $BM=\frac{11}{2}.$ Thus the perimeter is $11+\sqrt{605}$ , and the answer is $\boxed{616}$

Solution:
The triangle is symmetrical so we can split it in half ( $\triangle ABM$ and $\triangle ACM$ ).
Let $\angle BAM = y$ and $\angle BDM = 3y$ . By the Law of Sines on triangle $BAD$ $\frac{10}{\sin 2y} = \frac{BD}{\sin y}$ . Using $\sin 2y = 2\sin y\cos y$ we can get $BD = \frac{5}{\cos y}$ . We can use this information to relate $BD$ to $DM$ by using the Law of Sines on triangle $BMD$
\[\frac{\frac{5}{\cos y}}{\sin BMD} = \frac{1}{\sin 90^\circ - 3y}\]
$\sin BMD = 1$ (as $\angle BMD$ is a right angle), so $\frac{1}{\sin 90^\circ - 3y} = \frac{5}{\cos y}$ . Using the identity $\sin 90^\circ - x = \cos x$ , we can turn the equation into::
\[\frac{1}{\cos 3y} = \frac{5}{\cos y}\]
\[5\cos 3y = \cos y\]
\[5(4\cos ^3 y - 3\cos y) = \cos y\]
\[20\cos ^3 y = 16 \cos y\]
\[5\cos ^3 y = 4\cos y\]
\[5\cos ^2 y = 4\]
\[\cos ^2 y = \frac{4}{5}\]
Now that we've found $\cos y$ , we can look at the side lengths of $BM$ and $AB$ (since they are symmetrical, the perimeter of $\triangle ABC$ is $2(BM + AB)$
We note that $BM = 11\tan y$ and $AB = 11\sec y$
\[\sin ^2 y = 1 - \cos ^2 y\]
\[\sin ^2 y = \frac{1}{5}\]
\[\tan ^2 y = \frac{1}{4}\]
\[\tan y = \frac{1}{2}\]
(Note it is positive since $BM > 0$ ).
\[\sec ^2 y = \frac{5}{4}\]
\[\sec y = \frac{\sqrt{5}}{2}\]
\[BM + AB = 11\frac{\sqrt{5}+1}{2}\]
\[2(BM + AB) = 11(\sqrt{5} + 1)\]
\[2(BM + AB) = 11\sqrt{5} + 11\]
\[2(BM + AB) = \sqrt{605} + 11\]
The answer is $\boxed{616}$

Solution:
Suppose $\angle BAM=\angle CAM =x$ , since $\angle BDC=3\angle BAC$ , we have $\angle BDM=\angle MDC = 3x$ . Therefore, $\angle DBC=\angle DCB = 90^\circ -3x$ and $\angle ABD=\angle DCA=2x$ . As a result, $\triangle KAC$ is isosceles, $KC=KA$
Let $H$ be a point on the extension of $CD$ through $D$ such that $\overline{HB}\perp\overline{BC}$ and denote the intersection of $\overline{HC}$ and $\overline{AB}$ as $K$ . Then, $BH=2DM=2, \overline{HB}\parallel\overline{DM}$ , and $HD=DC$ by the Midpoint Theorem. So, $\angle HBA=x$ and $\angle CDM=\angle CHB=\angle HDA= 3x$
Consequently, $\triangle HBK\sim \triangle DAK$ \[\frac{BK}{KA}=\frac{HK}{KD}=\frac{1}{5}\] Assume $BK=a$ and $HK=b$ , then $KA=5a$ and $KD = 5b$ . Since $KC=KA, KC=5a$ , and since $HD=DC$ $KC=11b$ . Therefore, $a=\frac{11}{5}b$
In $\triangle BDM$ , by the Pythagorean Theorem $BM=\sqrt{36b^2-1}$ . Similarly in $\triangle BAM$ $BM=\sqrt{36a^2-121}$ . So \[\sqrt{36a^2-121}=\sqrt{36b^2-1}\] Since $a=\frac{11}{5}b$ , we have $b=\frac{5\sqrt{5}}{12}$ and $a=\frac{11\sqrt{5}}{12}$ . Consequently, $BM=\frac{11}{2}$ and $AB=\frac{11\sqrt{5}}{2}$ . Thus, the perimeter of $\triangle ABC$ is $11+\sqrt{605}$ , and the answer is $\boxed{616}$


Answer: 83.0

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution:
Take point $N$ inside $\triangle ABC$ such that $\angle CBN = 7^\circ$ and $\angle BCN = 23^\circ$
$\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ$ . Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM = CN$ . Hence $\triangle CMN$ is an equilateral triangle , so $\angle CNM = 60^\circ$
Then $\angle MNB = 360^\circ - \angle CNM - \angle CNB = 360^\circ - 60^\circ - 150^\circ = 150^\circ$ . We now see that $\triangle MNB$ and $\triangle CNB$ are congruent. Therefore, $CB = MB$ , so $\angle CMB = \angle MCB = \boxed{83}$

Solution:
From the givens, we have the following angle measures $m\angle AMC = 150^\circ$ $m\angle MCB = 83^\circ$ .  If we define $m\angle CMB = \theta$ then we also have $m\angle CBM = 97^\circ - \theta$ .  Then apply the Law of Sines to triangles $\triangle AMC$ and $\triangle BMC$ to get
\[\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin (97^\circ - \theta)}\]
Clearing denominators , evaluating $\sin 150^\circ = \frac 12$ and applying one of our trigonometric identities to the result gives
\[\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta\]
and multiplying through by 2 and applying the double angle formulas gives
\[\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta\]
and so $\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta \Longleftrightarrow \tan 7^{\circ} = \cot \theta$ ; since $0^\circ < \theta < 180^\circ$ , we must have $\theta = 83^\circ$ , so the answer is $\boxed{83}$

Solution:
Without loss of generality , let $AC = BC = 1$ .  Then, using the Law of Sines in triangle $AMC$ , we get $\frac {1}{\sin 150} = \frac {MC}{\sin 7}$ , and using the sine addition formula to evaluate $\sin 150 = \sin (90 + 60)$ , we get $MC = 2 \sin 7$
Then, using the Law of Cosines in triangle $MCB$ , we get $MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1$ , since $\cos 83 = \sin 7$ . So triangle $MCB$ is isosceles, and $\angle CMB = \boxed{83}$

Solution:
Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros
First, take point $E$ outside of $\triangle{ABC}$ so that $\triangle{CEB}$ is equilateral. Then, connect $A$ $C$ , and $M$ to $E$ . Also, let $ME$ intersect $AB$ at $F$ $\angle{MCE} = 83^\circ - 60^\circ = 23^\circ$ $CE = AB$ , and (trivially) $CM = CM$ , so $\triangle{MCE} \cong  \triangle{MCA}$ by SAS congruence. Also, $\angle{CMA} = \angle{CME} = 150^\circ$ , so $\angle{AME} = 60^\circ$ , and $AM = ME$ ,
making $\triangle{AME}$ also equilateral. (it is isosceles with a $60^\circ$ angle) $\triangle{MAF} \cong \triangle{EAF}$ by SAS ( $MA = AE$ $AF = AF$ , and $m\angle{MAF} = m\angle{EAF} = 30^\circ$ ), and $\triangle{MAB} \cong \triangle{EAB}$ by SAS ( $MA = AE$ $AB = AB$ , and $m\angle{MAB} = m\angle{EAB} = 30^\circ$ ). Thus, $\triangle{BME}$ is isosceles, with $m\angle{BME} = m\angle{BEM} = 60^\circ + 7^\circ = 67^\circ$ . Also, $\angle{EMB} + \angle {CMB} = \angle{CME} = 150^\circ$ , so $\angle{CME} = 150^\circ - 67^\circ  = \boxed{83}$

Solution:
Noticing that we have three concurrent cevians, we apply Ceva's theorem:
\[(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\sin \angle BCM)\] \[(\sin 23)(\sin 30)(\sin x) = (\sin 7)(\sin 37-x)(\sin 83)\]
using the fact that $\sin 83 = \cos 7$ and $(\sin 7)(\cos 7) = 1/2 (\sin 14)$ we have:
\[(\sin 23)(\sin x) = (\sin 14)(\sin 37-x)\]
By inspection, $x=14^\circ$ works, so the answer is $180-83-14= \boxed{083}$


Answer: 47.0

Triangle $ABC$ lies in the cartesian plane and has an area of $70$ . The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$

Solution:
The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$ . The equation of the median can be found by $-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}$ , so $q = -5p + 107$
Use determinants to find that the area of $\triangle ABC$ is $\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\  q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70$ (note that there is a missing absolute value ; we will assume that the other solution for the triangle will give a smaller value of $p+q$ , which is provable by following these steps over again) (alternatively, we could use the Shoelace Theorem ). We can calculate this determinant to become $140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}$ $\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$ . Thus, $q = \frac{1}{11}p + \frac{337}{11}$
Setting this equation equal to the equation of the median, we get that $\frac{1}{11}p + \frac{337}{11} = -5p + 107$ , so $\frac{56}{11}p = \frac{107 \cdot 11 - 337}{11}$ . Solving produces that $p = 15$ Substituting backwards yields that $q = 32$ ; the solution is $p + q = \boxed{047}$


Answer: 75.0

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution: Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \[\frac{50}{x} = \frac{10}{y},\] where $y = -x+a$ . Therefore substituting we have that $BG=\frac{5a}{6}$ . By similar triangles, we have that $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ . Then, we have that $\frac{ah}{2}=120$ . We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$ , and we have that it is $\boxed{75}$ by substituting.

Solution: For this problem, we have $\triangle{ADE}\sim\triangle{ABC}$ because of SAS and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$

Solution: The ratio of the $\overline{BG}$ to $\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\bigtriangleup ABG$ to the area of $\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\bigtriangleup ABG$ is $\frac{5}{5+1}\times120=100$ . Since $\overline{DE}$ is the midsegment of $\bigtriangleup ABC$ , so $\overline{DF}$ is the midsegment of $\bigtriangleup ABG$ . Thus, the ratio of the area of $\bigtriangleup ADF$ to the area of $\bigtriangleup ABG$ is $1:4$ , so the area of $\bigtriangleup ACG$ is $\frac{1}{4}\times100=25$ . Therefore, the area of quadrilateral $FDBG$ is $[ABG]-[ADF]=100-25=\boxed{75}$

Solution: The area of quadrilateral $FDBG$ is the area of $\bigtriangleup ABG$ minus the area of $\bigtriangleup ADF$ . Notice, $\overline{DE} || \overline{BC}$ , so $\bigtriangleup ABG \sim \bigtriangleup ADF$ , and since $\overline{AD}:\overline{AB}=1:2$ , the area of $\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4$ . Given that the area of $\bigtriangleup ABC$ is $120$ , using $\frac{bh}{2}$ on side $AB$ yields $\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}$ . Using the Angle Bisector Theorem, $\overline{BG}:\overline{BC}=50:(10+50)=5:6$ , so the height of $\bigtriangleup ABG: \bigtriangleup ACB=5:6$ . Therefore our answer is $\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}$

Solution: We try to find the area of quadrilateral $FDBG$ by subtracting the area outside the quadrilateral but inside triangle $ABC$ . Note that the area of $\triangle ADE$ is equal  to $\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}$ and the area of triangle $ABC$ is equal to $\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A$ . The ratio $\frac{[ADE]}{[ABC]}$ is thus equal to $\frac{1}{4}$ and the area of triangle $ADE$ is $\frac{1}{4} \cdot 120 = 30$ . Let side $BC$ be equal to $6x$ , then $BG = 5x, GC = x$ by the angle bisector theorem. Similarly, we find the area of triangle $AGC$ to be $\frac{1}{2} \cdot 10 \cdot x \cdot \sin C$ and the area of triangle $ABC$ to be $\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C$ . A ratio between these two triangles yields $\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}$ , so $[AGC] = 20$ . Now we just need to find the area of triangle $AFE$ and subtract it from the combined areas of $[ADE]$ and $[ACG]$ , since we count it twice. Note that the angle bisector theorem also applies for $\triangle ADE$ and $\frac{AE}{AD} = \frac{1}{5}$ , so thus $\frac{EF}{ED} = \frac{1}{6}$ and we find $[AFE] = \frac{1}{6} \cdot 30 = 5$ , and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$ , and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}$ , and we are done.


Answer: 75.0

Solution: Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \[\frac{50}{x} = \frac{10}{y},\] where $y = -x+a$ . Therefore substituting we have that $BG=\frac{5a}{6}$ . By similar triangles, we have that $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ . Then, we have that $\frac{ah}{2}=120$ . We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$ , and we have that it is $\boxed{75}$ by substituting.

Solution: For this problem, we have $\triangle{ADE}\sim\triangle{ABC}$ because of SAS and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$

Solution: The ratio of the $\overline{BG}$ to $\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\bigtriangleup ABG$ to the area of $\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\bigtriangleup ABG$ is $\frac{5}{5+1}\times120=100$ . Since $\overline{DE}$ is the midsegment of $\bigtriangleup ABC$ , so $\overline{DF}$ is the midsegment of $\bigtriangleup ABG$ . Thus, the ratio of the area of $\bigtriangleup ADF$ to the area of $\bigtriangleup ABG$ is $1:4$ , so the area of $\bigtriangleup ACG$ is $\frac{1}{4}\times100=25$ . Therefore, the area of quadrilateral $FDBG$ is $[ABG]-[ADF]=100-25=\boxed{75}$

Solution: The area of quadrilateral $FDBG$ is the area of $\bigtriangleup ABG$ minus the area of $\bigtriangleup ADF$ . Notice, $\overline{DE} || \overline{BC}$ , so $\bigtriangleup ABG \sim \bigtriangleup ADF$ , and since $\overline{AD}:\overline{AB}=1:2$ , the area of $\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4$ . Given that the area of $\bigtriangleup ABC$ is $120$ , using $\frac{bh}{2}$ on side $AB$ yields $\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}$ . Using the Angle Bisector Theorem, $\overline{BG}:\overline{BC}=50:(10+50)=5:6$ , so the height of $\bigtriangleup ABG: \bigtriangleup ACB=5:6$ . Therefore our answer is $\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}$

Solution: We try to find the area of quadrilateral $FDBG$ by subtracting the area outside the quadrilateral but inside triangle $ABC$ . Note that the area of $\triangle ADE$ is equal  to $\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}$ and the area of triangle $ABC$ is equal to $\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A$ . The ratio $\frac{[ADE]}{[ABC]}$ is thus equal to $\frac{1}{4}$ and the area of triangle $ADE$ is $\frac{1}{4} \cdot 120 = 30$ . Let side $BC$ be equal to $6x$ , then $BG = 5x, GC = x$ by the angle bisector theorem. Similarly, we find the area of triangle $AGC$ to be $\frac{1}{2} \cdot 10 \cdot x \cdot \sin C$ and the area of triangle $ABC$ to be $\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C$ . A ratio between these two triangles yields $\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}$ , so $[AGC] = 20$ . Now we just need to find the area of triangle $AFE$ and subtract it from the combined areas of $[ADE]$ and $[ACG]$ , since we count it twice. Note that the angle bisector theorem also applies for $\triangle ADE$ and $\frac{AE}{AD} = \frac{1}{5}$ , so thus $\frac{EF}{ED} = \frac{1}{6}$ and we find $[AFE] = \frac{1}{6} \cdot 30 = 5$ , and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$ , and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}$ , and we are done.


Answer: 7.0

Triangle $ABC$ with right angle at $C$ $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ $q$ $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$

Solution:
Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$ . It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$ . Hence $ODP$ is isosceles and $OD = DP = 2$
Denote $E$ the projection of $O$ onto $CD$ . Now $CD = CP + DP = 3$ . By the Pythagorean Theorem $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$ . Now note that $EP = \frac {1}{2}$ . By the Pythagorean Theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$ . Hence it now follows that,
\[\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}\]
This gives that the answer is $\boxed{007}$

Solution: Let $\angle{ACP}$ be equal to $x$ . Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$ . We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$ . Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$ . Plugging this in gives that $\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}$ . The answer is $\boxed{007}$

Solution:
Let $\alpha=\angle{ACP}$ $\beta=\angle{ABC}$ , and $x=BP$ . By Law of Sines,
$\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}$ (1), and
$\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}$ . (2)
Then, substituting (1) into (2), we get
$4-x=\frac{2}{x} \implies x^2-4x+2=0 \implies x=2-\sqrt{2} \implies \frac{4-x}{x}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$
The answer is $\boxed{007}$ .
~Rowechen

Solution:
Let $\angle{ACP}=x$ . Then, $\angle{APC}=2x$ and $\angle{A}=180-3x$ . Let the foot of the angle bisector of $\angle{APC}$ on side $AC$ be $D$ . Then,
$CD=DP$ and $\triangle{DAP}\sim{\triangle{APC}}$ due to the angles of these triangles.
Let $CD=a$ . By the Angle Bisector Theorem, $\frac{1}{a}=\frac{AP}{AD}$ , so $AD=a\cdot{AP}$ . Moreover, since $CD=DP=a$ , by similar triangle ratios, $\frac{AP}{a+a\cdot{AP}}=a$ . Therefore, $AP = \frac{a^2}{1-a^2}$
Construct the perpendicular from $D$ to $AP$ and denote it as $F$ . Denote the midpoint of $CP$ as $M$ . Since $PD$ is an angle bisector, $PF$ is congruent to $PM$ , so $PF=\frac{1}{2}$
Also, $\triangle{DFA}\sim{\triangle{BCA}}$ . Thus, $\frac{FA}{AC}=\frac{AD}{AB}\Longrightarrow\frac{\frac{a^2}{1-a^2}-\frac{1}{2}}{a+\frac{a^3}{1-a^2}}=\frac{\frac{a^3}{1-a^3}}{4}$ . After some major cancellation, we have $7a^4-8a^2+2=0$ , which is a quadratic in $a^2$ . Thus, $a^2 = \frac{4\pm\sqrt{2}}{7}$
Taking the negative root implies $AP<BP$ , contradiction. Thus, we take the positive root to find that $AP=2+\sqrt{2}$ . Thus, $BP=2-\sqrt{2}$ , and our desired ratio is $\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies{3+2\sqrt{2}}$
The answer is $\boxed{007}$

Solution:
Let $O$ be the circumcenter of $\triangle ABC$ . Since $\triangle ABC$ is a right triangle, $O$ will be on $\overline{AB}$ and $\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2$ . Let $\overline{OP} = x$
Next, let's do some angle chasing. Label $\angle ACP = \theta^{\circ}$ , and $\angle APC = 2\theta^{\circ}$ . Thus, $\angle PAC = (180-3\theta)^{\circ}$ , and by isosceles triangles, $\angle ACO = (180-3\theta)^{\circ}$ . Then, by angle subtraction, $\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}$
Using the Law of Sines: \[\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\] Using trigonometric identies, $\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}$ . Plugging this back into the Law of Sines formula gives us: \[\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\]
\[-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}\] \[-4\cos (2\theta)^{\circ}=x\] \[\cos(2\theta)^{\circ}=\frac{-x}4\]
Next, using the Law of Cosines: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}\] Substituting $\cos(2\theta)^{\circ}=\frac{-x}4$ gives us: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4\] \[4=1+x^2+\frac{x^2}{2}\]
Solving for x gives $x=\sqrt 2$
Finally: $\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2$ , which gives us an answer of $3+2+2=\boxed{007}$ . ~adyj


Answer: 182.0

Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: Do note that by counting the area in 2 ways, the first altitude is $x = \frac{ab}{c}$ . By similar triangles, the common ratio is $\rho = \frac{a}{c}$ for each height, so by the geometric series formula, we have \begin{align}     6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. \end{align} Writing $p=a+b+c$ and clearing denominators, we get \[13a=6p .\] Thus $p=13q$ $a=6q$ , and $b+c=7q$ , i.e. $c=7q-b$ . Plugging these into $(1)$ , we get $78q(q-b)=6bq$ , i.e., $14b=13q$ . Thus $q=14r$ and $p=182r$ $b=13r$ $a=84r$ $c=85r$ . Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=\boxed{182}$

Solution: Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$ . By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$ . Multiplying by the denominator and expanding, the equation becomes $\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a$ . Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$ , so $7ab = 6bc+6b^2$ and $7a=6b+6c$ . Checking for Pythagorean triples gives $13,84,$ and $85$ , so $p=13+84+85=\boxed{182}$

Solution:
We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.
First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$ . The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)$ . Thus,
$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)$ .     (1)
Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,
$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)$ .     (2)
Adding (1) and (2) together, we find that
$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$
Setting $a=C_{0}B$ $b=C_{0}A$ , and $c=AB$ , we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$

Solution:
Let $a = BC_0$ $b = AC_0$ , and $c = AB$ .
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$ , shown in red in the figure below.
Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\frac{a+c}{b}$ .
In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$
Guessing and checking Pythagorean triples reveals that $a = 84$ $b=13$ $c = 85$ , and $p = a + b + c = \boxed{182}$ satisfies this equation.

Solution: This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$ . Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$ . Thus $a = 168n^2, b = 26n^2, c = 170n^2$ . Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \boxed{182}$

Solution:
For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ \[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\] \[(x+y)xy=6(x^2+xy)(x-y)\] \[xy=6x(x-y)\] \[7xy=6x^2\] \[y=\frac{6}{7}x\]
Now this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $\boxed{182}$


Answer: 961.0

Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$

Solution:
Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$ . Also, the area we seek is simply the ratio $k=\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$ , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.
For ease, all ratios I will use to solve this problem are with respect to the area of $[AB_0C_0]$ . For example, if I say some area has ratio $\frac{1}{2}$ , that means its area is 45.
Now note that $k=$ 1 minus ratio of $[B_1C_1A]$ minus ratio $[B_0C_0C_1]$ . We see by similar triangles given that ratio $[B_0C_0C_1]$ is $\frac{17^2}{25^2}$ . Ratio $[B_1C_1A]$ is $(\frac{336}{625})^2$ , after seeing that $C_1C_0 = \frac{289}{625}$ , . Now it suffices to find 90 times ratio $[B_0B_1C_1]$ , which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find $k$ and clearing out the $5^8$ , we see that the answer is $90\cdot \frac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}$ , which gives $q= \boxed{961}$

Solution:
Let $k$ be the coefficient of the similarity of triangles \[\triangle B_0 C_1 C_0 \sim \triangle AB_0 C_0 \implies k = \frac {B_0 C_0}{AC_0} = \frac {17}{25}.\] Then area $\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \implies \frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$
The height of triangles $\triangle B_0C_1A$ and $\triangle AB_0C_0$ from $B_0$ is the same $\implies \frac {AC_1}{AC_0} = 1 – k^2.$
The coefficient of the similarity of triangles $\triangle AB_1C_1 \sim \triangle AB_0C_0$ is $\frac {AC_1}{AC_0} = 1 – k^2 \implies \frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$
Analogically the coefficient of the similarity of triangles $\triangle AB_2C_2 \sim \triangle AB_0C_0$ is $(1 – k^2)^2 \implies \frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on.
The yellow area $[Y]$ is $\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \frac {k^2}{1 – (1 – k^2)^2} = \frac{1}{2 – k^2}.$
The required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \cdot (1 –  \frac{1}{2 – k^2}) = [AB_0C_0 ] \cdot \frac  {1 –  k^2}{2 – k^2} = [AB_0C_0 ] \cdot \frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} =  [AB_0C_0 ] \cdot  \frac {336}{961}.$
The number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $\boxed{961}$