Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

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

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

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

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

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

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

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

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

Solution: [asy] size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); label("$E$",E,NE); label("$F$",F, S); label("$G$",G,SE); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); label("$5x$",midpoint(A--B),S); label("$6x$",midpoint(A--D),1.5*(-1,0)); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] Let the points formed by dropping altitudes from $A$ to the lines $BC$ $CD$ , and $BD$ be $E$ $F$ , and $G$ , respectively.
We have \[\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB\] and \[BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.\] For convenience, let $AB = 5x$ . By Heron's formula on $\triangle ABD$ , we have sides $5x,6x,9x$ and semiperimeter $10x$ , so \[\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},\] so $AB = 5x = \frac{45}{2\sqrt{2}}$
Then, \[BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}\] and \[\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.\] Finally, recalling that $ABCD$ is isosceles, \[K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},\] so $\sqrt{2}\cdot K = \boxed{567}$

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

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

Answer:  567
Problem: Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$

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

Answer:  567
Problem: Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then
\[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$

Solution: We know that $a_{1}=\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$ . Note also that $a_{2}=\tfrac{2038}{2040}=\tfrac{1019}{1020}=\tfrac{t}{t+1}$ for $t=1019$ . Then $a_{2+q}=\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ (by the Euclidean Algorithm), or $q+1$ divides $1001$ . Thus, the least value of $q$ is $6$ and $j=2+6=8$ . We know $a_{8}=\tfrac{1019+108}{1020+114}=\tfrac{1127}{1134}=\tfrac{161}{162}$ . Now $a_{8+q}=\tfrac{161+18q}{162+19q}$ unless $18q+161$ and $19q+162$ are not relatively prime which happens the first time $q+1$ divides $18q+161$ or $q+1$ divides $143$ or $q=10$ , and $j=8+10=18$ . We have $a_{18}=\tfrac{161+180}{162+190}=\tfrac{341}{352}=\tfrac{31}{32}$ . Now $a_{18+q}=\tfrac{31+18q}{32+19q}$ unless $18q+31$ and $19q+32$ are not relatively prime. This happens the first time $q+1$ divides $18q+31$ implying $q+1$ divides $13$ , which is prime so $q=12$ and $j=18+12=30$ . We have $a_{30}=\tfrac{31+216}{32+228}=\tfrac{247}{260}=\tfrac{19}{20}$ . We have $a_{30+q}=\tfrac{18q+19}{19q+20}$ , which is always reduced by EA, so the sum of all $j$ is $1+2+8+18+30=\boxed{059}$

Answer:  059
Problem: Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then
\[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$

Solution: Let $a_{j_1}, a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ be all terms in the form $\frac{t}{t+1},$ where $j_1<j_2<j_3<\cdots<j_u,$ and $t$ is some positive integer.
We wish to find $\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\frac{m}{m+1}$ for some positive integer $m.$
To find $\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\boldsymbol{k'}$ for which \[\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}}\] is reducible:
If $\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm , we have \begin{align*} d\mid m+18k' \text{ and } d\mid m+1+19k' &\implies d\mid m+18k' \text{ and } d\mid k'+1 \\ &\implies d\mid m-18 \text{ and } d\mid k'+1. \end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$
We will prove that for such value of $\boldsymbol{k',}$ the number $\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\boldsymbol{\frac{t}{t+1}:}$ \[a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}=\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\frac{\frac{m-18}{k'+1}+18}{\frac{m-18}{k'+1}+19}, \hspace{10mm} (*)\] where $t=\frac{m-18}{k'+1}+18$ must be a positive integer.
We start with $m=2020$ and $a_{j_1}=a_1=\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ by filling out the table below recursively: \[\begin{array}{c|c|c|c|c|c}  & & & & & \\ [-2ex] \boldsymbol{i} & \boldsymbol{m} & \boldsymbol{m-18} & \boldsymbol{k'+1} & \boldsymbol{k'} & \boldsymbol{a_{j_{i+1}} \left(\textbf{by } (*)\right)} \\ [0.5ex]  \hline   & & & & & \\ [-1.5ex]  1 & 2020 & 2002 & 2 & 1 & \hspace{4.25mm} a_2 = \frac{1019}{1020} \\ [1ex]      2 & 1019 & 1001 & 7 & 6 & \hspace{2.75mm} a_8 = \frac{161}{162} \\ [1ex]     3 & 161 & 143 & 11 & 10 & a_{18} = \frac{31}{32} \\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \frac{19}{20} \\ [1ex] 5 & 19 & 1 & \text{N/A} & \text{N/A} & \text{N/A} \\ [1ex] \end{array}\] As $\left(j_1,j_2,j_3,j_4,j_5\right)=(1,2,8,18,30),$ the answer is $\sum_{i=1}^{5}{j_i}=\boxed{059}.$

Answer:  059
Problem: Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

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

Answer:  301
Problem: Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

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

Answer:  301
Problem: Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

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

Answer:  301
Problem: Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

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

Answer:  301
Problem: Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Define the distance between two frogs as the number of sides between them that do not contain the third frog.
Let $E(a,b,c)$ denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are $a,b,$ and $c$ (in either clockwise or counterclockwise order). Without the loss of generality, assume that $a\leq b\leq c.$
We wish to find $E(4,4,4).$ Note that:
We have the following system of equations: \begin{align*} E(4,4,4)&=1+\frac{2}{8}E(4,4,4)+\frac{6}{8}E(2,4,6), \\ E(2,4,6)&=1+\frac{4}{8}E(2,4,6)+\frac{1}{8}E(4,4,4)+\frac{1}{8}E(2,2,8), \\ E(2,2,8)&=1+\frac{2}{8}E(2,2,8)+\frac{2}{8}E(2,4,6). \end{align*} Rearranging and simplifying each equation, we get \begin{align*} E(4,4,4)&=\frac{4}{3}+E(2,4,6), &(1) \\ E(2,4,6)&=2+\frac{1}{4}E(4,4,4)+\frac{1}{4}E(2,2,8), &\hspace{12.75mm}(2) \\ E(2,2,8)&=\frac{4}{3}+\frac{1}{3}E(2,4,6). &(3) \end{align*} Substituting $(1)$ and $(3)$ into $(2),$ we obtain \[E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],\] from which $E(2,4,6)=4.$ Substituting this into $(1)$ gives $E(4,4,4)=\frac{16}{3}.$
Therefore, the answer is $16+3=\boxed{019}.$

Answer:  019
Problem: Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We can solve the problem by removing $1$ frog, and calculate the expected time for the remaining $2$ frogs. In the original problem, when the movement stops, $2$ of the $3$ frogs meet. Because the $3$ frogs cannot meet at one vertex, the probability that those two specific frogs meet is $\tfrac13$ . If the expected time for the two frog problem is $E'$ , then the expected time for the original problem is $\tfrac 13 E'$
The distance between the two frogs can only be $0$ $2$ $4$ $6$ . We use the distances as the states to draw the following Markov Chain . This Markov Chain is much simpler than that of Solution 1 Supplement in the Remark section.
\begin{align*} E(2) &= 1 + \frac12 \cdot E(2) + \frac14 \cdot E(4)\\ E(4) &= 1 + \frac14 \cdot E(2) + \frac12 \cdot E(4) + \frac14 \cdot E(6)\\ E(6) &= 1 + \frac12 \cdot E(4) + \frac12 \cdot E(6) \end{align*}
By solving the above system of equations, $E(4) = 16$ . The answer for the original problem is $\frac{16}{3}$ $16 + 3 = \boxed{019}$

Answer:  019