Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$

Solution: Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$ . If one of the values in $S$ is $3$ or $4 \pmod{7}$ , observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \pmod{7}$ and $S$ contains a $0 \pmod{7}$ , then the values of $S$ must be $(b-3, b-2, b-1, b)$ . Similarly, if we are given a $b \equiv 4 \pmod{7}$ and we know that $0 \pmod{7}$ is in $S$ $S$ must be $(b, b+1, b+2, b+3).$ Hence, the only possibility for $a$ is $5, 6 \pmod{7}.$
In either case, we are guaranteed there is a $6, 0, 1 \pmod{7}$ value in $7$ . The difference comes down to if there is a $5 \pmod{7}$ value or a $1 \pmod{7}$ value. The person receiving such value will be able to determine all of $S$ but the $6, 0, 1 \pmod{7}$ people will not be able to differentiate the two cases ... yet.
Now consider which value among the consecutive integers is $c \equiv 3 \pmod{6}$ , if any. The person will know that $S$ is either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \pmod{6}$ value in $S$ . Neither the $6, 1 \pmod{7}$ person can be $3 \pmod{7}$ , else they can decipher what $S$ is right off the bat by considering which set has $0 \pmod{7}.$ This translates to the possible $5 \pmod{7}$ or $1 \pmod{7}$ person cannot be $0 \pmod{6}$ . We are given that $0 \pmod{7}$ and $0 \pmod{6}$ cannot be the same person. Hence we conclude one of the $6 \pmod{7}$ or $1 \pmod{7}$ must be the $0 \pmod{6}$ person.
Let the $6 \pmod{7}$ person be $0 \pmod{6}.$ Then--hypothetical-- $c \equiv 2 \pmod{7}$ person is $3 \pmod{6}.$ After the first round, the $6, 0, 1 \pmod{7}$ people realize that $2 \pmod{7}$ is not in $S$ else they would have deduced $S$ by noting $S$ was either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \pmod{6}$ and choosing the former based on where $0 \pmod{7}$ is. Hence they figure out $S$ by knowing $5 \pmod{7}.$ So $a \equiv 5 \pmod{7}$ and $a \equiv 5 \pmod{6}$ (from $6 \pmod{7}$ person being $0 \pmod{6}$ ).
Similarly, if $1 \pmod{7}$ person is $0 \pmod{6}$ we find that $a \equiv 6 \pmod{7}$ (so a $2 \pmod{7}$ is in $S$ ) and $a \equiv 4 \pmod{6}.$
By CRT, the possibilities are $a \equiv 5,  34 \pmod{42}.$ The sum of the greatest values of $S$ are the sum of $a + 3$ and so we get $(34 + 3) + (47 + 3) + (76 + 3) + (89 + 3) = \boxed{258}.$

Answer:  258
Problem: A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\overline{BD}.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(500);  pair A, B, C, D, P, A1, C1; B = origin; D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0); A = intersectionpoints(Circle(B,5),Circle(D,7))[0]; C = intersectionpoints(Circle(B,6),Circle(D,9))[1]; P = intersectionpoint(A--C,B--D); A1 = foot(A,B,D); C1 = foot(C,B,D); markscalefactor=3/160; draw(rightanglemark(A,A1,D),red); draw(rightanglemark(C,C1,B),red); dot("$A$",A,1.5*dir(aCos(7/sqrt(1649)))); dot("$B$",B,1.5*W); dot("$C$",C,1.5*dir(180+aCos(7/sqrt(1649)))); dot("$D$",D,1.5*E); dot("$E$",P,dir(180-(180-aCos(7/sqrt(1649)))/2)); dot("$A'$",A1,dir(-75)); dot("$C'$",C1,N); label("$\theta$",P,dir(180+aCos(7/sqrt(1649))/2),red); draw(A--A1^^C--C1,dashed); draw(A--B--C--D--cycle^^A--C^^B--D); [/asy] Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\triangle ABA',\triangle BCC',\triangle CDC',$ and $\triangle DAA',$ respectively: \[\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\hspace{36mm}(2) \\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\hspace{36mm}(3) \\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\hspace{36mm}(4) \end{array}\] Let the brackets denote areas. We get \begin{align*} [ABD]+[CBD]&=[ABCD] \\ \frac12(p+q+r+s)h_1+\frac12(p+q+r+s)h_2&=30 \\ \frac12(p+q+r+s)(h_1+h_2)&=30 \\ (p+q+r+s)(h_1+h_2)&=60. \hspace{49.25mm}(5) \end{align*} We subtract $(2)+(4)$ from $(1)+(3):$ \begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\ \left[(p+q+r)^2-s^2\right]+\left[(q+r+s)^2-p^2\right]&=21 \\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\ (p+q+r+s)(2q+2r)&=21 \\ 2(p+q+r+s)(q+r)&=21 \\ (p+q+r+s)(q+r)&=\frac{21}{2}. \hspace{9.5mm}(6) \end{align*} From right triangles $\triangle AEA'$ and $\triangle CEC',$ we have $\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.$ It follows that \begin{alignat*}{8} \tan\theta&=\frac{h_1}{r}\qquad&\implies\qquad h_1&=r\tan\theta, \hspace{64mm}&(1\star)\\ \tan\theta&=\frac{h_2}{q}\qquad&\implies\qquad h_2&=q\tan\theta. &(2\star) \end{alignat*} Finally, we divide $(5)$ by $(6):$ \begin{align*} \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ \frac{r\tan\theta+q\tan\theta}{q+r}&=\frac{40}{7} \hspace{15mm} &&\text{by }(1\star)\text{ and }(2\star)\\ \frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ \tan\theta&=\frac{40}{7}, \end{align*} from which the answer is $40+7=\boxed{047}.$

Answer:  047
Problem: A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: By Bretschneider's Formula, \[30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.\] Thus, $uv=3\sqrt{1649}$ . Also, \[[ABCD]=\tfrac 12 \cdot uv\sin{\theta};\] solving for $\sin{\theta}$ yields $\sin{\theta}=\tfrac{40}{\sqrt{1649}}$ . Since $\theta$ is acute, $\cos{\theta}$ is positive, from which $\cos{\theta}=\tfrac{7}{\sqrt{1649}}$ . Solving for $\tan{\theta}$ yields \[\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7},\] for a final answer of $\boxed{047}$

Answer:  047
Problem: Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$

Solution: Recall that $1000$ divides this expression if $8$ and $125$ both divide it. It should be fairly obvious that $n \geq 3$ ; so we may break up the initial condition into two sub-conditions.
(1) $5^n \equiv n \pmod{8}$ . Notice that the square of any odd integer is $1$ modulo $8$ (proof by plugging in $1^2,3^2,5^2,7^2$ into modulo $8$ ), so the LHS of this expression goes $5,1,5,1,\ldots$ , while the RHS goes $1,2,3,4,5,6,7,8,1,\ldots$ . The cycle length of the LHS is $2$ , RHS is $8$ , so the cycle length of the solution is $\operatorname{lcm}(2,8)=8$ . Indeed, the $n$ that solve this congruence are exactly those such that $n \equiv 5 \pmod{8}$
(2) $2^n \equiv n \pmod{125}$ . This is extremely computationally intensive if we try to calculate all $2^1,2^2,\ldots,2^{100} \pmod{125}$ , so we take a divide-and-conquer approach instead. In order for this expression to be true, $2^n \equiv n \pmod{5}$ is necessary; it shouldn't take too long for us to go through the $20$ possible LHS-RHS combinations, considering that $n$ must be odd. We only need to test $10$ values of $n$ and obtain $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$
With this in mind we consider $2^n \equiv n \pmod{25}$ . By the Generalized Fermat's Little Theorem, $2^{20} \equiv 1 \pmod{25}$ , but we already have $n$ modulo $20$ . Our calculation is greatly simplified. The LHS's cycle length is $20$ and the RHS's cycle length is $25$ , from which their least common multiple is $100$ . In this step we need to test all the numbers between $1$ to $100$ that $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$ . In the case that $n \equiv 3 \pmod{20}$ , the RHS goes $3,23,43,63,83$ , and we need $2^n \equiv n \equiv 2^3 \pmod{25}$ ; clearly $n \equiv 83 \pmod{100}$ . In the case that $n \equiv 17 \pmod{20}$ , by a similar argument, $n \equiv 97 \pmod{100}$
In the final step, we need to calculate $2^{97}$ and $2^{83}$ modulo $125$
Note that $2^{97}\equiv2^{-3}$ ; because $8\cdot47=376\equiv1\pmod{125},$ we get $2^{97} \equiv 47\pmod{125}$
Note that $2^{83}$ is $2^{-17}=2^{-16}\cdot2^{-1}$ . We have \begin{align*} 2^{-1}&\equiv63, \\ 2^{-2}&\equiv63^2=3969\equiv-31, \\ 2^{-4}&\equiv(-31)^2=961\equiv-39, \\ 2^{-8}&\equiv1521\equiv21, \\ 2^{-16}&\equiv441, \\ 2^{-17}&\equiv63\cdot441\equiv7\cdot(-31)=-217\equiv33. \end{align*} This time, LHS cycle is $100$ , RHS cycle is $125$ , so we need to figure out $n$ modulo $500$ . It should be $n \equiv 283,297 \pmod{500}$
Put everything together. By the second subcondition, the only candidates less than $1000$ are $283,297,783,797$ . Apply the first subcondition, $n=\boxed{797}$ is the desired answer.

Answer:  797
Problem: Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$

Solution: We have that $2^n + 5^n \equiv n\pmod{1000}$ , or $2^n + 5^n \equiv n \pmod{8}$ and $2^n + 5^n \equiv n \pmod{125}$ by CRT. It is easy to check $n < 3$ don't work, so we have that $n \geq 3$ . Then, $2^n \equiv 0 \pmod{8}$ and $5^n \equiv 0 \pmod{125}$ , so we just have $5^n \equiv n \pmod{8}$ and $2^n \equiv n \pmod{125}$ . Let us consider both of these congruences separately.
First, we look at $5^n \equiv n \pmod{8}$ . By Euler's Totient Theorem (ETT), we have $5^4 \equiv 1 \pmod{8}$ , so $5^5 \equiv 5 \pmod{8}$ . On the RHS of the congruence, the possible values of $n$ are all nonnegative integers less than $8$ and on the RHS the only possible values are $5$ and $1$ . However, for $5^n$ to be $1 \pmod{8}$ we must have $n \equiv 0 \pmod{4}$ , a contradiction. So, the only possible values of $n$ are when $n \equiv 5 \pmod{8} \implies n = 8k+5$
Now we look at $2^n \equiv n \pmod{125}$ . Plugging in $n = 8k+5$ , we get $2^{8k+5} \equiv 8k+5 \pmod{125} \implies 2^{8k} \cdot 32 \equiv 8k+5 \pmod{125}$ . Note, for $2^n \equiv n\pmod{125}$ to be satisfied, we must have $2^n \equiv n \pmod{5}$ and $2^n \equiv n\pmod{25}$ . Since $2^{8k} \equiv 1\pmod{5}$ as $8k \equiv 0\pmod{4}$ , we have $2 \equiv -2k \pmod{5} \implies k = 5m-1$ . Then, $n = 8(5m-1) + 5 = 40m-3$ . Now, we get $2^{40m-3} \equiv 40m-3 \pmod{125}$ . Using the fact that $2^n \equiv n\pmod{25}$ , we get $2^{-3} \equiv 15m-3 \pmod{25}$ . The inverse of $2$ modulo $25$ is obviously $13$ , so $2^{-3} \equiv 13^3 \equiv 22 \pmod{25}$ , so $15m \equiv 0 \pmod{25} \implies m = 5s$ . Plugging in $m = 5s$ , we get $n = 200s - 3$
Now, we are finally ready to plug $n$ into the congruence modulo $125$ . Plugging in, we get $2^{200s-3} \equiv 200s - 3 \pmod{125}$ . By ETT, we get $2^{100} \equiv 1 \pmod{125}$ , so $2^{200s- 3} \equiv 2^{-3} \equiv 47 \pmod{125}$ . Then, $200s \equiv 50 \pmod{125} \implies s \equiv 4 \pmod{5} \implies s = 5y+4$ . Plugging this in, we get $n = 200(5y+4) - 3 = 1000y+797$ , implying the smallest value of $n$ is simply $\boxed{797}$

Answer:  797
Problem: Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$

Solution: We wish to find the least positive integer $n$ for which $2^n+5^n-n\equiv0\pmod{1000}.$ Rearranging gives \[2^n+5^n\equiv n\pmod{1000}.\] Applying the Chinese Remainder Theorem, we get the following system of congruences: \begin{align*} 2^n+5^n &\equiv n \pmod{8}, \\ 2^n+5^n &\equiv n \pmod{125}. \end{align*} It is clear that $n\geq3,$ from which we simplify to \begin{align*} 5^n &\equiv n \pmod{8}, \hspace{15mm} &(1) \\ 2^n &\equiv n \pmod{125}. &(2) \end{align*} We solve each congruence separately:
Finally, combining the two results from above (bolded) generates the least such positive integer $n$ at $s=19:$ \begin{align*} n&=8r+5 \\ &=8(5s+4)+5 \\ &=40s+37 \\ &=\boxed{797} ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME II Discussion

Answer:  797
Problem: Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: In this solution, all angle measures are in degrees.
Let $M$ be the midpoint of $\overline{BC}$ so that $\overline{OM}\perp\overline{BC}$ and $A,G,M$ are collinear. Let $\angle ABC=13k,\angle BCA=2k$ and $\angle XOY=17k.$
Note that:
Together, we conclude that $\triangle OAM \sim \triangle OXY$ by AA, so $\angle AOM = \angle XOY = 17k.$
Next, we express $\angle BAC$ in terms of $k.$ By angle addition, we have \begin{align*} \angle AOM &= \angle AOB + \angle BOM \\ &= 2\angle BCA + \frac12\angle BOC \hspace{10mm} &&\text{by Inscribed Angle Theorem and Perpendicular Bisector Property} \\ &= 2\angle BCA + \angle BAC. &&\text{by Inscribed Angle Theorem} \end{align*} Substituting back gives $17k=2(2k)+\angle BAC,$ from which $\angle BAC=13k.$
For the sum of the interior angles of $\triangle ABC,$ we get \begin{align*} \angle ABC + \angle BCA + \angle BAC &= 180 \\ 13k+2k+13k&=180 \\ 28k&=180 \\ k&=\frac{45}{7}. \end{align*} Finally, we obtain $\angle BAC=13k=\frac{585}{7},$ from which the answer is $585+7=\boxed{592}.$

Answer:  592
Problem: Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $M$ be the midpoint of $BC$ . Because $\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o$ $AXOG$ and $OMYG$ are cyclic, so $O$ is the center of the spiral similarity sending $AM$ to $XY$ , and $\angle{XOY}=\angle{AOM}$ . Because $\angle{AOM}=2\angle{BCA}+\angle{BAC}$ , it's easy to get $\frac{585}{7} \implies \boxed{592}$ from here.

Answer:  592
Problem: Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Firstly, let $M$ be the midpoint of $BC$ . Then, $\angle OMB = 90^o$ . Now, note that since $\angle OGX = \angle XAO = 90^o$ , quadrilateral $AGOX$ is cyclic. Also, because $\angle OMY + \angle OGY = 180^o$ $OMYG$ is also cyclic. Now, we define some variables: let $\alpha$ be the constant such that $\angle ABC = 13\alpha, \angle ACB = 2\alpha,$ and $\angle XOY = 17\alpha$ . Also, let $\beta = \angle OMG = \angle OYG$ and $\theta = \angle OXG = \angle OAG$ (due to the fact that $AGOX$ and $OMYG$ are cyclic). Then, \[\angle XOY = 180 - \beta - \theta = 17\alpha \implies \beta + \theta = 180 - 17\alpha.\] Now, because $AX$ is tangent to the circumcircle at $A$ $\angle XAC = \angle CBA = 13\alpha$ , and $\angle CAO = \angle OAX - \angle CAX = 90 - 13\alpha$ . Finally, notice that $\angle AMB = \angle OMB - \angle OMG = 90 - \beta$ . Then, \[\angle BAM = 180 - \angle ABC - \angle AMB = 180 - 13\alpha - (90 - \beta) = 90 + \beta - 13\alpha.\] Thus, \[\angle BAC = \angle BAM + \angle MAO + \angle OAC = 90 + \beta - 13\alpha + \theta + 90 - 13\alpha = 180 - 26\alpha + (\beta + \theta),\] and \[180 = \angle BAC + 13\alpha + 2\alpha = 180 - 11\alpha + \beta + \theta \implies \beta + \theta = 11\alpha.\] However, from before, $\beta+\theta = 180 - 17 \alpha$ , so $11 \alpha = 180 - 17 \alpha \implies 180 = 28 \alpha \implies \alpha = \frac{180}{28}$ . To finish the problem, we simply compute \[\angle BAC = 180 - 15 \alpha = 180 \cdot \left(1 - \frac{15}{28}\right) = 180 \cdot \frac{13}{28} = \frac{585}{7},\] so our final answer is $585+7=\boxed{592}$

Answer:  592
Problem: Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] /* Made by MRENTHUSIASM */ size(375);  pair A, B, C, O, G, X, Y; A = origin; B = (1,0); C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); O = circumcenter(A,B,C); G = centroid(A,B,C); Y = intersectionpoint(G--G+(100,0),B--C); X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); pair O1=circumcenter(O,G,A); real r1=length(O1-O); markscalefactor=3/160; filldraw(O--X--Y--cycle, rgb(255,255,0)); draw(rightanglemark(O,G,X),red); draw(A--O--B,fuchsia+0.4); draw(Arc(O1,r1,-40,50),royalblue+0.5); draw(circumcircle(O,G,Y), heavygreen+0.5); dot("$A$",A,1.5*dir(180+585/7),linewidth(4)); dot("$B$",B,1.5*dir(-585/7),linewidth(4)); dot("$C$",C,1.5N,linewidth(4)); dot("$O$",O,1.5N,linewidth(4)); dot("$G$",G,1.5S,linewidth(4)); dot("$Y$",Y,1.5E,linewidth(4)); dot("$X$",X,1.5W,linewidth(4)); draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); [/asy] $\angle OAX = \angle OGX = 90^\circ \implies$ quadrilateral $XAGO$ is cyclic $\implies$
$\angle GXO =  \angle GAO,$ as they share the same intersept $\overset{\Large\frown} {GO}.$
$\angle OGY = \angle OMY =  90^\circ \implies$ quadrilateral $OGYM$ is cyclic $\implies$
$\angle GYO = \angle OMG,$ as they share the same intercept $\overset{\Large\frown} {GO}.$
In triangles $\triangle XOY$ and $\triangle AOM,$ two pairs of angles are equal, which means that the third angles $\angle XOY =  \angle AOM$ are also equal.
$\angle ABC :  \angle BCA :  \angle AOM = 13 : 2 : 17,$ so $\angle AOM =  \angle ABC + 2 \angle BCA.$
According to the Claim $\triangle ABC$ is isosceles, \[\angle ABC : \angle BCA : \angle BAC = 13 : 2 : 13.\] \[\angle BAC = \frac{13} {13 + 2 + 13} \cdot 180^\circ =  \frac {585^\circ}{7} \implies  585 + 7 = \boxed{592}.\]

Answer:  592
Problem: Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Extend $XA$ and meet line $CB$ at $P$ . Extend $AG$ to meet $BC$ at $F$ . Since $AF$ is the median from $A$ to $BC$ $A,G,F$ are collinear. Furthermore, $OF$ is perpendicular to $BC$
Draw the circumcircle of $\triangle{XPY}$ , as $OA\bot XP, OG\bot XY, OF\bot PY$ $A,G,F$ are collinear, $O$ lies on $(XYP)$ as $AGF$ is the Simson line of $O$ with respect to $\triangle{XPY}$ . Thus, $\angle{P}=180-17x, \angle{PAB}=\angle{C}=2x, 180-15x=13x, x=\frac{45}{7}$ , the answer is $180-15\cdot \frac{45}{7}=\frac{585}{7}$ which is $\boxed{592}$

Answer:  592
Problem: Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$

Solution: Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$ . Note that this formula also returns the correct value when $n=(k+1)^2$ , but not when $n=k^2$ . Thus $f(n)=k^2+3k+2-n$ for $k^2<n \leq (k+1)^2$
If $2 \mid (k+1)^2-n$ $g(n)$ returns the same value as $f(n)$ . This is because the recursion once again stops at $(k+1)^2$ . We seek a case in which $f(n)<g(n)$ , so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$
Write $7f(n)=4g(n)$ , which simplifies to $3k^2+k-10=3n$ . Notice that we want the LHS expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$ . We also want n to be strictly greater than $k^2$ , so $k-10>0, k>10$ . The LHS expression is always even (since $3k^2+k-10$ factors to $k(3k+1)-10$ , and one of $k$ and $3k+1$ will be even), so to ensure that $k$ and $n$ share the same parity, $k$ should be even. Then the least $k$ that satisfies these requirements is $k=16$ , giving $n=258$
Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$

Answer:  258
Problem: Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$

Solution: Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$ . If $k$ is odd, then $f(n)=g(n)$ . If $k$ is even, then \begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*} from which \begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*} Since $k$ is even, $m$ is even. Since $k\neq 0$ , the smallest $k$ is $2$ which produces the smallest $n$ \[k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.\] ~Afo

Answer:  258