Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$ . Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$ . The most likely explanation for this is that $Q$ is the midpoint of $XY$ , so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$ . Then our answer is $m+n=61+4=\boxed{065}$ . One can rigorously prove this using the methods above

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the tangents to $\omega$ at $A$ and $B$ intersect at $R$ . Then, since $RA^2=RB^2$ $R$ lies on the radical axis of $\omega_1$ and $\omega_2$ , which is $\overline{PQ}$ . It follows that \[-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).\] Let $Q'$ denote the midpoint of $\overline{XY}$ . By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), \[RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,\] whence $Q=Q'$ . Like above, $XP=\tfrac{11-\sqrt{61}}2$ . Since $XQ=\tfrac{11}2$ , we establish that $PQ=\tfrac{\sqrt{61}}2$ , from which $PQ^2=\tfrac{61}4$ , and the requested sum is $61+4=\boxed{065}$

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Firstly we need to notice that $Q$ is the middle point of $XY$ . Assume the center of circle $w, w_1, w_2$ are $O, O_1, O_2$ , respectively. Then $A, O_1, O$ are collinear and $O, O_2, B$ are collinear. Link $O_1P, O_2P, O_1Q, O_2Q$ . Notice that, $\angle B=\angle A=\angle APO_1=\angle BPO_2$ . As a result, $PO_1\parallel O_2O$ and $OO_1\parallel O_2P$ . So we have parallelogram $PO_2OO_1$ . So $\angle O_2PO_1=\angle O$ Notice that, $O_1O_2\bot PQ$ and $O_1O_2$ divides $PQ$ into two equal length pieces, So we have $\angle O_2QO_1=\angle O_2PO_1=\angle O$ . As a result, $O_2, Q, O, O_1,$ lie on one circle. So $\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P$ . Notice that since $\angle O_1PQ+\angle O_2O_1P=90^{\circ}$ , we have $\angle OQP=\angle OQO_1 + \angle O_1QP = \angle O_2O_1P + O_1PQ=90^{\circ}$ . As a result, $OQ\bot PQ$ . So $Q$ is the middle point of $XY$
Back to our problem. Assume $XP=x$ $PY=y$ and $x<y$ . Then we have $AP\cdot PB=XP\cdot PY$ , that is, $xy=15$ . Also, $XP+PY=x+y=XY=11$ . Solve these above, we have $x=\frac{11-\sqrt{61}}{2}=XP$ . As a result, we have $PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ . So, we have $PQ^2=\frac{61}{4}$ . As a result, our answer is $m+n=61+4=\boxed{065}$

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Note that the tangents to the circles at $A$ and $B$ intersect at a point $Z$ on $XY$ by radical axis theorem. Since $\angle ZAB = \angle ZQA$ and $\angle ZBA = \angle ZQB$ , we have \[\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},\] so $ZAQB$ is cyclic.
But if $O$ is the center of $\omega$ , clearly $ZAOB$ is cyclic with diameter $ZO$ , so $\angle ZQO = 90^{\circ}$ implies that $Q$ is the midpoint of $XY$ . Then, by power of point $P$ \[PY \cdot PX = PA \cdot PB = 15,\] whereas it is given that $PY+PX = 11$ . Thus \[PY, PX \in \left\{\tfrac 12 (11 \pm \sqrt{61})\right\}\] so $PQ = \frac{\sqrt{61}}{2}$ , i.e. $PQ^2 = \frac{61}{4}$ and the answer is $61+4 = \boxed{065}$

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Connect $AQ,QB$ , since $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}$ , so $\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}$ then, so $A,O,Q,B$ are concyclic
We let $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha$ , it is clear that $\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha$ , which leads to the conclusion $OQ\bot XY$ which tells $Q$ is the midpoint of $XY$
Then it is clear, $XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ , the answer is $\boxed{065}$

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));  draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B,red); draw(X -- Y,green);  dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70),blue); dot("$Q$", Q, dir(200)); dot("$X$", X, X); dot("$Y$", Y, Y); label("$3$", (A+P)/2, N, red); label("$5$", (B+P)/2, N, red); draw(brace(X,Y)); label("$11$",brace(X,Y),dir(20)); [/asy] $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$ , so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem would simply ask for $PQ$ . We can also figure out that since $PQ^2$ is rational, $PQ$ is $\sqrt{\text{[something]}}$ $PQ=QX-PX$ , and chances are low that $QX$ is some number with a square root plus or minus $\frac{\sqrt{61}}{2}$ to cancel out the $\frac{\sqrt{61}}{2}$ in $PX$ , so one can see that $PQ^2$ is most likely to be $\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}$ , and our answer is $61+4=\boxed{065}$

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We will show that $Q$ is the midpoint of $XY.$ To do this, let $Q^{\prime}$ be the altitude from $O$ to $XY$ or, equivalently, to $PQ.$ Notice that $O_{1}00_{2}P$ is a parallelogram. Thus, the height from $O$ to $O_{1}O_{2}$ is equal to the height from $P$ to $O_{1}O_{2}.$ Say that the line through $P$ perpendicular to $O_{1}O_{2}$ intersects $O_{1}O_{2}$ at $H.$ Then, $PQ$ is perpendicular to $O_{1}O_{2},$ so $H$ is on $PQ.$ Now, we have that the altitude from $O$ to $O_{1}O_{2}$ is equal to the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ (since $OQ^{\prime} \parallel O_{1}O_{2}$ ). However, the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ is just $Q^{\prime}H.$ Also, the altitude from $P$ to $O_{1}O_{2}$ is $PH$ , so $PH = Q^{\prime}H.$ Thus, $O_{1}O_{2}$ bisects $PQ^{\prime}.$ However, this is true for $Q,$ too, so $Q = Q^{\prime},$ and we are done. Now, by PoP, we have \[AP \cdot BP = XP \cdot YP = 15.\] Also, we have $XY = XP+YP = 11,$ so $XP = \frac{11 \pm \sqrt{61}}{2}$ . Notice that $XQ = \frac{XY}{2} = \frac{11}{2},$ so $PQ = \frac{\sqrt{61}}{2},$ giving us our answer of $\boxed{065}.$

Answer:  065
Problem: Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Like Solution 7, let $Q'$ be the altitude from $O$ to $XY$ . And, let $M$ be the intersection of $O_1O_2$ and $PQ$ . Construct $P'$ on line $AO$ such that $PP' \parallel O_2O_1$ . First, because of isosceles triangles $OAB$ $O_1AP$ , and $O_2BP$ , we have $\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}$ , which means $OO_1PO_2$ is a parallelogram. So, $O_2P = OO_1$ . It is also clear that $PP'O_1O_2$ is a parallelogram by virtue of our definition. Thus, $O_2P = O_1P' = OO_1$ . Since $OQ' \parallel O_1O_2 \parallel P'P$ (because of the right angles), $\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP$ . And, because $QM = MP$ $Q = Q'$ . From Power of a Point on $P$ , we have $XP(11-XP) = 15$ , giving us $XP = \frac{11 - \sqrt{61}}{2}$ . Since $OQ$ is perpendicular to $XY$ $Q$ is the midpoint of $XY$ , so $XQ = \frac{11}{2}$ . Thus, $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}$ . Therefore, our answer is $\boxed{65}$

Answer:  65
Problem: Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$E$",E,dir(-135)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120));  draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); [/asy] - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so 
that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$ , since $\triangle DEB$ is an $8-15-17$ .) The base $CD$ of the rectangle will be $9+6+6=21$ . Now, let $O$ be the intersection of $BD$ and $AC$ . This means that $\triangle ABO$ and $\triangle DCO$ are with ratio $\frac{21}{9}=\frac73$ . Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$ , and $x$ be the height of $\triangle ABO$ \[\frac{7}{3}=\frac{y}{x}\] \[\frac{7}{3}=\frac{8-x}{x}\] \[7x=24-3x\] \[10x=24\] \[x=\frac{12}{5}\]
This means that the area is $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$ . This gets us $54+5=\boxed{059}.$

Answer:  059
Problem: Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Using the diagram in Solution 1, let $E$ be the intersection of $BD$ and $AC$ . We can see that angle $C$ is in both $\triangle BCE$ and $\triangle ABC$ . Since $\triangle BCE$ and $\triangle ADE$ are congruent by AAS, we can then state $AE=BE$ and $DE=CE$ . It follows that $BE=AE$ and $CE=17-BE$ . We can now state that the area of $\triangle ABE$ is the area of $\triangle ABC-$ the area of $\triangle BCE$ . Using Heron's formula, we compute the area of $\triangle ABC=36$ . Using the Law of Cosines on angle $C$ , we obtain
\[9^2=17^2+10^2-2(17)(10)cosC\] \[-308=-340cosC\] \[cosC=\frac{308}{340}\] (For convenience, we're not going to simplify.)
Applying the Law of Cosines on $\triangle BCE$ yields \[BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC\] \[BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})\] \[0=389-34BE-(340-20BE)(\frac{308}{340})\] \[0=389-34BE+\frac{308BE}{17}\] \[0=81-\frac{270BE}{17}\] \[81=\frac{270BE}{17}\] \[BE=\frac{51}{10}\] This means $CE=17-BE=17-\frac{51}{10}=\frac{119}{10}$ . Next, apply Heron's formula to get the area of $\triangle BCE$ , which equals $\frac{126}{5}$ after simplifying. Subtracting the area of $\triangle BCE$ from the area of $\triangle ABC$ yields the area of $\triangle ABE$ , which is $\frac{54}{5}$ , giving us our answer, which is $54+5=\boxed{059}.$ -Solution by flobszemathguy

Answer:  059
Problem: Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120));  draw(D--(-6,0)--A,dotted); label("$8$",(D+(-6,0))/2,dir(180)); label("$6$",(A+(-6,0))/2,dir(-90));  draw((4.5,0)--(4.5,2.4),dotted); label("$h$", (4.5,1.2), dir(180)); label("$4.5$", (6,0), dir(90));  [/asy] - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$ . Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$ , so the other leg of the new triangle formed has length $4.5$ . Notice we have formed similar triangles, and we can solve for $h$
\[\frac{h}{4.5} = \frac{8}{15}\] \[h = \frac{36}{15} = \frac{12}{5}\]
So $\triangle ABE$ has area \[\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}\] And $54+5=\boxed{059}.$ - Solution by Duoquinquagintillion

Answer:  059
Problem: Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $a = \angle{CAB}$ . By Law of Cosines, \[\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}\] \[\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}\] \[\tan a = \frac{8}{15}\] \[A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}\] And $54+5=\boxed{059}.$

Answer:  059
Problem: Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $A=(0,0), B=(9,0)$ . Now consider $C$ , and if we find the coordinates of $C$ , by symmetry about $x=4.5$ , we can find the coordinates of D.
So let $C=(a,b)$ . So the following equations hold:
$\sqrt{(a-9)^2+(b)^2}=17$
$\sqrt{a^2+b^2}=10$
Solving by squaring both equations and then subtracting one from the other to eliminate $b^2$ , we get $C=(-6,8)$ because $C$ is in the second quadrant.
Now by symmetry, $D=(16, 8)$
So now you can proceed by finding the intersection and then calculating the area directly. We get $\boxed{059}$

Answer:  059
Problem: Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair O = (4.5,2.4); pair E = (-6,0); pair K = (1.993,3.737);  draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); label("$O$",O,dir(90));  label("$K$",K,dir(-120)); draw(A--K,dotted);  draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); label("$E$",E,dir(-135)); [/asy]
Since $\triangle ABC \cong \triangle BAD$ $\angle ADB = \angle BCA$ . Thus, $A$ $B$ $C$ $D$ are concyclic.
By Ptolemy's Theorem on $ABCD$ \[(AD)(BC) + (AB)(DC) = (BD)(AC)\] \[10^2 + 9(DC) = 17^2\] \[DC = 21\]
The altitudes dropped from $C$ and $D$ onto the extension of AB are equal, meaning that $DC \parallel AB$ . Therefore, $\triangle DCO \sim \triangle ABO$ . It follows that \[\frac{OB}{17 - OB} = \frac{AB}{DC} = \frac{9}{21} = \frac{3}{7}\] Solving yields $OB = \frac{51}{10}$
In $\triangle ABD$ , drop an altitude from $A$ to $BD$ . Call the intersection of this altitude and $BD$ $K$
The area of $\triangle ABD$ is $\frac{1}{2}(AB)(DE) = 36$ . Thus, $\frac{1}{2}(AK)(BD) = 36$ , and $AK = \frac{72}{17}$
Therefore, the area of $\triangle AOB$ is $\frac{1}{2}(OB)(AK) = \frac{1}{2}(\frac{51}{10})(\frac{72}{17}) = \frac{54}{5}$
The requested answer is $54 + 5 = \boxed{59}$

Answer:  59
Problem: Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$ . Then $P_7 = 1$ $P_6 = \frac12$ , and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$ . Working our way down, we find \[P_5 = \frac{3}{4}\] \[P_4 = \frac{5}{8}\] \[P_3 = \frac{11}{16}\] \[P_2 = \frac{21}{32}\] \[P_1 = \frac{43}{64}\] $43 + 64 = \boxed{107}$

Answer:  107
Problem: Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: Define a one jump to be a jump from $k$ to $k + 1$ and a two jump to be a jump from $k$ to $k + 2$
Case 1: (6 one jumps) $\left (\frac{1}{2} \right)^6 = \frac{1}{64}$
Case 2: (4 one jumps and 1 two jumps) $\binom{5}{1} \cdot \left(\frac{1}{2}\right)^5 = \frac{5}{32}$
Case 3: (2 one jumps and 2 two jumps) $\binom{4}{2} \cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$
Case 4: (3 two jumps) $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$
Summing the probabilities gives us $\frac{43}{64}$ so the answer is $\boxed{107}$

Answer:  107
Problem: Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: Let $P_n$ be the probability that the frog lands on lily pad $n$ . The probability that the frog never lands on pad $n$ is $\frac{1}{2}P_{n-1}$ , so $1-P_n=\frac{1}{2}P_{n-1}$ . This rearranges to $P_n=1-\frac{1}{2}P_{n-1}$ , and we know that $P_1=1$ , so we can compute $P_7$ \begin{align*} P_1&=1\\ P_2&=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}\\ P_3&=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}\\ P_4&=\dfrac{5}{8}\\ P_5&=\dfrac{11}{16}\\ P_6&=\dfrac{21}{32}\\ P_7&=\dfrac{43}{64}\\ \end{align*} We calculate $P_7$ to be $\frac{43}{64}$ , meaning that our answer is $\boxed{107}$

Answer:  107
Problem: Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: For any point $n$ , let the probability that the frog lands on lily pad $n$ be $P_n$ . The frog can land at lily pad $n$ with either a double jump from lily pad $n-2$ or a single jump from lily pad $n-1$ . Since the probability when the frog is at $n-2$ to make a double jump is $\frac{1}{2}$ and same for when it's at $n-1$ , the recursion is just $P_n = \frac{P_{n-2}+P_{n-1}}{2}$ . Using the fact that $P_1 = 1$ , and $P_2 = \frac{1}{2}$ , we find that $P_7 = \frac{43}{64}$ $43 + 64 = \boxed{107}$

Answer:  107
Problem: Find the number of $7$ -tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: \begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*}

Solution: As 71 is prime, $c$ $d$ , and $e$ must be 1, 1, and 71 (in some order). However, since $c$ and $e$ are divisors of 70 and 72 respectively, the only possibility is $(c,d,e) = (1,71,1)$ . Now we are left with finding the number of solutions $(a,b,f,g)$ satisfying $ab = 70$ and $fg = 72$ , which separates easily into two subproblems. The number of positive integer solutions to $ab = 70$ simply equals the number of divisors of 70 (as we can choose a divisor for $a$ , which uniquely determines $b$ ). As $70 = 2^1 \cdot 5^1 \cdot 7^1$ , we have $d(70) = (1+1)(1+1)(1+1) = 8$ solutions. Similarly, $72 = 2^3 \cdot 3^2$ , so $d(72) = 4 \times 3 = 12$
Then the answer is simply $8 \times 12 = \boxed{096}$

Answer:  096