Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$
[asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Solution: Let the center of the circle be $O$ . Fix $B,C,$ and $A$ . Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$ , it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$
Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$ . Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$ . Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$ . Clearly, $KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\frac{5}{2}$ . The same applies for $FO$ , which also equals $\frac{5}{2}$ . By the Pythagorean theorem, we deduce that $FL = 2$ , so $EK = 2$ . This is very important information! Now we know that $BE = 1$ , so by Power of a Point, $AE = ED = \sqrt{5}$
We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$ , we deduce that $EO = 2\sqrt{5}$ $EC=OC=5$ , so $\sin (CEO) = \frac{2\sqrt{5}}{5}$ . Furthermore, since $\sin (CEO) = \cos(DEC)$ , we know that $\cos (DEC) = \frac{2\sqrt{5}}{5}$ . By the law of cosines, \[DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10\] Therefore, $DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}$ . Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$ . Then, by the Pythagorean theorem, $OZ = \frac{7\sqrt{2}}{2}$ . Thus, $\sin (BOZ) = \frac{\sqrt{2}}{10}$ and $\cos (BOZ) = \frac{7\sqrt{2}}{10}$ . As a result, $\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}$ $7 \cdot 25 = \boxed{175}$

Answer:  175
Problem: The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$
[asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Solution: Let I be the intersection of AD and BC.
Lemma: $AI = ID$ if and only if $\angle AIO = 90$
Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$ , We can get $\triangle AIO \cong \triangle OID$
Let  be this the circle with diameter AO.
Thus, we get $\angle AIO = 90$ , implying I must lie on $\omega$ . I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.
Now, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C.
Let Z be (0,5).
Let Y be (-5,0).
Let X be the center of $\omega$ . Since $\omega$ 's radius is $\frac{5}{2}$ , the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \frac{3}{5}$ $sin(BOZ) =  \frac{3}{5}$ . If we let $sin(\theta) = \frac{3}{5}$ , we can find that what we are looking for is $sin(90 - 2\theta)$ , which we can evaluate and get $\frac{7}{25} \implies \boxed{175}$

Answer:  175
Problem: The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$
[asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Solution: Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$ . Additionally, this circle must be tangent to $\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ $N$ be the foot of the perpendicular from $P$ to $\overline{BM}$ , and $K$ be the foot of the perpendicular from $B$ to $\overline{AP}$ . Let $x=BK$
From right triangle $BKO$ , we get $KO = \sqrt{25-x^2}$ . Thus, $KP = \sqrt{25-x^2}-\frac52$
Since $BO = 5$ $BM = 3$ , and $\angle BMO$ is right, $MO=4$ . From quadrilateral $MNPO$ , we get $MN = \sqrt{PO^2 - (MO - NP)^2} = \sqrt{(5/2)^2 - (4 - 5/2)^2} = \sqrt{(5/2)^2 - (3/2)^2} = 2$ . Thus, $BN = 1$
Since angles $BNP$ and $BKP$ are right, we get \[BK^2+KP^2 = BN^2 + NP^2 \implies x^2 + \left(\sqrt{25-x^2}-\frac52\right)^2 = \left(\frac52\right)^2 + 1\] \[25 - 5\sqrt{25-x^2} = 1\] \[5\sqrt{25-x^2} = 24\] \[25(25-x^2) = 24^2\] \[25x^2 = 25^2 - 24^2 = 49\] \[x = \frac75\] Thus, $\sin \angle AOB = \frac{x}{5} = \frac{7}{25}\implies \boxed{175}$

Answer:  175