jmc

Let $E$ and $F$ be the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively, such that $\overline{BE}$ intersects $\overline{CF}$. Since $E$ and $F$ are midpoints, $BE=15$ and $CF=7$. $B$ and $C$ are located on the circumference of the circle, so $OB=OC=25$. The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so $△OEB$ and $△OFC$ are right triangles (with $\angle OEB$ and $\angle OFC$ being the right angles). By the Pythagorean Theorem, $OE=\sqrt{25^2-15^2}=20$, and $OF=\sqrt{25^2-7^2}=24$. Let $x$, $a$, and $b$ be lengths $OP$, $EP$, and $FP$, respectively. OEP and OFP are also right triangles, so $x^2=a^2+20^2\to a^2=x^2-400$, and $x^2=b^2+24^2\to b^2=x^2-576$ We are given that $EF$ has length 12, so, using the Law of Cosines with $△EPF$: $12^2=a^2+b^2-2ab\cos (\angle EPF)=a^2+b^2-2ab\cos (\angle EPO+\angle FPO)$ Substituting for $a$ and $b$, and applying the Cosine of Sum formula: $144=(x^2-400)+(x^2-576)-2\sqrt{x^2-400}\sqrt{x^2-576}\left(\cos \angle EPO\cos \angle FPO-\sin \angle EPO\sin \angle FPO\right)$ $\angle EPO$ and $\angle FPO$ are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines: $144=2x^2-976-2\sqrt{(x^2-400)(x^2-576)}\left(\frac{\sqrt{x^2-400}}{x}\frac{\sqrt{x^2-576}}{x}-\frac{20}{x}\frac{24}{x}\right)$ Combine terms and multiply both sides by $x^2$: $144x^2=2x^4-976x^2-2(x^2-400)(x^2-576)+960\sqrt{(x^2-400)(x^2-576)}$ Combine terms again, and divide both sides by 64: $13x^2=7200-15\sqrt{x^4-976x^2+230400}$ Square both sides: $169x^4-187000x^2+51,840,000=225x^4-219600x^2+51,840,000$ This reduces to $x^2=\frac{4050}{7}=(OP{)}^2$; $4050+7\equiv \boxed{57}(\mathrm{mod}1000)$.