Let $ABC$ be a triangle with $AB=34,BC=25,$ and $CA=39$. Let $O,H,$ and $ \omega$ be the circumcenter, orthocenter, and circumcircle of $\triangle ABC$, respectively. Let line $AH$ meet $\omega$ a second time at $A_1$ and let the reflection of $H$ over the perpendicular bisector of $BC$ be $H_1$. Suppose the line through $O$ perpendicular to $A_1O$ meets $\omega$ at two points $Q$ and $R$ with $Q$ on minor arc $AC$ and $R$ on minor arc $AB$. Denote $\mathcal H$ as the hyperbola passing through $A,B,C,H,H_1$, and suppose $HO$ meets $\mathcal H$ again at $P$. Let $X,Y$ be points with $XH \parallel AR \parallel YP, XP \parallel AQ \parallel YH$. Let $P_1,P_2$ be points on the tangent to $\mathcal H$ at $P$ with $XP_1 \parallel OH \parallel YP_2$ and let $P_3,P_4$ be points on the tangent to $\mathcal H$ at $H$ with $XP_3 \parallel OH \parallel YP_4$. If $P_1P_4$ and $P_2P_3$ meet at $N$, and $ON$ may be written in the form $\frac{a}{b}$ where $a,b$ are positive coprime integers, find $100a+b$.

[i]Proposed by Vincent Huang[/i]
1. **Understanding the Problem:**
   We are given a triangle \(ABC\) with sides \(AB = 34\), \(BC = 25\), and \(CA = 39\). We need to find the value of \(100a + b\) where \(ON\) is written in the form \(\frac{a}{b}\) with \(a\) and \(b\) being positive coprime integers. 

2. **Key Points and Definitions:**
   - \(O\) is the circumcenter.
   - \(H\) is the orthocenter.
   - \(\omega\) is the circumcircle.
   - \(A_1\) is the second intersection of line \(AH\) with \(\omega\).
   - \(H_1\) is the reflection of \(H\) over the perpendicular bisector of \(BC\).
   - \(Q\) and \(R\) are points where the line through \(O\) perpendicular to \(A_1O\) meets \(\omega\).
   - \(\mathcal{H}\) is the hyperbola passing through \(A, B, C, H, H_1\).
   - \(P\) is the second intersection of \(HO\) with \(\mathcal{H}\).
   - \(X\) and \(Y\) are points such that \(XH \parallel AR \parallel YP\) and \(XP \parallel AQ \parallel YH\).
   - \(P_1, P_2\) are points on the tangent to \(\mathcal{H}\) at \(P\) with \(XP_1 \parallel OH \parallel YP_2\).
   - \(P_3, P_4\) are points on the tangent to \(\mathcal{H}\) at \(H\) with \(XP_3 \parallel OH \parallel YP_4\).
   - \(N\) is the intersection of \(P_1P_4\) and \(P_2P_3\).

3. **Rectangular Hyperbola:**
   - Any hyperbola passing through \(A, B, C, H\) is a rectangular hyperbola.
   - The centers of these hyperbolas lie on the nine-point circle.

4. **Coordinate Setup:**
   - Let the hyperbola be \(xy = 1\).
   - Set \(A = (a, \frac{1}{a})\), \(B = (b, \frac{1}{b})\), \(C = (\frac{1}{b}, b)\), \(P = (p, \frac{1}{p})\) with \(a, p > 0\) and \(b < 0\).
   - The orthocenter \(H = (-\frac{1}{a}, -a)\).

5. **Tangents and Slopes:**
   - The tangent at \(H\) satisfies \(y = -a^2x - 2a\).
   - The tangent at \(P\) satisfies \(y = -\frac{1}{p^2}x + \frac{2}{p}\).
   - Line \(PH\) has slope \(\frac{a}{p}\).

6. **Finding \(X\) and \(Y\):**
   - \(HX\) is parallel to the \(x\)-axis and \(HY\) is parallel to the \(y\)-axis.
   - Circumcenter \((t, t)\) with \(t = \frac{(a+b)(ab+1)}{2ab}\).
   - Compute \(R = \left(\frac{(a+b)(ab+1)}{ab} - a, \frac{1}{a}\right)\) and \(S = \left(a, \frac{(a+b)(ab+1)}{ab} - \frac{1}{a}\right)\).
   - \(X = (p, -a)\) and \(Y = (-\frac{1}{a}, \frac{1}{p})\).

7. **Finding \(P_1, P_2, P_3, P_4\):**
   - \(P_1\) lies on the \(x\)-axis and \(P_4\) lies on the \(x\)-axis.
   - \(P_2\) lies on the \(y\)-axis and \(P_3\) lies on the \(y\)-axis.

8. **Intersection at Origin:**
   - \(P_1P_4\) and \(P_2P_3\) intersect at the origin, making \(N\) the center of the hyperbola.

9. **Finding \(ON\):**
   - Let \(M\) be the midpoint of \(BC\).
   - The power of \(O\) with respect to the nine-point circle is \(ON \cdot OM\).
   - \(ON \cdot OM = \frac{R^2 - OH^2}{4}\).
   - \(OH^2 = 9R^2 - a^2 - b^2 - c^2\).
   - \(ON \cdot OM = \frac{a^2 + b^2 + c^2 - 8R^2}{4}\).
   - \(OM = \sqrt{R^2 - \frac{a^2}{4}}\).
   - We need \(\frac{a^2 + b^2 + c^2 - 8R^2}{2\sqrt{4R^2 - a^2}}\).

10. **Plugging in Values:**
    - \(a = 25\), \(b = 39\), \(c = 34\), \(R = \frac{1105}{56}\).
    - \(\frac{429}{140}\).

The final answer is \(42900 + 140 = \boxed{43040}\).