Triangle $ABC$ has side lengths $AB=13, BC=14,$ and $CA=15$. Let $\Gamma$ denote the circumcircle of $\triangle ABC$. Let $H$ be the orthocenter of $\triangle ABC$. Let $AH$ intersect $\Gamma$ at a point $D$ other than $A$. Let $BH$ intersect $AC$ at $F$ and $\Gamma$ at point $G$ other than $B$. Suppose $DG$ intersects $AC$ at $X$. Compute the greatest integer less than or equal to the area of quadrilateral $HDXF$.

[i]Proposed by Kenan Hasanaliyev (claserken)[/i]
1. **Rename Points and Establish Reflections**:
   - Let \( D \) be the foot from \( A \) to \( BC \), \( E \) the foot from \( B \) to \( CA \), and \( F \) the foot from \( C \) to \( AB \).
   - Let the intersection of line \( AHD \) with \( \Gamma \) be \( H_A \), and the intersection of line \( BHE \) with \( \Gamma \) be \( H_B \).
   - Let \( X \) be the intersection of \( H_AH_B \) and \( AC \).

2. **Reflection Properties**:
   - \( H_A \) is the reflection of \( H \) over \( D \).
   - \( H_B \) is the reflection of \( H \) over \( E \).

3. **Angle Chasing**:
   - \(\measuredangle H_ABC = \measuredangle H_AAC = 90^\circ - \measuredangle ACB = \measuredangle CBH\).
   - This implies \( \triangle HDE \sim \triangle HH_AH_B \) by SAS similarity.

4. **Cyclic Quadrilateral**:
   - \( EHDC \) is cyclic, so \( \angle HED = \angle HCD = 90^\circ - \angle B \).

5. **Area Calculation**:
   - We need to compute the area of quadrilateral \( HDXF \).
   - \([HH_AXE] = [HH_AH_B] - [H_BEX]\).

6. **Using Heron's Formula**:
   - Semi-perimeter \( s = \frac{13 + 14 + 15}{2} = 21 \).
   - Area of \( \triangle ABC \):
     \[
     [ABC] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84.
     \]

7. **Sine Values**:
   - \(\sin C = \frac{2[ABC]}{ab} = \frac{2 \cdot 84}{13 \cdot 14} = \frac{12}{13}\).
   - \(\sin B = \frac{2[ABC]}{ac} = \frac{2 \cdot 84}{14 \cdot 15} = \frac{4}{5}\).
   - \(\sin A = \frac{2[ABC]}{bc} = \frac{2 \cdot 84}{13 \cdot 15} = \frac{56}{65}\).

8. **Calculating Heights**:
   - \( BD = AB \cos \angle B = 13 \cdot \frac{5}{13} = 5 \).
   - \( HD = \frac{BD}{\tan \angle C} = \frac{5}{4/3} = \frac{15}{4} \).
   - \( CE = BC \cos \angle C = 14 \cdot \frac{3}{5} = \frac{42}{5} \).
   - \( HE = \frac{CE}{\tan \angle A} = \frac{42/5}{56/33} = \frac{99}{20} \).

9. **Area of \( \triangle HH_AH_B \)**:
   - \[
   [HH_AH_B] = \frac{HH_A \cdot HH_B \cdot \sin \angle H_BHH_A}{2} = 2 \cdot HD \cdot HE \cdot \sin \angle C = 2 \cdot \frac{15}{4} \cdot \frac{99}{20} \cdot \frac{12}{13} = \frac{297}{10}.
   \]

10. **Area of \( \triangle H_BEX \)**:
    - \[
    [H_BEX] = \frac{H_BE \cdot EX}{2} = \frac{H_BE^2}{2 \tan \angle B} = \frac{HE^2}{2 \tan \angle B} = \frac{(\frac{99}{20})^2}{2 \cdot \frac{4}{3}} = \frac{3267}{640}.
    \]

11. **Final Area Calculation**:
    - \[
    [HH_AXE] = [HH_AH_B] - [H_BEX] = \frac{297}{10} - \frac{3267}{640} = 29.7 - 5.10546875 = 24.59453125.
    \]

12. **Greatest Integer Less Than or Equal to the Area**:
    - The greatest integer less than or equal to \( 24.59453125 \) is \( 24 \).

The final answer is \(\boxed{24}\).