The first question As shown in Figure 1, in $\angle ABC$, $\angle A=60^{\circ}, AB > AC$, point $O$ is the circumcenter, the two altitudes $BE$ and $CF$ intersect at point $H$, points $M$ and $N$ are on segments $BH$ and $HF$ respectively, and satisfy $BM = CN$. Find the value of $\frac{MH + NH}{OH}$.
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Solution 1: Connect $O B, O C$. Let the circumradius of $\triangle A B C$ be $R$.
By the properties of the circumcenter, we know that $\angle B O C$ is
By the properties of the circumcenter, we know that $\angle B H C=180^{\circ}-\angle A=120^{\circ}$.
    Therefore, points $B, C, H, O$ are concyclic. By Ptolemy's theorem,
    $B H \cdot O C=O H \cdot B C+C H \cdot O B$
    Also, $B C=2 R \sin 60^{\circ}=\sqrt{3} R$.
    Therefore, $(B H-C H) R=\sqrt{3} R \cdot O H$.
    Thus, $(B M+M H)-(C N-N H)=\sqrt{3} \cdot O H$.
    Given $B M=C N$, we have $\frac{M H+N H}{O H}=\sqrt{3}$.
    Solution 2: As shown in Figure 2, connect
$O B, O C, O N, O M$, and $M N$.
    $\angle B F C=\angle B E C=90^{\circ}$
    $\Rightarrow A, F, H, E$ are concyclic
    $\Rightarrow \angle B H F=\angle A=60^{\circ}$
        $\angle B H C=120^{\circ}$
        $\angle B O C=2 \angle A=120^{\circ}$
    $\Rightarrow \angle B O C=\angle B H C=120^{\circ}$
    $\Rightarrow B, O, H, C$ are concyclic
    $\Rightarrow \angle O B H=\angle O C H$
        $\left.\begin{array}{l}O B=O C \\ B M=C N\end{array}\right\} \Rightarrow \therefore \triangle B O M \cong \triangle C O N$
    $\Rightarrow\left\{\begin{array}{l}\angle O M B=\angle O N C \Rightarrow \\ O M=O N\end{array}\right.$
    $O M=O N$
$O, M, H, N$ are concyclic
    $\Rightarrow \angle M O N=\angle B H C=120^{\circ}$.
    In $\triangle M O N$, $M N=\sqrt{3} \cdot O N$
    In the cyclic quadrilateral $MONH$, by Ptolemy's theorem,
    $\left.\begin{array}{r}O N \cdot M H+N H \cdot O M=M N \cdot O H \\ O M=O N\end{array}\right\} \Rightarrow$
    $\frac{M H+N H}{O H}=\frac{M N}{O N}=\frac{\sqrt{3} \cdot O N}{O N}=\sqrt{3}$.
        Solution 3: As shown in Figure 1, it is easy to prove that $O, H, C, B$ are concyclic
        Let $\angle H O C=\theta$.
        $\theta=\angle H B C=90^{\circ}$
        $\frac{B H}{O H}-\frac{C H}{O H}=\frac{\sin \left(\theta+120^{\circ}\right)}{\sin \angle O B H}-\frac{\sin \theta}{\sin \angle O C H}$
            $=\frac{\sin \left(120^{\circ}+\theta\right)-\sin \theta}{\sin \left(\angle C-60^{\circ}\right)}=\frac{\sqrt{3} \sin \left(30^{\circ}-\theta\right)}{\sin \left(\angle C-60^{\circ}\right)}$
            $=\frac{\sqrt{3} \sin \left[30^{\circ}-\left(90^{\circ}-\angle C\right)\right]}{\sin \left(\angle C-60^{\circ}\right)}=\sqrt{3}$.
        Therefore, $\frac{M H+N H}{O H}=\frac{B H-C H}{O H}=\sqrt{3}$.
            (Zhang Weizhe, Class 007, Zhengding High School, Hebei, 050800)
        Solution 4: As shown in Figure 1, by $B M=C N$ it is easy to know $M H+N H=B H$
    $-C H$. Let the circumradius of $\triangle A B C$ be $R$, and $\angle A, \angle B, \angle C$
be the three interior angles. By the cosine theorem of the orthocenter,
        $B H-C H=2 R(\cos B-\cos C$
        $=-4 R \sin \frac{B+C}{2} \sin \frac{B-C}{2}$
        Given $\angle A=60^{\circ}$, we know
        $\frac{\angle B+\angle C}{2}=60^{\circ} . \frac{\angle B-\angle C}{2}=60^{\circ}-\angle C$.
        Therefore, $B H-C H=2 \sqrt{3} R \sin \left(\angle \dot{C}-60^{\circ}\right)$
        On the other hand, by $\angle B O C=\angle B H C=120^{\circ}$, we know $O, B, C$
$H$ are concyclic, $\angle O H B=\angle O C B=30^{\circ}, \angle O B H=\angle O B C-$
$\angle E B C=30^{\circ}-\left(90^{\circ}-\angle C\right)=\angle C-60^{\circ}$.
            By the sine theorem, $O H=2 R \sin \left(\angle C-60^{\circ}\right)$.
                Therefore, the required ratio is $\sqrt{3}$.
                Solution 5: As shown in Figure 3, construct
            the circumcircle of $A B C$, extend
        $B E$ to intersect $\odot O$ at $\lambda$. Extend
        $C F$ to intersect $O$ at $Y$. Connect
        $B Y$, draw $O Q \perp C Y$
at $O, O P \perp B X \mp P$. Let
        G. 