图 1 所示的均匀三角形薄片 $ABC$ 高为 $9y$，底边 $BC=6x$，且 $AB=AC$。

$P$、$Q$ 两点满足 $AP:PC=AQ:QB=2:1$。

将薄片沿 $PQ$ 折叠，形成折叠薄片 $F$。

$F$ 的质心到 $PQ$ 的距离为 $d$。

(a) 证明 $d=\frac{16}{9}y$。

将折叠薄片从 $P$ 点自由悬挂，使其处于平衡状态，$PQ$ 与竖直向下方向成 $\alpha$ 角。

已知 $\tan\alpha=\frac{64}{81}$，

(b) 用 $y$ 表示 $x$。

折叠只移动上方小三角形的质量。以折痕为轴，原大三角形质心恰在折痕上；扣掉原位置的小三角形面积矩，再加上镜像位置的面积矩。这样无须猜梯形质心。重叠处仍有两层质量，不能当成一层平面轮廓。

Use the unfolded orientation of Figure 1, measuring signed perpendicular distances from $PQ$ towards $BC$ as positive.

Since $AP:AC=AQ:AB=2:3$, triangles $APQ$ and $ACB$ are similar. Therefore

and the height of $APQ$ above $PQ$ is $(2/3)(9y)=6y$.

The original triangle has area

Its centroid is $9y/3=3y$ above $BC$. Since $PQ$ is also $3y$ above $BC$, its signed centroid distance from $PQ$ is zero.

The triangle $APQ$ has area

Its centroid is $6y/3=2y$ from its base $PQ$, so its signed distance before folding is $-2y$. After folding it is reflected across $PQ$, giving signed distance $+2y$.

The total mass is unchanged. Since the lamina is uniform, taking moments of areas, with the folded layer counted separately, gives

Thus

and hence

折叠后仍关于等腰三角形的对称轴对称，所以质心在 $PQ$ 中点的垂线上。悬挂时 $PG$ 竖直向下，所求角的正切为垂直折痕距离除以折痕半长。

Let $G$ be the centre of mass and let $M$ be the midpoint of $PQ$. By symmetry, $GM\perp PQ$. From (a),

When freely suspended from $P$, $PG$ is vertically downwards.

Thus the angle between $PQ$ and the downward vertical satisfies

Using the given value,

Hence

so