Question

在 $\triangle ABC$ 中， $AB = 360$ ， $BC = 507$ ，且 $CA = 780$ 。设 $M$ 为 $\overline{CA}$ 的中点， $D$ 为 $\overline{CA}$ 上使得 $\overline{BD}$ 平分角 $ABC$ 的点。设 $F$ 为 $\overline{BC}$ 上满足 $\overline{DF} \perp \overline{BD}$ 的点。若 $\overline{DF}$ 与 $\overline{BM}$ 交于 $E$ ，比值 $DE : EF$ 可写成 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

In $\triangle ABC,$ $AB = 360,$ $BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE : EF$ can be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：289

知识点：角平分线定理平行线相似

难度评级：3370

解答：

记 $c = AB = 360$ 、 $a = BC = 507$ 、 $b = CA = 780$ 。将 $\overline{FD}$ 过 $D$ 延长，与 $BA$ 射线在 $A$ 外侧交于 $G$ 。在三角形 $BGF$ 中，线段 $\overline{BD}$ 既是角平分线又是高，所以 $BG = BF = t$ 。角平分线还给出 $\frac{CD}{DA} = \frac{a}{c}$ ，因此对三角形 $ABC$ 中截线 $GDF$ 使用梅涅劳斯定理：所以 $\begin{aligned} &\frac{AG}{GB} \cdot \frac{BF}{FC} \cdot \frac{CD}{DA} \\ &= \frac{t - c}{t} \cdot \frac{t}{a - t} \cdot \frac{a}{c} = 1, \end{aligned}$ $t = \frac{2ac}{a + c}.$

现在取 $F_1$ 为 $\overline{AC}$ 上满足 $\overline{FF_1} \parallel \overline{BM}$ 的点。因为 $E$ 在 $\overline{BM}$ 上，有 $\overline{EM} \parallel \overline{FF_1}$ ，所以三角形 $DEM$ 与 $DFF_1$ 相似，且 $\frac{DE}{EF} = \frac{DM}{MF_1}$ 。角平分线比例给出 $AD = \frac{bc}{a + c}$ ，因此 $DM = \frac{b}{2} - \frac{bc}{a+c} = \frac{b(a - c)}{2(a + c)}$ 。又 $CF = a - t = \frac{a(a - c)}{a + c}$ ，所以 $CF_1 = CM \cdot \frac{CF}{CB} = \frac{b}{2} \cdot \frac{a - c}{a + c}$ ，并且 $MF_1 = \frac{b}{2}\left(1 - \frac{a - c}{a + c}\right) = \frac{bc}{a + c}$ 。

因此所以 $m + n = 49 + 240 = 289$ 。 $\begin{aligned} \frac{DE}{EF} &= \frac{DM}{MF_1} = \frac{a - c}{2c} \\ &= \frac{147}{720} = \frac{49}{240}, \end{aligned}$

Write $c = AB = 360,$ $a = BC = 507,$ $b = CA = 780.$ Extend $\overline{FD}$ beyond $D$ to meet ray $BA$ beyond $A$ at $G.$ In triangle $BGF,$ segment $\overline{BD}$ is both an angle bisector and an altitude, so $BG = BF = t.$ The bisector also gives $\frac{CD}{DA} = \frac{a}{c},$ so Menelaus' theorem for line $GDF$ crossing triangle $ABC$ says $\begin{aligned} &\frac{AG}{GB} \cdot \frac{BF}{FC} \cdot \frac{CD}{DA} \\ &= \frac{t - c}{t} \cdot \frac{t}{a - t} \cdot \frac{a}{c} = 1, \end{aligned}$ so $t = \frac{2ac}{a + c}.$

Now let $F_1$ be the point on $\overline{AC}$ with $\overline{FF_1} \parallel \overline{BM}.$ Since $E$ lies on $\overline{BM},$ we have $\overline{EM} \parallel \overline{FF_1},$ so triangles $DEM$ and $DFF_1$ are similar and $\frac{DE}{EF} = \frac{DM}{MF_1}.$ The bisector ratio gives $AD = \frac{bc}{a + c},$ so $DM = \frac{b}{2} - \frac{bc}{a+c} = \frac{b(a - c)}{2(a + c)}.$ Also $CF = a - t = \frac{a(a - c)}{a + c},$ so $CF_1 = CM \cdot \frac{CF}{CB} = \frac{b}{2} \cdot \frac{a - c}{a + c}$ and $MF_1 = \frac{b}{2}\left(1 - \frac{a - c}{a + c}\right) = \frac{bc}{a + c}.$

Therefore $\begin{aligned} \frac{DE}{EF} &= \frac{DM}{MF_1} = \frac{a - c}{2c} \\ &= \frac{147}{720} = \frac{49}{240}, \end{aligned}$ and $m + n = 49 + 240 = 289.$

2003 AIME I 第 15 题

答案：289

难度评级：3370

解答：

其他年份的第 15 题