AMC12 2018 B

AMC12 2018 B · Q12

AMC12 2018 B · Q12. It mainly tests Linear inequalities, Triangles (properties).

Side $\overline{AB}$ of $\triangle ABC$ has length 10. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m, n)$. What is $m + n$?

$\triangle ABC$的边$\overline{AB}$长为10。角$A$的角度平分线与$\overline{BC}$相交于$D$，且$CD = 3$。所有可能的$AC$值的集合是一个开区间$(m, n)$。求$m + n$？

(A) 16 16

(B) 17 17

(D) 19 19

(E) 20 20

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $q = AC$ and $r = BD$. By the Angle Bisector Theorem, $\dfrac{AC}{CD}=\dfrac{AB}{BD}$, which means $\dfrac{q}{3}=\dfrac{10}{r}$, so $r=\dfrac{30}{q}$. The possible values of $AC$ can be determined by considering the three Triangle Inequalities in $\triangle ABC$. • $AC+BC>AB$, which means $q+3+r>10$. Substituting for $r$ and simplifying gives $q^2-7q+30>0$, which always holds because $q^2-7q+30=\left(q-\dfrac{7}{2}\right)^2+\dfrac{71}{4}$. • $BC+AB>AC$, which means $3+r+10>q$. Substituting $r=\dfrac{30}{q}$, simplifying, and factoring gives $(q-15)(q+2)<0$, which holds if and only if $-2<q<15$. • $AB+AC>BC$, which means $10+q>3+r$. Substituting $r=\dfrac{30}{q}$, simplifying, and factoring gives $(q+10)(q-3)>0$, which holds if and only if $q>3$ or $q<-10$. Combining these inequalities shows that the set of possible values of $q$ is the open interval $(3,15)$, and the requested sum of the endpoints of the interval is $3+15=18$.

答案（C）：令 $q=AC$，$r=BD$。由角平分线定理， $\dfrac{AC}{CD}=\dfrac{AB}{BD}$，即 $\dfrac{q}{3}=\dfrac{10}{r}$，所以 $r=\dfrac{30}{q}$。 $AC$ 的可能取值可通过考虑 $\triangle ABC$ 的三个三角形不等式来确定。 • $AC+BC>AB$，即 $q+3+r>10$。代入 $r$ 并化简得 $q^2-7q+30>0$，该式恒成立，因为 $q^2-7q+30=\left(q-\dfrac{7}{2}\right)^2+\dfrac{71}{4}$。 • $BC+AB>AC$，即 $3+r+10>q$。代入 $r=\dfrac{30}{q}$，化简并因式分解得 $(q-15)(q+2)<0$，当且仅当 $-2<q<15$ 时成立。 • $AB+AC>BC$，即 $10+q>3+r$。代入 $r=\dfrac{30}{q}$，化简并因式分解得 $(q+10)(q-3)>0$，当且仅当 $q>3$ 或 $q<-10$ 时成立。综合这些不等式可得 $q$ 的取值范围为开区间 $(3,15)$，所求区间端点之和为 $3+15=18$。

Topics